Category: Class 12

Tamilnadu State Board New Syllabus Samacheer Kalvi 12th Business Maths Guide Pdf Chapter 7 Probability Distributions Ex 7.3 Text Book Back Questions and Answers, Notes.

Tamilnadu Samacheer Kalvi 12th Business Maths Solutions Chapter 7 Probability Distributions Ex 7.3

Question 1.
Define normal distribution.
Solution:
A random variable X is said to follow a normal distribution with parameters mean µ and varaince σ², if its probability density function is given by

Question 2.
Define standard normal variate.
Solution:
A random variable Z = (X – µ)/σ follows the standard normal distribution. Z is called the standard normal variate with mean 0 and standard deviation 1 i.e Z – N (0, 1). Its Probability density function is given by:
φ(z) = \(\frac { 1 }{\sqrt {2π}}\) e^-x²/2 -∞ < z < ∞

Question 3.
Write down the conditions in which the normal distribution is a limiting case of binomial distribution.
Solution:
The normal distribution of a variable when represented graphically, takes the shape of a symmetrical curve, known as the Normal Curve. The curve is asymptotic to x-axis on its either side.

Question 4. m
Write down any five characteristics of normal probability curve.
Solution:
Chief Characteristics or Properties of Normal Probability distribution and Normal probability Curve.
The normal probability curve with mean µ and standard deviation σ has the following properties:
(i) the curve is bell-shaped and symmetrical about the line x = u.
(ii) Mean, median and mode of the distribution coincide.
(iii) x-axis is an asymptote to the curve, (tails of the curve never touches the horizontal (x) axis)
(iv) No portion of the curve lies below the x-axis as f(x) being the probability function can never be negative.
(v) The points of inflexion of the curve are x = µ ± σ

Question 5.
In a test on 2,000 electric bulbs, it was found that bulbs of a particular make, was normally distributed with an average life of 2,040 hours and standard deviation of 60 hours. Estimate the number of bulbs likely to burn for
(i) more than 2,150 hours
(ii) less than 1,950 hours
(iii) more 1,920 hours but less than 2,100 hours.
Solution:
Let x denote the burning of the bulb follows normal distribution with mean 2,040 and standard deviation 60 hours.
Here m = 2040; σ = 60 and N = 2000
The standard normal variate
z = \(\frac { x-µ }{σ}\) = \(\frac { x-2040 }{60}\)
(i) p(morethan 2,150 hours)
p(x > 2150)
when x = 2150
z = \(\frac { 2150-2040 }{60}\) = \(\frac { 110 }{60}\)= 1.833
p(x > 2150) = p(z > 1.833)
= p(0 < z < ∞) – p(0 < z < 1.833)
= 0.5 – 0.4664
= 0.0336
∴ Number of bulbs whose burning time is more than 2150 hours=0.0336 × 2000
= 67.2 = 67(approximately)

(ii) p(less than 1950 hours)
p(x < 1950)
when x = 1950
z = \(\frac { 1950-3040 }{60}\) = \(\frac { -90 }{60}\)= -1.5
p(x < 1950) = p(z < -1.5) = p(z > 1.5)
= 0.5 – 0.4332
= 0.068

Numbers of bulbs whose burning time is less than
1950 = 0.0668 × 2000 = 133.6
= 134 (approximately)

(iii) p(more 1,920 hours but less than 2,100 hours)
= p(1920 < x < 2100)
when x = 1950
z = \(\frac { 1920-2040 }{60}\) = \(\frac { -120 }{60}\)= -2
when x = 2100
z = \(\frac { 2100-2040 }{60}\) = \(\frac { -60 }{60}\)= -2
∴ p(1920 < x < 2040) = p(-2 < z < 1)
= p(0 < z < 2) + p(0 < z < 1)
= 0.4772 + 0.3413
= 0.8185

∴ Number of bulbs whose burning time more than 1920 hours but less than 2100 hours) = 0.8185 × 2000
= 1637

Question 6.
In a distribution 30% of the items are under 50 and 10% are over 86. Find the mean and standard deviation of the distribution.
Solution:

z = \(\frac { x-µ }{σ}\)
Given that
p(x < 50) = 0.3 p(x > 86) = 0.1
p(z < -c) = 0.3
p(-c < z < 0) = 0.5 – 0.3
p(-c < z < 0) = 0.2 {from the table}
p(0 < z < c) = 0.2
c = 0.53
then -c = -0.53
∴ \(\frac { 50-µ }{σ}\) = -0.53
50 – µ = -0.53σ
µ – 0.53σ = 50 → 1
p(x < 50) = 0.1
p(0 < z < ∞) = -p(0 < z < c₁) = 0.1
p(0 < z < ∞) = p(0 < z < c₁) + 0.1
0.5 = p(0 < z < c₁) + 0.1
p(0 < z < c₁) = 0.5 – 0.1
p = (0 < z < c₁) = 0.4
c₁ = 1.29
∴ \(\frac { 86-µ }{σ}\) = 1.29
86 – µ = 1.29 σ
µ + 1.29σ = 86 → 2
solving eqn 1 & 2
eqn 2 ⇒ m + 1.29σ = 86
eqn 1 ⇒ m + 0.53σ = 50
– + – ………….
………… 1.82 ………… σ = 36 ………….
σ = \(\frac { 36 }{1.82}\) ∴ = 19.78
Substitute σ = 19.78 in eqn 1
µ – 0.53(19.78) = 50
µ -10.48 = 50
µ = 50 + 10.48
µ = 60.48
Mean = 60.48 and standard deviation = 19.78

Question 7.
X is normally distributed with mean 12 and sd 4. Find P (x ≤) 20 and P (0 ≤ x ≤ 12)
Solution:
x is normally distribution with mean 12 and sd 4
∴ µ = 12 and σ = 4
Standard normal variable
z = \(\frac { x-µ }{σ}\) = \(\frac { x-12 }{4}\)
(i) p(x ≤ 20)

when x = 20
z = \(\frac { 20-12 }{4}\) = \(\frac { 8 }{4}\) = 2
v(x ≤ 20) = \(\frac { 8 }{4}\) = 2
p(x ≤ 20) = p(z ≤ 2)
= 0.5 + p(0 < z < 2)
= 0.5 + 0.4772
= 0.9772

(ii) p(0 < x < 12 )

when x = 0
z = \(\frac { 0-12 }{4}\) = \(\frac { -12 }{4}\) = -3
when x = 12
z = \(\frac { 12-12 }{4}\) = \(\frac { 0 }{4}\) = 0
p(0 ≤ x ≤ 12) = p(-3 ≤ z ≤ 0)
= p(0 ≤ z ≤ 3)
= 0.4987

Question 8.
If the heights of 500 students are normally distributed with mean 68.0 inches and standard deviation 3.0 inches, how many students have height
(a) greater than 2 inches
(b) less than or equal to 64 inches
(c) between 65 and 71 inches.
Solution:
let x denote the height of a student N = 500; m = 68.0 inches and σ = 3.0 inches the standard normal variate
z = \(\frac { x-µ }{σ}\) = \(\frac { x-68 }{3}\)
a(greater than 72 inches)
p = p(x > 72)
when x = 72
z = \(\frac { 72-68 }{3}\) = \(\frac { 4 }{3}\) = 1.33
p(x > 72) = p(z > 1.33)
= 0.5 – 0.4082
= 0.0918

Number of students whose height are greater than 72 inches
= 0.0918 × 500
= 45.9
= 46 (approximately)

(b) p(less than or equal to 64 inches)
p(x ≤ 64)
when x = 64
z = \(\frac { 64-68 }{3}\) = \(\frac { -4 }{3}\) = -1.33
p(x ≤ 64) = p(z ≤ -1.33)
p(z ≥ -1.33)
= 0.5 – 0.4082
= 0.0918

∴ Number of heights whose ate less than or equal to 64 inches =0.0918 × 500
= 45.9
= 46 (approximately)

(c) p(between 65 and 71 inches)
p(65 ≤ x ≤ 71)
when x = 65

z = \(\frac { 65-68 }{3}\) = \(\frac { -3 }{3}\) = -1
when x = 71
z = \(\frac { 71-68 }{3}\) = \(\frac { 3 }{3}\) = 1
p(65 ≤ x ≤ 71) = p(-1 < z < 1)
= p(-1 < z < 0) + p(0 < z < 1)
= p(0 < z < 1) + p(0 < z < 1)
= 2 × [p(0 < z < 1)]
= 2 × 0.3413
= 0.6826
∴ Number of students whose height between 65 and 7 inches = 0.6826 × 500
= 341.3
= 342 (approximately)

Question 9.
In a photographic process, the developing time of prints may be looked upon as a random variable having the normal distribution with a mean of 16.28 seconds and a standard deviation of 0.12 second. Find the probability that it will take less than 16.35 seconds to develop prints.
Solution:
let x be the random variable have long the normal distribution with a mean of 16.28 seconds and a standard deviation of 0.12 second

µ = 16,28 and σ = 0.12
The standard normal variate
z = \(\frac { x-µ }{σ}\) = \(\frac { x-16.28 }{0.12}\) = 1
p(less than 16.35 seconds) = p(x < 16.35)
when x = 16.35
z = \(\frac { 16.35-16.28 }{0.12}\) = \(\frac { 0.07 }{0.12}\) = 0.583
p(x< 16.35) = p(z < 0.583)
= p(—∞ < z < 0) + p(0 < z < 0.583)
= 0.5 + 0.2190
= 0.7190

Question 10.
If the heights of 500 students are normally distributed with mean 68.0 inches and standard deviation 3.0 inches, how many students have height (a) greater than 2 inches (b) less than or equal to 64 inches (c) between 65 and 71 inches.
Solution:
Let x be a normal variate with mean 400 labour days and standard deviation of 100 labour days
m = 400 and σ = 100
The construction work should be completed within 450 days.
The standard normal variate
\(\frac { x-µ }{6}\) = \(\frac { x-400 }{100}\)
personality for 1 labour day = Rs 10,000
If personality amount is = 2,00,000 than No of excess

days = \(\frac { 200000 }{10000}\) = 20
∴ x = 450 + 20 = 470
when x = 470
z = \(\frac { 470-400 }{100}\) = \(\frac { 70 }{100}\) = 0.7
= p(x ≥ 470) = p(z ≥ 0.7)
= 0.5 – 0.2580
= 0.2420

(ii) p(at most 500 days) = p(x ≤ 500 )
when x = 500

z = \(\frac { 500-400 }{100}\) = \(\frac { 100 }{100}\) = 1
p(x ≤ 500) = p(z ≤ 1)
= p(∞ < z < 0) -r- p(0 < z < 1)
= 0.5 + 0.3415
= 0.8413

Tamilnadu State Board New Syllabus Samacheer Kalvi 12th Business Maths Guide Pdf Chapter 7 Probability Distributions Ex 7.2 Text Book Back Questions and Answers, Notes.

Tamilnadu Samacheer Kalvi 12th Business Maths Solutions Chapter 7 Probability Distributions Ex 7.2

Question 1.
Define possion distribution.
Solution:
Poisson distribution was derived in 1837 by a French Mathematician Simeon D. Poisson.
A random variable X is said to follow a Poisson distribution with parameter X if it assumes only non-negative values and its probability mass function is given by

Question 2.
Write any 2 examples for possion distribution
Solution:
1. The number of alpha particles emitted by a radioactive substance in a fraction of a second.
2. Number of road accidents occurring at a particular interval of time per day.

Question 3.
Write the condition for which the possion distribution is limiting case of binomial distribution
Solution:
Poisson distribution is a limiting case of binomial distribution under the following conditions:
(i) n, the number of trials is indefinitely large i.e n → ∞
(ii) p, the constant probability of success in each trial is very small, i.e. p → 0.
(iii) np = λ is finite. Thus p = \(\frac { λ }{n}\) and q = 1 – (\(\frac { λ }{n}\))
where λ, is a positive real number.

Question 4.
Derive the mean and variance of possion distribution.
Solution:
Derivation of Mean and variance of Poisson distribution

Variance (X) = E(X²) – E(X)²

Variance (X) = E(X²) – E(X)²
= λ² + λ – (λ)²
= λ

Question 5.
Mention the properties of possion distribution.
Solution:
Poisson distribution is the only distribution in which the mean and variance are equal.

Question 6.
The mortality rate for a certain disease is 7 in 1000. What is the probability for just 2 deaths on account of this disease in a group of 400? Give e^(-2.8) = 0.06
Solution:
Since the mortality rate for a certain disease in 7 in loop
∴ p = \(\frac { 7 }{1000}\) and n = 400
The value of mean A = λp = 400 × \(\frac { 7 }{1000}\)
∴ λ = 2.8
Let x be a random variable following
distribution with p(x) = \(\frac { e^{-λ}λ^x }{x!}\)
∴ the distribution is p(x = 2) = \(\frac { e^{-2-8}(2.8)^2 }{2!}\)
\(\frac { 0.06×7.84 }{2}\)
= 0.2352

Question 7.
Mention the properties of possion distribution.
Solution:
p(defective bulbs) = \(\frac { 5 }{100}\)
n = 120
The value of mean λ = np = 120 × \(\frac { 5 }{100}\)
λ = 6
Hence, x follows possion distribution with
P(x) = \(\frac { e^{-λ}λ^x }{x!}\)
(x = 0) = \(\frac { e^{-6}(6)^0 }{0!}\) = e^-6
= 0.0025

Question 8.
A car hiring firm has two cars. The demand for cars on each day is distributed as a possion variate, with mean 1.5. Calculate the proportion of days on which
(i) Neither car is used
(ii) Some demand is refused
Solution:
In a possion distribution n=2
mean λ = 1.5
x follows poison distribution
With in p(x) = \(\frac { e^{-λ}λ^x }{x!}\)
(i) p(neither car is used) = p(x = 0)
\(\frac { e^{-1.5}(1.5)^0 }{0!}\) = e^-1.5
= 0.2231

(ii) p(some demand is refused) = p(x > 2)

= 1 – 0.2231 [1 + 1.5 + 1.125]
= 1 – 0.2231 [3.625]
= 1-0.8087
= 0.1913

Question 9.
The average number of phone calls per minute into the switch board of a company between 10.00 am and 2.30 pm is 2.5. Find the probability that during one particular minute there will be (i) no phone at all (ii) exactly 3 calls (iii) atleast 5 calls
Solution:
The average number of phone cells per minute into the switch board of a company is λ = 2.5
x follows poisson distribution with

(iii) p(atleast 5 calls) = p(x ≥ 5)
= p(x = 5) + p(x = 6) + …………..
= 1 – p(x < 5)

Question 10.
The distribution of the number of road accidents pre day in a city is possion with mean 4. Find the number of days out of 100 days when there will be (i) no accident (ii) atleast 2 accidents and (iii) at most 3 accidents.
Solution:
In a possion distribution
mean λ = 4
n = 100
x follows possion distribution with
p(x) = \(\frac { e^{-λ}λ^x }{x!}\) = \(\frac { e^{-4}(4)^x }{x!}\)
(i) p(no accident) = p(x = 0)
= \(\frac { e^{-4}(4)^0 }{0!}\) = e^-4 = 0.0183
out of 100 days there will be no accident
= n × p(x = 0)
= 100 × 0.0183 = 1.83
= 2 days (approximately)

(ii) p(atleast 2 accidents)
= p(x ≥ 2)
= p(x = 2) + p(x = 3) + p(x = 4) + …………
= 1 – p(x < 2)
= 1 – [p(x = 0) + p(x = 1)]
= 1 – [\(\frac { e^{-4}(4)^0 }{0!}\) + \(\frac { e^{-4}(4)^1 }{1!}\)]
= 1 – e^-4 [l + 4]
= 1 – 0.0183(5) = 1 – 0.0915
= 0.9085
= Out of 100 days there will be atleast 2 accidents = n × p(x ≥ 2)
= 100 × 0.9085
= 90.85
= 91 days (approximately)

(iii) p(atmost 3 accident) = p(x ≤ 3)
= p(x = 0) + p(x = 1) + p(x = 2) + p(x = 3)

out of 100 days there will be at most 3 acccident = n × p(x ≤ 3)
= 100 × 0.4331
= 43.31
= 43 days(approximately)

Question 11.
Assuming that a fatal accident in a factory during the year is 1/1200/ calculate the probability that in a factory employing 300 workers there will be atleast two fatal accidents in year, (given e^-0.25 = 0.7788 Solution:
Let p be the probability of a fatal accident in a factory during the yeart
p = \(\frac { 1 }{1200}\) and n = 300 1200
λ = np = 300 × \(\frac { 1 }{1200}\) = \(\frac { 1 }{4}\)
λ = 0.25
x follows poison distribution with
p(x) = \(\frac { e^{-λ}λ^x }{x!}\) + \(\frac { e^{-0.25}(0.25) }{x!}\)
p(atleasttwo fatal accidents) = p(x ≥ 2)
= p(x = 2) + p(x = 3) + p(x = 3) + p(x = 4) + ……….
= 1 – p(x < 2)
= 1 – {p(x = 0) + p(x = 1)}
= 1 – 0.7788 [1 + 0.25]
= 1 – 0.7788(1.25)
= 1 – 0.9735 = 0.0265
∴ p(x ≥ 2) = 0.0265

Question 12.
The average number of customers, who appear in a counter of a certain bank per minute is two. Find the probability that during a given minute (i) No customer appears (ii) three or more customers appear.
Solution:
The average number of customers ,who appear in a counter of a certain bank per minute = 2
∴ λ = 2
x follows poisson distribution with
p(x) = \(\frac { e^{-λ}λ^x }{x!}\)
x follows poisson distribution with p(x) = \(\frac { e^{-λ}λ^x }{x!}\)
(i) p(no customber appears) = p(x = 0)
= \(\frac { e^{-2}(2)^0 }{0!}\) = e^-2
= 0.1353

(ii) p(three or more customers appears) = p(x ≥ 3)
= p(x = 3) + p{x = 4) + p(x = 5) + ……
= 1 – p(x < 3)
= 1 – {p(x = 0) + p(x = 1) + p(x = 2)}

= 1 – 2^-2 [1 + 2 + 2]
= 1 – 0.1353(5)
= 1 – 0.6765
= 0.3235

Tamilnadu State Board New Syllabus Samacheer Kalvi 12th Business Maths Guide Pdf Chapter 7 Probability Distributions Ex 7.1 Text Book Back Questions and Answers, Notes.

Tamilnadu Samacheer Kalvi 12th Business Maths Solutions Chapter 7 Probability Distributions Ex 7.1

Question 1.
Define Biomial distribution.
Solution:
Binomial distribution was discovered by James Bernoulli (1654-1705) in the year 1700.
A random variable X is said to follow binomial distribution with parameter n and p, if it assumes only non-negative value and its probability mass function in given by

Question 2.
Define Bernouli trials.
Solution:
A random experiment whose outcomes are of two types namely success S and failure F, occurring with probabilities p and q respectively, is called a Bernoulli trial.
Some examples of Bernoulli trials are:
(i) Tossing of a coin (Head or tail)
(ii) Throwing of a die (getting even or odd number)

Question 3.
Derive the mean and variance of bionomial distribution
Solution:
Derivation of the Mean and Variance of Binomial distribution:
The mean of the binomial distribution

= np(q + p)n – 1 [since p + q = 1]
= np
E(X) = np
The mean of the binomial distribution is np.
Var(X) = E(X²) – E(X²)

= n(n – 1)p²(q + p)(n – 2) + np
n(n – 1 )p² + np
Variance = E(X²) – [E(X)]²
= n²p² – np² + np – n²p²
= np(1 – p) = npq
Hence, mean of the BD is np and the Variance is npq.

Question 4.
Write down the condition for which the bionomial distribution can be used.
Solution:
The Binomial distribution can be used under the following conditions:
1. The number of trials ‘n finite
2. The trials are independent of each other.
3. The probability of success ‘p’ is constant for each trial.
4. In every trial there are only two possible outcomes – success or failure.

Question 5.
Mention the properties of bionomial distribution.
Solution:
Properties of Binomial distribution
1. Binomial distribution is symmetrical if p = q = 0.5, It is skew symmetric if p ≠ q. It is positively skewed if p < 0.5 and it is negatively skewed it p > 0.5.
2. For Binomial distribution, variance is less than mean
Variance npq = (np) q < np

Question 6.
If 5% of the items produced turn out to be defective, then find out the probability that out of 10 items selected at random there are
(i) exactly three defectives
(ii) atleast two defectives
(iii) exactly 4 defectives
(iv) find the mean and variance.
Solution:
Probability of getting a defective item

In binomial distribution
p(X = x) = nC_xp^xq^n-x

Let x = (19)²
log x = 7 log(19)⁷
= 7 log 19
= 7 × 1.2788
= 8.9516
x = Antilog 8.956
= 8.945 × 10⁸
ut y = (20)¹⁰
log y = (10) log 20
= 10 × 1.3010
= 13.010
y = Anti log 13.010
= 1.023 × 10³

(ii) p(atleast two defectives)
= p(x ≥ 2)
= p(x = 2) = p(x = 3) + ………….. + p(x = 10)
= 1 – p(x < 2)
= 1 – {p(x = 0) + p(x = 1)}

Let x = (19)⁹
log x = 9 log 19
= 9 × 1.2788
= 11.5092
x = Antilog 11.5092
= 3.229 × 1011

(iii) p(extactly 4 defectives) = p(X = 4)

Let x = (19)⁶
log x = 6 log 19
= 6 × 1.2788
log x = 7.66728
Antilog (7.66728)
x = 4.648 × 10⁷

(iv) mean E(x) = np
= 10 × \(\frac { 1 }{20}\) = \(\frac { 1 }{2}\) = 0.5
Varaince = npq
= 10 × \(\frac { 1 }{20}\) = \(\frac { 19 }{20}\) = \(\frac { 19 }{40}\) = 0.475

Question 7.
In a particular university 40% of the students are having news paper reading habit. Nine university students are selected to find their views on reading habit. Find the probability that
(i) none of those selected have news paper reading habit
(ii) all those selected have news paper reading habit
(iii) atleast two third have news paper reading habit.
Solution:
let p to the probability of having newspaper reading habit
p = \(\frac { 40 }{100}\) = \(\frac { 2 }{5}\)
q = 1 – p = 1\(\frac { 2 }{5}\) = \(\frac { 5-2 }{5}\) = \(\frac { 3 }{5}\) and n = 9
In the binomial distribution p(x = 4) = nc_x p^xq^n-r
The binomial distribution P(x) = 9c_x (\(\frac { 2 }{5}\))^x (\(\frac { 3 }{5}\))^9-x

(i) p(none of those selected have newspaper reading

(ii) P(all those selected have newspaper reading habit)

(iii) p(at least two third have newspaper reading habit)

Question 8.
In a family of 3 children, what is the probability that there will be exactly 2 girls?
Solution:
let p be the probability of getting a girl child= 1/2 q = 1 – p ⇒ = 1 – 1/2
∴ q = \(\frac { 1 }{2}\) and n = 3
In a binomial distibution
p(x = x) nc_x p^xq^n-x
p(exactly 2 girls) p(x = 2)

Question 9.
Defects in yarn manufactured by a local mill can be approximated by a distribution with a mean of 1.2 defects for every 6 metres of length. If lengths of 6 metres are to be inspected, find the probability of less than 2 defects.
Solution:
Given n = 6
mean np = 1.2
⇒ 6p = 1.2
p = \(\frac { 1.2 }{6}\) p = 0.2 (or) p = \(\frac { 1 }{5}\)
q = 1 – p = 1 – \(\frac { 1 }{5}\)
∴ q = \(\frac { 4 }{5}\)
The binomial distribution
p(x) = 6C_x (0.2)^x(0.8)^6-x
p(x < 2) = p(x = 0) + p(x = 1)

Question 10.
If 18% of the bolts produced by a machine are defective, determine the probability that out of the 4 bolts chosen at random
Solution:
n = 4
probability of defective bolts p = 18/100
9 = 1 – p = 1.018 = 0.82
The binoial distribution p(x) = 4C_x(0.18)^x(0.82)^4-x
= 4C₁(0.18)¹(0.82)^4-1
= 4 × 0.18 × (0.82)³
= 0.72 × 0.551368
= 0.39698496
p(X = 1)= 0.3969 approximately

(ii) p(none will be defective)p(x = 0)
= 4C₀(0.18)°(0.82)^4-0
= (1)(1)(0.45212176)
p(x = 0) = 0.45212

(iii) p(almost 2 will be defective) = p(x ≤ 2)
= p(x = 0) + (p(x = 1) + p(x = 2)
= 4C₀(0.18)°(0.82)^4-0 + 4C₁(0.18)₁(0.82)^4-1 + 4C₂ (0.18)² (0.82)^4-2
= (0.82)⁴ + 4 × (0.18) × (0.82)³ + \(\frac { 4×3 }{1×2}\) × (0.18)² (0.82)²
= 0.45212176 + (0.72 × 0.551368) + (6 × 0.0324 × 0.6724)
= 0.45212176 + 0.39698496 + 013071456
= 0.97982128
= 0.9798

Question 11.
If 18% of the bolts produced by a machine are defective, determine the probability that out of the 4 bolts chosen at random
(i) exactly one will be defective
(ii) none will be defective
(iii) atmost 2 will be defective
Solution:
p = 0.09 = \(\frac { 9 }{100}\)
q = 1 – p ⇒ q – 1 – \(\frac { 9 }{100}\)
q = \(\frac { 100-9 }{100}\)
q = \(\frac { 91 }{100}\)
In a binomial distribution p(x = x) = nc_xp^xq^n-x
p(atleast one success) = p(x ≥ 1)

Taking log on both sides log ≤ (0.66) ≤ n log(0.91)
∴ n ≥ log \(\frac { log(0.66) }{log(0.91)}\)
n ≥ 5
∴ 5 or more trials are needed

Question 12.
Among 28 professors of a certain department, 18 drive foreign cars and 10 drive local made cars. If 5 of these professors are selected at random, what is the probability that atleast 3 of them drive foreign cars?
Solution:
Let p be the probability of foreign cars driven by the professors
p = \(\frac { 18 }{28}\) = \(\frac { 9 }{14}\)
Let q be the probability of local made cars driv¬en by the professors
q = \(\frac { 10 }{28}\) = \(\frac { 5 }{14}\)
and n = 5
The binomial distribution p(x = x) = nc_xp^xq^n-x

Question 13.
Out of 750 families with 4 children each, how many families would be expected to have (i) atleast one boy (ii) atmost 2 girls (iii) and children of both sexes? Assume equal probabilities for boys and girls?
Solution:
Assume equal probabilities for boys girls let p be the probability of having a boy Let x be the random variable for getting either a boy or a girl
∴ p = \(\frac { 1 }{2}\) and q \(\frac { 1 }{2}\) and n = 4
In binomial distribution p(x = 4) = nc_xp^xq^n-x
Here the binomial distribution is p(X= x)

= 1 – 0.0625
= 0.9375
for 750 families (p ≥ 1) = 750 × 0.9375
= 703.125
= 703(approximately)

(ii) p(almost 2 girls) = p(x ≤ 2)
= p(x = 0) = p(x = 1) + p(x = 2)

For 750 families p(x ≤ 2) = 0.6875 × 750
= 515.625
= 516 (approximately)

(iii) p(children of both sexes) = p(x = 1) + p(x = 2)p(x = 3)

= 0.875 x 750
For 750 families p(x = 2) = 0.875 × 750
= 656.25
= 656 (approximately)

Question 14.
Forty percent of business travellers carry a laptop. In a sample of 15 business travelers,
Solution:
Given n = 5
p = \(\frac { 40 }{100}\) = \(\frac { 2 }{5}\)
q = 1 – p = 1 – \(\frac { 2 }{5}\) \(\frac { 5-2 }{5}\) = \(\frac { 3 }{5}\)
The binomial distributionp (X = x) = 15c_x (\(\frac { 2 }{5}\))^x (\(\frac { 3 }{5}\))^15-x
(i) p(probability that 3 will have a laptop) = p(x = 3)

(ii) p(12 of the traels will not have a laptop)
= 1 – p(x = 12)
= 1 – 15c₁₂ (\(\frac { 2 }{5}\))¹² (\(\frac { 3 }{5}\))^15-12

(iii) p(at least three of the travelers have a laptop)
= p(x ≥ 3)
p(x ≥ 3) = 1 – p (x < 3)
= 1 – [p(x = 0) + p(x = 1) + p(x = 2)]
p(x < 3) = p(x = 0) + p(x = 1) + p(x = 2)

Question 15.
A pair 5f dice is thrown 4 times. If getting a doublet is considered a success, find the probability of 2 successes.
Solution:
In a throw of a pair dice the doublets are (1, 1),(1, 2) (3, 3),(4, 4),(5, 5),(6, 6)
probability of getting a doublet p = \(\frac { 6 }{36}\) = \(\frac { 1 }{6}\)
⇒ q = 1 – q = \(\frac { 5 }{6}\) and n = 4 is given
The probability of success = \(\left[\begin{array}{l}
4 \\
x
\end{array}\right]\) [latex]\frac { 1 }{6}[/latex]^x [latex]\frac { 5 }{6}[/latex]^4-x
Therefore the probability of 2 success are

Question 16.
The mean of a binomial distribution is 5 and standard deviation is 2. Determine the distribution.
Solution:
In a binomial distribution
mean np = 5 → (1)
Standard deviation \(\sqrt { npq}\) = 2
squaring on body sides
npq = 4 → (2)

n = 5 × 5 ⇒ n = 25
∴ the binomial distribution is
P(X = x) = nc_xp^xq^n-x
(i.e) p(X = x) = \(\left[\begin{array}{l}
25 \\
x
\end{array}\right]\) (\(\frac { 1 }{5}\))^x (\(\frac { 4 }{5}\))^(25-x)

Question 17.
Determine the binomial distribution for which the mean is 4 and variance 3. Also find P(X = 15)
Solution:
In a binomial distribution
mean np = 4 → (1)
variance npq = 3 → (2)

n = 4 × 4 ⇒ n = 16
The binomial distribution is p(X = x)nc_xp^xq^n-x

Question 18.
Assume that a drug causes a serious side effect at a rate of three patients per one hundred. What is the probability that atleast one person will have side effects in a random sample of ten patients taking the drug?
Solution:
Here n = 10

= 0.5344

Question 19.
Consider five mice from the same litter, all suffering from Vitamin A deficiency. They are fed a certain dose of carrots. The positive reaction means recovery from the disease.
Assume that the probability of recovery is 0.73. What is the probability that atleast 3 of the 5 mice recover.
Solution:
n = 5
Let probability of recovery p = 0.73
q = 1 – p = 1 – 0.73
∴ q = 0.27
The binomial distribution is
p (x = x) = nC_xp^xq^n-x
p(x = x) = 5C_x(0.73)^x(0.27)^5-x
p(atleast 3 of the 5 mice recover) = p(x ≥ 3)
= p(x = 3) + p(x = 4) + p(x = 5)
= 5C₃ (0.73)³(0.27)^5-3 + 5C₄ (0.73)⁴ (0.27)^5-4 + 5C₅ (0.73)⁵ (0.27)^5-5
5C₂ (0.73)³ (0.27)² + 5c₁ (0.73)⁴ (0.27)¹ + 5c₀(0.73)⁵(0.27)°
[\(\frac { 5×4 }{1×2}\) × 0.389017 × 0.0729] + [5 × 0.28398241 × 0.27] + (1 × 0.2073071593 × 1)
= 0.283593393 + 0.3833762535 + 0.2073071593
= 0.2836 + 0.3834 + 0.2073
= 0.8743

Question 20.
Consider five mice from the same litter, all suffering from Vitamin A deficiency. They are fed a certain dose of carrots. The positive reaction means recovery from the disease. Assume that the probability of recovery is 0.73. What is the probability that atleast 3 of the 5 mice recover.
Solution:
Success = 2 × fails
p = 2q ⇒ p = 2(1 – p)
p = 2 – 2p ⇒ p + 2p = 2
3p = 2 and p = 2/3
q = 1 – p = 1 – 2/3
q = 1/3 and n = 5
The binomial destribution is
p (X = x) = nC_xp^xq^n-x
= 5C(2/3)^x (1/3)
(i) p(three successes) = p(x = 3)

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Samacheer Kalvi 12th Business Maths Book Solutions Answers Guide

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Samacheer Kalvi 12th Business Maths Guide Chapter 1 Applications of Matrices and Determinants

• Chapter 1 Applications of Matrices and Determinants Ex 1.1
• Chapter 1 Applications of Matrices and Determinants Ex 1.2
• Chapter 1 Applications of Matrices and Determinants Ex 1.3
• Chapter 1 Applications of Matrices and Determinants Ex 1.4
• Chapter 1 Applications of Matrices and Determinants Miscellaneous Problems

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• Chapter 2 Integral Calculus I Ex 2.1
• Chapter 2 Integral Calculus I Ex 2.2
• Chapter 2 Integral Calculus I Ex 2.3
• Chapter 2 Integral Calculus I Ex 2.4
• Chapter 2 Integral Calculus I Ex 2.5
• Chapter 2 Integral Calculus I Ex 2.6
• Chapter 2 Integral Calculus I Ex 2.7
• Chapter 2 Integral Calculus I Ex 2.8
• Chapter 2 Integral Calculus I Ex 2.9
• Chapter 2 Integral Calculus I Ex 2.10
• Chapter 2 Integral Calculus I Ex 2.11
• Chapter 2 Integral Calculus I Ex 2.12
• Chapter 2 Integral Calculus I Miscellaneous Problems

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• Chapter 3 Integral Calculus II Ex 3.1
• Chapter 3 Integral Calculus II Ex 3.2
• Chapter 3 Integral Calculus II Ex 3.3
• Chapter 3 Integral Calculus II Ex 3.4
• Chapter 3 Integral Calculus II Miscellaneous Problems

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• Chapter 4 Differential Equations Ex 4.1
• Chapter 4 Differential Equations Ex 4.2
• Chapter 4 Differential Equations Ex 4.3
• Chapter 4 Differential Equations Ex 4.4
• Chapter 4 Differential Equations Ex 4.5
• Chapter 4 Differential Equations Ex 4.6
• Chapter 4 Differential Equations Miscellaneous Problems

12th Business Maths Guide Volume 1 Chapter 5 Numerical Methods

• Chapter 5 Numerical Methods Ex 5.1
• Chapter 5 Numerical Methods Ex 5.2
• Chapter 5 Numerical Methods Ex 5.3
• Chapter 5 Numerical Methods Miscellaneous Problems

Tamilnadu State Board Samacheer Kalvi 12th Business Maths Book Volume 2 Solutions

12th Business Maths Guide Volume 2 Chapter 6 Random Variable and Mathematical Expectation

• Chapter 6 Random Variable and Mathematical Expectation Ex 6.1
• Chapter 6 Random Variable and Mathematical Expectation Ex 6.2
• Chapter 6 Random Variable and Mathematical Expectation Ex 6.3
• Chapter 6 Random Variable and Mathematical Expectation Miscellaneous Problems

TN 12th Business Maths Solution Book Chapter 7 Probability Distributions

• Chapter 7 Probability Distributions Ex 7.1
• Chapter 7 Probability Distributions Ex 7.2
• Chapter 7 Probability Distributions Ex 7.3
• Chapter 7 Probability Distributions Ex 7.4
• Chapter 7 Probability Distributions Miscellaneous Problems

12th Std Business Maths Guide Pdf Chapter 8 Sampling Techniques and Statistical Inference

• Chapter 8 Sampling Techniques and Statistical Inference Ex 8.1
• Chapter 8 Sampling Techniques and Statistical Inference Ex 8.2
• Chapter 8 Sampling Techniques and Statistical Inference Ex 8.3
• Chapter 8 Sampling Techniques and Statistical Inference Miscellaneous Problems

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• Chapter 9 Applied Statistics Ex 9.1
• Chapter 9 Applied Statistics Ex 9.2
• Chapter 9 Applied Statistics Ex 9.3
• Chapter 9 Applied Statistics Ex 9.4
• Chapter 9 Applied Statistics Miscellaneous Problems

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• Chapter 10 Operations Research Ex 10.1
• Chapter 10 Operations Research Ex 10.2
• Chapter 10 Operations Research Ex 10.3
• Chapter 10 Operations Research Ex 10.4
• Chapter 10 Operations Research Miscellaneous Problems

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Tamilnadu State Board New Syllabus Samacheer Kalvi 12th Business Maths Guide Pdf Chapter 6 Random Variable and Mathematical Expectation Miscellaneous Problems Text Book Back Questions and Answers, Notes.

Tamilnadu Samacheer Kalvi 12th Business Maths Solutions Chapter 6 Random Variable and Mathematical Expectation Miscellaneous Problems

Question 1.
The probability function of a random variable X is given by

Evaluate the following probabilities.
(i) P(X ≤ 0)
(ii) P(X ≤ 0)
(iii) P( X ≤ 2) and
(iv) P(0 ≤ X ≤ 10)
Solution:
(i) From the data

x	-2	0	10
P(x = x)	1/4	1/4	1/2

(i) P(x ≤ 0) = P( x = -2) + P(x = 0)
= \(\frac { 1 }{4}\) + \(\frac { 1 }{4}\) = \(\frac { 2 }{4}\) = \(\frac { 1 }{2}\)

(ii) P(x < 0) = P(x = -2) = \(\frac { 1 }{4}\)

(iii) P(|x| ≤ 2) = P(-2 ≤ X ≤ 2)
= P(X = -2) + P(X = 0) = \(\frac { 1 }{4}\) + \(\frac { 1 }{4}\)
= \(\frac { 2 }{4}\) = \(\frac { 1 }{2}\)

(iv) p(0 ≤ X ≤ 10) = P (X = 0) + P(X = 10)
= \(\frac { 1 }{4}\) + \(\frac { 1 }{2}\) = \(\frac { 1+2 }{4}\) = \(\frac { 3 }{4}\)

Question 2.
The probability function of a random variable X is given by

(a) Compute: (i) P(1 ≤ X ≤ 2) and (ii) P(X = 3).
(b) Is X a discrete random variable? Justify your answer.
Solution:
W.K.T Probability density Function
f(x) = \(\frac { d[F(x)] }{dx}\)

(ii) P(X = 3) = 0
(b) X is not discrete since f is not a step function.

Question 3.
The p.d.f. of X is defined as
f(x) = \(\left\{\begin{array}{l}
\mathrm{k}, \text { for } 0<x \leq 4 \\
0, \text { otherwise }
\end{array}\right.\)
Solution:
Let X and a random variable if a Probability density function

Question 4.
The probability distribution function of a discrete random variable X is

where k is some constant. Find (a) k and (b) P(X > 2).
Solution:
(a) Let X be the random variable of a probability distribution function
W.K.T Σpi = 1
P(x = 1) + P(x = 3) + P(x = 5) = 1
2k + 3k + 4k = 1
9k – 1 ⇒ k = 1/9

(b) P(x > 1) = P(x = 3) + P(x = 5)
= 3k + 4k = 7k
= 7(1/9) = 7/9

Question 5.
The probability distribution function of a discrete random variable X is
f(x) = \(\left\{\begin{array}{l}
\mathrm{a}+\mathrm{b} x^{2}, 0 \leq x \leq 1 \\
0, \text { otherwise }
\end{array}\right.\)
where a and b are some constants. Find (i) a and b if E(X) = 3/5 (ii) Var(X).
Solution:
Let X be due continuous variable of density function

Question 6.
Prove that if E(X) = 0, then V(X) = E(X²)
Solution:
V(X) = E(X²) – [E(X)]²
= E(X²) – 0 {Given that E(X) = 0}
Var(X) = E(X²)
Hence proved.

Question 7.
What is the expected value of a game that works as follows: I flip a coin and, if tails pay you ₹ 2; if heads pay you ₹ 1. In either case I also pay you ₹ 50.
Solution:
Let x be the remain variable denoting the amount paying for a game of flip coin then x and takes 2 and 1
P(X = 2) = \(\frac { 1 }{2}\) (getting a head)
p(X = 1) = \(\frac { 1 }{2}\) (getting a tail)
Hence the probability of X is

x	2	1
P(x = x)	1/2	1/2

Expected value E(X) = \(\sum_{ x }\)x P(x)
= (2)(\(\frac { 1 }{2}\)) + 1 (\(\frac { 1 }{2}\))
= 1 + \(\frac { 1 }{2}\) = \(\frac { 3 }{2}\)
Since I pay you ₹ 50 in either case
E(X) = 50 × 3/2 = ₹ 75

Question 8.
Prove that, (i) V(aX) = a²V(X) , and (ii) V(X + b) = V(X).
Solution:
LHS = V(ax)
= E(ax)² – [E(ax)]²
= a² E(x²) – [aE(x)]²
= a²E(x²) – a²E(x)]²
= a²E(x²) – E(x)²]
= a²v(x)
= RHS
Hence proved

(ii) LHS = V(x + b)
= E (x + b)² – [E(x + b)]²
= E(x² + 2bx + b²) – [E(x) + b]² –
= E(x²) + 2bE(x) + b² – [E(x)]² + b² + 2bE(x)]
= E(x²) + 2bE(x) + b² -[E(x)]² + b² – 2bE(x)]
= E(x²) – [E(x)]²
= V(x)
= RHS
LHS = RHS Hence proved.

Question 9.
Consider a random variable X with p.d.f.
f(x) = \(\left\{\begin{array}{l}
3 x^{2}, \text { if } 00 \\
0, \text { otherwise }
\end{array}\right.\)
Find the expected life of this piece of equipment.
Solution:
Let X be the random variable

Question 10.
The time to failure in thousands of hours of an important piece of electronic equipment used in a manufactured DVD player has the density function
f(x) = \(\left\{\begin{array}{l}
2 e^{-2 x}, x>0 \\
0, \text { otherwise }
\end{array}\right.\)
Find the expected life of this piece of equipment.
Solution:
Let X be the random variable

Tamilnadu State Board New Syllabus Samacheer Kalvi 12th Business Maths Guide Pdf Chapter 6 Random Variable and Mathematical Expectation Ex 6.2 Text Book Back Questions and Answers, Notes.

Tamilnadu Samacheer Kalvi 12th Business Maths Solutions Chapter 6 Random Variable and Mathematical Expectation Ex 6.2

Question 1.
Find the expected value for the random variable of an unbiased die
Solution:
When a un based die is thrown , any one of the number 1, 2, 3, 4, 5, 6, can turn up that x denote the random variable taking the values from 1 to 6

Question 2.
Let X be a random variable defining number of students getting A grade. Find the expected value of X from the given table

Solution:
Let X be a random variable taking values 0, 1, 2, 3,
Expected value E(x) = Σp₁x₁
= (0.2 × 0) + (0.1 × 1) + (0.4 × 2) + (0.3 × 3)
= 0 + 0.1 + 0.8 + 0.9
∴ E(x) = 0.18

Question 3.
The following table is describing about the probability mass function of the random variable X

Find the standard deviation of x.
Solution:
Let x be the random variable taking the values 3, 4, 5
E(x) = Σpixi
= (0.1 × 3) + (0.1 × 4) + (0.2 × 5)
0.3 + 0.4 + 1.0
E(x) = 1.7
E(x²) = Σpixi²
= (0.1 × 3²)+ (0.1 × 4²) + (0.2 × 5²)
= (0.1 × 9) + (0.1 × 16) + (0.2 × 25)
E(x²) = 7.5
Var(x) = E(x²) – (E(x)]²
= 7.5 – (1.7)²
= 7.5 – 2.89
Var(x) = 4.61
Standard deviation(S.D) = \(\sqrt { var (x)}\)
= \(\sqrt { 4.61}\)
σ = 2.15

Question 4.
Let X be a continuous random variable with probability density function
f_x (x) = \(\left\{\begin{array}{l}
2 x, 0 \leq x \leq 1 \\
0, \text { otherwise }
\end{array}\right.\)
Find the expected value of X.
Solution:
Let x be a continuous random variable. In the probability density function, Expected

Question 5.
Let X be a continuous random variable with probability density function
f_x (x) = \(\left\{\begin{array}{l}
2 x, 0 \leq x \leq 1 \\
0, \text { otherwise }
\end{array}\right.\)
Find the mean and variance of X.
Solution:
Let x be a continuous random variable. In the probability density function,

Question 6.
In an investment, a man can make a profit I of ₹ 5,000 with a probability of 0.62 or a loss of ₹ 8,000 with a probability of 0.38 . Find the expected gain.
Solution:

X	5000	-8000
P(x = x)	0.62	0.38

Let x be the random variable of getting gain in an Investment
E(x) be the random variable of getting gain in an Investment
E(x) = ΣPixi
= (0.62 × 5000) + [0.38 × (-8000)]
= 3100 – 3040
E(x) = 60
∴ Expected gain = ₹ 60

Question 7.
What are the properties of mathematical expectation?
Properties of Mathematical expectation
Solution:
i. E(a) = a, where ‘a’ is a constant.
ii. E(aX) = aE(X)
iii. E(aX + b) = aE(X) + b, where ‘a’ and ‘b’ are constants.
iv. If X ≥ 0, then E(X) ≥ 0
v. V(a) = 0
vi. If X is random variable, then V(aX + b) = a²V(X)

Question 8.
What do you understand by mathematical expectation?
Solution:
The average value of a random Phenomenon is termed as mathematical expectation or expected value.
The expected value is weighted average of the values of a random variable may assume

Question 9.
How do you define variance in terms of Mathematical expectation?
Solution:
The variance of X is defined by
Var(X) = Σ[x – E(X)]² p(x)
If X is discrete random variable with probability mass function p(x).
Var(X) = \(\int_{ -∞ }^{∞}\) [x- E(X)]² f_x (x) dx
If X is continuous random variable with probability density function f_x (x).

Question 10.
Define mathematical expectation in tears of discrete random variable?
Solution:
Let X be a discrete random variable with probability mass function (p.m.f.) p(x). Then, its expected value is defined by
E(X) = \(\sum_{ x }\) x p(x) ……(1)

Question 11.
State the definition of mathematical expectation using continous random variable.
Solution:
If X is a continuous random variable and fix) is the value of its probability density function at x, the expected value of X is
E(X) = \(\int_{ -∞ }^{∞}\) x f(x) dx …….(2)

Question 12.
In a business venture a man can make a profit of ₹ 2,000 with a probability of 0.4 or have a loss of ₹ 1,000 with a probability of 0.6. What is his expected, variance and standard deviation of profit?
Solution:
Let X be the random variable of getting profit in a business

X	2000	-1000
P(x = x)	0.4	0.6

E(x) = Σ_xxp_x(x)
= (0.4 × 2000) +[0.6 × (-1000)]
= 800 – 600
E(X) = 200
∴ Expected value of profit = ₹ 200
E(X²) = Σx² P_x(x)
= [(2000)² × 0.4] + [(-1000)² × 0.6]
= (4000000 × 0.4) + (1000000 × 0.6)
E(X²) = 2200000
Var(X) = E(X²) – [E(X)]²
= 22000000 – (200)²
= 2200000 – 40000
Var(X) = 21,60,000
Variance of his profit = ₹ 21,60,000
Standard deviation(S.D) = \(\sqrt { var (x)}\)
σ = \(\sqrt { 2160000}\)
σ = 1469.69
Standard deviation of his profit is ₹ 1,469.69

Question 13.
The number of miles an automobile tire lasts before it reaches a critical point in tread wear can be represented by a p.d.f.

Find the expected number of miles (in thousands) a tire would last until it reaches the critical tread wear point.
Solution:
We know that,

Question 14.
A person tosses a coin and is to receive ₹ 4 for a head and is to pay ₹ 2 for a tail. Find the expectation and variance of his gains.
Solution:
Let X denote the amount the person receives in a game
Then X takes values 4,-2 and
So P(X = 4) = P (of getting a head)
= \(\frac { 1 }{2}\)
P(X = – 2) = P (of getting a tail)
= \(\frac { 1 }{2}\)
Hence the Probability distribution is

X	4	-2
P(x = x)	1/2	1/2

E(x²) = 10
Var(x) = E(x²) – E(x²)
= 10 – (1)²
Var(x) = 9
∴ His expected gain = ₹ 1
His variance of gain = ₹ 9

Question 15.
Let X be a random variable and Y = 2X + 1. What is the variance of Y if variance of X is 5?
Solution:
Var(X) = 5
Var(Y) = var (2x + 1) {∴ v(ax + b) = a²v(x)}
= (2)² var(X)
= 4(5)
Var() = 20

Tamilnadu State Board New Syllabus Samacheer Kalvi 12th Business Maths Guide Pdf Chapter 6 Random Variable and Mathematical Expectation Ex 6.1 Text Book Back Questions and Answers, Notes.

Tamilnadu Samacheer Kalvi 12th Business Maths Solutions Chapter 6 Random Variable and Mathematical Expectation Ex 6.1

Question 1.
Construct cumulative distribution function for the given probability distribution.

Solution:
F(0) = P(x ≤ 0) = p(0) = 0.3
F(1) = P(x ≤ 1) = p(0) + p(1)
= 0.3 + 0.2 = 0.5
F(2) = P(x ≤ 2) = P(0) + P(1) + P(2)
= 0.3 + 0.2 + 0.4 = 0.9
F(3) = P(x ≤ 3) = P(0) + P(2) + P(3) + P(4)
= 0.3 + 0.2 + 0.4 + 0.1 = 1

Question 2.
Let X be a discrete random variable with the following p.m.f

Find and plot the c.d.f. of X.
Solution:
F(3) = P(x ≤ 3) = P(3) = 0.3
F(5) = P(x ≤ 5) = P(x = 3) + (x = 5)
= 0.3 + 0.2 = 0.5
F(8) = P(x ≤ 8) = P(3) + P(5) + P(8)
= 0.3 + 0.2 + 0.3
= 0.8
F(10) = P(x ≤ 10)
= P(3) + P(5) + P(8) + P(10)
= 0.3 + 0.2 + 0.3 + 0.2
= 1

Question 3.
The discrete random variable X has the following probability function.

where k is a constant. Show that k = \(\frac { 1 }{18}\)
Solution:
From the data
P(x = 2) = kx
= 2k
P(x = 4) = kx
= 4k
P(x = 6) = kx
= 6k
P(x = 8) = k(x – 2)
= k(8 – 2) = 6k
Since P(X = x) is a probability mass function
\(\sum_{x=2}^{8}\) P(X = x) = 1
\(\sum_{i=2}^{∞}\) P(xi) = 1
(i.e) P(x = 2) + P(x = 4) + P(x = 6) + P(x = 8) = 1
2k + 4k + 6k + 6k = 1
18k = 1
∴ k = \(\frac { 1 }{18}\)

Question 4.
The discrete random variable X has the probability function

Solution:
From the data
P(x = 1) = k
p(x = 2) = 2k
p(x = 3) = 3k
P(x = 4) = 4k
Since P(X = x) is a probability Mass function
\(\sum_{x=1}^{4}\) P(X = x) = 1
\(\sum_{i=1}^{∞}\) P(x_i) = 1
p(x = 1) + P(x = 2) + P(x = 3) + P(x = 4) = 1
P(x = 1) + P(x = = 2) + P(x = 3) + P(x = 4) = 1
k + 2k + 3k + 4k = 1
10k = 1
k = \(\frac { 1 }{10}\)
∴ k = 0.1

Question 5.
Two coins are tossed simultaneously. Getting a head is termed as success. Find the t probability distribution of the number of successes.
Solution:
Let X is the random variable which counts the Number of Heads when the coins are tossed the outcomes are stated below

Question 6.
The discrete random variable X has the probability function.

(i) Find k
(ii) Evaluate p( x < 6), p(x ≥ 6)and p(0 < x < 5) (iii) If P(X ≤ x) > 1, 2 then find the minimum value of x.
Solution:
(i) Since the condition of Probability Mass function
\(\sum_{i=2}^{∞}\) P(x_i) = 1
\(\sum_{i=0}^{7}\) P(x_i) = 1
P(x = 0) + P(x = 1) + P(x = 2) + P(x = 3) + P(x = 4) + P P(x = 6) + p(x = 7) = 1
0 + k + 2k + 2k + 3k + k² + 2k² + 7k² + k = 1
10k² + 9k – 1 = 0
10k² + 10k – k – 1 = 0
10k(k + 1) -1(k + 1) = 0
(k + 1)(10k – 1) = 0
k + 1 = 0; 10k – 1 = 0
k = -1 10k = 1
k = \(\frac { 1 }{10}\)
k = -1 is not possible

(ii) P(x < 6)
P(x < 6) = P(x = 0) + P(x = 1) + P(x = 2)+ P(x = 3) + p(x = 4) + P(x = 5)
= 0 + k + 2k +2k + 3k + k²
= 8k + k²

P(x ≥ 6) = P(x = 6) + p(x = 7)
= 2k² + 7k² + k
= 9k² + k

P(0 < x < 5)
= P(x = 1) + P(x = 2) + p(x = 3) + P(x = 4)
= k + 2k + 2k + 3k
= 8k = 8(\(\frac { 1 }{10}\))
∴ P(0 < x < 5) = \(\frac { 8 }{10}\)

(iii) We want the minimum value of x for which P(X ≥ x) > \(\frac { 1 }{2}\)
Now P(X ≤ 0) = 0 < \(\frac { 1 }{2}\)
P(X ≤ 1) = P(x = 0) + P(X = 1) = 0 + k = k
\(\frac { 1 }{10}\) < \(\frac { 1 }{2}\)
P( X ≤ 2) = P(X = 0) + P(X = 1) + P(X = 2)
= 0 + k + 2k = 3k
\(\frac { 3 }{10}\) \(\frac { 1 }{2}\)
P(X ≤ 3) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 2) + P(X = 3)
= 0 + k + 2k + 2k = 5k
= \(\frac { 5 }{10}\) = \(\frac { 1 }{2}\)
P(X ≤ 4) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 2) + P(X = 3) + P(X = 4)
= 0 + k + 2k + 2k + 3k = 8k
= \(\frac { 8 }{10}\) > \(\frac { 1 }{2}\)
This Shows that the minimum value of X for which P(X ≤ x) > \(\frac { 1 }{2}\) is 4

Question 7.
The distribution of a continuous random variable X in range (-3, 3) is given by p.d.f.

Verify that the area under the curve is unity.
Solution:
Since(-3, 3) in the range of given p.d.f
Area under the curve A = \(\int_{-∞}^{∞}\) f(x)dx = \(\int_{-3}^{3}\) f(x)dx

Hence the Area under the curve is unity

Question 8.
A continuous random variable X has the following distribution function:

Find (i) k and (ii) the probability density function.
Solution:
(i) Here F(3) – F(1) = 1
K(3 – 1)⁴ – O = 1
K(2)⁴ = 1
K(16) = 1
k = \(\frac { 1 }{16}\)

(ii) The Probability density function
f(x) = \(\frac { d(F(x)) }{dx}\) = \(\left\{\begin{array}{l}
4 k(x-1)^{3}, 1<x \leq 3 \\
0 \text { else where }
\end{array}\right.\)

Question 9.
The length of time (in minutes) that a certain person speaks on the telephone is found to be random phenomenon, with a probability function specified by the probability density function f(x) as

(a) Find the value of A that makes f (x) a p.d.f.
(b) What is the probability that the number of minutes that person will talk over the phone is (i) more than 10 minutes, (ii) less than 5 minutes and (iii) between 5 and 10 minutes.
Solution:
(a) Since f(x) is a probability density Function

b (i) more than 10 minutes

Question 10.
Suppose that the time in minutes that a person has to wait at a certain station for a train is found to be a random phenomenon with a probability function specified by the distribution function

(a) Is the distribution function continuous? If so, give its probability density function?
(b) What is the probability that a person will have to wait (i) more than 3 minutes, (ii) less than 3 minutes and (iii) between 1 and 3 minutes?
Solution:
(i) Yes, the distribution function is continuous on [0, 4]
The probability density function

(b) The probability that a person will have to wait
(i) more than 3 minutes

(iii) between 1 and 3 minutes

Question 11.
Define random variable.
Solution:
A random variable (r.v.) is a real valued function defined on a sample space S and taking values in (-∞, ∞) or whose possible values are numerical outcomes of a random experiment.

Question 12.
Explain What are the types of random variable.
Solution:
Random variables are classified into two types namely discrete and continuous random variables these are important for practical applications in the field of Mathematics and Statistics.

Question 13.
Define dicrete random Variable
Solution:
A variable which can assume finite number of possible values or an infinite sequence of countable real numbers is called a discrete random variable.

Question 14.
What do you understand by continuous random variable?
Solution:
A random variable X which can take on any value (integral as well as fraction) in the interval is called continuous random variable.

Question 15.
Describe what is meant by a random variable
Solution:
If S is a sample space with a probability measure and x is a real valued function defined over the elements of S then x is called a random variable A random variable is also called a change variable.

Question 16.
Distinguish between discrete and continuous random variable
Solution:

Question 17.
Explain the distribution function of random variable
Solution:
The discrete cumulative distribution function or distribution function of a real valued discrete random variable X takes the countable number of points x₁, x₂, …. with corresponding probabilities p(x₁), p(x₂),… and then the cumulative distribution function is defined by
F_x(x) = P(X ≤ x), for all x ∈ R
i.e. F_x (x) = \(\sum_{x \leq x}\) p(x_i)

Question 18.
Explain the terms (i) probability Mass function (ii) probability density function and (iii) probability
Solution:
(i) If X is a discrete random variable with distinct values x₁, x₂, …. x_n, …, then the function, denoted by P_x(x) and defined by

This is defined to be the probability mass function or discrete probability function of X.

(ii) The probability that a random variable X takes a value in the interval [t₁, t₂] (open or closed) is given by the integral of a function called the probability density function f_x(x):
P(t₁ ≤ X ≤ t₂) = \(\int_{t_{1}}^{t_{2}}\) f_x(x)dx.

(iii)The probability distribution of a random variable X is defined only when we have the various values oft the various values of the random variable e.g. x₁, x₂ …… x_n togather with respective probabilities p₁, p₂, p₃ …… p₄ satisfying
\(\sum_{i=1}^{n}\) Pi = 1

Question 19.
What are the properties of (i) discrete random variable and (ii) Continuous random variable
Solution:
(i) The probability mass function p(x) must satisfy the following conditions
(i) p(x_i) ≥ 0 ∀ i,
(ii) \(\sum_{i=1}^{∞}\) p(x_i) = 1

(ii) The probability density functions/)(x) or simply ; by/(x) must satisfy the following conditions.
(i) f(x) ≥ 0 ∀ x and
(ii) \(\int_{-∞}^{∞}\) f(x) dx = 1

Question 20.
State the Properties of distribution function.
Solution:
The function F_x(x) or simply F(x) has the following properties.
(i) 0 ≤ F(x) ≤ 1, -∞ < x < ∞
(ii) F(-∞) = \(\lim _{x \rightarrow-\infty}\) F(x) = 0 and F(+∞) = \(\lim _{x \rightarrow ∞}\) F(x) = 1.
(iii) F(.) is a monotone, non-decreasing function; ; that is, F(a) < F(b) for a < b.
(iv)F(.) is continuous from the right; that is, \(\lim _{h \rightarrow 0}\) F(x + h) = F(x).
(v) F(x) = \(\frac { d }{dx}\) F(x) = f(x) ≥ 0;
(vi) F'(x) = \(\frac { d }{dx}\) F(x) = f(x) ⇒ dF(x) = f(x)dx
dF(x) is known as probability differential of X.
(vii) P(a ≤ x ≤ b) = \(\int_{a}^{b}\) f(x)dx = \(\int_{-∞}^{b}\) f(x)dx – \(\int_{-∞}^{a}\) f(x)dx
= P(X ≤ b) – P(X ≤ a)
= F(b) – F(a)

Tamilnadu State Board New Syllabus Samacheer Kalvi 12th Physics Guide Pdf Chapter 11 Recent Developments in Physics Text Book Back Questions and Answers, Notes.

Tamilnadu Samacheer Kalvi 12th Physics Solutions Chapter 11 Recent Developments in Physics

12th Physics Guide Recent Developments in Physics Text Book Back Questions and Answers

Part – I

TextBook Evaluation:

I. Multiple Choice Questions:

Question 1.
The particle size of ZnO material is 30 nm. Based on the dimension it is classified as
a) Bulk material
b) Nanomaterial
c) Soft material
d) Magnetic material
Answer:
b) Nanomaterial

Question 2.
Which one of the following is the natural nanomaterial.
a) Peacock feather
b) Peacock beak
c) Grain of sand
d) Skin of the Whale
Answer:
a) Peacock feather

Question 3.
The blue print for making ultra-durable synthetic material is mimicked from ________.
a) Lotus leaf
b) Morpho butterfly
c) Parrot fish
d) Peacock feather
Answer:
c) Parrot fish

Question 4.
The method of making nanomaterial by assembling the atoms is called ______.
a) Top down approach
b) Bottom up approach
c) Cross down approach
d) Diagonal approach
Answer:
b) Bottom up approach

Question 5.
“Sky wax” is an application of nano product in the field of ________.
a) Medicine
b) Textile
c) Sports
d) Automotive industry
Answer:
c) Sports

Question 6.
The materials used in Robotics are ________.
a) Aluminium and silver
b) Silver and gold
c) Copper and gold
d) Steel and aluminium
Answer:
d) Steel and aluminium

Question 7.
The alloys used for muscle wires in Robots are _______.
a) Shape memory alloys
b) Gold copper alloys
c) Gold silver alloys
d) Two-dimensional alloys
Answer:
a) Shape memory alloys

Question 8.
The technology used for stopping the brain from processing pain is ______.
a) Precision medicine
b) Wireless brain sensor
c) Virtual reality
d) Radiology
Answer:
c) Virtual reality

Question 9.
The particle which gives mass to protons and neutrons are _______.
a) Higgs particle
b) Einstein particle
c) Nanoparticle
d) Bulk particle
Answer:
a) Higgs particle

Question 10.
The gravitational waves were theoretically proposed by _______.
a) Conrad Rontgen
b) Marie Curie
c) Albert Einstein
d) Edward Purcell
Answer:
c) Albert Einstein

II. Short Answers Questions:

Question 1.
Distinguish between Nanoscience and Nanotechnology.
Answer:

Nanoscience	Nanotechnology
1. Nanoscience is the science of objects with a typical size of 1-100 nm. Nano means one-billionth of a metre that is 10-9m.	Nanotechnology is a technology involving design, production.
2. If matter is divided into such small objects the mechanical, electrical, optical magnetic and other properties change.	Characterization and application of nano structured materials.

Question 2.
What is the difference between Nano materials and Bulk materials?
Answer:

Nano Materials	Bulk Mateerials
1. Nano materials are particle that have their size in 1-100 nm range atleast in one dimension.	Bulk materials are particle that have their size above lOOnm in all dimensions.
2. We cannot see particles of nanomaterials from the naked eye.	We can see particle of most of the bulk materials from the naked eye.
3. The example of nanomaterials include nano zymes, titanium dioxide, nano particles, graphene, etc.	The example of bulk materials include plaster sand, gravel, cement, ore, slag, salts, etc.

Question 3.
Give any two examples for “Nano” in nature.
Answer:
1.  Single-strand DNA:
A single strand of DNA, the building block of all living things, is about three nanometers wide.

2. Morpho Butterfly:
The scales on the wings of a morpho butterfly contain nanostructures that change the way light waves interact with each other, giving the wings brilliant metallic blue and green hues. Mimic in laboratories – Manipulation of colours by adjusting the size of nanoparticles with which the materials are made.

Question 4.
Mention any two advantages and disadvantages of Robotics.
Answer:
Advantage:

1. The robots are much cheaper than humans.
2. Robots never get tired like humans.
3. Stronger and faster than humans.

Disadvantage:

1. Robots have no sense of emotions or conscience.
2. They lack empathy and hence create an emotionless workplace.
3. Unemployment problem will increase.

Question 5.
Why steel is preferred in making Robots?
Answer:
Steel is several times stronger. In any case, because of the inherent strength of metal, robot bodies are made using a sheet, bar, rod, channel, and other shapes.

Question 6.
What are black holes?
Answer:

1. Black holes are the end-stage of stars which are the highly dense massive objects. Its mass ranges from 20 times mass of the sun to 1 million times mass of the sun.
2. It has very strong gravitational force such that no particle or even light can escape from it.
3. The existence of black holes is studied when the stars orbiting the black hole behave differently from the other stars. Every galaxy has a black hole at its center.

Question 7.
What are sub atomic particles?
Answer:

1. The three main subatomic particles that form an atom are protons, neutrons, and electrons.
2. Subatomic particles are particles that are smaller than the atom, proton and neutron are made up of quarks which is interact through gluons.
3. A subatomic particle having two types of particles, they are elementary particles and composite particles.

III. Long Answer Questions:

Question 1.
Discuss the applications of Nanomaterials in various fields.
Answer:

(i) Automotive industry:

• Lightweight construction
• Painting (fillers, base coat, clear coat)
• Catalysts
• Tires (fillers)
• Sensors
• Coatings for window screen and car bodies

(ii) Chemical industry:

• Fillers for paint systems
• Coating systems based on nanocomposites
• Impregnation of papers
• Switchable adhesives
• Magnetic fluids

(iii) Engineering

• Wear protection for tools and machines (anti blocking coatings, scratch resistant coatings on plastic parts, etc.)
• Lubricant – free bearings

(iv) Electronic industry

• Data memory
• Displays
• Laser diodes
• Glass fibres
• Optical switches
• Filters (IR-blocking)
• Conductive, antistatic coatings

(v) Construction:

• Construction materials
• Thermal insulation
• Flame retardants
• Surface – functionalised building materials for wood, floors, stone, facades, tiles, roof tiles, etc.
• Facade coatings
• Groove mortar

(vi) Medicine:

• Drug delivery systems
• Contrast medium
• Prostheses and implants
• Agents in cancer therapy
• Active agents
• Medical rapid tests
• Antimicrobial agents and coatings

(vii) Textile / fabrics / non – wovens:

• Surface – processed textiles
• Smart clothes

(viii) Energy:

• Fuel cells
• Solar cells
• Batteries
• Capacitors

(ix) Cosmetics:

• Sun protection
• Lipsticks
• Skin creams
• Tooth paste

(x) Food and drinks:

• Package materials
• Additives
• Storage life sensors
• Clarification of fruit juices

(xi) Household:

• Ceramic coatings for irons
• Odors catalyst
• Cleaner for glass, ceramic, floor, windows

(xii) Sports / outdoor:

• Ski wax
• Antifogging of glasses/goggles
• Antifouling coatings for ships/boats
• Reinforced tennis rackets and balls.

Question 2.
What are the possible harmful effects of the usage of Nanoparticles? Why?
Answer:

1. The research on the harmful impact of the application of nanotechnology is also equally important and fast developing.
2. The major concern here is that the nanoparticles have the dimensions same as biological molecules such as protein.
3. They may easily get absorbed onto the surface of living organisms and they might enter the tissues and fluids of the body.
4. The interaction with living systems is also affected by the dimensions of the nanoparticles.
5. For instance, nanoparticles of a few nanometers size may reach well inside biomolecules.
6. It is also possible for the inhaled nanoparticles to reach the blood to reach other sites such as the liver, heart, or blood cells.
7. Researchers are trying to understand the response of living organisms to the presence of nanoparticles of varying size, shape, chemical composition, and surface characteristics.

Question 3.
Discuss the functions of key components in Robots?
Answer:

Power conversion unit:
Robots are powered by batteries, solar power, and hydraulics.

Actuators:

1. Converts energy into movement.
2. The majority of the actuators produce rotational (or) linear motion.

Most robots are composed of 3 main parts.

The Controller:

1. This is also known as the “brain” which is run by a computer program.
2. It gives commands for the moving parts to perform the job.

Most robots are composed of 3 main parts.

Mechanical parts:

1. Motors
2. Pistons
3. Grippers
4. Wheels and gears that make the robot move, grab, turn and lift.

Sensors:

1. To tell the robot about its surroundings.
2. It helps to determine the sizes and shapes of the objects around, the distance between the objects, and the directions as well.

Question 4.
Elaborate any two types of Robots with relevant examples.
Answer:
(i) Human-Robot: Certain robots are made to resemble humans in appearance and replicate human activities like walking, lifting, and sensing, etc.

1. Power conversion unit: Robots are powered by batteries, solar power, and hydraulics.
2. Actuators: Converts energy into movement. The majority of the actuators produce rotational or linear motion.
3. Electric motors: They are used to actuate the parts of the robots like wheels, arms, fingers,
   legs, sensors, camera, weapon systems etc. Different types of electric motors are used. The most often used ones are AC motor, Brushed DC motor, Brushless DC motor, Geared DC motor, etc.
4. Pneumatic Air Muscles: They are devices that can contract and expand when air is pumped inside. It can replicate the function of a human muscle. The contract almost 40% when the air is sucked inside them.
5. Muscle wires: They are thin strands of wire made of shape memory alloys. They can contract by 5% when an electric current is passed through them.
6. Piezo Motors and Ultrasonic Motors: Basically, we use it for industrial robots.
7. Sensors: Generally used in task environments as it provides information of real-time knowledge.
8. Robot locomotion: Provides the types of movements to a robot.
   The different types are:
   • Legged
   • Wheeled
   • Combination of Legged and Wheeled Locomotion
   • Tracked slip/skid.

(ii) Industrial Robots:
Six main types of industrial robots:

1. Cartesian
2. SCARA (Selective Compliance Assembly Robot Arm)
3. Cylindrical
4. Delta
5.  Polar
6. Vertically articulated

Six-axis robots are ideal for:

1. Arc Welding
2. Spot Welding
3. Material Handling
4. Machine Tending
5. Other Applications

Question 5.
Comment on the recent advancement in medical diagnosis and therapy.
Answer:
The recent advancement in medical diagnosis and therapy:

1. Virtual reality
2. Precision medicine
3. Health wearables
4. Artificial organs
5.  3 – D printing
6. Wireless brain sensors
7. Robotic surgery
8. Smart inhalers

1. Virtual reality:
Medical virtual reality is effectively used to stop the brain from processing pain and cure soreness in the hospitalized patients. Virtual reality has enhanced surgeries by the use of 3D models by surgeons to plan operations. It helps in the treatment of Autism, Memory loss, and Mental illness.

2. Precision medicine:
Precision medicine is an emerging approach for disease treatment and prevention that takes into account individual variability in genes, environment, and lifestyle for each person. In this medical model, it is possible to customise healthcare, with medical decisions, treatments, practices, or products which are tailored to the individual patient.

3. Health wearables:
A health wearable is a device used for tracking a wearer’s vital signs or health and fitness-related data, location, etc. Medical wearables with artificial intelligence and big data provide an added value to healthcare with a focus on diagnosis, treatment, patient monitoring, and prevention.

Note Big Data: Extremely large data sets that may be analysed computationally to reveal patterns, trends, and associations, especially relating to human behavior and interactions.

4. Artificial organs:
An artificial organ is an engineered device or tissue that is implanted or integrated into a human. It is possible to interface it with living tissue or to replace a natural organ. It duplicates or augments a specific function or functions of human organs so that the patient may return to normal life as soon as possible.

5. 3D printing:
Advanced 3D printer systems and materials assist physicians in a range of operations in the medical field from audiology, dentistry, orthopedics, and other applications.

6. Wireless brain sensors:
Wireless brain sensors monitor intracranial pressure and temperature and then are
absorbed by the body. Hence there is no need for surgery to remove these devices.

7. Robotic surgery:
Robotic surgery is a type of surgical procedure that is done using robotic systems. Robotically-assisted surgery helps to overcome the limitations of pre – existing minimally invasive surgical procedures and to enhance the capabilities of surgeons performing open
surgery.

8. Smart inhalers:
Inhalers are the main treatment option for asthma. Smart inhalers are designed with health systems and patients in mind so that they can offer maximum benefit. Smart inhalers use Bluetooth technology to detect inhaler use, remind patients when to take their medication, and gather data to help guide care.

Part – II:

12th Physics Guide Recent Developments in Physics Additional Questions and Answers

I. Match the following:

Question 1.

I	II
1. Molecule	a. Located in the center of an atom
2. Nucleon	b. Electrons revolving around the atomic nucleus
3. Atom	c. Protons & Neutrons comprising the nucleus of an atom
4. Nucleus	d. Composed of two (or) more

Answer:

1. d
2. c
3. b
4. a

Question 2.

I	II
1. Geroge Devol	a. Audiology
2. Nano	b. No need surgery
3. Wireless brain sensors	c. One billionth of a meter (109 m)
4. 3D Printing	d. Unimate

Answer:

1. d
2. c
3. b
4. a

Question 3.

I	II
1. Higgs Particle	a. Slavic word
2. Albert Einstein	b. Mass of particle
3. Black Holes	c.Gravitational waves
4. Robot	d. End stage of stars

Answer:

1. b
2. c
3. d
4. a

Question 4.

I	II
1. Stephen Hawking	a. Rossum universal robots
2. Einstein’s theory	b. Twin mars rovers
3. Karl Capek	c. Field of black holes
4. Outer space	d. General relativity

Answer:

1. c
2. d
3. a
4. b

Question 5.

I	II
1. Inhalers	a. Aluminium and steel
2. 3D Printing	b. Mars pathfinder mission
3. Make robots	c. Dentistry
4. Outer Space	d. Asthma

Answer:

1. d
2. c
3. a
4. b

II. Fill in the blanks:

Question 1.
_____ existed in nature long before scientists began studying them in laboratories.
Answer:
Nanoscale structures

Question 2.
_____ and _____ is the interdisciplinary area covering its applications in various fields.
Answer:
Nanoscience, technology

Question 3.
Chinese scientists have created the world’s first autonomous DNA robots to combat _______.
Answer:
cancer tumours

Question 4.
________ is the fundamental entity of matter.
Answer:
Atom

Question 5.
There are two ways of preparing the nanomaterials _____ and _____ approaches.
Answer:
top-down, bottom-up

III. Choose the odd man out:

Question 1.
The robotic system mainly consists of ________.
a) sensors
b) power supplies
c) blood cells
d) control system
Answer:
c) blood cells

Question 2.
The key components of a robot are _______.
a)Power conversion unit
b) Actuators
c) Electric motors
d) Muscle wires
e) Delta
f) Sensors
Answer:
e) Delta

Question 3.
Six main types of industrial robots are _______.
a) Cartesian
b) SCARA
c) Cylindrical
d) Delta
e) Polar
f) Ultrasonic motors
Answer:
f) Ultrasonic motors

Question 4.
The recent advancement in medical technology includes ________.
a) Artificial organs
b) Precision medicine
c) Virtual reality
d) Pool cleaning
e) 3D printing etc
Answer:
d) Pool cleaning

Question 5.
Household robots are used as _______.
a) floor cleaners
b) gutter cleaners
c) Pool cleaning
d) Investigation of the rocks
e) Lawn mowing
Answer:
d) Investigation of the rocks

IV. Find the correct pair:

Question 1.
a) SCARA – Spot welding
b) George Devol and – First robot Joseph Engalberger company
c) Found rocks and soils – Pool cleaning
d) Size less than 100 nm – Bulk solid
Answer:
b) George Devol and Joseph Engalberger – First robot company

Question 2.
a) Top-down approach – Plasma etching
b) Chemical Industry – Glass Fibres
c) Medicine – Active agents
d) Human-Robot – Machine Tending
Answer:
c) Medicine – Active agents

V. Find the incorrect pair:

Question 1.
a) 3D Printing – Audiology
b) Karel Capek – Rossum universal robots
c) Inhalers – Asthma
d) Higgs particle – End stage of stars
Answer:
d) Higgs particle – End stage of stars

Question 2.
a) Higgs particles – God particles
b) Stephen Hawking – Field of Black holes
c) Molecule – Fundamental entity of matter
d) George Devol – Unimate
Answer:
c) Molecule – Fundamental entity of matter

VI. Choose the incorrect statement:

Question 1.
Statement 1: Nanotechnology is a technology involving the design, production, characterization, and applications of nanostructured materials.
Statement 2: Nanoparticles of a few micrometers size may reach well inside biomolecules, which is not possible for larger nanoparticles.
a) statement 1
b) statement 2
c) statement 1 and 2
d) None of these
Answer:
b) Statement 2:
Correct Sentence:
Nanoparticles of a few nanometer-size may reach well inside biomolecules, which is not possible for larger nanoparticles.

Question 2.
a) Statement 1: Five major fields of robotics are human-robot interface, mobility, manipulation, programming, and sensors.
b) Statement 2: Aluminum is a softer metal and is, therefore, easier to work with.
c) Statement 3: Industrial robots are used for exploring stars, planets, etc.
d) Statement 4: The robotic system mainly consists of sensors, power supplies, control systems, manipulators and
necessary software.
Answer:
c) Statement 3
Correct Sentence:
Industrial robots are used for welding, cutting, robotic water jet cutting, lifting, etc.

Question 3.
a) Statement 1: Accelerated mass emits gravitational force which is very week.
b) Statement 2: Black holes are the strongest source of gravitational waves.
c) Statement 3: Cosmology is the branch that involves the origin and evolution of the universe.
Answer:
a) Statement 1:
Correct Sentence:
Accelerated mass emits gravitational wave which is very week.

VII. Choose the correct statement:

Question 1.
a) Statement 1: Human is a mechanical device.
b) Statement 2: Nanoparticles can also cross cell membranes.
c) Statement 3: Top up and Bottom down are the two ways to preparing the nanomaterials.
d) Statement 4: It is possible to deliver a drug directly to a specific cell in the body by designing the surface of bulk particles.
Answer:
b) Statement 2: Nanoparticles can also cross cell membranes.

Question 2.
a) Statement 1: Manipulation of colours is found in laboratories by Morpho butterfly in nature.
b) Statement 2: Similar nanostructures are made in a lab to glow in different colors from peacock feathers in nature.
c) Statement 3: Water repellent nano paints are made from a lotus leaf surface idea.
d) Statement 4: All statements are correct.
Answer:
d) Statement 4: All statements are correct

Question 3.
Nanomaterial-based products in different fields.
a) Statement 1: Lightweight construction in medicine.
b) Statement 2: Displays nanomaterial application in the engineering field.
c) Statement 3: Antimicrobial agents and coating in the medical field.
d) Statement 4: None of these
Answer:
c) Statement 3: Antimicrobial agents and coating in the medical field.

VIII. Assertion and Reason:

a) Assertion is correct, the reason is correct; the reason is a correct explanation for the assertion.
b) Assertion is correct, the reason is correct; the reason is not a correct explanation for the assertion.
c) Assertion is correct, reason is incorrect.
d) Assertion is incorrect, reason is correct.

Question 1.
Assertion (A):
Nanoparticles can also cross cell membranes.
Reason (R):
It is also possible for the inhaled nanoparticles to reach the blood, other sites as the liver, heart.
Answer:
a) Assertion is correct, the reason is correct; the reason is a correct explanation for the assertion.

Question 2.
Assertion (A):
A drug delivery system is a medical application of a nano-based product.
Reason (R):
The mechanical parts of Robotics are motors, pistons, grippers, wheels, and gears.
Answer:
b) Assertion is correct, the reason is correct; the reason is not a correct explanation for the assertion.

Question 3.
Assertion (A):
Human robots replicate human activities like walking, lifting, and sensing, etc.
Reason (R):
Electric motors are not used in robots.
Answer:
c) Assertion is correct, reason is incorrect.

Question 4.
Assertion (A):
Face recognition is a natural intelligence of robots.
Reason (R):
Robot can translate words from one language to another.
Answer:
d) Assertion is incorrect, reason is correct.

IX. Choose the correct answer:

Question 1.
An automatic apparatus or device that performs functions ordinarily ascribed to human or operate with what appears to be almost human intelligence is called ……………..
(a) Robot
(b) Human
(c) Animals
(d) Reptiles.
Answer:
(a) Robot.

Question 2.
If the particle of a solid is of size less than 100 nm, it is said to be a .
a) nano solid
b) bulk solid
c) nano & bulk solids
d) None of these
Answer:
a) nano solid

Question 3.
The basic components of the robot are ……………..
(a) mechanical linkage
(b) sensors and controllers
(c) user interface and power conversion unit
(d) All the above.
Answer:
(d) All the above.

Question 4.
Nano means ________.
a) 10^-33 m
b) 10^-6 m
c) 10^-9 m
d) 10^-12 m
Answer:
c) 10^-9 m

Question 5.
_______ and _______ are the two important phenomena that govern nano properties.
a) Quantum confinement effects
b) Surface effects
c) Quantum confinement (or) Surface light
d) Quantum confinement effect and Surface confinement effects
Answer:
d) Quantum confinement effect and Surface confinement effects

Question 6.
Which of the following atoms do not move from each other ……………..
(a) Shape memory alloys
(b) Nanomaterials
(c) Dielectrics
(d) Static materials.
Answer:
(b) Nanomaterials

Question 7.
A health wearable is a device used for ________.
a) health and fitness-related data, location, etc.
b) to replace a natural organ
c) (a) and (b)
d) (a) or (b)
Answer:
a) health and fitness-related data, location, etc.

Question 8.
Atom is made up of ________.
a) electrons
b) protons
c) neutrons
d) above all
Answer:
d) above all

Question 9.
For nanometers whose diameters less than …………….are used for welding purposes.
(a) 10 nm
(b) 20 nm
(c) 30 nm
(d) 40 nm.
Answer:
(a) 10 nm

Question 10.
The Strongest source of gravitational waves is _______.
a) black holes
b) accelerated mass
c) sun
d) stars
Answer:
a) black holes

Question 11.
This is the black hole at the centre of the milky way galaxy.
a) Sagittarius A*
b) Sagittarius B*
c) Sagittarius C*
d) Sagittarius D*
Answer:
a) Sagittarius A*

Question 12.
Who worked in the field of black holes.
a) Marie curie
b) Stephen Hawking
c) Conrad Rontgen
d) Edward Purcell
Answer:
b) Stephen Hawking

Question 13.
Who is the father of the modem robotics industry formed the world’s first robotic company in 1956 ……………..
(a) Joliot
(b) Cormark
(c) Engelberger
(d) Edward purcell.
Answer:
(c) Engelberger

Question 14.
Slavic word robota means ________.
a) labour
b) work
c) labour (or) work
d) None of these
Answer:
c) labour (or) work

Question 15.
These robots are used for vacuum cleaners, floor cleaners, pool cleaning, etc.,
a) Household robots
b) Industrial robots
c) Space robots
d) None of these
Answer:
a) Household robots

Question 16.
Similar nanostructures are made in a lab to glow in different colors.
a) Morpho butterfly
b) Peacock feathers
c) DNA
d) Parrotfish
Answer:
b) Peacock feathers

Question 17.
Manipulation of colours by adjusting the size of nanoparticles from this idea is _______.
a) Morpho butterfly
b) Peacock feathers
c) Parrotfish
d) Lotus leaf surface
Answer:
a) Morpho butterfly

Question 18.
The phenomenon of artificial radioactivity was invented by ……………..
(a) Joliot and Irene curie
(b) Felix Bloch and Edward Purcell
(c ) Connick and Hounsfield
(d) Wilhelm Conrad – Rontgen.
Answer:
(a) Joliot and Irene curie

Question 19.
Bottom up approach example is _______.
a) ball milling
b) sol-gel
c) lithography
d) plasma etching
Answer:
d) plasma etching

Question 20.
Coatings for wind-screen and car bodies
a) Chemical industry
b) Electronic industry
c) Automotive industry
d) Medicine
Answer:
c) Automotive industry

Question 21.
This is effectively used to stop the brain from processing pain and cure soreness in the hospitalized patients.
a) Virtual reality
b) Precision medicine
c) Health wearables
d) Artificial organs
Answer:
a) Virtual reality

Question 22.
This is an emerging approach for disease treatment.
a) Artificial organs
b) Health wearables
c) Precision medicine
d) Virtual reality
Answer:
c) Precision medicine

Question 23.
This is possible to interface it with living tissue (or) to replace a natural organ.
a) Precision medicine
b) Health wearables
c) Virtual reality
d) Artificial organs
Answer:
d) Artificial organs

Question 24.
This device used for tracking a wearer’s vital signs.
a) 3D printing
b) Health wearables
c) Artificial organs
d) Robotic surgery
Answer:
b) Health wearables

Question 25.
This monitor intracranial pressure and temperature and then are absorbed by the body.
a) 3D printing
b) Health wearables
c) wireless brain sensors
d) Robotic surgery
Answer:
c) wireless brain sensors

Question 26.
Composed of two or more atoms.
a) Molecule
b) Atom
c) Nucleus
d) Nucleon
Answer:
a) Molecule

Question 27.
Electrons revolving around the atomic nucleus.
a) Molecule
b) Atom
c) Nucleus
d) Nucleon
Answer:
b) Atom

Question 28.
Composed of protons and neutrons located in the center of an atom.
a) Molecule
b) Atom
c) Nucleus
d) Nucleon
Answer:
c) Nucleus

Question 29.
Protons and neutrons comprising the nucleus of an atom.
a) Molecule
b) Atom
c) Nucleus
d) Nucleon
Answer:
d) Nucleon

Question 30.
Smart inhaler uses this technology to detect inhaler use.
a) system technology
b) Bluetooth technology
c) both (a) and (b)
d) None of these
Answer:
b) Bluetooth technology

X. Two mark questions:

Question 1.
What is physics?
Answer:
Physics is the basic building block for Science, Engineering, Technology, and Medicine.

Question 2.
What is Robotics?
Answer:
Robotics is an integrated study of mechanical engineering, electronic engineering, computer engineering, and science.

Question 3.
What are Robots?
Answer:
The robot is a mechanical device designed with electronic circuitry and programmed to perform a specific task.

Question 4.
What is meant by ‘Robot’? Write its uses?
Answer:
Robot is a mechanical device designed with electronic circuitry and programmed to perform a specific task. These automated machines are highly significant in this robotic era where they can take up the role of humans in certain dangerous environments that are hazardous to people like defusing bombs, finding survivors in unstable ruins, and exploring mines and shipwrecks.

Question 5.
Name main types of industrial robots.
Answer:
Six main types of industrial robots are,

1. Cartesian
2. SCARA
3. Cylindrical
4. Delta
5. Polar
6. Vertically articulated

Question 6.
Name the axis robots.
Answer:
Six-axis robots are ideal for

1. Arc Welding
2. Spot Welding
3. Material Handling
4. Machine Tending
5. Other applications

Question 7.
What is the aim of artificial intelligence?
Answer:
The aim of artificial intelligence is to bring in human-like behavior in robots.

Question 8.
Define cosmology?
Answer:
Cosmology is the branch that involves the origin and evolution of the universe. It deals with the formation of stars, galaxy, etc.

Question 9.
What are the uses of outer space robots?
Answer:
In outer space, robots are used for exploring stars, planets, etc., investigation of the mineralogy of the rocks and soils on Mars, analysis of elements found in rocks and soils.

Question 10.
Name some outer space robots.
Answer:

1. Mars Rovers of NASA
2. Twin Mars Rovers
3. Mars Pathfinder Mission

Question 11.
What are the uses of household robots?
Answer:
Household robots are used as,

1. Vacuum cleaners
2. Floor cleaners
3. Gutter cleaners
4. Lawn mowing
5. Pool cleaning
6. To open and close doors

Question 12.
What are the developments in Nano-robots?
Answer:
Nano-robots are being developed to be in the bloodstream to perform small surgical procedures, to fight against bacteria, repairing individual cells in the body.

Question 13.
Define Particle physics.
Answer:
Particle physics deals with fundamental particles of nature. Protons and neutrons are made of quarks.

Question 14.
What is cosmology?
Answer:
Cosmology is the branch that involves the origin and evolution of the universe.

Question 15.
What is physics?
Answer:

1. Physics is the basic building block for Science, Engineering, Technology and Medicine.
2. Nanoscience is the science of objects with typical sizes of 1-100 nm.

Question 16.
What is Nano?
Answer:
Nano means one-billionth of a meter that is 10^-9 m.

Question 17.
What is nano solid?
Answer:
If the particle of a solid is of size less than lOOnm, it is said to be a nano solid.

Question 18.
What is bulk solid?
Answer:
When the particle size exceeds 100 nm, it forms a bulk solid.

Question 19.
Name the two ways of preparing the nanomaterials.
Answer:

• Top-down approaches
• Bottom-up approaches

Question 20.
What are the major fields of robotics?
Answer:
Five major fields of robotics are

1. Human-robot interface
2. Mobility
3. Manipulation
4. Programming
5. Sensors

Question 21.
Is the accelerated mass emits gravitational waves?
Answer:
Yes, the accelerated mass emits gravitational waves which are very weak.

Question 22.
What is the important phenomena of nano properties?
Answer:
Quantum confinement effects and surface effects are the two important phenomena that govern nano properties.

Question 23.
Is the nano form of the material the same as its bulk counterpart?
Answer:
No, The nano form of the material shows strikingly different properties when compared to its bulk counterpart.

Question 24.
Can nanoparticles get absorbed?
Answer:
Yes, Nanoparticles can easily get absorbed onto the surface of living organisms and they might enter the tissues and fluids of the body.

Question 25.
What is the function of inhaled nanoparticles?
Answer:
The inhaled nanoparticles reach the blood and that may also reach other sites such as the liver, heart, and blood cells.

Question 26.
Can nanoparticles cross the cell membrane?
Answer:
Yes, nanoparticles can also cross cell membranes.

Question 27.
Is the larger nanoparticles can reach inside the biomolecules?
Answer:

1. The absorbing nature depends on the surface of the nanoparticle.
2. Nanoparticles of a few nanometers size may reach well inside biomolecules, which is not possible for larger nanoparticles.

Question 28.
What is Unimate?
Answer:
In 1954, George Devol invented the first digitally operated programmable robot called “Unimate”.

Question 29.
Write a short note on Human-Robot.
Answer:
Certain robots are made to resemble humans in appearance and replicate human activities like walking, lifting and sensing, etc.

XI. Three mark questions:

Question 1.
List out the nanomaterial-based products in the automotive industry.
Answer:

1. Lightweight construction
2. Painting (fillers, base coat, clear coat)
3. Catalysts
4. Tires (fillers)
5. Sensors
6. Coatings for wind-screen and car bodies

Question 2.
Give five examples of nanomaterial-based products in the chemical industry.
Answer:

1. Fillers for paint systems
2. Coating systems based on nanocomposites
3. Impregnation of papers
4. Switchable adhesives
5. Magnetic fluids

Question 3.
Write the application of nanomaterial-based products in construction.
Answer:

1. Construction materials
2. Thermal insulation
3. Flame retardants
4. Surface-functionalized building materials for wood, floors, stone, facades, tiles, roof tiles, etc.
5. Facade coatings
6. Groove mortar

Question 4.
What are the applications of nanomaterial-based products in the following areas?
i) Textile
ii) Energy
iii) cosmetics
Answer:

Textile/fabrics/ non-wovens	Energy	Cosmetics
1. Surface Processed textiles 2. Smart clothes	1. Fuel cells 2. Solar cells 3. Batteries 4. Capacitors	1. Sun protection 2. Lipsticks 3. Skin creams 4. toothpaste

Question 5.
List the nanomaterial-based products in Engineering.
Answer:

1. Wear protection for tools and machines (anti-blocking coatings, scratch-resistant coatings on plastic parts, etc.)
2. Lubricant-free bearings.

Question 6.
Write about the top-down approach of nanomaterials synthesis.
Answer:
Top-down approach:

• Nanomaterials are synthesised by breaking down bulk solids into nano sizes.
• Ex: Ball milling, sol-gel, lithography.

Question 7.
What is bottom-up approach?
Answer:

1. Nanomaterials are synthesised by assembling the atoms/molecules together.
2. Selectively atoms are added to create structures.
3. Example: Plasma etching and chemical vapour deposition.

Question 8.
Short note on nano in laboratories.
Answer:

1. The nanostructures made in the laboratory mimic some of nature’s amazing nanostructures.
2. As the nanostructures are so small, specialized methods are needed to manufacture objects in this size range.
3. There are two ways of preparing the nanomaterials, top-down and bottom-up approaches.

Question 9.
Mention the uses of wireless brain sensors.
Answer:

1. Wireless brain sensors monitor intracranial pressure and temperature and then are absorbed by the body.
2. Hence there is no need for surgery to remove these devices.

Question 10.
What is robotic surgery?
Answer:

1. Robotic surgery is a type of surgical procedure that is done using robotic systems.
2. Robotically assisted surgery helps to overcome the limitations of pre-existing minimally-invasive surgical procedures and to enhance the capabilities of surgeons performing open surgery.

Question 11.
What do you know about smart inhalers?
Answer:

1. Inhalers are the main treatment option for asthma.
2. Smart inhalers are designed with health systems and patients in mind so that they can offer maximum benefit.
3. Smart inhalers use BlueTooth technology to detect inhaler use, remind patients when to take their medication, and gather data to help guide care.

Question 12.
What is the role of physics in medical diagnosis?
Answer:

1. Medical science very much revolves around physics principles.
2. Medical instrumentation has widened the life span due to the technology integrated diagnosis and treatment of most of the diseases.
3. This modernisation in all fields is possible due to the efficient application of fundamental physics.

Question 13.
What is gravitational waves? How are they produced?
Answer:

1. Gravitational waves are the disturbances in the curvature of space-time and it travels with the speed of light.
2. Any accelerated charge emits an electromagnetic wave. Any accelerated mass emits gravitational waves but these waves are very weak even for masses like earth. The strongest source of gravitational waves is black holes.
3. The recent discoveries of gravitational waves are emitted by two black holes when they merge into a single black hole.

Question 14.
Can we completely replace humans with robots? Give any three reasons.
Answer:
No, we cannot completely replace humans with robots. Following are the reasons:

1. Robots have no sense of emotions or conscience.
2. Robots cannot handle unexpected situations.
3. Humans cannot be replaced by robots in decision-making.

Question 15.
What is a quark?
Answer:

1. A quark is a type of elementary particle and a fundamental constituent of matter.
2. Quarks combine to form composite particles called hadrons, the most stable of which are protons and neutrons, the components of atomic nuclei.

Question 16.
How nano are found in mimic in laboratories from i) Morpho butterfly ii) Peacock feathers.
Answer:

1. Morpho butterfly: Manipulation of colours by adjusting the size of nanoparticles with which the materials are made.
2. Peacock feathers: Similar nanostructures are made in a lab to glow in different colors.

Question 17.
From parrotfish, how nano are found in laboratories?
Answer:
The natural structure provides a blueprint for creating ultra-durable synthetic materials that could be useful for mechanical components in electronics and in other devices that undergo repetitive movement, abrasion, and contact stress.

Question 18.
How the lotus leaf surface (Nano in nature) found in laboratories?
Answer:

1. Water repellant nano paints are made.
2. Coating with such nano paints gives durability, protection against stains and dirt also enhances fuel efficiency when coated on ships.

Question 19.
What do you know about animate?
Answer:

1. In 1954, George Devol invented the first digitally operated programmable robot called Unimate.
2. George Devol and Joseph Engelberger, the father of the modern robotics industry formed the world’s first robot company in 1956.
3. In 1961, Unimate was operated in a General Motors automobile factory for moving car parts around in New Jersey.

XII. Five mark questions:

Question 1.
Write the advantage of Robotics.
Answer:

1. Robots are much cheaper than humans.
2. Robots never get tired like humans. It can work 24 x 7. Hence absenteeism in the workplace can be reduced.
3. Robots are more precise and error-free in performing the task.
4. Stronger and faster than humans.
5. Robots can work in extreme environmental conditions: extreme hot (or) cold, space (or) underwater.
6. In dangerous situations like bomb detection and bomb deactivation.
7. In warfare, robots can save human lives.
8. Robots are significantly used in handling materials in chemical industries especially in nuclear plants which can lead to health hazards in humans.

Question 2.
List out the disadvantage of Robotics.
Answer:

1. Robots have no sense of emotions or conscience.
2. They lack empathy and hence create an emotionless workplace.
3. If ultimately robots would do all the work, and the humans will just sit and monitor them, health hazards will increase rapidly.
4. The unemployment problem will increase.
5. Robots can perform defined tasks and can’t handle unexpected situations
6. The robots are well programmed to do a job and if a small thing goes wrong it ends up in a big loss to the company.
7. If a robot malfunctions, it takes time to identify the problem, rectify it, and even program if necessary. This process requires significant time.
8. Humans can’t be replaced by robots in decision-making.
9. Till the robot reaches the level of human intelligence, the humans in the workplace will exit.

Question 3.
Give an example of “Nano” in nature.
Answer:

1. A single strand of DNA the building block of all living things is about three nanometers wide.
2. The scales on the wings of a morpho butterfly contain nanostructures that change the way light waves interact with each other, giving the wings brilliant metallic blue and green hues.
3. Peacock feathers get their iridescent coloration from light interacting with 2-dimensional photonic crystal structures just tens of nanometers thick.
4. Parrotfish crunches up coral all day. The source of the parrot fish’s powerful bite is the interwoven fiber nanostructure.
5. Crystals of a mineral called fluorapatite are woven together in a chain mail-like arrangement. This structure gives parrotfish teeth incredible durability.
6. Lotus leaf surface scanning electron micrograph (SEM) showing the nanostructures on the surface of a leaf from a lotus plant. This is the reason for the self-cleaning process in lotus leaf.

Question 4.
List out the applications of Nanotechnology in various fields.
Answer:

1. Optical engineering and communication
2. Electronics
3. Metallurgy and materials
4. Defense and security
5. Energy storage
6. Biomedical and drug delivery
7. Agriculture and food
8. Cosmetics and paints
9. Biotechnology
10. Textile

Question 5.
Explain about Nanorobots.
Answer:

1. The size of the nanorobots is reduced to a microscopic level to perform a task in very small spaces.
2. However, it is only in the developmental stage.
3. The future prospects of it are much expected in the medical field:
   • Nano-robots in the bloodstream to perform small surgical procedures.
   • To fight against bacteria
   • Repairing individual cells in the body.
4. It can travel into the body and once the job is performed it can find its way out.
5. Chinese scientists have created the world’s first autonomous DNA robots to combat cancer tumours.

Question 6.
List out the recent advancement in medical technology.
Answer:

1. Virtual reality
2. Precision medicine
3. Health wearables
4. Artificial organs
5. 3D printing
6. Wireless brain sensors
7. Robotic surgery
8. Smart inhalers

Question 7.
Discuss the applications of nanomaterial-based products in
i) Electronic Industry
ii) Medicine.
Answer:

Electronic industry	Medicine
1. Data memory 2. Displays 3. Laser diodes 4. Glass fibers 5. Optical switches 6. Filters 7. Conductive, antistatic coatings	1. Drug delivery system 2. Active agents 3. Contrast medium 4. Medical rapid tests 5. Prostheses and implants 6. Antimicrobial agents and coatings 7. Agents in cancer therapy

Question 8.
Write about artificial intelligence briefly.
Answer:
The aim of artificial intelligence is to bring in human-like behaviour in robots. It works on

1. Face recognition
2. Providing a response to player’s actions in computer games
3. Taking decisions based on previous actions
4. To regulate the traffic by analyzing the density of traffic on roads.
5. Translate words from one language to another

Question 9.
What are the uses of industrial robots?
Answer:
Industrial robots are used for

1. Welding
2. Cutting
3. Lifting
4. Packing
5. Transport
6. Sorting
7. Bending
8. Assembling
9. Manufacturing
10. Weaponry
11. Industrial goods
12. Laboratory research
13. Mass production of consumer
14. Robotic water jet cutting
15. Robotic laser cutting
16. Handling hazardous materials like nuclear waste.

Tamilnadu State Board New Syllabus Samacheer Kalvi 12th Physics Guide Pdf Chapter 10 Communication Systems Text Book Back Questions and Answers, Notes.

Tamilnadu Samacheer Kalvi 12th Physics Solutions Chapter 10 Communication Systems

12th Physics Guide Communication Systems Text Book Back Questions and Answers

Part – I:

Textbook Evaluation:

I. Multiple Choice questions:

Question 1.
The output transducer of the communication system converts the radio signal into
a) Sound
b) Mechanical energy
c) Kinetic energy
d) None of the above
Answer:
a) Sound

Question 2.
The signal is affected by noise in a communication system
a) At the transmitter
b) At the modulator
c) In the channel
d) At the receiver
Answer:
c) In the channel

Question 3.
The variation of frequency of carrier wave with respect to the amplitude of the modulating signal is called
a) Amplitude modulation
b) Frequency modulation
c) Phase modulation
d) Pulse width modulation
Answer:
b) Frequency modulation

Question 4.
The interntionally accepted frequency deviation for the purpose of FM broadcasts.
a) 75 kHz
b) 68 kHz
c) 80 kHz
d) 70 kHz
Answer:
a) 75 kHz

Question 5.
The frequency range of 3 MHz to 30 MHz is used for
a) Ground wave propagation
b) Space wave propagation
c) Sky wave propagation
d) Satellite communication
Answer:
c) Sky wave propagation

II. Short Answers Questions:

Question 1.
Give the factors that are responsible for transmission impairments.
Answer:
Attenuation:
The strength of signal decreases with increasing distance which causes loss of energy.
Distortion:
Change in the shape of the signal
Noise:
Random or unwanted signal mixes up with the original sound.

Question 2.
Distinguish between wireline and wireless communication? Specify the range of electromagnetic waves in which it is used.
Answer:

Wireline Communication	Wireless Communication
It uses wires, cables and optical fibres as a medium	It uses free space as a medium
It cannot be used for long distance transmission	It can be used for long distance transmission
Ex. Telephone, intercom and cable TV	Ex. Mobile, radio or TV broadcasting and satellite communication

Question 3.
Explain centre frequency or resting frequency in frequency modulation.
Answer:
When the frequency of the baseband signal is zero (no input signal), there is no change in the frequency of the carrier wave. It is at its normal frequency and is called center frequency or resting frequency.

Question 4.
What does RADAR stand for?
Answer:

• RADAR stands for Radio Detection and Ranging
• It is used to sense, detect and locate distant objects like aircraft, ships, spacecraft, etc.

Question 5.
What do you mean by the Internet of Things?
Answer:
Internet of Things (IoT), it is made possible to control various devices from a single device. Example: home automation using a mobile phone.

III. Long Answers Questions:

Question 1.
What is modulation? Explain the types of modulation with necessary diagrams.
Answer:
Modulation:
For long-distance transmission, the low-frequency baseband signal (input signal) is superimposed onto a high-frequency radio signal by a process called modulation.

i) Amplitude modulation:
1. If the amplitude of the carrier signal is modified according to the instantaneous amplitude of the baseband signal.
2. The frequency and the phase of the carrier signal remains constant.
3. It is used in radio and TV broadcasting.

Advantages of AM:
1. Easy transmission and reception
2. Lesser bandwith
3. Low cost

Disadvantages of AM:
1. High noise level
2. Low efficiency
3. Small operating range

ii) Frequency modulation:
1. Frequency of the carrier signal is modified according to the instantaneous amplitude of the baseband signal.
2. The amplitude and the phase of the carrier signal remain constant.
3. Increase in the amplitude of the baseband signal increases the carrier signal frequency.
4. Louder signals form compressions and weaker signals form rarefactions.
5. When the frequency of the baseband signal is zero, there is no change in the frequency of the carrier wave. This is known as centre frequency or resting frequency.

Advantages:
1. Decrease in noise
2. Efficiency is high
3. Operating range is quite large
4. Better quality

Disadvantages:
1. Requires wider channel
2. Less area is covered.
3. Transmitters and receivers are complex and costly.

Phase Modulation:

1. The instantaneous amplitude of the baseband signal modifies the phase of the carrier signal keeping the amplitude and frequency constant.
2. The frequency shift depends on

• the amplitude of the modulating signal and
• the frequency of the signal.

3. When the modulating signal goes positive, the amount of phase lead increases with the amplitude of the modulating signal.
4. The carrier signal is compressed or its frequency is increased.
5. The negative half cycle of the baseband signal produces lag in the carrier signal.
6. Phase modulated wave also comprises of compressions and rarefactions.
7. When the signal voltage is zero, (A, C and E) the carrier frequency is unchanged.

Advantages:
1. FM Signal produced from PM signal is very stable.
2. The center frequency called resting frequency is stable.

Question 2.
Elaborate on the basic elements of a communication system with the necessary block diagram.
Answer:
1. Information (Baseband or input signal):
i. Information can be in the form of a sound signal like speech, music, pictures or computer data.
ii. These can be given as input to the input transducer.

 

2. Input transducer:
i. It converts variation in a physical quantity such as pressure, temperature, sound into equivalent electrical signals or vice versa.
ii. It converts the information into corresponding electrical signals.
iii. The electrical equivalent of the original information is called baseband signal.

3. Transmitter:
It is located at the broadcasting station.

a) Amplifier:
Very weak transducer output is amplified

b) Oscillator:
i. For long-distance, the high-frequency carrier wave is transmitted into space.
ii. Energy of the wave is proportional to its frequency, so the carrier wave has very high energy.

c) Modulator:
It superimposes the baseband signal onto the carrier signal and generates the modulated signal.

d) Power amplifier:
To cover a large distance, the power level of the electrical signal is increased.

4. Transmitting antenna:
i. It travels in the form of electromagnetic waves with the velocity of light
ii. Used for long-distance transmission
iii. Ex: Mobile, radio or TV broadcasting, satellite communication.

5. Communication channel:
To carry the electrical signal from transmitter to receiver with less noise or distortion.

Wireline Communication:
i. Medium – Wires, Cables, and optical fibers
ii. Cannot be used for long-distance transmission
iii. Ex: Telephone, intercom and Cable TV

Wireless communication:
i. Medium – free space
ii. Signals are transmitted in the form of electromagnetic waves.

6. Noise:
i. Undesirable electrical signal
ii. Reduces the quality of the transmitted signal.
iii. It may be man-made or natural.
iv. It cannot be completely eliminated but it can be reduced.

7. Receiver:
i. The transmitted signals are fed into the receiver.
ii. Receiver consists of the demodulator, amplifier, detector etc.
iii. Demodulator extracts the baseband signal from the carrier.
iv. Then it is amplified and detected, then fed to the output transducer.

8. Repeaters:
i. Combination of transmitter and receiver
ii. Increase the range or distance through which the signals are sent.
iii. Signals are received, amplified, and retransmitted with a different frequency to the destination.
iv. Ex. Communication satellite in space.

9. Output transducer:
i. Converts the electrical signal back to its original form such as sound, music, pictures or data.
ii. Ex : Loudspeakers, Picture tubes, Computer monitor etc.

10. Attenuation:
The loss of strength of a signal while propagating through a medium.

11. Range:
i. Maximum distance between the source and the destination.
ii. The signal is received with sufficient strength.

Question 3.
Explain the three inodes of propagation of electromagnetic waves through space.
Answer:
According to the frequency range, the electromagnetic wave transmitted by the transmitter travels in three different modes.

1. Ground Wave propagation (or) Surface Wave propagation:
i. The electromagnetic waves are transmitted by the transmitter glide over the surface of the earth to reach the receiver.
ii. Both transmitting and receiving antennas must be close to the earth.
iii. The size of the antenna decides the efficiency of the radiation of signals.
iv. Some reasons for attenuation are

a) Increasing distance:
Attenuation of the signal depends on
(i) Power of the transmitter
(ii) frequency of the transmitter
(iii) Condition of the earth surface.

b) Absorption of energy by the Earth:
i. Transmitted signal is in contact with the Earth, it induces charges in the Earth and constitutes a current.
ii. The Earth behaves like a leaky capacitor which leads to the attenuation of the wave.

c) Tilting of the wave:
i. The wavefront starts gradually tilting according to the curvature of the Earth.
ii. When the tilt increases, the electric field strength decreases.
iii. At some distance, the surface wave dies out due to energy loss.
iv. The frequency of the ground waves is mostly less than 2 MHz
v. It is used in local broadcasting, radio navigation, for ship – to ship, ship – to shore communication and mobile communication.

2. Skywave propagation (or) Ionospheric propagation:
i. The electromagnetic waves are radiated from an antenna, directed upwards at large angles gets reflected by the ionosphere back to earth.
ii. Due to the absorption of ultraviolet rays, cosmic ray and other high energy radiations like α, β rays from sun get ionized.
iii. These charged ions provide a reflecting medium for the reflection of radio waves.
iv. The phenomenon of bending the radio waves back to earth is due to total internal reflection.
v. The shortest distance between the transmitter and the point of reception of the skywave along the surface is called as the skip distance.
vi. When the angle of emission increases, the reception of ground waves decreases.
vii. A zone where there is no reception of electromagnetic waves neither ground nor sky is called skip zone or skip area.
viii. In this mode of propagation, the frequency range of EM waves is 3 to 30 MHz.
ix. It is used for short wave broadcast services.
x. Extremely long-distance communication is possible.
xi. A single reflection helps to travel a distance of approximately 4000 km.

3. Space Wave propagation:
i. The process of sending and receiving information signal through space.
ii. EM waves of very high frequencies above 30 MHz are used.
iii. Waves travel in a straight line from the transmitter to the receiver.
iv. The transmission towers must high.
v. For high frequencies, the signals will not encounter the curvature of the earth.
vi. It travels with less attenuation and loss of signal strength.
vii. TV broadcast, satellite communication and RADAR are based on this propagation
viii. It has some advantages, larger bandwith, high data rates, better directivity, small antenna size, low power consumption etc.
ix. Range depends on the height of the antenna (h).
x. Range or distance (d) = \(\sqrt{2 \mathrm{Rh}}\)
R – Radius of the earth
R = 6400 Km.

Question 4.
What do you know about GPS? Write a few applications of GPS.
Answer:
GPS:

1. GPS stands for Global Positioning System.
2. It offers geolocation and time information anywhere on or near the earth.
3. It works with the assistance of a satellite network.
4. It broadcasts a precise signal like an ordinary radio signal.
5. It convey the location data which is translated by the GPS software.
6. This software recognize the satellite, its location ahd time taken to travel from each satellite.
7. Then the data from each satellite is processed to estimate the location of the receiver.

Applications:

1. It is used in fleet vehicle management for tracking cars, trucks and buses.
2. In wildlife management for counting of wild animals.
3. In engineering for making tunnels, bridges etc.

Question 5.
Give the applications of ICT is mining and agriculture sectors.
Answer:
ICT in mining:

1. Improves operational efficiency, remote monitoring and disaster locating system.
2. Provides audio – visual warning to the trapped underground miners.
3. Helps to connect remote sites.

ICT in agriculture:

1. Increases food productivity and farm management.
2. Optimize the use of water, seeds and fertilizers etc.
3. It is used in farming to decides the suitable place for the species to be planted.
4. It can be used in sophisticated technologies like robots, temperature, moisture sensors, aerial images and GPS technology.

Question 6.
Modulation helps to reduce the antenna size in wireless communication Explain.
Answer:

1. Antenna is used at both transmitter and receiver.
2. Antenna height is important in wireless communication.
3. The height of the antenna must be a muiltiple of \(\frac{\lambda}{4}\).
   i.e h = \(\frac{\lambda}{4}\)
   Where, λ = \(\frac{C}{\gamma}\), C = Velocity of light
   γ – frequency λ – wavelength
4. Consider two base band signals.
5. One is modulated and the other is not modulated
6. Frequency of the original baseband signal, γ = 10 KHz
7. Frequency of the modulated signal, γ = 1 MHz
8. Height required to transmit the original baseband signal is,
   h₁ = \(\frac{\lambda}{4}=\frac{C}{4 \gamma}\)
   = \(\frac{3 \times 10^{8}}{4 \times 10 \times 10^{3}}\) = 7.5 Km
   Height required to transmit the modulated signal is
   h₂ = \(\frac{\lambda}{4}=\frac{C}{4 \gamma}\)
   = \(\frac{3 \times 10^{8}}{4 \times 10 \times 10^{6}}\) = 75 m
9. From the above example, it is clear that modulated signals reduce the height of the antenna.

Question 7.
Fiber-optic communication is gaining popularity among the various transmission media – justify.
Answer:
The method of transmitting information from one place to another in terms of light pulses through an optical fiber is called fiber optic communication. It is in the process of replacing wire transmission in communication systems. Light has very high frequency (400 THz – 790 THz) than microwave radio systems.  The fibers are made up of silica glass or silicon dioxide which is highly abundant on Earth.

Now it has been replaced with materials such as chalcogenide glasses, fluoro aluminate crystalline materials because they provide larger infrared wavelength and better transmission capability. As fibers are not electrically conductive, it is preferred in places where multiple channels are to be laid and isolation is required from electrical and electromagnetic interference.

Applications:
Optical fiber system has a number of applications namely, international communication, inter-city communication, data links, plant and traffic control and defense applications.

Merits:

1. Fiber cables are very thin and weight lesser than copper cables.
2. This system has much larger bandwidth. This means that its information-carrying capacity is larger.
3. Fiber optic system is immune to electrical interferences.
4. Fiber optic cables are cheaper than copper cables.

Demerits:

1. Fiber optic cables are more fragile when compared to copper wires.
2. It is an expensive technology.

PART – II:

12th Physics Guide Communication Systems Additional Questions and Answers

I. Matching Type Questions:

Question 1.

Column 1	Column 2
1. Wireless communication	a. 30 MHz to 400 GHz
2. Ground wave propagation	b. 3 MHz to 30 GHz
3. Sky wave propagation	c. 2 KHz to 2 MHz
4. Space wave propagation	d. 2 KHz to 400 GHz

Answer:

1. d
2. c
3. b
4. a

Question 2.

Column 1	Column 2
1. GPS	a. Fisheries
2. GSM	b. Military and Navigation systems
3. ICT	c. Mobile communication
4. RADAR	d. Counting of wild animals

Answer:

1. d
2. c
3. a
4. b

Question 3.

Column 1	Column 2
1. Uplink Frequency band	a. 20 – 20,000 Hz
2. Downlink Frequency band	b. 400 THz – 790 THz
3. Very High frequency of light	c. 6 GHz
4. Audio frequency	d. 4 GHz

Answer:

1. c
2. d
3. b
4. a

II. Fill in the blanks:

Question 1.
_______ gives the difference between the upper and lower frequency limits of the signal.
Answer:
Bandwidth

Question 2.
For a frequencv less than the critical frequencv, skip distance is _______.
Answer:
Zero

Question 3.
Multimode fibers operate at the speed of _______.
Answer:
10 Mbps

Question 4.
Radar uses _______ for communication.
Answer:
electromagnetic waves.

III. Choose the Odd man out:

Question 1.
a) Input transducer
b) Amplifier
c) Oscillator
d) Demodulator
Answer:
d) Demodulator

Question 2.
a) Telephone
b) mobile
c) intercom
d) Cable TV
Answer:
b) Mobile

Question 3.
a) Weather satellites
b) Communication
c) satellites
d) Navigation satellites
e) RADAR
Answer:
e) RADAR

Question 4.
a) Total internal reflection
b) skip distance
c) amplitude modulation
d) skip zone
Answer:
c) Amplitude modulation

IV. Choose the incorrect pair:

Question 1.
i) Space wave propagation – LOS
ii) Tracking cars – ICT
iii) Counting of wild animals – GPS
iv) Home automation using
a mobile phone – IoT
Answer:
ii) Tracking cars – ICT

Question 2.
i) Input transducer – Microphone
ii) Output transducer – Loudspeakers
iii) Transmitter – Broadcasting station
iv) Oscillator – Low-frequency carrier wave
Answer:
iv) Oscillator – Low-frequency carrier wave

Question 3.
i) Bandwidth, BW – γ₁ – γ₂
ii) Amplitude modulation system – 10 KHz
iii) Single side – band system – 5 KHz
iv) Height of the Antenna – \(\frac{\lambda}{4}\)
Answer:
i) Bandwidth, BW – γ₁ – γ₂

Question 4.
i) Amplitude – High-efficiency modulation
ii) Baseband signal – input signal
iii) Carrier signal – a radio signal
iv) Resting frequency – centre frequency
Answer:
i) Amplitude modulation – High efficiency

V. Choose the correct pair:

Question 1.
i) Attenuation – Amplitude of the transmitter
ii) High frequency – High skip distance
iii) Skip distance – The longest distance
iv) RADAR – Ground wave propagation
Answer:
ii) High frequency – High skip distance

Question 2.
i) Audio frequency – 200 to 2000 Hz
ii) Carrier wave – Cosine wave
iii) Louder signal – Compressions
iv) Frequency shift – Rarefaction
Answer:
iii) Louder signal- Compressions

Question 3.
i) Velocity of light – 3 × 10⁶ m/ s
ii) Radius of the earth – 6800 Km
iii) Date speed for homes – 2 Gbps
iv) Antenna height – d = \(\sqrt{2 \mathrm{Rh}}\)
Answer:
iv) Antenna height – d = \(\sqrt{2 \mathrm{Rh}}\)

Question 4.
i) Skywave propagation – 3 MHz to 30 MHz
ii) Space wave propagation – 30 MHz to 40 MHz
iii) Ground wave propagation – 20 KHz to 20 MHz
iv) Wireless communication – 1 KHz to 2 KHz
Answer:
i) Skywave propagation – 3 MHz to 30 MHz

VI. Assertion and Reason:

Question 1.
Assertion:
Short wave bands are used for the transmission of radio waves to a large distance.
Reason:
Short waves are reflected by the ionosphere.
a) Assertion and the reason are true and the reason is the correct explanation of the assertion.
b) Assertion and the reason are true and the reason is not the correct explanation of the assertion.
c) Assertion is true but reason is false
d) Assertion is false but reason is true.
Answer:
a) Assertion and the reason are true and the reason is the correct explanation of the assertion.

Question 2.
Assertion:
Television broadcasting becomes weaker with increasing distance.
Reason:
The power transmitted from the TV transmitter varies inversely as the distance of the receiver.
a) Assertion and the reason are true and the reason is the correct explanation of the assertion.
b) Assertion and the reason are true and the reason is not the correct explanation of the assertion.
c) Assertion is true but reason is false
d) Assertion is false but reason is true.
Answer:
c) Assertion is true but reason is false

Question 3.
Assertion:
Optical fibre communication has immunity to cross talk.
Reason:
Optical interference between fibres is zero.
a) Assertion and the reason are true and the reason is the correct explanation of the assertion.
b) Assertion and the reason are true and the reason is not the correct explanation of the assertion.
c) Assertion is true but reason is false
d) Assertion is false but reason is true.
Answer:
a) Assertion and the reason are true and the reason is the correct explanation of the assertion.

Question 4.
Assertion:
A dish antenna is highly directionals
Reason:
This is because a dipole antenna is omnidirectional.
a) Assertion and the reason are true and the reason is the correct explanation of the assertion.
b) Assertion and the reason are true and the reason is not the correct explanation of the assertion.
c) Assertion is true but reason is false
d) Assertion is false but reason is true.
Answer:
b) Assertion and the reason are true and the reason is not the correct explanation of the assertion.

VII. Choose the correct statement:

Question 1.
i) A transducer is a device which converts an electrical signal into variations in a physical quantity.
ii) Transmitting antenna radiates the radio signal into space in all directions.
iii) The power amplifier decreases the power level of the signal
iv) The minimum distance between the source and the destination is called a range
Answer:
Statement (ii) is correct.

Question 2.
i) Attenuation depends on the frequency of the transmitter.
ii) Increase in the tilt increases the electric field strength.
iii) Transmitting and receiving antenna must be close to the earth.
iv) The frequency of the ground waves is most less than 1 MHz.
Answer:
Statements (i) and (iii) are correct

Question 3.
i) The higher the frequency, lower is the skip distance.
ii) Bending of radio waves back to earth is called refraction.
iii)The frequency range of EM wave in skywave propagation is 30 to 300 MHz.
iv) When the angle of emission increases, the reception of ground waves decreases.
Answer:
Statement (iv) is correct.

Question 4.
i) The transducer output is very weak
ii)The electrical equivalent of the original information is called the carrier signal.
iii) Microphone converts electrical energy into sound energy.
iv) Wireline communication uses free space as a communication medium.
Answer:
Statement (i) is correct.

VIII. Choose the incorrect statement:

Question 1.
i) Noise is the undesirable electrical signal.
ii) It attenautes or reduces the quality of the signal
iii) It may be man – made or natural
iv) Noise can be completely eliminated.
Answer:
Statement (iv) is correct.

Question 2.
i) Operating range is quite large in FM
ii) In FM, there is a large increase in noise
iii) Am radio has better quality compared to FM radio.
iv) FM requires a much under channel
Answer:
Statement (ii) & (iii) are incorrect

Question 3.
i) Noise level is low in AM
ii) There is a lesser landwidth in AM
iii) In A.M the efficiency is low
iv) Small operating range in AM
Answer:
Statement (i) is incorrect

Question 4.
i) ICT in mining helps to connect remote sites.
ii) ICT is widely used in decreasing food productivity
iii) ICT improves operational efficiency
iv) ICT are also used in fisheries
Answer:
Statement (ii) is incorrect

IX. Choose the best answer:

Question 1.
Which of the following frequencies will be suitable for beyond the horizon communication using sky waves?
(a) 10 kHz
(b) 10 MHz
(c) 1 GHz
(d) 1000 GHz
Answer:
(b) 10 MHz
Hint:
A frequency of 10 kHz will require a very large radiating antenna while frequencies 1GHz and 1000 GHz will penetrate the ionosphere and cannot be reflected by it.

Question 2.
Cellular phones use radio waves in ______ band.
a) long-wave
b) short wave
c) medium wave
d) ultra-high frequency
Answer:
d) ultra-high frequency

Question 3.
Antenna is
(a) inductive
(b) capacitive
(c) resistive above its resonant frequency
(d) resistive at the resonant frequency
Answer:
(d) resistive at the resonant frequency.
Hint:
An antenna is a tuned circuit consisting of an inductance and a capacitance. At the resonant frequency, it is resistive.

Question 4.
In space wave propagation, the range of the propagation depends on the height (h) of the antenna given by the equation.
a) \(\sqrt{2 \mathrm{Rh}}\)
b) \(\sqrt{\frac{R h}{2}}\)
c) \(\sqrt{2 R^{2} h}\)
d) \(\frac{h}{2 R}\)
Answer:
a) \(\sqrt{2 \mathrm{Rh}}\)

Question 5.
_______ modulation is used in radio and TV broadcasting.
a) Frequency
b) Phase
c) Amplitude
d) Carrier
Answer:
c) Amplitude

Question 6.
If a ________ wave is used as the baseband signal, then phase reversal takes place in the modulated signal.
a) Square
b) Sine
c) Cosine
d) triangular
Answer:
a) Square

Question 7.
An oscillator is producing FM waves of frequency 2 kHz with a variation of 10 kHz. What is the modulation index?
(a) 0.67
(b) 5.00
(c) 0.20
(d) 1.5
Answer:
(b) 5.00
Hint:
m_f = \(\frac { ∆f }{ f }\) = \(\frac { 10kHz }{ 2kHz }\) = 5

Question 8.
The best example of transducer is________
a) Loudspeaker
b) picture tubes
c) computer monitor
d) microphone
Answer:
d) microphone

Question 9.
________ generates high – frequency carrier wave.
a) Modulator
b) Oscillator
c) Amplifier
d) Power amplifier
Answer:
b) Oscillator

Question 10.
The maximum distance upto which TV transmission from a TV tower of height h can be received is proportional to
(a) h^1/2
(b) h
(c) ^3/2
(d) h²
Answer:
(a) h^1/2
Hint:
d = \(\sqrt { 2Rh } \) ; d ∝ h^1/2

Question 11.
The height of the antenna must be a multiple of _______.
a) \(\frac{\lambda}{4}\)
b) \(\frac{\lambda}{8}\)
c) 4λ
d) 2λ
Answer:
a) \(\frac{\lambda}{4}\)

Question 12.
The relation between the frequency, the velocity of light, and wavelength is given by ________.
a) C = γλ
b) C = \(\frac{\lambda}{\gamma}\)
c) λ = \(\frac{C}{\gamma}\)
d) both (a) and (c)
Answer:
d) both (a) and (c)

Question 13.
In AM, the channel bandwidth is ______ the signal frequency
a) thrice
b) twice
c) equal to
d) half
Answer:
b) twice

Question 14.
The height of the transmitting and receiving antenna must be a multiple of
a) 2λ
b) λ/4
c) λ/2
d) 4λ
Answer:
b) λ/4

Question 15.
A laser beam is used for carrying out surgery because it
(a) is highly monochromatic
(b) is highly coherent
(e) is highly directional
(d) can be sharply focused
Answer:
(d) can be sharply focused
Hint:
A laser beam is highly monochromatic, directional, and coherent and hence it can be sharply focused for carrying out surgery.

Question 16.
The _______ of the antenna plays a major role in deciding the efficiency of the radiation of the signals.
a) shape
b) distance
c) direction
d) size
Answer:
d) size

Question 17.
A single reflection helps the radio waves to travel a distance of approximately
a) 40 km
b) 400 km
c) 4000 km
d) 4 km
Answer:
c) 4000 km

Question 18.
When the angle of emission increases, the reception of ground waves _______.
a) increases
b) decreases
c) remains constant
d) none of the above
Answer:
b) decreases

Question 19.
The principle used in the transmission of signals through an optical fiber is
(a) total internal reflection
(b) refraction
(c) dispersion
(d) interference
Answer:
(a) total internal reflection
Hint:
Signals propagate through an optical fiber by suffering repeated total internal reflections.

Question 20.
The fibres are made up of _______.
a) glass
b) silicon
c) silica glass
d) both a & c
Answer:
d) both a & c

Question 21.
_______ provide the fastest transmission rate compared to any other form of transmission
a) Cable wires
b) Fibers
c) Fiber optic cables
d) Copper wire
Answer:
c) Fiber optic cables

Question 22.
________ is a fast-growing technology in the field communication system.
a) Mobile
b) Satellite
c) Fiber optics
d) Internet
Answer:
d) Internet

Question 23.
The skywave propagation is suitable for radio waves of frequency
(a) from 2 MHz to 50
(b) upto 2MHz
(c) from 2 MHz to 30 MHz
(d) from 2MHz to 20 MHz
Answer:
(c) from 2 MHz to 30 MHz

Question 24.
To store all the information available on the internet, we need over
a) 2 billion DVDs
b) 1 billion DVDs
c) 3 billion DVDs
d) 4 billion DVDs
Answer:
b) 1 billion DVDs

Question 25.
_______ is used to search for information on the World Wide Web
a) Communication
b) Satellite
c) E-Commerce
d) Search engine
Answer:
d) Search engine

Question 26.
_______ is used for making tunnels, bridges etc.
a) GSM
b) TDMA
c) GPS
d) GPRS
Answer:
c) GPS

Question 27.
Fiber optic cables provide data speed of ____ for homes and businesses.
a) 1 Gbps
b) 2 Gbps
c) 10 Mbps
d) 5 Gbps
Answer:
a) 1 Gbps

Question 28.
A modem is a device which performs
(a) modulation
(b) demodulation
(c) rectification
(d) modulation and demodulation
Answer:
(d) modulation and demodulation

X. Two Marks Questions:

Question 1.
What is a communication system?
Answer:
The setup used to transmit information from one point to another is called a communication system.

Question 2.
What is called a baseband signal?
Answer:
The electrical equivalent of the original information.

Question 3.
What is a transducer?
Answer:
A device that converts variations in a physical quantity into an equivalent electrical signal.

Question 4.
Define bandwidth.
Answer:
The frequency range over which the baseband signals or the information signals such as voice, music, the picture is transmitted.

Question 5.
Define bandwidth?
Answer:
The frequency range over which the baseband signals or the information signals such as voice, music, picture, etc. is transmitted is known as bandwidth.

Question 6.
What is skip distance?
Answer:
The shortest distance between the transmission and the point of reception of the skywave along the surface.

Question 7.
What is skip zone or skip area?
Answer:

• The zone in which there is no reception of electromagnetic waves.
• It is also called as skip area.

Question 8.
What is called fibre optic communication?
Answer:
The method of transmitting information from one place to another in terms of light pulses through an optical fiber.

Question 9.
What is mean by fibre optic communication?
Answer:
The method of transmitting information from one place to another in terms of light pulses through an optical fiber is called fiber optic communication.

Question 10.
Expand GPS and write a note on it.
Answer:

• GPS stands for Global Positioning System.
• It is a global navigation satellite system that offers geolocation and time anywhere on or near the earth.

Question 11.
Why do we need carrier waves of very high frequency in the modulation of signals?
Answer:
High-frequency carrier waves are used

1. to increase operating range
2. to reduce antenna length
3. to convert the wideband signal into the narrowband signal.

Question 12.
Why is modulation needed at all?
Answer:
Modulation is needed.

1. to transmit a low-frequency signal to a distant place.
2. for protecting the waveform of the signal
3. to keep the height of antenna small.

Question 13.
Which is better for high fidelity reception FM or AM?
Answer:
FM transmission gives high fidelity due to the presence of a large number of sidebands.

Question 14.
Why is the transmission of signals through a coaxial cable not possible for frequencies greater than 20 MHz?
Answer:
For frequencies greater than 20 MHz, dielectric loss becomes quite high.

Question 15.
Why are short wave bands used for long-distance transmission of signals?
Answer:
Radio waves of short wave bands can be easily reflected bv the ionosphere. So they are used in long-distance transmission.

Question 16.
Why microphone is used as an input transducer?
Answer:
A transducer is a device that converts sound into an equivalent electrical signal So, a microphone which converts sound into electrical energy is used as a transducer.

Question 17.
What are the elements in the transmitter?
Answer:

1. Amplifier
2. Modulator
3. Oscillator
4. Power amplifier

Question 18.
Draw the block diagram of Transmission.
Answer:

Question 19.
Define range.
Answer:
The maximum distance between the source and the destination upto which the signal is received with sufficient strength.

Question 20.
Mention the application of ICT is fisheries.
Answer:

• Satellite vessel monitoring system helps to identify fishing zones.
• Use of barcodes helps to identify time and date of catch, species name, quality of fish.

Question 21.
What is satellite communication?
Answer:
A mode of communication of signal between transmitter and receiver via satellite.

Question 22.
What are the basic elements required for the transmission and reception of a signal?
Answer:
Transducer, amplifier, carrier signal, modulator, power amplifier, medium of transmission, tansmitting and receiving antenna, demodulator and detector.

Question 23.
Give the reasons for attenuation in the Ground wave propagation.
Answer:

1. Increasing distance
2. Absorption of energy by the earth
3. Tilting of the wave.

XI. Three Marks questions:

Question 1.
What are the advantages of FM?
Answer:

1. Large decrease in noise. An increase in signal noise ratio
2. The operating range is quite large
3. The transmission efficiency is very high
4. FM bandwidth covers the entire frequency range.
5. FM radio has better quality compared to AM radio.

Question 2.
What are the disadvantages of FM?
Answer:

1. FM requires a much wider channel
2. FM transmitters and receivers are more complex and costly.
3. In FM reception, less area is covered compared to AM.

Question 3.
Give the reason why transmission of TV signals via sky wave is not possible.
Answer:

1. Television frequencies lie in the range 100 – 220 MHz which cannot be reflected by the ionosphere.
2. So, sky wave propgation is not used in TV transmission.

Question 4.
Show in diagram, the skip distance and the skip zone.
Answer:

Question 5.
It is necessary to use satellites for long distance TV transmission. Why?
Answer:

1. TV signals being of high frequency are not reflected by the ionosphere.
2. Ground wave transmission is possible only up to a limited range.
3. So, satellites are used for long distance TV transmission.

Question 6.
Which is more efficient mode of transmission FM or AM?
Answer:

1. FM transmission is more efficient because all the transmitted power is useful.
2. In AM transmission, most of the power goes waste in transmitting the carrier alone.

Question 7.
Give an account of E-commerce and search engine.
Answer:
E-Commerce:
Buying and selling of goods and services, transfer of funds are done by an electronic network.
Search engine:
A web-based service tool used to search for information on World Wide Web.

Question 8.
Expand GSM. Give a note on it.
Answer:

• GSM – Global System for Mobile Communication.
• Increases the utilization of bandwidth of the network, sharing of the networks, error detections etc.

Question 9.
Mention some of the applications of optical fiber system.
Answer:

1. International Communication
2. Intercity communication
3. data links
4. plant and traffic control
5. defence applications.

Question 10.
What are repeaters in an electronic communication system?
Answer:

1. Repeaters are a combination of transmitter and receivers.
2. Used to increase the range or distance through which the signals are sent.
3. The received signals are transmitted with a carrier signal of different frequencies to the destination.

Question 11.
State two factors by which the range of transmission of T.V. signals can be increased.
Answer:
The range of TV transmission can be increased by using

• tall antenna and
• geostationary satellites.

Question 12.
Compare the difference between PM and FM.
Answer:

Phase Modulation	Frequency Modulation
1. Smaller bandwidth	Larger bandwidth
2. High transmission speed	Low transmission speed
3. More information can be sent.	Less information can be sent

Question 13.
Why should transmitters broadcasting programmes use different carrier frequencies?
Answer:

1. Different audio signals fall in the same spectral range.
2. Different transmitting stations are allowed different slots in radio frequency range.
3. A single receiver can tune into these frequencies without any confusion or overlap.

Question 14.
Give one example each of a system that uses the (i) sky – wave (ii) space wave mode of propagation.
Answer:

1. Short broadcast services use skywave propagation.
2. TV broadcast, microwaves links, satellite communication use space wave propagation.

Question 15.
A transmitting antenna has a height of 40m and the height of the receiving antenna is 30m. What is the maximum distance between them for line-of-sight communication?
[The radius of the earth is 6.4 × 10⁶ m]
Solution:
The total distance d between the transmitting and receiving antennas will be the sum of the individual distances of coverage,
d = d₁ + d₂
= \(\sqrt{2 R}\left(\sqrt{h_{1}}+\sqrt{h_{2}}\right)\)
= \(\sqrt{2 \times 6.4 \times 10^{6}}\)
= 16 × 10²√5 × (6.32+ 5.48)
= 42217m
d = 42.217 km

XII. Five Mark Questions:

Question 1.
Explain mobile communication? Write its applications.
Answer:
Mobile communication is used to communicate with others in different locations without the use of any physical connection like wires or cables. It allows the transmission over a wide range of area without the use of the physical link. It enables people to communicate with each other regardless of a particular location like office, house, etc. It also provides communication access to remote areas.

It provides the facility of roaming:
that is. the user may move from one place to another without the need of compromising on the communication. The maintenance and cost of installation of this communication network are also cheap.

Applications:

1. It is used for personal communication and cellular phones offer voice and data connectivity with high speed.
2. Transmission of news across the globe is done within a few seconds.
3. Using the Internet of Things (IoT), it is made possible to control various devices from a single device.
   Example: home automation using a mobile phone.
4. It enables smart classrooms, online availability of notes, monitoring student activities etc. in the field of education.

Question 2.
Give a brief account on RADAR and its applications.
Answer:

1. RADAR stands for Radio Detection and Ranging system.
2. The angle, range or velocity of the objects which are invisible to the human eye can be determined.
3. Electromagnetic waves are used for communication.
4. Signal is initially radiated into space by an antenna in all direction.
5. Then it gets reflected or reradiated in many directions as it strikes the object.
6. Radar antenna receives the reflected radio signal and it is delivered to the receiver.
7. These signals are processed and amplified to determine the geographical statistics of the object.
8. the range is determined by calculating the time taken by the signal to travel from RADAR to the target and back.

Applications:
RADAR is used

1. to sense, detect and distant objects like aircraft, ships, spacecraft etc.
2. in military for locating and detecting the targets.
3. in navigation systems such as ship-borne surface search, air search and weapons guidance system.
4. to measure precipitation rate and wind speed in meteorological observation
5. to locate and rescue people in emergency situations.

Question 3.
What do you know about mobile communication? Give its applications.
Answer:

1. Mobile communication is used to communicate with others in different locations without the use of any physical connection like wires or cables.
2. It allows the transmission over a wide range of areas without the use of the physical link.
3. It also provides communication access to remote areas.
4. It provides the facility of roaming.
5. User may move from one place to another without the need of compromising on the communication.
6. The maintenance and cost of installation are cheap.

Applications:

1. It is used to communicate with each other regardless of a particular location like office, house etc.
2. It is used for personal communication.
3. Cellular phones offer voice and data connectivity with high speed.
4. Transmission of news across the globe is done within a few seconds.
5. IoT is used to control various devices from a single device
   Ex. home automation using a mobile phone.
6. It enables smart classrooms, online availability of notes, monitoring student activities etc in the field of education.

Question 4.
What is bandwidth? Explain the bandwidth of the transmission system.
Answer:

1. The frequency range over which the baseband signals or the information signals such as voice, music, picture, etc is transmitted.
2. Each of these signals has different frequencies. The communication system depends on the nature of the frequency band for a given signal.
3. Bandwidth gives the difference between the upper and lower frequency limits df the signal.
4. The portion of the electromagnetic spectrum occupied by the signal.
5. If γ₁ and γ₂ are the lower and upper-frequency limits of a signal.
6. Bandwidth, BW = γ₂ – γ₁

Bandwidth of Transmission system:

1. The range of frequencies required to transmit a piece of specified information in a particular channel.
2. An amplitude modulation system requires a channel bandwidth of 10 kHz to transmit a 5 kHz signal.
3. A single side-band system requires only 5 kHz channel bandwidth for a 5 kHz signal.
4. Because in amplitude modulation, the channel bandwidth is twice the signal frequency.
5. It is required to reduce the channel bandwidth to accommodate more channels in the available electromagnetic spectrum.

Question 5.
Explain the process of modulation.
Answer:

1. For long-distance transmission, the low-frequency baseband signal is superimposed onto a high-frequency radio signal.
2. The energy of the information signal is sufficient to send directly.
3. A very high-frequency signal called carrier signal is used to carry the baseband signal.
4. As the frequency of the carrier signal is very high, it can be transmitted to long distances with less attenuation.
5. The carrier wave is a sine wave signal.
6. The carrier signal will be more compatible.
7. It propagates in free space with greater efficiency.
8. A sine wave can be represented as
   e_c = E_c sin (2πγ_ct + φ)
   Where, E_c – amplitude γ_c – frequency
   φ – the initial phase of the carrier wave at time ‘t’
9. The characteristics of the carrier signal are modified in three types.
   • Amplitude modulation
   • Frequency modulation
   • Phase modulation.

Question 6.
What do you know about INTERNET? Write its few applications?
Answer:
Internet is a fast-growing technology in the field of a communication system with multifaceted tools. It provides new ways and means to interact and connect with people. Internet is the largest computer network recognized globally that connects millions of people through computers. It finds extensive applications in all walks of life.

Applications:

1. Search engine: The search engine is basically a web-based service tool used to search for information on World Wide Web.
2. Communication: It helps millions of people to connect with the use of social networking: emails, instant messaging services, and social networking tools.
3. E-Commerce: Buying and selling of goods and services, transfer of funds are done over an electronic network.

Tamilnadu State Board New Syllabus Samacheer Kalvi 12th Physics Guide Pdf Chapter 9 Semiconductor Electronics Text Book Back Questions and Answers, Notes.

Tamilnadu Samacheer Kalvi 12th Physics Solutions Chapter 9 Semiconductor Electronics

12th Physics Guide Semiconductor Electronics Text Book Back Questions and Answers

Part I:

Textbook Evaluation:

I. Multiple choice questions:

Question 1.
The barrier potential of a silicon diode is approximately
a) 0.7 V
b) 0.3V
c) 2.0 V
d) 2.2V
Answer:
a) 0.7 V

Question 2.
Doping a semiconductor results in
a) The decrease in mobile charge carriers
b) The change in chemical properties
c) The change in the crystal structure
d) The breaking of the covalent bond
Answer:
c) The change in the crystal structure

Question 3.
A forward-biased diode is treated as
a) An open switch with infinite resistance
b) A closed switch with a voltage drop of 0V
c) A closed switch in series with a battery voltage of 0.7V
d) A closed switch in series with a small resistance and a battery.
Answer:
d) A closed switch in series with a small resistance and a battery.

Question 4.
If a half -wave rectified voltage is fed to a load resistor, which part of a cycle the load current will flow?
a) 0° – 90°
b) 90° – 180°
c) 0° – 180°
d) 0° – 360°
Answer:
c) 0°-180°

Question 5.
The primary use of a zener diode is
a) Rectifier
b) Amplifier
c) Oscillator
d) Voltage regulator
Answer:
d) Voltage regulator

Question 6.
The principle in which a solar cell operates
a) Diffusion
b) Recombination
c) Photovoltaic action
d) Carrier flow
Answer:
c) Photovoltaic action

Question 7.
The light emitted in an LED is due to
a) Recombination of charge carriers
b) Reflection of light due to lens action
c) Amplification of light falling at the junction
d) Large current capacity.
Answer:
a) Recombination of charge carriers

Question 8.
When a transistor is fully switched on, it is said to be
a) Shorted
b) Saturated
c) Cut-off
d) Open
Answer:
b) Saturated

Question 9.
The specific characteristic of a common emitter amplifier is
a) High input resistance
b) Low power gain
c) Signal phase reversal
d) Low current gain
Answer:
c) Signal phase reversal

Question 10.
To obtain sustained oscillation in an oscillator,
a) Feedback should be positive
b) Feedback factor must be unity
c) Phase shift must be 0 or 2π
d) All the above
Answer:
d) All the above

Question 11.
If the input to the NOT gate is A = 1011, its output is
a) 0100
b) 1000
c) 1100
d) 0011
Answer:
a) 0100

Question 12.
The electrical series circuit in digital form is
a) AND
b) OR
c) NOR
d) NAND
Answer:
a) AND

Question 13.
Which one of the following represents a forward bias diode? (NEET)

Answer:
a) A node must have a high potential

Question 14.
The given electrical network is equivalent to

a) AND gate
b) OR gate
c) NOR gate
d) NOT gate
Answer:
c) NOR gate

Question 15.
The output of the following circuit is 1 when the input ABC is

a) 101
b) 100
c) 110
d) 010
Answer:
a) 101

II. Short Answer Questions:

Question 1.
Define electron motion in a semiconductor.
Answer:
To move the hole in a given direction, the valence electrons move in the opposite direction. Electron flow in an N-type semiconductor is similar to electrons moving in a metallic wire. The N-type dopant atoms will yield electrons available for conduction.

Question 2.
Distinguish between intrinsic and extrinsic semiconductors.
Answer:

Intrinsic Semiconductor	Extrinsic Semiconductor
1. It is pure form of semiconductor	Small amount of impurity is added
2. No doping takes place in intrinsic semiconductor	Doping takes place
3. Number of free electrons in conduction is equal to number of holes in valence band.	Number of free electrons and holes are not equal
4. It has bad electrical characteristics	It has good electrical conductivity

Question 3.
What do you mean by doping?
Answer:
The process of adding impurities to the intrinsic semiconductor is called doping.

Question 4.
How electron-hole pairs are created in a semiconductor material?
Answer:
When external voltage is applied in semiconductor free electrons from valence band moves to conduction band and recombined with holes. So electron-hole pairs are created.

Question 5.
A diode is called a unidirectional device. Explain
Answer:
A diode is called a unidirectional device, i.e., current flows in only one direction (anode to cathode internally) when a forward voltage is applied, the diode conducts and when a reverse voltage is applied, there is no conduction. A mechanical analogy is a rat chat, which allows motion in one direction only.

Question 6.
What do you mean by leakage current in a diode?
Answer:

1. Under biasing very small current (μA) flows across the junction in a diode.
2. It is due to minority charge carriers.
3. This current is called leakage current or reverse saturation current.

Question 7.
Draw the output waveform of a full-wave rectifier.
Answer:

Input and output waveforms

Question 8.
Distinguish between avalanche and Zener breakdown.
Answer:

Avalanche breakdown	Zener breakdown
1. The phenomenon of increasing the free electrons or current in the semiconductor by applying a higher voltage	The process in which electrons are moving across the barrier from valence band of P-type material to conduction band of lightly filled n -material
2. Depletion layer is thick	Depletion layer is thin
3. Electric field is weak	Electric field is strong
4. Produces pairs of electrons and holes	Produces electrons
5. Doping is low	Doping is heavy
6. Breakdown voltage vary	Voltage remains constant

Question 9.
Discuss the biasing polarities in NPN and PNP transistors.
Answer:
In a PNP transistor, base and collector will be negative with respect to emitter indicated by the middle letter N whereas base and collector will be positive in an NPN transistor [indicated by the middle letter P.

Question 10.
Explain the current flow in an NPN transistor.
Answer:

1. In NPN transistor electron flow from Emitter to collector. So conventional current flow from collector to emitter.
2. Electrons from emitter region flow towards base region constitute emitter current (I_E). Electrons after reaching base region recombine with holes – Most of electrons reach collector region. This constitute collector current (I_c). After recombination of holes in base region by bias voltage V_EE constitute base current I_B.
   I_E = I_B + I_C [Since base current is very low I_c ≈ I_C]

Question 11.
What is the phase relationship between the AC input and output voltages in a common emitter amplifier? What is the reason for the phase reversal?
Answer:
In a common emitter amplifier, the input and output voltages are 180° out of phase or in, opposite phases. The reason for this can be seen from the fact that as the input voltage rises, so the current increases through the base circuit.

Question 12.
Explain the need for a feedback circuit in a transistor oscillator.
Answer:
If the portion of the output fed to the input is in phase with the input, then the magnitude of the input increases. It is necessary for sustained oscillations, so feed back circuit is needed for transistor oscillator.

Question 13.
Give circuit symbol, logical operation, truth table, and Boolean expression of AND, OR, NOT, NAND, NOR, and EX-OR gates.
Answer:

Question 14.
State De Morgan’s first and second theorems.
Answer:
First Theorem:
The complement of the sum is equal to the product of its complements
\(\overline{\mathrm{A}+\mathrm{B}}=\overline{\mathrm{A}} \cdot \overline{\mathrm{B}}\)
Second Theorem:
The complement of the product is equal to sum of its complements
\(\overline{A \cdot B}=\bar{A}+\bar{B}\)

III. Long Answer Questions:

Question 1.
Elucidate the formation of a N – type and P – type semiconductors.
Answer:
1. N – type semiconductor:
N – type semiconductor is obtained by doping a pure Germanium or silicon with dopant from group V pentavalent elements like phosphorous, Arsenic, Antimony.

n-type extrinsic semiconductor:
(a) Free electron which is loosely attached to the lattice
(b) Representation of donar energy level.

1. Dopants has 5 valence electrons, Germanium has 4 valence electrons.
2. During the process of doping few Ge atoms’ are replaced by group V dopants.
3. Four of the five valence electrons of impurities bound with 4 valence electrons of neighbouring Ge atoms.
4. The 5th valence electron of impurity atom is loosely attached, it has not formed covalent bond.
5. Energy level loosely attached electron is just below the conduction band edge which is called donor energy level.
6. At room temperature electrons easily moves to conduction band with absorption of thermal energy.
7. These thermally generated electrons leave holes in valance band.
8. Such a semiconductor doped with a pentavalent impurity is called n-type semiconductor.

2. P-type semiconductor.
1. Trivalent atom from group III elements such as Boron, Aluminium, Gallium, Indium is added with Germanium or silicon is a p-type semiconductor.
2. The dopants with 3 valence electrons bound with neighbouring Germanium atom one electron position of the dopant in Ge lattice will remain vacant.
3. The missing electron position in covalent band is denoted as a hole.
4. To make complete covalent bond with all 4 neighbouring atom, dopant needs one more electrons.
5. Dopants accept electron from neighbouring atoms. This impurity is called acceptor impurity.
6. Energy level of hole created by each impurity atom is just above valence band, which is called acceptor energy level.
7. In such extrinsic semiconductor holes are majority carriers and thermally generated electrons are minority carriers. This semiconductor is called a p-type semiconductor.

P-type extrinsic semiconductor
(a) Hole generated by the dopant
(b) Representation of acceptor energy level.

Question 2.
Explain the formation of PN junction diode. Discuss its V-I characteristics.
Answer:

p-n junction diode
(a) Schematic representation
(b) Circuit symbol

i. A p-n junction diode is formed when p-type semiconductor is fused with N-type semiconductor.
ii. Depending on polarity of external source to the p-n junction there are two types of biasing
1. Forward bias
2. Reverse bias

1. Forward bias:
i. If positive terminal of external voltage source is connected to p-side and negative terminal to n-side forward bias takes place.
ii. Electron moves to n-side holes move to p side Recombination takes place near junction and reduce depletion region.
iii. Electron from n-side accelerate towards p side it experience reduced potential barrier at junction.
iv. Applied voltage is increased, width of depletion region and barrier potential further reduced.
v. So large number of electrons pass through junction.

2. Reverse bias:
i. If positive terminal of external voltage source is connected to p-side and negative terminal to n-side reverse bias takes place.

ii. Depletion region is increased potential barrier is also increased.
iii. Majority charge carriers from both sides experience a great barrier to cross the junction. So diffusion current reduces.
iv. The current flows under reverse bias is called reverse saturation current I_S

V-I Characteristics:

1. Forward characteristics:
i. A graph is plotted by taking forward bias voltage in X axis and current i Y axis.
ii. Current flow is negligible when applied voltage is less than threshold voltage beyond that increase in current is significant even for small increase in voltage.
iii. Graph shows clearly current flow is non linear. It doesnot obey ohm’s law.
iv. Forward resistance r_f = \(\frac{\Delta \mathrm{V}}{\Delta \mathrm{I}}\)
v. Diode behaves as conductor when it is forward bias.

p-n junction diode
(a) diode under forward bias
(b) forward characteristics

2. Reverse characteristics:
i. In reverse bias p-region is connected to negative terminal and n-region to positive terminal, Graph is drawn for reverse bias characteristics.
ii. Very small current in pA flows across junction. This is due to leakage current or reverse saturation current.
iii. The reverse bias voltage can be increased only up to rated value otherwise diode will enter into breakdown region.

p-n junction diode
(a) diode under reverse bias
(b) reverse characteristics

Forward and reverse characteristics of a diode

Question 3.
Draw the circuit diagram of a half wave rectifier and explain its working.
Answer:
i. The circuit consist of transformer, p-n junction diode and resistor.
ii. In half wave rectifier either a positive half or the negative half of AC input is passed through it, other half is blocked.

(a) Input signal
(b) half wave rectifier circuit
(c) input and output waveforms

During positive half cycle:
1. When positive half cycle of input signal passes. A becomes positive with respect to B.
2. Diode is forward bias, current flows through R_L output voltage V₀ is developed.
3. Waveform is shown in figure.

During negative half cycle:
1. When negative half cycle of input signal passes A is negative with respect to B.
2. Diode is reverse bias, does not conduct No current is passed through R_L. No voltage drop across R_L.
3. Negative half cycle of ac supply is suppressed at output waveform is shown in figure.
4. Efficiency of half wave rectifier is 40.67%.

Question 4.
Explain the construction and working of a full wave rectifier.
Answer:
i. Full wave rectifier consist two p-n junction diodes a center tapped transformer and load resistor (R_L)
ii. Due to centre tap transformer output voltage rectified by each diode is only one half of total secondary voltage.

(a) Full wave rectifier circuit
(b) Input and output waveforms

During positive half cycle:
1. Input signal passes through the circuit M is positive, G is zero, N is negative potential.
2. D₁ forward bias, D₂ reverse bias.
3. Current flows through MD₁ AGC
4. Positive half cycle of voltage appears across R_L in direction G to C.

During negative half cycle:
1. Terminal N is positive, G is at zero, M is negative potential.
2. D₂ is forward bias, D₁ is reverse bias.
3. Current flows through ND₂BGC
4. Negative half cycle of voltage across R_L from G to C.
5. Hence in full wave rectifiers both postive and negative half cycle of input signals pass through the circuit in same direction.
6. Efficiency of full wave rectifier is 31.2%.

Question 5.
What is an LED? Give the principle of operation with a diagrajm?
Answer:
1. LED is a p-n junction diode which emit visible or invisible light when it is forward bias.
2. It converts electrical energy to light energy.
3. It consists of p-layer, n-layer and substrate. External resistance in series with biasing source is required to limit forward current through LED. It has anode and cathode.

(a) Circuit symbol of LED
(b) Inside view of LED
(c) Schematic diagram to explain recombination process

4. When p-n junction is forward bias, conduction band electron on n-side and valence band hole on p side diffuse across the junction.
5. When they cross junction, they become excess minortiy carriers. These excess minority carriers recombine with opposite charged majority carriers in respective region.
6. During recombination process energy is released is the form of light or heat.
7. The colour of the light is determined by energy band gap of the material.
8. LED’s with wide range colours such as blue (Sic), green (AlGaP), red(GaAsP), white(GaInN) is also available.

Question 6.
Write notes on Photodiode.
Answer:
                     
(a) Circuit symbol
(b) Schematic view of photodiode

1. Photodiode is a p-n junction diode which converts an optical signal into electric current.
2. It works in reverse bias.
3. It consists of F p-n junction semiconductor made of photosensitive material kept safely inside a plastic case.
4. It has small transparent window to allow light.
5. When photon (hυ) strikes depletion region of diode some valence band electrons elevated to conduction band. In turns holes are developed in valence band. This creates electron hole pair.
6. Amount of electron – hole pair generated depends on intensity of light.
7. These electrons and hole swept across the junction by electric field. Thus holes moves to -n side, electrons to p side.
8. When external circuit is made, electrons flow through external circuit and constitute the photo current.

Applications:
Alarm system, count items in conveyer belt, photo conductors, CD players, smoke detectors, detectors for computed tomography etc.

Question 7.
Explain the working principle of a solar cell. Mention its applications.
Answer:
11. Solar cell converts light energy directly into electricity or electric potential difference by photovoltaic effect.

Cross-sectional view of a solar cell

2. It generates emf when radiations falls on p-n junction. A solar cell is of two type p-type and n-type.
3. Both types use a combinations of p-type and n-type silicon which together forms the p-n junction.
4. In solar cell, electron – hole pairs are generated due to absorption of light near junction.
5. Electrons move towards n-type silicon, holes moves towards p-type silicon layer.
6. Electrons reaching n-side are collected by front contact and holes reaching p-side are collected by back electrical contact.
7. Thus potential difference is developed across solar cell. When external load is connected, photocurrent flows through it.
8. Many solar cells are connected in series or parallel to form solar panel or module.

Applications:
1. Widely used in calculators, watches, toys, portable power supplies etc.
2. Used in satellites and space stations.
3. Solar panels are used to generate electricity.

Question 8.
Sketch the static characteristics of a common emitter transistor and bring out the essence of input and output characteristics.
Answer:

Static characteristics of a NPN transistor in common emitter configuration

i. The circuit to study the static characteristics of NPN transistor is given in figure
ii. Bias supply voltages V_BB and V_CC bias, base – emitter junction and collector – emitter junction
iii. Junction potentials are V_BE and Y_CE
iv. R₁ and R₂ are used to vary base and collector currents respectively.

1. Input Characteristics:
i. Input Characteristics curve gives relationship between I_B and V_BE at constant V_CE
ii. For constant collector emitter voltage V_CE, Base emitter voltage V_BE increases for corresponding Base current I_B which is recorded and graph is ploted.
iii. The curve looks like forward characteristics

Input characteristics of a NPN transistor in common emitter configuration

iv. Beyond knee voltage base current increases with increase in base emitter voltage for silicon 0.7 V & for Germanium 0.3 V
v. Increase is V_CE decreases I_B. This shift the curves outward.
vi. Input resistance R_i = \(\left(\frac{\Delta V_{B E}}{\Delta I_{B}}\right)_{V_{C B}}\)

2. Output characteristics:

Output characteristics of a NPN transistor in common emitter configuration

i. The output characteristics gives the relation between ∆I_C with respect to ∆V_CE at constant I_B
ii. Initially base current I_B is set to particular value. Increasing collector emitter voltage V_CE corresponding collector current I_C increases. A graph is plotted.

iii. Saturation region:
When V_CE increased above 0V, I_C increases rapidly almost independent of I_B called knee voltage Transistor operated above this knee voltage

iv. Cut off region:
A small I_C exist even after I_B is reduced to zero. This current is due to presence of minority carriers across collector – base junction and the surface leakage current (I_CEO). This region is called cut-off region.

v. Active region:
In this region emitter – base junction is forward bias, collector – base junction is reverse bias. Transistor in this region can be used for voltage, current and power amplification.

vi. Breakdown region:
If V_CE is increased beyond rated value, given I_c increases enormously leading to junction breakdown of transistor. This avalanche breakdown can damages the transistor.
vii. Output resistance R_o = \(\left(\frac{\Delta V_{C E}}{\Delta I_{C}}\right)_{I_{B}}\)

Question 9.
Describe the function of a transistor as an amplifier with the neat circuit diagram. Sketch the input and output wave form.
Answer:
i. Amplification is the process of increasing the signal strength (increase in the amplitude)
ii. NPN transistor is connected in CE configuration
iii. To start with, Q point is fixed to get maximum signal swing at the output.
iv. R_C – to measure output voltage
C₁ allows only AC signals to pass,
Bypass capacitor C_E provides a low resistance path
Coupling capacitor C_C is used to couples next stage amplifier.
V_s input source signal
I_C = BI_B (∴ B = \(\frac{I_{C}}{I_{B}}\))
V_CE = V_CC – I_CR_C

                       
(a)Transistor as an amplifier
(b) Input and output waveform showing 180° phase reversal.

Working of the amplifier:

During positive half cycle:
1. Input signal V_s increases the forward voltage across emitter base. I_B increases. also increases I_C also increases β times.
2. This increase the voltage drop across R_c which decreaseV_CE.
3. Therefore input signal in the positive direction produces an amplified signal in negative direction at the output. Hence output signal is reversed by 180°

During negative half cycle:
1. Input signal V_s decreases the forwarded voltage across emitter – base. As a result I_B decreases, I_C increases.
2. Increase in I_C decreases potential drop across R_c and increases V_CE.
3. Input signal in negative direction produces amplified signal in the positive direction at the output.
4. Therefore 180° phase reversal is observed during negative half signal.

Question 10.
Transistor functions as a switch. Explain.
Answer:
1. In saturation and cut-off region transistor works as an electronic switch that helps to turn ON or OFF a circuit.
2. presence of dc source at the input (saturation region):

Transistor as a switch

3. When high input voltage (V_in = +5V) is applied, I_B increases and in turn increases I_C.
4. Transistor will move into saturation region (turned ON)
5. Increase in I_C increases voltage drop across R_c lowering the output voltage close to zero.
6. Transistor acts like a closed switch (ON condition)

Absence of dc source at the input (cut-off region) :
1. At low input voltage (V_in = 0V), decreases the I_B and in turn decreases I_C.
2. Transistor moves to cut off region (turned ON)
3. Decrease in I_C decreases voltage drop across R_c, increasing output voltage +5V transistor acts as open switch (OFF condition)
4. It is manifested that high input gives low output and low input gives high output.

Question 11.
State Boolean laws. Elucidate how they are used to simplify Boolean expressions with suitable example.
Answer:
1. Complement law:
\(\overline{\mathrm{A}}\) = A
A Y = \(\overline{\mathrm{A}}\)
O Y = \(\overline{\mathrm{O}}\) = 1
I Y = \(\overline{\mathrm{I}}\) = \(\overline{\mathrm{O}}\)

2. OR laws:

3. AND laws:

4. Commutative laws:
A + B = B + A
A.B = B.A

5. Associative laws
A + (B + C) = (A + B) + C
A.(B.C) = (A.B).C

6. Distributive laws
A (B + C) = AB + AC
A + BC = (A + B) (A + C)

Question 12.
State and prove De Morgan’s First and Second theorems.
Answer:
First Theorem:
The complement of the sum is equal to the product of its complements
\(\overline{A+B}=\bar{A} \cdot \bar{B}\)

Second Theorem:
The complement of the product of two inputs is equal to the sum of its complements.
\(\overline{\mathrm{A} \cdot \mathrm{B}}=\overline{\mathrm{A}}+\overline{\mathrm{B}}\)

IV. Numerical Problems:

Question 1.
The given circuit has two ideal diodes connected as shown in figure below. Calculate the current flowing through the resistance R₁

Solution:
D₁ will act as Reverse bias so current will not pass through it
Total Resistance = 4Ω
V = 10 V
Current flow through R₁I = \(\frac{\mathrm{V}}{\mathrm{R}}\)
= \(\frac{10}{4}\) = 2.5 A

Question 2.
Four silicon diodes and a 10Ω resistor are connected as shown in figure below. Each diode has a resistance of 1Ω. Find the current flows through the 18Ω resistor.

Solution:
Current flow through D₁ and D₃ = 2Ω
Current flow through D₂ and D₄ = 2Ω
These two are in parallel so
Net resistants = \(\frac{1}{2}\) + \(\frac{1}{2}\) = 1Ω
Total Resistance = 1Ω+ 18Ω = 19Ω
V = 2.5Ω
I = \(\frac{\mathrm{V}}{\mathrm{R}}\) = \(\frac{2.5}{19}\)
I = 0.13 A

Question 3.
Assuming V_CESat = 0.2 V and β = 50, find the minimum base current (I_B) required to drive the transistor given in the figure to saturation.

Solution:
V_CC = 3V
β = 50
I_B =?
Collector current
V_CC = 3 – 0.2 = 2.8 V
I_C = \(\frac{\mathrm{V}_{\mathrm{CC}}}{\mathrm{R}_{\mathrm{c}}}\)
= \(\frac{2.8}{1 \times 10^{3}}\) = 2.8 × 10^-3 A

β = \(\frac{\mathrm{I}_{\mathrm{c}}}{\mathrm{I}_{\mathrm{B}}}\)
I_B = \(\frac{\mathrm{I}_{\mathrm{c}}}{\beta}\) = \(\frac{2.8 \times 10^{-3}}{50}\)
= 56 × 10^-6 A
I_B = 56 µA

Question 4.
A transistor having α = 0.99 and V_BE = 0.7 V, is given in the circuit. Find the value of the collector current.

Solution:
V_cc = 12 V
α = 0.99
V_BE = 0.7 V
I_C =?
Applying Kirchoff’s voltage law,
V_BE + (I_C + I_B) 1k + 10k.I_B + (I_C + I_B) 1k = 12 ………………(1)
I_B = \(\frac{\mathrm{I}_{\mathrm{C}}}{\beta}\)

β = \(\frac{\alpha}{1-\alpha}\) = \(\frac{0.99}{1-0.99}\) = 0.99

I_B = \(\frac{\mathrm{I}_{\mathrm{C}}}{99}\)

I_C = β I_B
I_C = 99 I_B
I_C = \(\frac{\mathrm{I}_{\mathrm{C}}}{99}\)
V_BE = 0.7 V
Substitute in equation (1)
\(\frac{0.7}{10^{3}}\) + (99 I_B + I_B) + 10 I_B + (99 I_B + I_B) = \(\frac{12}{10^{3}}\)
100 I_B + 10 I_B + 100 I_B = \(\frac{12-0.7}{10^{3}}\)
210 I_B = \(\frac{11.3}{10^{3}}\) ;
I_B = 0.054 × 10^-3 A
I_C = 99 I_B = 99 × 0.054 × 10^-3 A
I_C = 5.34 × 10^-3 A
I_C = 5.34 mA

Question 5.
In the circuit shown in the figure, the BJT has a current gain (β) of 50. For an emitter-base voltage V_EB = 600 mV, calculate the emitter-collector voltage V_EC (in volts)

Solution:
β = 50
V_EB = 600 × 10^-3V
R_B = 60 kΩ
R_C = 500 kΩ
V_B = V_E – V_EB
= 3 – 0.6 = 2.4 V
I_B = \(\frac{\mathrm{V}_{\mathrm{B}}}{\mathrm{R}_{\mathrm{B}}}\)
= \(\frac{2.4}{60 \times 10^{3}}\)
= 0.04 × 10^-3 A
I_B = 40 µA
I_C = β I_B
= 50 × 40 × 10^-6
I_C = 2mA
V_C = I_C R_C = 500I_C
= 500 × 2 × 10^-3
V_EC = V_E – V_C
= 3 – 1 = 2V

Part II:

12th Physics Guide Semiconductor Electronics Additional Questions and Answers

I. Choose the correct answer:

Question 1.
A transistor has α = 0.95, it has change in emitter current of 100 milliampere then the change in collector current is
a) 95 mA
b) 99.05 mA
c) 100.95 mA
d) 100 mA
Answer:
a) 95 mA
Solution:
α = \(\frac{\Delta \mathrm{I}_{\mathrm{c}}}{\Delta \mathrm{I}_{\mathrm{e}}}\)
∆I_C = ∆I_e × α
= 100 × 0.95
= 95 mA

Question 2.
Diode can works as
a) Demodulator
b) Modulator
c) Amplifier
d) Rectifier
Answer:
a) Demodulator

Question 3.
Three amplifiers of each with gain 10 volt are connected in series then the overall amplification is ______.
a) 10/3
b) 13
c) 1000
d) none of these
Answer:
c) 1000
Solution:
10 × 10 × 10 = 1000

Question 4.
A Transistor has an α = 0.95 then β is
a) 1/19
b) 19
c) 1.5
d) 0.95
Answer:
b) 19
Solution:
β = \(\frac{\alpha}{1-\alpha}\) = \(\frac{0.95}{1-0.95}\)

Question 5.
In p – type semiconductor germanium is doped with _______.
a) aluminium
b) boron
c) gallium
d) all of these
Answer:
d) all of these

Question 6.
For an npn transistor, the collector current is 10 mA. If 90% of electron emitted reach the collector then
a) base current will be 1 mA
b) base current will be 10 mA
c) emitter current will be 11 mA
d) emitter current will be 9 mA
Answer:
a) & c)
Solution:
0.9 I_e10 = I_C
I_e = \(\frac{10}{0.9}\) = 11 mA
I_B = 11 – 10 = 1 mA

Question 7.
The depletion layer in p-n junction region is caused by _______.
a) drift of holes
b) diffusion of charge carriers
c) drift
d) drift of electrons
Answer:
b) diffusion of charge carriers

Question 8.
When – n – type semiconductor is heated ______.
a) number of electrons only increase
b) number of holes increase
c) number of electrons and holes remains same
d) number of electrons and holes increase equally
Answer:
d) number of electrons and holes increase equally

Question 9.
The device that can act as complete electronic circuit is _______.
a) junction diode
b) IC
c) transistor
d) diode
Answer:
b) IC

Question 10.
In following circuit the output Y for all possible input A and B is expressed by truth table

a)

b)

c)

d)

Answer:
c)

Y’ = \(\overline{A+B}\)
Y = \(\overline{Y^{\prime}}\) = \(\overline{\overline{A+B}}\)
Y = A + B

Question 11.
Which one is correct for reverse bias is applied in junction diode
a) Increase minority career
b) lower potential barrier
c) raise in potential barrier
d) increase in majority carrier
Answer:
c) raise in potential barrier

Question 12.
In common emitter configuration V_c = 1.5 V_c change in base current from 100 µA to 150 µA produce change in collector current from 5 mA to 10 mA. The current gain (β) is
a) 75
b) 100
c) 50
d) 67
Answer:
b) 100
Solution:
∆I_B = 50 µA
∆I_C = 5 mA
β = \(\frac{\Delta \mathrm{I}_{\mathrm{B}}}{\Delta \mathrm{I}_{C}}\)
= \(\frac{5 \mathrm{~mA}}{50 \mu \mathrm{A}}\)
= 100

Question 13.
To use a transistor as an amplifier
a) the emitter base junction is forward biased and base collector junction is reverse biased
b) no bias voltage is required
c) both junction are forward bias
d) both junctions are reverse bias
Answer:
a) the emitter base junction is forward biased and base collector junction is reverse biased

Question 14.
The circuit shown in figure contains two diodes each with forward resistance of 50Ω and with infinite backward resistance. If battery voltage is 6V the current through 100Ω resistance is

a) Zero
b) 0.02 mA
c) 0.03 mA
d) 0.36 mA
Answer:
b) 0.02 mA
Solution:
Current will not pass through D₂
Total resisatnce = 50 + 150 + 100 = 300Ω
Current = \(\frac{6}{300}\) = 0.02 V

Question 15.
Type of material which emits white light in LED
a) GaInN
b) Sic
c) AlGaP
d) GaInP
Answer:
a) GalnN

Question 16.
Blue colour LED is made up of
a) Sic
b) AIGaP
c) GaAsP
d) GaInP
Answer:
a) Sic

Question 17.
The energy band gap is maximum in
a) metals
b) super conductors
c) insulators
d) semiconductors
Answer:
c) insulators

Question 18.
The part of transistor which is mostly heavily doped to produce large majority carriers
a) emitter
b) base
c) collector
d) None of these
Answer:
a) emitter

Question 19.
In the middle of depletion layer of reverse – biased p-n junction
a) electric field is zero
b) potential is maximum
c) electric field is maximum
d) potential is zero
Answer:
c) electric field is maximum

Question 20.
What is the current flowing in the circuit?

a) 1.71 A
b) 2.0 A
c) 2.31 A
d) 1.33 A
Answer:
b) 2.0 A
Solution:
D₁ is reverse bias will not conduct = \(\frac{12}{6}\) = 2 A

Question 21.
A diode as a rectifier converts _______.
a) a.c to d.c
b) d.c to a.c
c) AC only
d) dc only
Answer:
a) ac to d.c

Question 22.
An oscillator is nothing but an amplifier with
a) positive feedback
b) large gain
c) no feed back
d) negative feedback
Answer:
a) positive feedback

Question 23.
In Intrinsic semiconductor at room temperature number of electrons and holes are
a) equal
b) zero
c) unequal
d) infinite
Answer:
a) equal

Question 24.
The level formed due to impurity atom in forbidden energy gap, very near to valence band is p-type semiconductor is called
a) An acceptor level
b) A donor level
c) Conduction level
d) A forbidden level
Answer:
a) An acceptor level

Question 25.
The symbol represents _______ gate.

a) NAND gate
b) OR gate y
c) AND gate
d) NOT gate
Answer:
d) NOT gate

Question 26.
In p-n junction, avalanche current flows is circuit when biasing
a) Forward
b) Reverse
c) Zero
d) Excess
Answer:
b) Reverse

Question 27.
For detecting light, which is correct?
a) Photodiode has to be forward biased
b) Photodiode has to be reversed biased
c) LED has to be connected in forwarded bias
d) LED in reverse bias
Answer:
b) Photodiode has to be reversed biased

Question 28.
What is the output Y in the above circuit, when all the three inputs A, B and C are 0?

a) 0
b) 1
c) 10
d) 11
Answer:
a) 0

Question 29.
Which type of semiconductor device does not need any bias voltage
a) Photodiode
b) Zero diode
c) Solar cell
d) Transistor
Answer:
c) Solar cell

Question 30.
The frequency of oscillator is_____ 2 1
a) f = \(\frac{1}{2 \pi \mathrm{LC}}\)

b) ω² = \(\frac{1}{\mathrm{LC}}\)

c) ω = \(\frac{1}{2 \pi \sqrt{\mathrm{LC}}}\)

d) f = \(\frac{1}{2 \pi \sqrt{\mathrm{LC}}}\)
Answer:
d) \(\frac{1}{2 \pi \sqrt{\mathrm{LC}}}\)

Question 31.
The logic circuit shown in figure represents ______ gate.

a) OR
b) AND
c) NOR
d) NAND
Answer:
d) NAND

Question 32.
Maximum efficiency of full wave rectifier
a) 100%
b) 81.2%
c) 40.6%
d) 95%
Answer:
b) 81.2%

Question 33.
The diffusion current in P-N junction is from
a) p side to n side
b) n side top side
c) Both (a) and (b)
d) None
Answer:
a) p side to n side

Question 34.
The current through an ideal PN junction shown in figure

a) 5 mA
b) 10 mA
c) 70 mA
d) 100 mA
Answer:
b) 10 mA
Solution:
V = 5-(-2)
V = 7V
I = \(\frac{7}{100}\)
I = 10 MA

Question 35.
Efficiency of half wave rectifier is
a) 81.2%
b) 100%
c) 40.6%
d) 95%
Answer:
c) 40.6%

Question 36.
In common base amplifier, phase difference between input voltage and output voltage
a) 0
b) π/4
c) π/2
d) π
Answer:
a) 0

Question 37.
What will be the input of A and B for Boolean expression \(\overline{(A+B)}\) \(\overline{(\mathrm{A} \cdot \mathrm{B})}\) = 1
a) 0,0
b) 0,1
c) 1,0
d) 1,1
Answer:
a) 0,0
Solution:
\(\overline{0} \cdot \overline{0}\) = 1
1.1 = 1

Question 38.
The value of current flowing through AB in this circuit

a) 10 mA
b) 20 mA
c) 15 mA
d) 11 mA
Solution:
PD = 3 -(-7) = 10V
I = \(\frac{10}{1000}\)
= 10^-2 A = 10 mA

Question 39.
To get output, 1 for following circuit. The correct choice for input is ________.

a) A = 0, B = 1, C = 0
b) A = 1, B = 0, C = 0
c) A = 1, B = 1, C = 0
d) A = 1, B = 0, C = 1
Answer:
c) A = 1, B = 1, C = 0
Solution:
Y’ = A+B
Y = (A + B) C = 1
(1 + 0).1 = 1

Question 40.
The given electrical network is equal to

a) AND gate
b) OR gate
c) NOR gate
d) Ex-OR gate
Answer:
c) NOR gate
Y₁ = \(\overline{A+B}\) = \(\overline{\mathrm{A}} \cdot \overline{\mathrm{B}}\)
Y₂ = \(\overline{\mathrm{Y}}_{1}\) = \(\overline{\mathrm{A}} \cdot \overline{\mathrm{B}}\) = A + B
Y = \(\overline{A+B}\)

Question 41.
Which gate will give high input when odd numbers of input are high
a) NAND gate
b) OR gate
c) NOR gate
d) EX-OR gate
Answer:
d) EX-OR gate

Question 42.
The current gain (β) of a transistor in common emitter mode is 40. To change the collector current by 160 mA at constant V_CE, The necessary change in the base current is
a) 0.2 mA
b) 4 µA
c) 4 mA
d) 40 mA
Answer:
c) 4 mA

Question 43.
Which one is Barkhausen condition for sustained Oscillation
a) phase shift 0°
b) loop gain is unity
c) \(|A B|\) = 0
d) Both (a) and (b)
Answer:
d) Both (a) and (b)

Question 44.
Tank circuit consists of ________.
a) Capacitor
b) Inductor
c) Both (a) and (b)
d) Diode
Answer:
c) Both (a) and (b)

Question 45.
The given electrical network is equivalent to:

(a) NAND gate
(b) OR gate
(c) NOT gate
(d) Ex-OR gate
Answer:
b) OR gate

Question 46.
The maximum reverse bias that can be applied before entering into zener region is referred as
a) PIV rating
b) PAV rating
c) RIM rating
d) None
Answer:
a) PIV rating

Question 47.
For reverse voltage between 4 and 6V ___________ are present
a) Zener effect
b) Avalanche effect
c) Both effects
d) None
Answer:
c) Both effects

Question 48.
Solar cell works on _______ effect.
a) Photo emissive
b) Photo electric
c) Photo conducting
d) Photo voltaic
Answer:
d) Photo voltaic

Question 49.
Which one is the symbol for photodiode
a)

b)

c)

d)
Answer:
c)

II. Choose the Correct Statement:

Question 1.
Which of the statement is not true?
a) The resistance of intrinsic semiconductor decreases with increase of temperature
b)Doping pure Si with trivalent impurities give p-type semiconductor.
c) The majority carriers in n type semiconductors are holes
d) A p-n type junction can act as a semiconductor diode
Answer:
c) The majority carriers in n type semiconductors are holes

Question 2.
Which one is correct statement A — p type semiconductor is
i) A silicon crystal is doped with arsenic impurity
ii) Si doped with aluminium impurity
iii) Ge doped with phosphorous impurity
iv) Ge doped with boron impurity
a) (i) and (ii) are correct
b) (ii) and (iii) are correct
c) (i) and (iv) are correct
d) (i) only correct
Answer:
b) (ii) and (iii) are correct

Question 3.
Read the following statement carefully
Y: The resistivity of semiconductor decreases with increase of temperature
Z: The resistivity value of semiconductor is 10^-5 to 10⁶ Ωm
a) Y is true but Z is false
b) Y is false but Z is true
c) Both Y and Z are true
d) Both Y and Z are false
Answer:
c) Both Y and Z are true

Question 4.
Which one of the following statement is correct?
a) The depletion region of P-N junction diode increase with forward bias
b) The depletion region of P-N junction diode decrease with reverse bias
c) The depletion region of PN junction diode does not change with biasing
d) The depletion region of PN junction diode decreases with forward biasing.
Answer:
d) The depletion region of PN junction diode decreases with forward biasing.

Question 5.
Assertion:
The size, capacity of chips are progressed enormously with advancement of technology
Reason:
Computers, mobile phones are possible in small size and low cost of ICs
a) Assertion and Reason are true and Reason is correct explanation of Assertion
b) Assertion and Reason are true but Reason is not correct explanation of Assertion
c) Assertion is true but Reason is false
d) Assertion is false but Reason is true
Answer:
a) Assertion and Reason are true and Reason is correct explanation of Assertion

III. Fill in the blanks:

Question 1.
The reverse saturation current of silicon diode doubles for every _______ rise in temperature.
Answer:
10°C

Question 2.
The external voltage applied to p-n junction is called _______.
Answer:
bias voltage

Question 3.
A semiconductor has ________ resistance coefficient.
Answer:
negative

Question 4.
The most commonly used semiconductor is ____________.
Answer:
silicon

Question 5.
Find the odd one out. Trivalent atoms used in doping
a) Boron
b) Aluminium
c) Indium
d) Arsenic
Answer:
d) Arsenic

Question 6.
Find the odd one out Pentavalent dopant used in n-type semiconductor
a) Phosphorus
b) Antimony
c) Galium
d) Arsenic
Answer:
c) Galium

IV. Match the following:

Question 1.

a) 1 – d 2 – b 3 – a 4 – c
b) 1 – d 2 – a 3 – b 4 – c
c) 1 – d 2 – a 3 – c 4 – b
d) 1- c 2 – a 3 – b 4 – d
Answer:
b) 1 – d 2 – a 3 – b 4 – c

Question 2.

1. Silicon	a. 6 ev
2. Germanium	b. holes
3. Insulator	c. 1.1 ev
4. Deficiency of electron	d. 0.7 ev

a) 1 – c 2 – d 3 – b 4 – a
b) 1 – a 2 – b 3 – c 4 – d
c) 1 – c 2 – d 3 – a 4 – b
d) 1 – b 2 – c 3 – a 4 – d
Answer:
c) 1 – c 2 – d 3 – a 4 – b

V. Two Mark Questions:

Question 1.
Define Energy band.
Answer:
Band of very large number of closely spaced energy levels in a very small energy range is known as energy band.

Question 2.
What is valence band?
Answer:
The energy band formed due to the valence orbits is called valence band.

Question 3.
What is called conduction band?
Answer:
The energy band formed by unoccupied orbits is called conduction band.

Question 4.
What is forbidden energy gap?
Answer:
The energy gap between valence band and conduction band is called forbidden energy gap.

Question 5.
What are passive and active components?
Answer:
Passive components:
Components that cannot generate power in a circuit.
Active components:
components that can generate power in a circuit.

Question 6.
Give example for Pentavalent elements.
Answer:
Phosphorus, Arsenic and Antimony

Question 7.
Write – the examples of trivalent impurities.
Answer:
Boron, Aluminum, Gallium and Indium

Question 8.
Differntiate Donor impurities, Acceptor impurities.
Answer:

Donor Impurity	Acceptor Impurity
Impurity atoms which donate electrons to the conduction band are called Donor impurities. Ex: Pentavalent elements like Phosphorous, Arsenic and Antimony	Dopants (or) Impurities which accept electron from neighbouring atoms are called acceptor impurity. Ex: Trivalent elements like Boron, Aluminium, Gallium, Indium

Question 9.
What is barrier potential?
Answer:
The difference in potential across the depletion layer is called barrier potential.

Question 10.
What are extrinsic semiconductors?
Answer:
An extrinsic semiconductor is a semiconductor doped by a specific impurity which is able to deeply modify its electrical properties, making it suitable for electronic applications (diodes, transistors etc.) or optoelectronic applications (light emitters and detectors).

Question 11.
What is called bias voltage?
Answer:
The external voltage applied to the p-n junction is called bias voltage.

Question 12.
What is called Forward bias and reverse bias in p-n junction diode?
Answer:

1. If positive terminal of external voltage source is connected to p-side and negative terminal to n-side is called forward bias.
2. If positive terminal of battery is connected to n-side and negative terminal to p-side. The junction is said
   to be reverse bias.

Question 13.
What is reverse saturation current?
Answer:
The current that flows under a reverse bias is called the reverse saturation current. It is represented as I_s.

Question 14.
Write down the applications of Zener diode.
Answer:
The Zener diode can be used as:

1. Voltage regulators
2. Peak clippers
3. Calibrating voltages
4. Provide fixed reference voltage in a network for biasing
5. Meter protection against damage from accidental application of excessive voltage.

Question 15.
In the combination of the following gates, write the Boolean equation for output Y in terms of inputs A and B

Answer:
The output at the 1st AND gate : \(\mathrm{A} \overline{\mathrm{B}}\)
The output at the 2nd AND gate : \(\overline{\mathrm{A}} \mathrm{B}\)
The output at the OR gate : Y = \(\mathrm{A.} \overline{\mathrm{B}}+\overline{\mathrm{A}} \cdot \mathrm{B}\)

Question 16.
What are the characteristics of ideal diode?
Answer:

1. It acts like conductor when it is forward biased.
2. When it is reverse bias it acts like an insulator
3. The barrier potential is assumed to be zero and hence it behaves like a resistor.

Question 17.
What are Optoelectronic devices?
Answer:
Optoelectronics deals with devices which convert electrical energy into light and light into electrical energy through semiconductors.

Question 18.
Define efficiency of rectifier.
Answer:
Efficiency (η) is the ratio of the output DC power to the ac input power supplied to the circuit.

Question 19.
What is Zener diode. Draw its symbol
Answer:
Zener diode is a reverse biased heavily doped silicon diode. It is specially designed to operate in breakdown region

Question 20.
What is called optoelectronics?
Answer:

1. 1. Optoelectronics deals with devices which convert electrical energy into light and light into electrical energy through semiconductors.
2. Optoelectronic devices are LED, photodiode and solar cells.

Question 21.
What is light emitting diode?
Answer:
Light Emitting Diode (LED) is a p-n junction diode which emits visible or invisible light when it is forward biased. Symbol is

Question 22.
What is photodiode? Draw the circuit symbol.
Answer:
A p-n junction – diode which converts an optical signal into electric current is known as photodiode. Circuit symbol is

Question 23.
What is solar cell?
Answer:
A solar cell is photo voltaic cell converts light energy directly into electricity or electric potential difference by photovoltaic effect.

Question 21.
Write down the applications of Oscillators.
Answer:
Applications of oscillators:

• to generate periodic sinusoidal or nonsinusoidal waveforms.
• to generate RF carriers.
• to generate audio tones
• to generate clock signals in digital circuits.
• as sweep circuits in TV sets and CRO.

Question 25.
What is Input resistance?
Answer:
The ratio of the change in base emitter voltage (∆V_BE) to change in base current (∆I_B) at constant collector emitter voltage (V_CE) is called input resistance (R_i)
R_i = \(\left(\frac{\Delta V_{B E}}{\Delta I_{B}}\right)_{V_{C E}}\)

Question 26.
What is output resistance?
Answer:
The ratio of change in collector – emitter voltage (∆V_CE) to the corresponding change is collector current (∆I_C) at constant base current.
R₀ = \(\left(\frac{\Delta \mathrm{V}_{\mathrm{CE}}}{\Delta \mathrm{I}_{\mathrm{C}}}\right)_{\mathrm{I}_{\mathrm{B}}}\)

Question 27.
What is forward current gain in common emifter mode?
Answer:
The ratio of change in collector current (∆I_C) to change in base current (I_B) at constant collector – emitter voltage (V_CE) is called forward current gain (B)
B = \(\left(\frac{\Delta \mathrm{I}_{\mathrm{C}}}{\Delta \mathrm{I}_{\mathrm{B}}}\right)_{\mathrm{V}_{\mathrm{CE}}}\)
The value is very high and range from 50 to 200.

Question 28.
What is amplification?
Answer:

1.  Amplification is the process of increasing the signal strength (increase the amplitude)
2. If large amplification is required, the transistors are cascaded.

Question 29.
Write the Barkhausen conditions for sustained oscillations?
Answer:

1. The loop phase shift is 0° or integral multiple of 2n.
2. The loop again must be unity \(|A B|\) = 1
   A – Voltage gain,
   B – Feedback ratio.

Question 30.
Draw the circuit diagrams of transistor in CB and CC modes.
Answer:

CB Mode	CC Mode

Question 31.
Differentiate damped and undamped Oscillations.
Answer:

Damped Oscillation	Undamped Oscillation
1. If amplitude of electrical oscillations decreases with time due to energy loss is damped oscillation.	1. The amplitude of electrical oscillation remains constant with time is damped oscillation.

VI. Three Mark Questions:

Question 1.
Differentiate Intrinsic semiconductor and Extrinsic semiconductors.
Answer:

Intrinsic Semiconductor	Extrinsic Semiconductor
1. It is in pure form	It formed by adding trivalent or pentavalent impurity.
2. Holes and electrons are equal	No. of holes are more in p-type. No of electrons are more in n-type.
3. Conductivity increases with raise in temperature	Conductivity depends on the amount of impurity added
4. Conductivity is slightly increased	Conductivity is greatly increased.

Question 2.
Find the differences between N-type and p-type semiconductors.
Answer:

N – type	P – type
1. Pentavalent impurities are added	Trivalent impurities are added.
2. Majority carriers are electrons.	Majority carriers are holes.
3. Minority carriers are holes	Minority carriers are electrons
4. They are negatively charged materials	They are positively charged materials

Question 3.
Write briefly about common — base configuration. Draw the circuit symbol for it.
Answer:
NPN transistor in common base configuration

(a) Schematic circuit diagram

(b) Circuit Symbol

1. Base is common to both input and output
2. Emitter current I_E is input collector current is output current I_C.
3. Input signal is applied between base and emitter output is measured between collector and base.

Question 4.
What is Zener diode? Write the applications.
Answer:
Zener diode is reverse biased with heavily doped silicon diode specially works in breakdown region. Zener diode can be used as

1. Voltage regulators
2. Peak Clippers
3. Calibrating Voltages
4. Provide fixed reference voltage in-network for biasing
5. Meter protection against damage from accidental application of excessive

Question 5.
What are the applications of LED?
Answer:

1. Indicator lamps on front panel of scientific and lab equipments
2. Seven – segment displays
3. Traffic – signals, exit signals in emergency vehicle lighting.
4. Industrial process control, position encoders, bar graph readers.

Question 6.
List the applications of photodiodes
Answer:

1. Alarm system
2. Count items on a conveyer belt
3. Photo conductors
4. CD players smoke detectors
5. Detectors for computed demography

Question 7.
Explain the different modes of transistor biasing.
Answer:
1. Forward Active:

• In this bias emitter – base junction is forward bias, collector base junction is reverse bias. .
• Transistor is in active mode of operation.
• Transistor functions as amplifier.

2. Saturation:

• Emitter base junction and collector-base junction are forward biased.
• Very large flow of current across junction.
• Transistor is used as closed switch.

3. Cut-Off:

• Emitter – base and collector-base junction are reverse biased.
• Transistor is acts as an open switch.

Question 8.
Write a note on common – Emitter configuration Draw the circuit diagram.
Answer:
NPN transistor in common emitter configuration

(a) Schematic circuit diagram

(b) Circuit Symbol
1. Emitter is common to both input and output.
2. Base current I_B is input current I_C is output current.
3. Input signal is applied between base and emitter. Output is measured between collector and emitter.

Question 9.
Write a note on common – collector configuration.
Answer:
1. Collector is common to both the inputs and outputs.
2. In is input current and Ig is output current.
3. Input signal is applied between base and collector. Output is measured between emitter and collector.
NPN transistor in common collector configuration

(a) schematic circuit diagram

(b) Circuit diagram

Question 10.
Write the applications of Oscillators.
Answer:

1. To generate a periodic sinusoidal or non-sinusoidal wave forms.
2. To generate RF carriers.
3. To generate audio tones .
4. To generate clock signals in digital circuits.
5. As sweep circuits in TV sets and CRO.

Question 11.
Write a short note on NAND gate.
Answer:
The circuit symbol of NAND gate is

Boolean Equation:
Y = \(\overline{A \cdot B}\)

Logic operation:
1. The output Y equals the complement of AND
2. The output is logic zero only when all inputs are high

Question 12.
Write a not on NOR gate.
Answer:
Circuit symbol:

A, B inputs, Y – output

Boolean equation:
Y = \(\overline{A+B}\)

Logic operation:
1. Y equals to complement of OR
2. The output is high when all inputs are low.

Question 13.
Write a note on Ex – OR gate.
Answer:
The circuit symbol is

A, B inputs, Y – output

Boolean equation is
Y = \(A \cdot \bar{B}+\bar{A} \cdot B\)
Y = \(A \oplus B\)
Logic operation:
The output is high only when either of two inputs is high.

Question 14.
Write down the concept in details of Integrated Chips (IC’s) Integrated Chips
Answer:
An integrated circuit is also referred as an IC or a chip or a microchip. It consists of thousands to millions of transistors, resistors, capacitors, etc. integrated on a small flat piece of semiconductor material that is normally Silicon. Integrated circuits (ICs) are the keystone of modem electronics. With the advancement in technology and the emergence of Very Large Scale Integration (VLSI) era it is possible to fit more aind more transistors on chips of same piece.

ICs have two main advantages over ordinary circuits: cost and performance. The size, speed, and capacity of chips have progressed enormously with the advancement in technology. Computers, mobile phones, and other digital home appliances are now made possible by the small size and low cost of ICs. ICs can function as an amplifier, oscillator, timer, microprocessor and computer memory.

These extremely small ICs can perform calculations and store data using either digital or analog technology. Digital ICs use logic gates, which work only with values of ones and zeros. A low signal sent to a component on a digital IC will result in a value of 0, while a high signal creates a value of 1.

Question 15.
In the following diagrams, indicate which of the diodes are forward biased and which are reverse biased?

Answer:
a) Forward biased
b) Reverse biased
c) Reverse biased

Question 16.
Find the current through the Zener diode when the load resistance is 1 kΩ. Use diode approximation.

Answer:
Voltage across AB, V_z = 9V
Voltage drop across R = 15 – 9 = 6V
Current through the resistor
I = \(\frac{\mathrm{V}}{\mathrm{R}}\)
= \(\frac{6}{1 \times 10^{3}}\) = 6 mA
Voltage across load resistor,
V_AB = 9V
Current through load resistor,
I_L = \(\frac{\mathrm{V}_{\mathrm{AB}}}{\mathrm{R}_{\mathrm{L}}}=\frac{9}{2 \times 10^{3}}\)
= 4.5 mA
Current through the zener diode,
I_z = I – I_p = 6 mA = 1.5 mA

VII. Five Mark Questions:

Question 1.
Explain the forward and reverse bias of Zener diode. Discuss its V-I Characterestics.
Answer:
1. The forward characteristics of Zener diode is similar to ordinary p-n junction diode.
2. It starts conducting around 0.7 H. However reverse characteristics is highly significant.
3. The increase in reverse voltage normally generate reverse current. While in Zener diode when reverse voltage (V₂) increase in current is very sharp the voltage remains constant.
4. If reverse current is increased further diode will be damaged V_z – Zener breakdown voltage
5. I_z(min) minimum current to sustain breakdown
6. I_z(max) maximum current limited by maximum power dissipation.
7. Zener diode is operated is reverse bias having voltage greater than V_z and current less than I_z(max)
8. The reverse characteristics is not exactly vertical so diode posses zener dynamic impedence.
9. Zener resistance in inverse of slope in breakdown region.
10. Increase in the Zener current produces only a very small increase in reverse voltage.
11. Voltage of ideal Zener diode doesn’t change. V_z remains almost constant even when I_z increases.

(a) forward bias

(b) reverse bias

(c) V-I characteristics

Question 2.
How zener diode works as a voltage regulator? Explain.
Answer:

Circuit to study voltage regulation by Zener diode

1. A Zener diode working in breakdown region can serve as a voltage regulator.
2. It maintains constant output voltage V₀ even when input voltage V_i or load current I_L varies.
3. In this circuit, V_i is regulated to constant voltage Zener voltage V_z at output represented as V₀ using Zener diode.
4. The output is maintained constant as long as V_i. does not fall below V₂.
5. When potential developed across the diode is greater than V_z diode moves into Zener breakdown region. It conducts, large current through R_i.
6. The total current is always less than the maximum Zener diode current. Under all conditions V₀ = V_z Thus output voltage is regulated.

Question 3.
Transistor works as oscillator Explain.
Answer:

Block diagram of an Oscillator

1. An electronic oscillator converts dc energy into ac energy of high frequency
2. An oscillator circuit consist of tank circuit, an amplifier and feed back circuit.
3. Tank circuit generates electrical oscillations and acts as ac input source to transistor amplifier.
4. Amplifier amplifies input ac signal Feed back circuit produces Block diagram of an oscillator a portion of output to tank circuit to sustain the oscillations without energy loss.
5. So an oscillator does not require an external output signal. It is self – sustained.
6. Feedback Network Portion of output is fed to input is in phase with input the magnitude of input increases. It is necessary for sustained oscillations.
Tank Circuit:

tank circuit

1. The LC tank circuit consist of inductance and capacitor connected in parallel
2. Whenever energy is supplied to tank circuit from DC source, energy is stored in inductor and capacitor alternatively This produces oscillations of definite frequency
3. In order to produce undamped oscillations, positive feedback is provided from output to frequency f = \(\frac{1}{2 \pi \sqrt{\mathrm{LC}}}\)

VIII. Additional Problems:

Question 2.
In PNP transistor circuit, the collector current is 10 mA. If 90% of the holes reach the collector, find emitter and base currents.
Solution:
Here, I_E = 10 mA
As 90% of the holes reach the collector, so the collector current.
I_C = 90% of I_E
I_C = \(\frac { 90 }{ 100 }\) I_E
I_E = \(\frac { 100 }{ 90 }\) I_C = \(\frac { 100 }{ 90 }\) x 10
I_E ≃ 11mA
Base current, I_B = I_E – I_C = 11 – 10
I_B = 1 mA.

Question 2.
An NPN BJT having reverse saturation current I_S = 10^-15 A is biased in the forward active region with V_BE = 700 mV and the current gain (β) may vary from 50 to 150 due to manufacturing variations. What is the maximum emitter current (in μA)
Solution:
I_S = 10^-15 A
V_BE = 700
V_T = 25 mV
β range from 50 to 150
I_C = I₀ e^(V_BE/V_T)
I_E = \(\frac { β+1 }{ β }\) I_C
I_E = \(\frac { β+1 }{ β }\) I_S e^V_BE/V_T
I_E will be maximum when β is 50
= 1.02 × 10^-15 × e^{700 × 10-3/25 × 10-3}
I_E = 1475 μA

Question 3.
The current gain β of silicon transistor used in circuit shown in figure is 50. (Barrier potential of silicon is 0.69 V) find I_B, I_E and I.

Solution:
V_BB = 2V
V_CC = 10V
β = 50
R_B = 10 kΩ
Barrier potential of Si V_BE = 0.69 V
R_C = 1 kΩ
V_BB = I_B R_B + V_BE
I_B = \(\frac{\mathrm{V}_{\mathrm{BB}}-\mathrm{V}_{\mathrm{BE}}}{\mathrm{R}_{\mathrm{B}}}\)
= \(\frac{2-0.69}{10 \times 10^{3}}\) = 131 µA
β = \(\frac{\mathrm{I}_{\mathrm{c}}}{\mathrm{I}_{\mathrm{B}}}\)
I_C = I_Bβ
= 131 × 10^-6 × 50 = 6.55 mA
∴ I_E = I_C + I_B
I_E = 6.55 mA + 131 µA
= 6.55 mA + 0.131 mA
I_E = 6.681 mA

Question 4.
You are given the two circuits shown in figure show that circuit
(a) acts as OR gate while the circuit
(b) acts as AND gate
Solution:
(a) A, B input Y output

(a) A, B inputs Y output

output of NOT gate = \(\overline{A+B}\)
This will be input for NOT gate.
Its output will be
\(\overline{\overline{A+B}}\) = A + B
∴ Y = A + B
Hence the circuit functions as OR gate.

(b) A, B are inputs Y – output

Y = \(\overline{\mathrm{A}}+\overline{\mathrm{B}}\)
= \(\overline{\overline{\bar{A}} \cdot \overline{\bar{B}}}\)
Y = A.B
Hence this circuit functions as AND gate.

Question 5.
In the circuit shown in the figure, the input voltage V_i is 20V, V_BE = 0V and V_CE = 0V, what are the values of I_B I_C and β?

Solution:

Question 6.
Name the logic gate equivalent of the output Y. If the same waveform I / P is given to both the inputs A and B, draw the corresponding output waveform.

Solution:

b) The output waveform will be the same as that of input waveform as it is OR (Addition) operation and same input is given to both A and B.