Category: Class 12

Tamilnadu State Board New Syllabus Samacheer Kalvi 12th Maths Guide Pdf Chapter 11 Probability Distributions Ex 11.5 Textbook Questions and Answers, Notes.

Tamilnadu Samacheer Kalvi 12th Maths Solutions Chapter 11 Probability Distributions Ex 11.5

Question 1.
Compute P(X = k) for the binomial distribution, B(n, p) where
(i) n = 6, p = \(\frac { 1 }{ 3 }\), k = 3
(ii) n = 10, p = \(\frac { 1 }{ 5 }\), k = 4
(iii) n = 9, p = \(\frac { 1 }{ 2 }\), k = 7
Solution:

Question 2.
The probability that Mr. Q hits a target ait any trial is \(\frac { 1 }{ 4 }\). Suppose he tries at the target 10 times. Find the probability that he hits the target (i) exactly 4 times (ii) atleast one time
Solution:
Let p be the probability of hitting the target
n = 10, p = \(\frac { 1 }{ 4 }\), X ~ B(n, p)
P(X = x) = ⁿC_x p^x q^n-x, x = 0, 1, 2, …. n
q = 1 – p = \(\frac { 3 }{ 4 }\)
(i) Probability of hitting the target exactly 4 times

(ii) Probability of hitting atleast one time
P(X ≥ 1) = 1 – P(X < 1) = 1 – P(X = 0)
= 1 – ¹⁰C₀ × (\(\frac { 1 }{ 4 }\))° × (\(\frac { 3 }{ 4 }\))¹⁰
= 1 – \(\frac { 3^{10} }{ 4^{10} }\)

Question 3.
Using binomial distribution find the mean and variance of X for the following experiments
(i) A fair coin is tossed 100 times and X denotes the number of heads.
(ii) A fair die is tossed 240 times and X denotes the number of times that four appeared.
Solution:
(i) Let p be the probability of getting heads
⇒ p = \(\frac { 1 }{ 2 }\)
n = 100, p = \(\frac { 1 }{ 2 }\), X ~ B(n, p)
q = 1 – p = \(\frac { 1 }{ 2 }\)
Mean = np
= 100 × \(\frac { 1 }{ 2 }\)
= 50
Variance = npq
= 50 × \(\frac { 1 }{ 2 }\)
= 25

(ii) Let p be the probability of getting 4 when a die is thrown
n = 240, p = \(\frac { 1 }{ 6 }\), X ~ B(n, p)
q = 1 – p = \(\frac { 5 }{ 6 }\)
Mean = np
= 240 × \(\frac { 1 }{ 6 }\) = 40
Variance = npq
= 40 × \(\frac { 5 }{ 6 }\) = \(\frac { 100}{ 3 }\)

Question 4.
The probability that a certain kind of component will survive a electrical test is \(\frac { 3 }{ 4 }\). Find the probability that exactly 3 of the 5 components tested survive.
Solution:
Let p be the probability of a certain component survive in an electrical test.
n = 5, p =\(\frac { 3 }{ 4 }\), X ~ B(n, p), k = 3
q = 1 – p = \(\frac { 1 }{ 4 }\)
P(X = x) = ⁿC_x p^x q^n-x, x = 0, 1, 2, …. n

Question 5.
A retailer purchases a certain kind of electronic, device from a manufacturer. The manufacturer indicates that the defective rate of the device 5 is 5%. The inspector of the retailer randomly picks 10 items from a shipment. What is the probability that there will be
(i) atleast one defective item
(ii) exactly two defective items?
Solution:
Let p be the probability that indicates the defective rate of an electronic device
p = 5%
= 0.05
q = 1 – p = 0.95
n = 10, p = 0.05, X ~ B(n, p)
P(X = x) = ⁿC_x p^x q^n-x, x = 0, 1, 2, …. n
(i) Atleast 1 defective item
P(X ≥ 1) = 1 – (X < 1)
= 1 – P(X = 0)
= 1 – ¹⁰C₀ (0.05)° (0.95)¹⁰
= 1 – (0.95)¹⁰

(ii) Exactly two defective items
P(X = 2) = ¹⁰C₂ (0.05)² (0.95)⁸

Question 6.
If the probability that fluorescent light has a useful life of atleast 600 hours is 0.9, find the probabilities that among 12 such lights
(i) exactly 10 will have a useful life of atleast 600 hours
(ii) At atleast 11 will have a useful life of atleast 600 hours
(iii) At atleast 2 will not have a useful life of at least 600 hours.
Solution:
Let p be the probability of the useful life hours of a fluorescent light.
n = 12, p = 0.9, X ~ B(n, p)
q – 1 – p = 0.1
P(X = x) = ⁿC_x p^x q^n-x, x = 0, 1, 2, …. n
(i) Exactly 10
P(X = 10) = ¹²C₁₀ (0.9)¹⁰ (0.1)²

(ii) Atleast 11
P(X ≥ 11) = P(X = 11) + P(X = 12)
= ¹²C₁₁ (0.9)¹¹ (0.1)¹ + ¹²C₁₂ (0.9)¹² (0.1)°
= 12 × (0.9)¹¹ × 0.1 + 1 × (0.9)¹² × 1
= (0.9)¹¹ (12 × 0.1 × 0.9)
= (0.9)¹¹ (1.2 + 0.9)
= (2.1) (0.9)¹¹

(iii) Atleast 2 will not have a useful
P(X ≤ 10) = 1 – P(X > 10)
= 1 – [P(X = 11) + P(X = 12)]
= 1 – (2.1)(0.9)¹¹

Question 7.
The mean and standard deviation of a binomial variate X are respectively 6 and 2. Find
(i) The probability mass function
(ii) P(X = 3)
(iii) P(X ≥ 2).
Solution:
X ~ B(n, p)
Given, Mean = 6, Standard deviation = 2
np = 6, \(\sqrt { npq }\) = 2
npq = 4
\(\frac { npq }{ np }\) = \(\frac { 4 }{ 6 }\)
q = \(\frac { 2 }{ 3 }\)
p = 1 – q = \(\frac { 1 }{ 3 }\)
npq = 4
n × \(\frac { 1 }{ 3 }\) × \(\frac { 2 }{ 3 }\) = 4
n = 18
(i) Probability mass function
P(X = x) = ⁿC_x p^x q^n-x, x = 0, 1, 2, …. n

Question 8.
If X ~ B(n, p) such that 4P(X = 4) = P(x = 2) and n = 6. Find the distribution, mean and standard deviation of X.
Solution:
X ~ B(n, p)
Given, 4P(X = 4) = P(X = 2), n = 6
P(X = x) = ⁿC_x p^x q^n-x, x = 0, 1, 2, …. n
⇒ 4 ⁶C₄ p⁴ q² = ⁶C₂ p² q⁴
⇒ 4 p² = q²
4 (1 – q)² = q²
4(1 – 2q + q²) = q²
3q² – 8q + 4 = 0
(q – 2) (3q – 2) = 0
q = \(\frac { 2 }{ 3 }\) (q ≠ 2)
Distribution
P(X = x) = ⁶C_x (\(\frac { 1 }{ 3 }\))^x (\(\frac { 2 }{ 3 }\))^6-x
x = 0, 1, 2, 3, 4, 5, 6
Mean = np = 6 × \(\frac { 1 }{ 3 }\) = 2
Variance = npq = 2 × \(\frac { 2 }{ 3 }\) = \(\frac { 4 }{ 3 }\)

Question 9.
In a binomial distribution consisting of 5 independent trials, the probability of 1 and 2 successes are 0.4096 and 0.2048 respectively. Find the mean and variance of the random variable.
Solution:
n = 5, X ~ B(n, p)
Given, P(X = 1) = 0.4096
P(X = 2) = 0.2048
P(X = x) = ⁿC_x p^x q^n-x, x = 0, 1, 2, …. n
⇒ ⁵C₁ p q⁴ = 0.4096
⇒ ⁵C₂ p² q³ = 0.2048
5 p q⁴ = 0.4096 …….. (1)
10 p² q³ = 0.2048 ………. (2)
(1) divided by (2),
⇒ \(\frac { 5pq^4 }{ 10p^2q^3 }\) = 2
\(\frac { p }{ q }\) = 4
q = 4p
q = 4 (1 – q)
q = 4 – 4q
5q = 4
q = \(\frac { 4 }{ 5 }\)
p = 1 – q = \(\frac { 1 }{ 5 }\)
Mean = np
= 5 × \(\frac { 1 }{ 5 }\) = 1
Variance = npq
= 1 × \(\frac { 4 }{ 5 }\) = \(\frac { 4 }{ 5 }\)
Distribution
P(X = x) = ⁵C_x (\(\frac { 1 }{ 5 }\))^x (\(\frac { 4 }{ 5 }\))^5-x
x = 0, 1, 2, 3, 4, 5.

Tamilnadu State Board New Syllabus Samacheer Kalvi 12th Maths Guide Pdf Chapter 11 Probability Distributions Ex 11.4 Textbook Questions and Answers, Notes.

Tamilnadu Samacheer Kalvi 12th Maths Solutions Chapter 11 Probability Distributions Ex 11.4

Question 1.
For the random variable X with the given prob-ability mass function as below, find the mean and variance.

Solution:


Var(X)= E(X²) – [E(X)]²
= 8.1 – (2.3)²
= 8.1 – 5.29
= 2.81

Var (X) = E(X²) – [E(X)]²
= 3.33 – (1.67)²
= 3.33 – 2.79
= 0.54




Var (X) = E(X²) – [E(X)]²
= 8 – 4
= 4

Question 2.
Two balls are drawn in succession without replacement from an urn containing four red balls and three black balls. Let X be the possible outcomes drawing red balls. Find the probability mass function and mean for X.
Solution:
Let X be the random variable that denotes the number of red balls.
X = {0, 1, 2}
Sample space consist of ⁷C₂ = 21
X= 0, X^-1 (BB) = ³C₂
X= 1, X^-1 (BR) = ³C₁ × ⁴C₁ = 12
X = 2, X^-1 (RR) ⁴C₂ = 6

Question 3.
If µ and σ² are the mean and variance of the discrete random variable X and E(X + 3) = 10 and E(X + 3)² = 116, find µ and σ²
Solution:
Given E(X + 3) = 10
E(aX + b) = aE(X) + b
E(X + 3) = 10 ⇒ E(X) + 3 = 10
⇒ µ = E(X) = 7
E(X + 3)² = 116
E(X² + 6X + 9) = 116
E(X²) + 6E(X) + 9 = 116
E(X²) + 6(7) + 9 = 116
E(X²) = 65
σ² = Var (X) = E(X²) – [E(X)]²
= 65 – 49
= 16

Question 4.
Four fair coins are tossed once. Find the probability mass function, mean and variance for a number of heads that occurred.
Solution:
Let X be the random variable that denotes the number of heads when four coins are tossed once.
X ={0, 1, 2, 3}
n(S) = 16

Var (X) = E(X²) – [E(X)]²
= 5 – 4
= 1

Question 5.
A commuter train arrives punctually at a station every half hour. Each morning, a student leaves his house to the train station. Let X denote the amount of time, in minutes, that the student waits for the train from the time he reaches the train station. It is known that the pdf of X is

Obtain and interpret the expected value of the random variable X.
Solution:

The average waiting time for the student is 15 minutes.

Question 6.
The time to failure in thousands of hours of an electronic equipment used in a manufactured computer has the density function

Find the expected life of this electronic equipment.
Solution:

Expected life of the electronic equipment = \(\frac { 1 }{ 3 }\)

Question 7.
The probability density function of the random variable X is given by

find the mean and variance of X.
Solution:
Mean:

Var (X) = E(X²) – [E(X)]²
\(\frac { 3 }{ 8 }\) – \(\frac { 1 }{ 4 }\)
= \(\frac { 1 }{ 8 }\)

Question 8.
A lottery with 600 tickets gives one prize of Rs 200, four prizes of Rs 100 and six prizes of Rs 50. If the ticket costs is Rs 2, find the expected winning amount of a ticket.
Solution:
Let X be the random variable denotes the amount win in the lottery

= 0.5
Loss = Rs 0.5

Tamilnadu State Board New Syllabus Samacheer Kalvi 12th Commerce Guide Pdf Chapter 9 Human Resource Management Text Book Back Questions and Answers, Notes.

Tamilnadu Samacheer Kalvi 12th Commerce Solutions Chapter 9 Human Resource Management

12th Commerce Guide Human Resource Management Text Book Back Questions and Answers

I. Choose The Correct Answer.

Question. 1
Human resource is a …………. asset.
a) Tangible
b) Intangible
c) Fixed
d) Current
Answer:
b) Intangible

Question 2.
Human Resource management is both ………………….. and ………………..
a) Science and art
b) Theory and practice
c) History and Geography
d) None of the above
Answer:
a) Science and art

Question 3.
Planning is a …………. Function.
a) selective
b) pervasive
c) both a and b
d) none of the above
Answer:
b) pervasive

Question 4.
Human resource management determines the ……………………….. relationship.
a) internal, external
b) employer, employee
c) Owner, Servant
d) Principle, Agent
Answer:
b) employer, employee

Question 5.
Labour turnover is the rate at which employees ………….. the organisation.
a) enter
b)leave
c) salary
d)None of the above
Answer:
b) leave

II. Very Short Answer Questions.

Question 1.
Give the meaning of Human Resource.
Answer:
In an organisation, the human resource is the employees who are inevitable for the survival and success of the enterprise.

Question 2.
What is Human Resource Management?
Answer:
The Branch of Management that deals with managing resource is known as “Human Resource Management (HRM)

Question 3.
State two features of HRM.
Answer:
Features of Human Resource Management.
Universally relevant: Human Resource Management has universal relevance.
Goal-oriented: The accomplishment of organisational goals is made possible through the best utilisation of human resources in an organisation.

Question 4.
Mention two characteristics of Human Resource. [OWE]
Answer:

• HR is Only factor of production that lives.
• HR can be Work as a team.
• HR – Exhibits innovations and creativity.

Question 5.
What are the managerial functions of HRM.
Answer:
The functions of human resource management may be classified as under: I Managerial function – Planning, Organising, Directing, Controlling.
II Operative function – Procurement, Development, Compensation, Retention, Integration, Maintenance.

III. Short Answer Questions.

Question 1.
Define the term Human Resource Management.
Answer:
HRM – Definition:
“HRM as that part of management process which is primarily concerned with the human constituents of an organisation”. E.F.L. BRECH

Question 2.
What are the characteristics of Human resources? [OWE] [MOM]
Answer:

1. Human resource exhibits innovation and creativity.
2. Human resources alone can think, act, analyse and interpret.
3. Human resources are emotional beings.
4. Human resources can be motivated either financially or non-financially.
5. The behaviour of human resources is unpredictable.
6. Over the years human resources gain value and appreciation.
7. Human resources are movable.
8. Human resources can work as a team.

Question 3.
What is the significance of Human resource?
Answer:

• HR is possible only through human relations.
• HR managers all other factors of production.
• HR can be utilised at all levels of management.
• HR is helpful to utilise all other resources.
• HR helps in industrial relation.
• HR well protected by legal frame works.

Question 4.
State the functions of Human Resource Management.
Answer:
The functions of human resource management may be classified as under: I Managerial function – Planning, Organising, Directing, Controlling.
II Operative function – Procurement, Development, Compensation, Retention, Integration, Maintenance.

1. Planning: Planning is deciding in advance what to do, how to do and who is to do it. It bridges the gap between where we are and where we want to go.
2. Organising: It includes division of work among employees by assigning each employee their duties, delegation of authority as required.
3. Procurement: Acquisition deals with job analysis, human resource planning, recruitment, selection, placement, transfer and promotion.
4. Development: Development includes performance appraisal, training, executive development, career planning and development, organisational development.
5. Compensation: It deals with job evaluation, wage and salary administration, incentives, bonuses, fringe benefits and social security schemes.

IV. Long Answer Questions.

Question 1.
Explain the characteristics of Human Resource. [OWE] [MOM] [LACE] (any 5) Characteristics
Answer:

•  HR is the Only factor of production that lives.
•  HR can Work as a team.
•  HR Exhibits innovation and creativity.
•  HR is Movable.
•  HR Over years – gain values.
•  HR can Motivated either in cash or in kind.
•  HR the Labour of employees that is hired and not the employee himself.
•  HR alone Act, Analyse, Think and interpret.
•  HR Create all other resources.
•  HR are Emotional beings.

Question 2.
Describe the significance of Human Resource Management. (MEERA) (JCE) (any 5)
Answer:
The role of human resource management is the process of acquiring, training, appraising, and compensating employees. The significance of human resource management is given below:

1. To identify manpower needs: Determination of manpower needs in an organisation is very important as it is a form of investment.
2. To ensure the correct requirement of manpower: At any time the organisation should
   not suffer from shortage or surplus manpower which is made possible through human resource management
3. To select the right man for the right job: Human resource management ensures the right talent to select the right employee for the right job.
4. To update the skill and knowledge: Human resource management enables employees to remain up-to-date through training and development programmes.
5. To appraise the performance of employees: A periodical appraisal of the performance of employees through human resource management activities boosts up good performers and motivates slow performers.

Question 3.
Elaborate on the Managerial functions of Human Resource Management.
Answer:
Managerial Functions : [COP] [D]
Controlling”

• It is comparing the actuals with the standards and to check whether activities are going on as per plan and rectify deviations.
• The control process includes fixing the standards, measuring actual performance, comparing the actual with standard laid down, measuring deviations and taking corrective decisions.
• This is made possible through observation, supervision, reports, records and audit.

Organising:

• It includes division of work among employees assigning duties to each employee.
• Delegation of authority as required, creation of accountability to make employees responsible.

Planning:

• Planning is thinking before doing.
• Planning is deciding in advance What to do, How to do, Who is to do it.
• It bridges the gap between where we are and where we want to go.
• It involves determination of objectives, policies, procedures, rules, strategies, programmes and budgets.

Directing:

• It involves issue of orders and instructions along with supervision, guidance and motivation to get the best out of employees.
• This reduces waste of time, energy and money and early attainment of organisational objectives.

Question 4.
Discuss the Operative functions HRM.
Answer:
Operating functions of HRM:

1. Procurement: Acquisition deals with job analysis, human resource planning, recruitment, selection, placement and promotion.
2. Development: It includes performance appraisal, training, executive development, and organizational development.
3. Compensation: It deals with job evaluation, wage and salary administration, incentives, bonus schemes.
4. Retention: This is made possible through health and safety, social security, job satisfaction and quality of work life.
5. Maintenance: This encourages employees to work with job satisfaction, reducing labour turnover, for human resource.

12th Commerce Guide Human Resource Management Additional Important Questions and Answers

I. Choose the correct answer.

Question 1.
create all other resources.
a) Skill
b) workforce
c) Natural resources
d) Human resource
Answer:
d) Human resources

Question 2.
The function of HRM concerned with …………………………
a) Motivating
b) Maintaining
c) Hiring
d) All of these
Answer:
d) All of those

Question 3.
How to encourage the best performer in an organisation?
a) Bonus
b) incentives
c) Both (a) and (b)
d) NOTA
Answer:
c) Both (a) and (b)

Question 4.
Human Resource Exbhits ……………………
a) Talent
b) Creativity
c) Innovation
d) All of those
Answer:
d) All of those

Question 5.
Who are inevitable for the survival and success of the enterprise?
a) Employee
b) Proprietors
c) Debtors
d) Creditors
Answer:
a) Employee

Question 6.
HR is ………………
a) Im-morable
b) In-tangble
c) Movable
d) NOTA
Answer:
c) Movable

II. Match the following

Question 1.
Match List I with List – II

Answer:
a) (i) 4,(ii) 1,(iii) 2,(iv) 3

III. Very short answer questions.

Question 1.
Define the term Human Resource.
Answer:
Man of all sources available to him, can grow and develop”.

Question 2.
Give two points of difference between HR and HRM .[MO]
Answer:

IV. Long Answer Questions.

Question 1.
Explain the features of HRM? [CUGAI]
Answer:
Continuous Process:
As long as there is HR in the running of an organisation, the activities relating to managing , HR exists.

Universal Relevent:

• HRM has universal relevance.
• The approach and style varies depending the nature of organisation structure and is applicable at all levels.

Goal Oriented :
The accomplishment of organisational goals is made possible through best utilisation of HR is an organisation.

A Systematic Approach:
HRM lavs emphasis on a systematic approaching in managing the tasks performed by HR of an organisation.

Intangible:
HRM is an intangible function which can be measured only by results.

Question 2.
Differentiate HR from HRM.[MOPET]
Answer:

Tamilnadu State Board New Syllabus Samacheer Kalvi 12th Maths Guide Pdf Chapter 11 Probability Distributions Ex 11.3 Textbook Questions and Answers, Notes.

Tamilnadu Samacheer Kalvi 12th Maths Solutions Chapter 11 Probability Distributions Ex 11.3

Question 1.
The probability density function of X is given by

Find the value of k.
Solution:
Since f(x) is a probability density function

Question 2.
The probability density function of X is

Find
(i) P(0.2 ≤ X < 0.6)
(ii) P(1.2 ≤ X < 1.8)
(iii) P(0.5 ≤ X < 1.5)
Solution:

Question 3.
Suppose the amount of milk sold daily at a milk booth is distributed with a minimum of 200 litres and a maximum of 600 litres with probability density function

Find
(i) the value of k
(ii) the distribution function
(iii) the probability that daily sales will fall between 300 litres and 500 litres?
Solution:

(i) Since f(x) is a probability density function

Question 4.
The probability density function of X is given by

Find
(i) the value of k
(ii) the distribution function
(iii) P(X < 3)
(iv) P(5 ≤ X)
(v) P(X ≤ 4)
Solution:
(i) Since f is a probability density function

(ii) The distribution function F(x) = \(\int_{-\infty}^{x}\) f(u) du

Question 5.
If X is the random variable with probability density function f(x) given by,

then find
(i) the distribution function F(x)
(ii) P(-0.5 ≤ x ≤ 0.5)
Solution:
Distribution function F(x) = \(\int_{-\infty}^{x}\) f(u) du
case 1.
x < -1
F(x) = \(\int_{-\infty}^{x}\) f(u) du = 0

Question 6.
If X is the random variable with distribution function F(x) given by,

Find
(i) the probability density function f(x)
(ii) P(0.3 ≤ X ≤ 0.6)
Solution:
(i) Probability density function

(ii) P(0.3 ≤ X ≤ 0.6)
P(a ≤ X ≤ b) = F(b) – F(a)
P(0.3 ≤ X ≤ 0.6) = F(0.6) – F(0.3)
= \(\frac { 1 }{ 2 }\) [(0.6)² + (0.6)] – \(\frac { 1 }{ 2 }\) [(0.3)² + (0.3)]
= \(\frac { 1 }{ 2 }\) (0.96 – 0.39)
= 0.285

Tamilnadu State Board New Syllabus Samacheer Kalvi 12th Maths Guide Pdf Chapter 11 Probability Distributions Ex 11.2 Textbook Questions and Answers, Notes.

Tamilnadu Samacheer Kalvi 12th Maths Solutions Chapter 11 Probability Distributions Ex 11.2

Question 1.
Three fair coins are tossed simultaneously. Find the probability mass function for a number of heads that occurred.
Solution:
When three coins are tossed, the sample space is
S = {HHH, HHT, HTH, HTT, THH, THT, TTH, TTT}
‘X’ is the random variable that denotes the number of heads.
∴ ‘X’ can take the values of 0, 1, 2 and 3Sample space S = {HHH, HHT, HTH, THH, HTT, THT, TTH, TTT}

Probability mass function

Question 2.
A six sided die is marked ‘1’ on one face, ‘3’ on two of its faces and ‘5’ on remaining three faces. The die is thrown twice. If X denotes the total score in two throws, find
(i) the probability mass function
(ii) the cumulative distribution function
(iii) P(4 ≤ X < 10)
(iv) P(X ≥ 6)
Solution:
Let X be the random variable denotes the total score in two thrown of a die.
Sample space S

n (S) = 36
X = {2, 4, 6, 8, 10}

(i) Probability mass function

(ii) Cumulative distribution function

= \(\frac { 36 }{ 36 }\)
= 1

Question 3.
Find the probability mass function and cumulative distribution function of a number of girl children in families with 4 children, assuming equal probabilities for boys and girls.
Solution:
Let X be the random variable denotes the number of girl child among 4 children
X = {0, 1, 2, 3, 4}

(i) Probability mass function

(ii) Cumulative distribution

Question 4.
Suppose a discrete random variable can only take the values 0, 1, and 2. The probability mass function is defined by

Find (i) the value of k.
(ii) cumulative distribution function
(iii) P(X ≥ 1)
Solution:

(i) Given f is a probability mass function
\(\sum_{x}\) f(x) = 1
Probability mass function is

(ii) Cumulative distribution function
F(0) = P(x ≤ 0)
= P(x = 0)
= \(\frac { 1 }{ 8 }\)
F(1) = P(x ≤ 1)
= P(x = 0) + P(x = 1)
= \(\frac { 1 }{ 8 }\) + \(\frac { 2 }{ 8 }\) = \(\frac { 3 }{ 8 }\)
F(2) = P(x ≤ 2)
= P(x = 0) + P(x = 1) + P(x = 2)
= \(\frac { 1 }{ 8 }\) + \(\frac { 2 }{ 8 }\) + \(\frac { 5 }{ 8 }\) = 1

(iii) P(X ≥ 1)
= P(X = 1) + P(X = 2)
= \(\frac { 2 }{ 8 }\) + \(\frac { 5 }{ 8 }\) + \(\frac { 7 }{ 8 }\)

Question 5.
The cumulative distribution function of a discrete random variable is given by

Find
(i) the probability mass function
(ii) P(X < 1)
(iii) P(X ≥ 2)
Solution:

f(-1) = P(X = -1) = F(-1) – F(-1) = 0.15 – 0 = 0.15
f(0) = P(X = 0) = F(0) – F(-1) = 0.35 – 0.15 = 0.20
f(1) = P(X = 1) = F(1) – F(0) = 0.60 – 0.35 = 0.25
f(2) = P(X = 2) = F(2) – F(1) = 0.85 – 0.60 = 0.25
f(3) = P(X = 3) = F(3) – F(2) = 1 – 0.85 = 0.15

(ii) P(X < 1)
P(X < 1) = P(X = -1) + P(X = 0)
= 0.15 + 0.20
= 0.35

(iii) P(X ≥ 2)
P(X ≥ 2) = P(X = 2) + P(X = 3)
= 0.25 + 0.15
= 0.40

Question 6.
A random variable X has the following probability mass function.

Find
(i) the value of k
(ii) P(2 ≤ X < 5)
(iii) P(3 < X)
Solution:
(i) Given f(x) in a probability mass function
\(\sum_{x}\) f(x) = 1
k² + 2k² + 3k² + 2k + 3k = 1
6k² + 5k = 1
6k² + 5k – 1 = 0
(k + 1) (6k – 1) = 0
k = \(\frac { 1 }{ 6 }\)
(k ≠ -1 neglecting negative terms)
Probability mass function

(ii) P(2 ≤ X < 5)
P(2 ≤ X < 5) = P(X = 2) + P(X = 3) + P(X = 4)
= \(\frac { 2 }{ 36 }\) + \(\frac { 3 }{ 36 }\) + \(\frac { 2 }{ 6 }\)
= \(\frac { 2+3+12 }{ 36 }\) + \(\frac { 17 }{ 36 }\)

(iii) P(X > 3) = P(X = 4) + P(X = 5)
= \(\frac { 2 }{ 6 }\) + \(\frac { 3 }{ 6 }\)
= \(\frac { 5 }{ 6 }\)

Question 7.
The cumulative distribution function of a discrete random variable is given by

Find
(i) the probability mass function
(ii) P(X < 3) and (iii) P(X ≥ 2)
Solution:
(i) Probability mass function

(ii) P(X < 3)
P(X < 3) = P(X = 0) + P(X = 1) + P(X = 2)
= \(\frac { 1 }{ 2 }\) + \(\frac { 1 }{ 10 }\) + \(\frac { 1 }{ 5 }\)
= \(\frac { 4 }{ 5 }\)

(ii) P(X ≥ 2)
P(X ≥ 2) = P(X = 2) + P(X = 3) + P(X = 4)
= \(\frac { 1 }{ 5 }\) + \(\frac { 1 }{ 10 }\) + \(\frac { 1 }{ 10 }\)
= \(\frac { 4 }{ 10 }\) = \(\frac { 2 }{ 5 }\)

Tamilnadu State Board New Syllabus Samacheer Kalvi 12th Maths Guide Pdf Chapter 11 Probability Distributions Ex 11.1 Textbook Questions and Answers, Notes.

Tamilnadu Samacheer Kalvi 12th Maths Solutions Chapter 11 Probability Distributions Ex 11.1

Question 1.
Suppose X is the number of tails that occurred when three fair coins are tossed once simultaneously. Find the values of the random variable X and number of points in it. inverse – images.
Solution:
Let X is the random variable that denotes the number of tails when three coins are tossed simultaneously.
Sample space S = {HHH, HHT, HTH, HTT, THH, THT, TTH, TTT}
∴ ‘X’ takes the values 0,1, 2, 3
i.e., X (HHH) = 0 ; X (HHT, HTH, THH) = 1 ; X (HTT, THT, TTH) = 2 ; X (TTT) = 3

Question 2.
In a pack of 52 playing cards, two cards are drawn at random simultaneously. If the number of black cards drawn is a random variable, find the values of the random variable and number of points in its inverse images.
Solution:
Let X be the random variable of number of black cards occur.
∴ X = {0, 1, 2}
The sample space consists of 52C₂ = 1326
X = 0, X (both are white cards) = 26C₂ = 325
X = 1, X (one black and one white cards)
= 26C₁ × 26C₁
= 676
X = 2, X (both are black cards) = 26C₂ = 325

Question 3.
An urn contains 5 mangoes and 4 apples. Three fruits are taken at random. If the number of apples taken is a random variable, then find the values of the random variable and number of points in its inverse images.
Solution:
Number of mangoes = 5
Number of Apples = 4
Total number of fruits = 9
Let ‘X’ be the random variable that denotes the number of apples taken, then it takes the values 0, 1, 2, 3
X (MMM) = 0
X (AMM (or) MAM (or) MMA) = 1
X (AAM (or) AMA (or) MAA) = 2
X (AAA) = 3

Question 4.
Two balls are chosen randomly from an urn containing 6 red and 8 black balls. Suppose that we win Rs 15 for each red ball selected and we lose Rs 10 for each black ball selected. If X denotes the winning amount, then find the values of X and the number of points in its inverse images.
Solution:
Let X be the random variable that denotes the j winning amount.
X (Both are black balls) = Rs 2 (-10) = -Rs 20
X (one red and one black ball) = Rs 15 – Rs 10 = Rs 5
X (both are red ball) = Rs 2 (15) = Rs 30
∴ X = {-20, 5, 30}
The sample space consists of ¹⁴C₂ = 91
X = – 20, Both black balls = ⁸C₂ = 28
X = 5, One black, one red ball = ⁸C₁ × 6C₁
= 8 × 6 = 48
X = 30, Both are white balls = ⁶C₂ = 15

Question 5.
A six-sided die is marked ‘2’ on one face, ‘3’ on two of its faces, and ‘4’ on the remaining three faces. The die is thrown twice. If X denotes the total score in two throws, find the values of the random variable and number of points in its inverse images.
Solution:
Let X be the random variable denotes the total score is two throws of a die.
Sample Space S

n (S) = 36
X = {4, 5, 6, 7, 8}
From the sample space

Tamilnadu State Board New Syllabus Samacheer Kalvi 12th Commerce Guide Pdf 8 Securities Exchange Board of India Text Book Back Questions and Answers, Notes.

Tamilnadu Samacheer Kalvi 12th Commerce Solutions Chapter 8 Securities Exchange Board of India

I. Choose The Correct Answer.

Question 1.
Securities Exchange Board of India was first established in the year…………………
a) 1988
b) 1992
c) 1995
d) 1998
Answer:
a) 1988

Question 2.
The headquarters of SEBI is…………..
a) Calcutta
b) Bombay
c) Chennai
d) Delhi
Answer:
b) Bombay

Question 3.
In which year SEBI was constituted as the regulator of capital markets in India?
a) 1988
b) 1992
c) 2014
d) 2013
Answer:
a) 1988

Question 4.
Registering and controlling the functioning of collective investment schemes as………….
a) Mutual Funds
b) Listing
c) Rematerialisation
d) Dematerialization
Answer:
d) Dematerialization

Question 5.
SEBI is empowered by the Finance ministry to nominate ………………….. members on the Governing body of every stock exchange.
a) 5
b) 3
c) 6
d) 7
Answer:
b) 3

Question 6.
The process of converting physical shares into electronic form is called ………………
a) Dematerialisation
b) Delisting
c) Materialisation
d) Debarring
Answer:
a) Dematerialisation

Question 7.
Trading is dematerialized shares commenced on the NSE is ………………………
a) January 1996
b) June 1998
c) December 1996
d) December 1998
Answer:
c) December 1996

Question 8.
…………..was the first company to trade its shares in Demat form.
a) Tata Industries
b) Reliance Industries
c) Infosys
d) Birla Industries
Answer :
b) Reliance Industries

Question 9.
…………….. enables small investors to participate in the investment on the share capital of large companies.
a) Mutual Funds
b) Shares
c) Debentures
d) Fixed deposits
Answer:
a) Mutual Funds

Question 10.
PAN stands for …………………
a) Permanent Amount Number
b) Primary Account Number
c) Permanent Account Number
d)Permanent Account Nominee
Answer:
c) Permanent Account Number

II. Very Short Answer Questions.

Question 1.
Write short notes on SEBI.
Answer:
Securities and Exchange Board of India (SEBI) was first established in the year 1988 as a non-statutory body for regulating the securities market.

Question 2.
Write any two objectives of SEBI.
Answer:
Control Over Brokers:
The important object is to supervise or check the activities of the Brokers, and other intermediaries in order to control the capital market.

Regulation of Stock Exchange :

• The first objective of SEBI is to regulate the Stock Exchange.
• So that efficient services may be provided to all the parties operating there.

Question 3.
What is a Demat Account?
Answer:
A Demat account holds all the shares that are purchased in electronic or dematerialized form.

Question 4.
Mention the headquarters of SEBI.
Answer:
Head Quarters – Mumbai [Bandra Kurla – Complex]
.
Question 5.
What are the various ID proofs?
Answer:
Proof of identity: PAN card, voter’s ID, passport, driver’s license, bank attestation, IT returns, electricity bill, telephone bill, ID cards with applicant’s photo issued by the central or state government are the ID proofs.

III. Short Answer Questions.

Question 1.
What is meant by Dematerialization?
Answer:

•  Dematerialization [DEMAT] is the process by which physical share certificates of an investor are taken back by the Company or Register and destroyed.
• Then an equivalent number of securities in the Electronic form is credited to the investor’s account with his Depository Participant. [DP]
•  DEMAT is done at the request of the Investor.
•  He has to open an account with a DP.
•  In DEMAT A/c – Purchase – Credited his account.
•  In DEMAT A/ c – Sales – Debited his account

Question 2.
What are the documents required for a Demat account?
Answer:

• For opening a Demat account, we submit proof of identity and address along with a passport size photo and the account opening form.
• Documents for Identity: Pan card, Voters ID, Passport, Driver’s License, IT Returns, Electricity and Telephone Bills are the Identity documents.
• Documents for Address: Ration card, Passport, Voter’s ID card, Driving License, Telephone Bills, Electricity bills are the address documents.

Question 3.
What is the power of SEBI under the Securities Contract Act?
Answer:

•  Power to grant License to the new stock exchange.
•  Power to direct any stock exchange to amend the rules.
•  Power to supersede governing body of any stock exchange.
•  Power to ask for information and accounts of the stock exchange.
•  Power to suspend the business of the stock exchange.
•  Power to prohibit contracts of the stock exchange.

Question 4.
What is meant by Insider trading?
Answer:
Insider trading means the buying and selling of securities by directors, promoters, etc., who have access to some confidential information about the company. This affects the interests of the general investors and is essential to check this tendency.

Question 5.
Draw the organizational structure of SEBI.
Answer:
Organization Structure of SEBI

IV. Long Answer Questions.

Question 1.
What are the functions of SEBI?
Answer:
functions:

•  SEBI is the NODAL agency which safeguards the interests of an investor in the Indian Financial Market.
  SEBI performs:
  
• Safeguarding the interest of the investors by means of adequate education and guidance.
• SEBI makes Rules and Regulations that must be strictly followed by the participants of the financial market.
• Regulating and Controlling the business on Stock Exchange.
• Registration of Brokers and sub-brokers is made mandatory.
• Conduct inspection and inquiries of stock exchanges, intermediaries, and self-regulating organizations.
• Barring insider trading in securities.
• Prohibiting deceptive and unfair methods by intermediaries.
• Registering and Controlling share agents, bankers, trustees, registrars, merchant bankers, underwriters, managers, etc.
• SEBI regulates mergers and amalgamation as a way to protect the interest of the investors. Registering and Controlling the function of collective investment schemes such as mutual funds.
• Promoting self-regulatory organizations intermediaries.
  Carrying out steps in order to develop the capital markets by having an accommodating approach.

Question 2.
Explain the powers of SEBI.
Answer:

The various powers of a SEBI are explained below:

1. Powers Relating to Stock Exchanges and Intermediaries: SEBI has wide powers to get the information from the stock exchange and intermediaries regarding their business transactions for inspection.
2. Power to Impose Monetary Penalties: SEBI has the power to impose monetary penalties on capital market intermediaries for violations.
3. Power to Initiate Actions in Functions Assigned
4. Power to Regulate Insider Trading: SEBI has the power to regulate insider trading or can regulate the functions of merchant bankers.
5. Powers Under Securities Contracts Act: For the regulation of the stock exchange, the Ministry of Finance issued a notification, for delegating several of its powers under the securities contract Act.

Question 3.
What are the benefits of Dematerialisation? (advantages)
Answer:
Benefits of Dematerialisation:

1. The risks relating to physical certificates like loss, theft, forgery are eliminated completely with a Demat Account.
2. The risk of paperwork enables quicker transactions and higher efficiency in trading.
3. The shares which are created through mergers and consolidation of companies are credited automatically in the Demat account.
4. There is no stamp duty for the transfer of securities.
5. Certain banks also permit holding of both equity and debt securities in a single account
6. A Demat account holder can buy or sell any amount of shares.

12th Commerce Guide Securities Exchange Board of India Additional Important Questions and Answers

I.Choose the correct answer.

Question 1.
In which year is SEBI Act being passed by the Indian Parliament?
a) 1990
b) 1991
c) 1992
d) 1993
Answer:
c) 1992

Question 2.
…………….. is the foremost objective of SEBI.
a) Security
b) Regulate
c) Control
d) NOTA
Answer:
a) Security

Question 1.
SEBI got the statutory powers in the year _____
(a) 1988
(b) 1992
(c) 1969
(d) 1980
Answer:
(b) 1992

Question 4.
Which is an Apex body that maintains and Regulates the capital market?
a) Stock Exchange
b) SEBI
c) OTCEI
d) NSE
Answer:
b) SEBI

Question 5.
Name the three key functions of SEBI.
Answer:
SEBI performs three key functions.
They are:

1. quasi-legislative
2. quasi-judicial
3. quasi-executive

Question 6.
SEBI has the following number of members including the chairman.
a) 4
b) 5
c) 6
d) 7
Answer:
c) 6

Question 7.
Pick the odd one out:
a) Voter ID
b) PAN
c) ID card
d) PostCard
Answer:
d) Post Card

Question 8.
which one of the following is not correctly matched?
a) DEMAT – Dematerialisation
b) PAN – Proof
c) SEBI -12 members
d) HQ – Bendrakurla
Answer :
c) SEBI -12 members

Question 9.
Choose the correct statement.
(i) SEBI Act was passed in the year 1992 in Indian Parliament.
(ii) It protects the interest of the Investors.
(iii) It Regulates and controls the stock exchange.
a) (i) is correct
b) (ii) is correct
c) (iii) is correct
d) All (i), (ii) and (iii) are correct
Answer :
d) All (i), (ii) and (iii) are correct

II. Match the following

Question 1.

Answer :
a) (i) – 2, (ii) – 1, (iii) – 4, (iv) – 3

III. Assertion and Reason :

Question 1. Assertion (A): DEMAT is done at the request of the investor.
Reason (R): A DEMAT A/c holder can buy or sell any amount of shares.
a) (A) is True (R) is True
b) (A) and (R) are False
c) (A) is True (R) is False
d) (A) is False (R) is True
Answer :
a) (A) is True (R) is True

Question 2.
Assertion (A): SEBI is a supervisory (Apex) Body.
Reason (R): So, It cannot regulates and control the stock exchange.
a) (A) is True (R) is False
b) (A) is False (R) is True
c) Both (A) and (R) are True
d) Both (A) and (R) are False
Answer :
a) (A) is True (R) is False|

IV. Very Short answer questions.

Question 1.
Write a note on PAN.
Answer:
PAN, or permanent account number, is a unique 10-digit alphanumeric identity allotted to each taxpayer by the Income Tax Department. It also serves as identity proof.

Tamilnadu State Board New Syllabus Samacheer Kalvi 12th Maths Guide Pdf Chapter 10 Ordinary Differential Equations Ex 10.9 Textbook Questions and Answers, Notes.

Tamilnadu Samacheer Kalvi 12th Maths Solutions Chapter 10 Ordinary Differential Equations Ex 10.9

Choose the most suitable answer from the given four alternatives:

Question 1.
The order and degree of the differential equation \(\frac { d^2y }{ dx^2 }\) + (\(\frac { dy }{ dx }\))^1/3 + x^1/4 = o are respectively
(a) 2, 3
(b) 3, 3
(c) 2, 6
(d) 2, 4
Solution:
(a) 2, 3
Hint:

In this equation, the highest order derivative is \(\frac { d^2y }{ dx^2 }\) & its power is 3.
∴ Its order = 2 & degree = 3

Question 2.
The differential equation representing the family of curves y = A cos (x + B), where A and B are parameters, is
(a) \(\frac { d^2y }{ dx^2 }\) – y = 0
(b) \(\frac { d^2y }{ dx^2 }\) + y = 0
(c) \(\frac { d^2y }{ dx^2 }\) = 0
(d) \(\frac { d^2x }{ dy^2 }\) = 0
Solution:
(b) \(\frac { d^2y }{ dx^2 }\) + y = 0
Hint:
Given equation is y = A cos (x + B) …….. (1)
where A & B are parameters.
Differentiating equation (1) twice successively, because we have two arbitrary constants.
\(\frac { dy }{ dx }\) = A sin (x + B)
Again differentiating \(\frac { d^2y }{ dx^2 }\) = -A cos(x + B) = -y
∵ y = A cos (x + B
∵ \(\frac { d^2y }{ dx^2 }\) + y = 0 as the required differential equation.

Question 3.
The order and degree of the differential equation, \(\sqrt { sin x }\) (dx + dy) = \(\sqrt { cos x }\) (dx – dy) is
(a) 1, 2
(b) 2, 2
(c) 1, 1
(d) 2, 1
Solution:
(c) 1, 1
Hint:
Given \(\sqrt { sin x }\) (dx + dy) = \(\sqrt { cos x }\) (dx – dy)
divide by dx on both sides, we get

In this equation, the highest, order derivative is \(\frac { dy }{ dx }\) & its power is 1.
∴ Its order = 1 & degree = 1

Question 4.
The order of the differential equation of all circles with centre at (h, k) and radius ‘a’ is
(a) 2
(b) 3
(c) 4
(d) 1
Solution:
(a) 2
Hint:
We know that equation of a circle be (x – h)² + (y – k)² = a².
Here we have two constants. Therefore, Its order is 2.

Question 5.
The differential equation of the family of curves y = Ae^x + Be^-x, where A and B are arbitrary constants is
(a) \(\frac { d^2y }{ dx^2 }\) + y = 0
(b) \(\frac { d^2y }{ dx^2 }\) – y = 0
(c) \(\frac { dy }{ dx }\) + y = 0
(d) \(\frac { dy }{ dx }\) – y = 0
Solution:
(b) \(\frac { d^2y }{ dx^2 }\) – y = 0
Hint:
Given y = Ae^x + Be^-x ……… (1)
where A & B are arbitrary constants.
Differentiate equation(1) twice continuously, we get
(∵ Two constants so differentiate twice)

\(\frac { d^2y }{ dx^2 }\) – y = 0 as the required differential equation.

Question 6.
The general solution pf the differential equation \(\frac { dy }{ dx }\) = \(\frac { y }{ x }\) is
(a) xy = k
(b) y = k log x
(c) y = kx
(d) log y = kr
Solution:
(c) y = kx
Hint:
Given \(\frac { dy }{ dx }\) = \(\frac { y }{ x }\)
The equation can be written as
\(\frac { dy }{ y }\) = \(\frac { dx }{ x }\)
Integrating on both sides, we get
∫\(\frac { dy }{ y }\) = ∫\(\frac { dy }{ x }\)
log y = log x + log k
log y = log k x [∵ log m + log n = log mn]
Remove log, we get
y = kx is a required differential equation

Question 7.
The solution of the differential equation 2x\(\frac { dy }{ dx }\) – y = 3 represents
(a) straight lines
(b) circles
(c) parabola
(d) ellipse
Solution:
(c) parabola
Hint:

2 log (3 + y) = log x + log k
log (3 + y)² = log kx
Remove log, we get
(3 + y)² = kx is a solution of the differential equation which is a Parabola.

Question 8.
The solution of \(\frac { dy }{ dx }\) + p(x) y = 0 is
(a) y = ce^∫pdx
(b) y = ce^-∫pdx
(c) x = ce^-∫pdy
(d) x = ce^∫pdy
Solution:
(b) y = ce^-∫pdx
Hint:
Given \(\frac { dy }{ dx }\) + p(x) y = 0
\(\frac { dy }{ dx }\) = -p(x)y
The equation can be written as
\(\frac { dy }{ y }\) = -p(x) dx
Taking integration on both sides, we get

∴ y = ce^-∫pdx is a solution of the given differential equation.

Question 9.
The integrating factor of the differential equation \(\frac { dy }{ dx }\) + y = \(\frac { 1+y }{ x }\) is
(a) \(\frac { x }{ e^x }\) + y = 0
(b) \(\frac { e^x }{ x }\) – y = 0
(c) λe^x
(d) e^x
Solution:
(b) \(\frac { e^x }{ x }\) – y = 0
Hint:
Given differential equation is

Question 10.
The Integrating factor of the differential equation \(\frac { dy }{ dx }\) + p(x) y = Q(x) is x, then p(x)
(a) x
(b) \(\frac { x^2 }{ 2 }\)
(c) \(\frac { 1 }{ x }\)
(d) \(\frac { 1 }{ x^2 }\)
Solution:
(c) \(\frac { 1 }{ x }\)
Hint:
The given differential equation is
\(\frac { dy }{ dx }\) + p(x) y = Q(x)
Integrating factor e^∫pdx = x
Taking log on both sides, we get ∫p dx = log x
Now differentiating, we get I
\(\frac { d }{ dx }\) [∫p dx]= \(\frac { d }{ dx }\) [log x]
∴ p = \(\frac { 1 }{ x }\)

Question 11.
The degree of the differential equation
y(x) = 1 + \(\frac { dy }{ dx }\) + \(\frac { 1 }{ 1.2 }\) (\(\frac { dy }{ dx }\))² + \(\frac { 1 }{ 1.2.3 }\) (\(\frac { dy }{ dx }\))³ + ……. is
(a) 2
(b) 3
(c) 1
(d) 4
Solution:
(c) 1
Hint:
Given
y(x) = 1 + \(\frac { dy }{ dx }\) + \(\frac { 1 }{ 1.2 }\) (\(\frac { dy }{ dx }\))² + \(\frac { 1 }{ 1.2.3 }\) (\(\frac { dy }{ dx }\))³ + ……
Here the highest derivative term is \(\frac { dy }{ dx }\) and its power is 1.
∴ Degree of the differential equation is 1.

Question 12.
If p and q are the order and degree of the differential equation y \(\frac { dy }{ dx }\) +x³ (\(\frac { d^2y }{ dx^2 }\)) + xy = cos x, when
(a) p < q
(b) p = q
(c) p > q
(d) p exists and q does not exist
Solution:
(c) p > q
Hint:
Given equation is y\(\frac { dy }{ dx }\) +x³ (\(\frac { d^2y }{ dx^2 }\)) + xy = cos x
The highest order derivative of the differential equation is \(\frac { d^2y }{ dx^2 }\) and its degree is 1.
∴ Its order is p = 2 and degree q = 1
∴ p > q

Question 13.
The solution of the differential equation \(\frac { dy }{ dx }\) + \(\frac { 1 }{ \sqrt{1-x^2} }\) = 0 is
(a) y + sin^-1 x = c
(b) x + sin^-1 y = 0
(c) y² + 2sin^-1 x = c
(d) x² + 2sin^-1 y = 0
Solution:
(a) y + sin^-1 x = c
Hint:
The given equation is \(\frac { dy }{ dx }\) = –\(\frac { 1 }{ \sqrt{1-x^2} }\)
The equation can be written as
dy = \(\frac { 1 }{ \sqrt{1-x^2} }\)
Integrating on both sides, we get
∫dy = -∫\(\frac { dx }{ \sqrt{1-x^2} }\)
y = -sin^-1 (x) + C
y + sin^-1 (x) = C
∴ y + sin^-1 (x) = C is a solution of the given differential equation.

Question 14.
The solution of the differential equation \(\frac { dy }{ dx }\) = 2xy is
(a) y = Ce^x²
(b) y = 2x² + C
(c) y = Ce^-x² + C
(d) y = x² + C
Solution:
(a) y = Ce^x²
Hint:
Given
\(\frac { dy }{ dx }\) = 2xy
The equation can be written as
\(\frac { dy }{ y }\) = 2xdx
Taking integration on both sides, we get

∴ y = Ce^x² is a solution to the differential equation.

Question 15.
The general solution of the differential equation log(\(\frac { dy }{ dx }\)) = x + y is
(a) e^x + e^y = C
(b) e^x + e^-y = C
(c) e^-x + e^y = C
(d) e^-x + e^-y = C
Solution:
(b) e^x + e^-y = C
Hint:
Given differential equation is log \(\frac { dy }{ dx }\) = x + y,
log \(\frac { dy }{ dx }\) = x + y
\(\frac { dy }{ dx }\) = e^x+y = e^x e^y
\(\frac { dy }{ dx }\) = e^x e^y
The given equation can be written as
\(\frac { dy }{ dx }\) = e^x e^y
e^-y dy = e^x dx
Taking integration on both sides, we get
∫ e^-y dy = ∫ e^x dx
\(\frac { e^{-y} }{ -1 }\) = e^x + C
-e^-y = e^x + C
-e^x – e^-y = C
-(e^x + e^-y) = C
e^x + e^-y = -C
∴ e^x + e^-y = C Where – C = C
∴ e^x + e^-y = C is a solution of the given differential equation.

Question 16.
The solution of \(\frac { dy }{ dx }\) = 2^y-x is
(a) 2^x + 2^y = C
(b) 2^x – 2^y = C
(c) \(\frac { 1 }{ 2^x }\) – \(\frac { 1 }{ 2^y }\) = C
(d) x + y = C
Solution:
(c) \(\frac { 1 }{ 2^x }\) – \(\frac { 1 }{ 2^y }\) = C
Hint:
Given
\(\frac { dy }{ dx }\) = 2^y-x
The equation can be written as

\(\frac { 1 }{ 2^x }\) – \(\frac { 1 }{ 2^y }\) = C is a solution of the given differential equation.

Question 17.
The solution of the differential equation
\(\frac { dy }{ dx }\) = \(\frac { y }{ x }\) + \(\frac { ∅(\frac { y }{ x }) }{ ∅(\frac { y }{ x }) }\) is
(a) x∅(\(\frac { y }{ x }\)) = k
(b) ∅(\(\frac { y }{ x }\)) = kx
(c) y∅(\(\frac { y }{ x }\)) = k
(d) ∅(\(\frac { y }{ x }\)) = ky
Solution:
(b) ∅\(\frac { y }{ x }\) = kx
Hint:

log ∅(v) = log x + log k
log ∅(v) = log xk
∅(v) = kx
∅(y/x) = kx

Question 18.
If sin x is the integrating factor of the linear differential equation \(\frac { dy }{ dx }\) + Py = Q, then P is
(a) log sin x
(b) cos x
(c) tan x
(d) cot x
Solution:
(d) cot x
Hint:
Given integrating factor
e^∫pdx = sin x
∫pdx = log sin x ……. (1)
Differential equation (1) with respect to x, we get

∴ The value of P is cot x

Question 19.
The number of arbitrary constants in the general solutions of order n and n + 1 is respectively.
(a) n – 1, n
(b) n, n + 1
(c) n + 1, n + 2
(d) n + 1, n
Solution:
(b) n, n + 1
Hint:
If one arbitrary constant, Differentiate one time, so order is 1.
If two arbitrary constants, Differentiate two times, so order is 2.
.
.
.
If n arbitrary constants, Differentiate n times, so the order is n.
If n + 1 arbitrary constants, Differentiate n + 1 times, so order is n + 1.

Question 20.
The number of arbitrary constants in the particular solution of a differential equation of third order is
(a) 3
(b) 2
(c) 1
(d) 0
Solution:
(d) 0

Question 21.
Integrating factor of the differential equation
\(\frac { dy }{ dx }\) = \(\frac { x+y+1 }{ x+1 }\) is
(a) \(\frac { 1 }{ x+1 }\)
(b) x+ 1
(c) \(\frac { 1 }{ \sqrt{x+1} }\)
(d) \(\sqrt { x+1 }\)
Solution:
(a) \(\frac { 1 }{ x+1 }\)
Hint:

Question 22.
The population P in any year t is such that the rate of increase in the population is proportional to the population. Then
(a) P = Ce^kt
(b) P = Ce^-kt
(c) P = Ckt
(d) Pt = C
Solution:
(a) \(\frac { 1 }{ x+1 }\)
Hint:

Question 23.
P is the amount of certain substance left in after time t. If the rate of evaporation of the substance is proportional to the amount remaining, then
(a) P = Ce^kt
(b) P = Ce^-kt
(c) P = Ckt
(d) Pt = C
Solution:
(b) P = Ce^-kt
Hint:

Question 24.
If the solution of the differential equation \(\frac { dy }{ dx }\) = \(\frac { ax+3 }{ 2y+f }\) represents a circle, then the value of a is
(a) 2
(b) -2
(c) 1
(d) -1
Solution:
(b) -2
Hint:
Given \(\frac { dy }{ dx }\) = \(\frac { ax+3 }{ 2y+f }\)
The equation can be written as
(2y + f) dy = (ax + 3) dx ……. (1)
Integrating equation (1) on both sides, we get

It is solution of the given differential equation.
Since this solution represents a circle,
co-efficient of x² = co-efficient of y²
i.e) \(\frac { a }{ 2 }\) = -1
a = -2

Question 25.
The slope at any point of a curve y = f(x) is given by \(\frac { dy }{ dx }\) = 3x² and it passes through (-1, 1). Then the equation of the curve is
(a) y = x³ + 2
(b) y = 3x² + 4
(c) y = 3x³ + 4
(d) y = x³ + 5
Solution:
(a) y = x³ + 2
Hint:
Given differential equation is \(\frac { dy }{ dx }\) = 3x²
The equation can be written as dy = 3x² dx ……… (1)
Integrating equation (1) on both sides, we get
∫dy = ∫ 3x²dx
y = \(\frac { 3x^3 }{ 3 }\) + C
y = x³ + C …….. (2)
Since it passes through (-1, 1)
So, y = x³ + C becomes
1 =(-1)³ + C
1 = -1 + C
1 + 1 = C
∴ C = 2
Substituting C value in equation (2), We get The equation of the curve is y = x³ + 2

Tamilnadu State Board New Syllabus Samacheer Kalvi 12th Chemistry Guide Pdf Chapter 14 Biomolecules Text Book Back Questions and Answers, Notes.

Tamilnadu Samacheer Kalvi 12th Chemistry Solutions Chapter 14 Biomolecules

12th Chemistry Guide Biomolecules Text Book Back Questions and Answers

Part – I Text Book Evaluation

I. Choose the Correct Answer

Question 1.
Which one of the following rotates the plane polarized light towards left? (NEET Phase – II)
a) D(+) Glucose
b) L(+) Glucose
c) D(-) Fructose
d) D(+) Galactose
Answer:
c) D(-) Fructose

Question 2.
The correct corresponding order of names of four aldoses with configuration given below Respectively is, (NEET Phase I)1551
a) L-Erythrose, L-Threose, L-Erythrose, D-Threose
b) D-Threose,D-Ervthrose, L-Threose, L-Erythrose
c) L-Erythrose, L-Threose, D-Erythrose, D-Threose
d) D-Erythrose, D-Threose, L-Erythrose, L-Threose
Answer:
d) D-Erythrose, D-Threose, I-Erythrose, L-Threose

Question 3.
Which one given below is a non-reducing sugar? (NEET Phase – I)
a) Glucose
b) Sucrose
c) maltose
d) Lactose
Answer:
b) Sucrose

Question 4.
Glucose(HCN) → Product (hydroIysis) → Product (HI + Heat) A, the compoundA is
a) Heptanoic acid
b) 2-lodohexane
c)Heptane
d) Hep tanol

Answer:
a) Heptanoic acid

Question 5.
Assertion: A solution of sucrose in water is dextrorotatory. But on hydrolysis in the presence of little hydrochloric acid, it becomes levorotatory. (AIIMS)
Reason: Sucrose hydrolysis gives unequal amounts of glucose and fructose. As a result of this change in sign of rotation is observed.
a) If both assertion and reason are true and reason is the correct explanation of assertion
b) If both assertion and reason are true but reason is not the correct explanation of assertion
c) If assertion is true but reason is false
d) If both assertion and reason is false
Answer:
a) If both assertion and reason are true and reason is the correct explanation of assertion

Question 6.
The central dogma of molecular genetics states that the genetic information flows from (NEET Phase-II)
a) Amino acids Protein DNA
b) DNA Carbohydrates Proteins
c) DNA RNA Proteins
d) DNA RNA Carbohydrates
Answer:
c) DNA RNA Proteins

Question 7.
In a protein, various amino acids linked together by (NEET Phase – I)
a) Peptide bond
b) Dative bond
c) α – Glycosidic bond
d) β- Glycosidic bond
Answer:
a) Peptide bond

Question 8.
Among the following, the achiral amino acid is (AIIMS)
a) 2-ethylalanine
b) 2-methyl glycine
c) 2-hydroxymethylserine
d) Tryptophan
Solution:

Answer:
c) 2-hydroxymethylserine

Question 9.
The correct statement regarding RNA and DNA respectively is (NEET Phase – I)
a) the sugar component in RNA is arabinose and the sugar component in DNA is ribose
b) the sugar component in RNA is 2′- deoxyribose and the sugar component in DNA is arabinose
c) the sugar component in RN A is arabinose and the sugar component in DNA is 2‘- deoxyribose
d) the sugar component in RNA is ribose and the sugar component in DNA is 2 – deoxyribose
Answer:
d) the sugar component in RNA is ribose and the sugar component in DNA is 2′-deoxyribose

Question 10.
In aqueous solution amino acids mostly exist in,
a) NH₂-CH(R)-COOH
b) NH₂-CH(R)-COO-
c) H₃N⁺-CH(R)-COOH
d) H₃N⁺-CH(R)-COO-
Answer:
d) H₃N⁺-CH(R)-COO-

Question 11.
Which one of the following is not produced by the body?
a) DNA
b) Enzymes
c) Hormones
d) Vitamins
Answer:
d) Vitamins

Question 12.
The number of sp² and sp ³ hybridised carbon in fructose are respectively
a) 1 and 4
b) 4 and 2
c) 5 and 1
d) 1 and 5
Answer:
d) 1 and 5

Question 13.
Vitamin B2 is also known as
a) Riboflavin
b) Thiamine
c) Nicotinamide
d) Pyridoxine
Answer:
a) Riboflavin

Question 14.
The pyrimidine bases present in DNA are
a) Cytosine and Adenine
b) Cytosine and Guanine
c) Cytosine and Thiamine
d) Cytosine and Uracil
Answer:
c) Cytosine and Thiamine

Question 15.
Among the following L-serine is


Answer:

Question 16.
The secondary structure of a protein refers to
a) fixed configuration of the polypeptide backbone
b) hydrophobic interaction
c) sequence of a-amino acids
d) α -the helical backbone
Answer:
d) α helical backbone

Question 17.
Which of the following vitamins is water-soluble?
a) Vitamin E
b) Vitamin K
c) Vitamin A
d) Vitamin B
Answer:
d) Vitamin B

Question 18.
Complete hydrolysis of cellulose gives
a) L-Glucose
b) D-Fructose
c) D-Ribose
d) D-Glucose
Answer:
d) D-Glucose

Question 19.
Which of the following statement is incorrect?
a) Ovalbumin is a simple food reserve in the egg- white
b) Blood proteins thrombin and fibrinogen are involved in blood clotting
c) Denaturation makes the protein more active
d) Insulin maintains the sugar level in the human body.
Answer:
c) Denaturation makes the protein more active

Question 20.
Glucose is an aldose. Which one of the following reactions is not expected with glucose?
a) It does not form an oxime
h) It does not react with the Grignard reagent
c) It does not form osazones
d) It does not reduce tollens reagent
Answer:
b) It does not react with the Grignard reagent

Question 21.
If one strand of the DNA has the sequence ‘ATGCTTGA’, then the sequence of complementary strand would be
a) TACGAACT
b) TCCGAACT
c) TACGTACT
d) TACGRAGT
Answer:
a) TACGAACT

Question 22.
Insulin, a hormone chemically is
a) Fat
b) Steroid
c) Protein
d) Carbohydrates
Answer:
c) Protein

Question 23.
α -D (+) Glucose and β -D (+) glucose are
a) Epimers
b) Anomers
c) Enantiomers
d) Conformational isomers
Answer:
b) Anomers

Question 24.
Which of the following are epimers
a) D(+)-Glucose and D(+)-Galactose
b) D(+)-Glucose and D(+)-Mannose
c) Neither (a) nor (b)
d) Both (a) and (b)
Answer:
d) Both (a) and (b)

Question 25.
Which of the following amino acids is achiral?
a) Alanine
b) Leucine
c) Proline
d) Glycine
Answer:
d) Glycine

II. Short Answer

Question 1.
What type of linkages holds together monomers of DNA?
Answer:

1. Phospho diester linkages hold together monomers of DNA
2. Phosphoric acid forms phospho diester bond between neucleotides
3. The sugar – phosphate linkage forms the backbone of each strand of DNA.

Question 2.
Give the differences between primary and secondary structure of proteins
Answer:

Question 3.
Name the Vitamins whose deficiency cause

1. rickets
2. scurvy

Answer:

1. Rickets – deficiency of Vitamin D
2. Scurvy – deficiency of Vitamin C

Question 4.
Write the Zwitter ion structure of alanine
Answer:

Question 5.
Give any three difference between DNA and RNA
Answer:

Question 6.
Write a short note on peptide bond
Answer:

• Amino acids are linked covalently by peptide bonds.
• The carboxyl group of the first amino acid reacts with the amino group of the second amino acid to give an amide linkage between these two amino acids.
• This amide linkage is called a peptide bond.
• Peptide bond is –

• The resulting compound is called a dipeptide

• If the number of amino acids attached is less it is called a polypeptide.
• If the number of amino acids attached is large it is called a protein.

Question 7.
Give two difference between Hormones and vitamins
Answer:

Question 8.
Write a note on the denaturation of proteins.
Answer:
Denaturation of proteins.
1. In general, protein has a unique three – dimensional structure formed by interactions such as disuiphide bond, hydrogen bond, hydrophobic and electrostatic interactions.

2. These interactions can be disturbed when the protein is exposed to a higher temperature in certain chemicals such urea, alternation of pH, ionic strength, etc. It leads to the loss of the three – dimensional structure.

3. The process of a protein-losing its higher-order structure without losing the primary structure, It is called denaturation of the protein. When a protein denatures, its biological function is also lost.

4. Since the primary structure is intact, this process can be reversed in certain proteins. This can happen spontaneously upon restoring the original conditions or with the help of special enzymes called chaperons.

5. Example: Coagulation of egg white by the action of heat.

Question 9.
What is reducing and non – reducing sugars?
Answer:

Question 10.
Why carbohydrates are generally optically active?
Answer:

• Almost all carbohydrates are optically active because the’ have one or more chiral carbon.
• The number of optical isomers = 2ⁿ
  where n = the total number of chiral carbons.

Question 11.
Classify the following into monosaccharides, oligosaccharides and polysaccharides.
i) starch
ii) fructose
iii) sucrose
iv) lactose
v) maltose
Answer:

Question 12.
How are vitamins classified?
Answer:
Vitamins are classified into two groups based on their solubility in water and in fat. They are,

1. Water-soluble vitamins
2. Oil or fat-soluble vitamins

Water-soluble vitamins – Vitamins which dissolve in water are called water-soluble vitamins. Examples – Vitamins of B group and Vitamin C Oil or fat-soluble vitamins: Vitamins which dissolve in oils or fat are called oil or fat-soluble vitamins. Examples – Vitamin A, D, E and K.

Question 13.
What are hormones? Give Examples.
Answer:
The hormone is an organic substance that is sëcreted by one tissue into the bloodstream and induces a physiological response in other tissues. It is an intercellular signaling molecule. Virtually every process is a complex organism is regulated by one or more hormones. Examples: insulin, epinephrine, estrogen, androgen etc.

Question 14.
Write the structure of all possible dipeptides which can be obtained from glycine and alanine.
Answer:

Question 15.
Define enzymes.
Answer:
There are many biochemical reactions that occur in our living cells. Digestion of food and harvesting the energy from them, and synthesis of necessary molecules required for various – cellular functions are examples for such reactions. All these reactions are catalysed by special proteins called enzymes.
(or)
Enzymes are biocatalysts produced by the living cells which catalyse many biochemical reactions in animal and plant bodies. They are more specific in their action.

Question 16.
Write the structure of α – D (+) glucophyranos
Answer:

Question 17.
What are different types of RNA which are found in cell?
Answer:

• Ribosomal RNA (rRNA)
• Messenger RNA (rnRNA)
• Transfer RNA (tRNA)

Question 18.
Write a note on formation of α – helix
Answer:

• In the α – helix sub – structure, the aminoacids are arranged in a right handed helical (spiral) structure.
• This structure is stabiised by hydrogen bond between the carhonyl oxygen of one amino acid (nth residue) with amino hydrogen of the fifth residue (n + 4^th residue).
• The side chains of the residues protrude outside of the helix.
• Each turn of an helix contains about 3.6 residues and in about 5.4 A long.
• The amino acid proline produces a kink in the helical structure and often called as a helix breaker due to its rigid cyclic structure.

Question 19.
What are the functions of lipids in living organisms?
Answer:

• Lipids are the integral component of cell membrane. They are necessary for the structural integrity of the cell.
• The main function of triglycerides in animals is as an energy reserve.
• They yield more energy than carbohydrates and proteins.
• They act as a protective coating in aquatic organisms.
• Lipids of connective tissue give protection to internal organs.
• Lipids help in the absorption and transport of fat-soluble vitamins.
• They are essential for the activation of enzymes such as lipases.
• Lipids act as emulsifiers in fat metabolism.

Question 20.
Is the following sugar, D – sugar or L – sugar?
Answer:

Because the H and OH on C₄ carbon are in the same configuration as the H and OH on C₂
carbon in L — Glyceraldehyde.

12th Chemistry Guide Biomolecules Additional Questions and Answers

Part – II – Additional Questions

Question 1.
Polyhydroxy aldehydes or ketones are called
a) Vitamins
b) Enzymes
c) Carbohydrates
d) Lipids
Answer:
c) Carbohydrates

Question 2.
The general molecular formula of Carbohydrates is
a) Cn(H₂)O_2n
b) Cn(H₂O)n
c) Cn(H₂O)_{2n + 2}
d) Cn(H₂O)_{2n -2}
Answer:
b) Cn(H₂O)n

Question 3.
Green leaves of plants, during photosynthesis synthesise
a) Vitamins
b) Enzymes
c) Carbohydrates
d) Lipids
Answer:
c) Carbohydrates

Question 4.
During photosynthesis, carbondioxide and water are converted into
a) Sucrose
b) Fructose
c) Maltose
d) Glucose
Answer:
d) Glucose

Question 5.
The number of optical isomers of a carbohydrate is given by the formula 2n, where n is the number of
a) Carbon atoms
b) Chiral carbon atoms
c) achiral carbon atom
d) hydroxy group
Answer:
b) Chiral Carbon atoms

Question 6.
Which among the following is not a monosaccharide?
a) Fructose
b) Erythrose
c) Maltose
d) Ribose
Answer:
c) Maltose – Disaccharide

Question 7.
The medicinal value of a drug is measured in terms of its
a) Deoxyribose
b) Goldnumber
c) Therapeutic index
d) Equilibrium constant
Answer:
c) Therapeutic index

Question 8.
Fructose is a
a) aldopentose
b) ketopentose
c) aldohexose
d) ketohexose
Answer:
d) ketohexose

Question 9.
Which is known as blood sugar?
a) Glucose
b) fructose
c) sucrose
d) maltose
Answer:
a) Glucose

Question 10.
Sucrose undergoes hydrolysis when boiled with dil.H₂So₄ to give
a) Glucose
b) fructose
c) both (a) & (b)
d) maltose
Answer:
c) both (a) & (b)

Question 11.
The number of asymmetric carbon atoms present in glucose are
a) 3
b) 4
c) 5
d) 6
Answer:
b) 4

Question 12.
Glucose is
a) dextro rotatory
b) laevorotatory
c) optically inactive
d) non – reducing sugar
Answer:
a) dextro rotatory

Question 13.
Glucose
a) Saccharic acid, Gluconic acid
b) Gluconic acid, Saccharic acid
c) Glycollic acid, Tartaric acid
d) Tartaric acid, Glycollic acid
Answer:
b) Gluconic acid, Saccharic acid

Question 14.
The isomers called anomers differ in the configuration of
a) C1 carbon
b) C2 carbon
c) C4 carbon
d) C5 carbon
Answer:
a) C1 carbon

Question 15.
The cyclic structure of glucose is similar to
a) furan
b) pyran
c) pyridine
d) pyrrole
Answer:
(b) pyran

Question 16.
Sugars differing in configuration at an asymmetric centre are known as
a) anomers
b) epimers
c) enantimoers
d) diasteromers
Answer:
(b) epimers

Question 17.
D – mannose and D – glucose are
a) C – 2 epimers
b) C – 3 epimers
c) C – 4 epimers
d) C – 5 epimers
Answer:
a) C – 2 epimers

Question 18.
D – glucose and D – galactose are
a) C – 2 epimers
b) C – 3 epimers
c) C – 4 epimers
d) C- 5 epimers
Answer:
c) C – 4 epimers

Question 19.
In our body galactose is coverted into glucose by
a) hydrolysis
b) mutrarotation
c) fermentation
d) epimerisation
Answer:
d) epimerisation

Question 20.
Fructose is .
a) dexto rotatory
b) laevo rotatory
c) optically inactive
d) disaccharide
Answer:
b) laevo rotatory

Question 21.
Glucose and fructose are
a) Chain isomers
b) Position isomers
c) Functional isomers
d) tautomers
Answer:
c) Functional isomers
Hint: Glucose – Aldohexose
Fructose – Ketohexose

Question 22.
Fruit sugar is
a) glucose
b) fructose
c) ribose
d) erythrose
Answer:
b) fructose

Question 23.
Equal amount of glucose and fructose is termed as
a) Fruit sugar
b) blood sugar
c) invert sugar
d) cane sugar
Answer:
c) invert sugar

Question 24.
Sucrose is converted into glucose and fructose by the enzyme
a) Zymase
b) invertase
c) lactase
d) fructase
Answer:
b) invertase

Question 25.
Partial reduction of fructose with sodium amalgam and water produces mixture of
a) D – Glucose and D – mannose
b) Sorbitol and Mannitol
c) GIvcollic acid and Tartaric acid
d) D – Glucose and D – Galactose
Answer:
b) Sorbitol and Mannitol

Question 26.
Sorbitol and Mannitol are
a) enantiomers
b) tautomers
c) epimers
d) functional isomers
Answer:
c) epimers

Question 27.
Which of the following statement incorrect?
a) Nucleoside + Phosphate → Nucleotide
b) Nucleoside + Base → Nucleotide
c) Sugar + Base → Nucleoside
d) n Nucleoside → Polynucleotide
Answer:
b) Nucleoside + Base → Nucleotide

Question 28.
The number of asymmetric carbon atom present in fructose is
a) 2
b) 3
c) 4
d) 5
Answer:
b) 3

Question 29.
Fructose forms a five membered ring similar to
a) Furan
b) Pyran
c) Pyridime
d) Pyrrole
Answer:
a) Furan

Question 30.
The general formula of disaccharides is
a) Cn(H₂O)_2n
b) Cn(H₂O)_n
c) Cn(H₂O)_n-1
d) Cn(H₂O)_{2n + 2}
Answer:
c) Cn(H₂O)_n-1

Question 31.
In disaccharides, two monosaccharides are linked by
a) hydrogen bond
b) ionic bond
c) peptide linkage
d) glvcosidic linkage
Answer:
d) glycosidic linkage

Question 32.
Honey is a mixture of
i) glucose
ii) fructose
iii) sucrose
iv) Maltose
a) (i) & (ii)
b) (ii) & (iii)
c) (i (ii) & (iii)
d) (i), (ii), (iv)
Answer:
c) (i), (ii), (iii)

Question 33.
Which carbon atoms of α -D – glucose and β – D – fructose are joined together in sucrose?
a) C1 and C2
b) C1 and C4
c) C1 and C5
d) C2 and C4
Answer:
a) C1 and C2

Question 34.
The disaccharide present in milk of mammals is
a) glucose
b) lactose
c) maltose
d) sucrose
Answer:
b) lactose

Question 35.
Which among the following is a non-reducing sugar
a) Lactose
b) Maltose
c) Sucrose
d) Glucose
Answer:
c) sucrose

Question 36.
On hydrolysis lactose gives
i) galactose
ii) glucose
iii) fructose
iv) maltose
a) (i) & (ii) h) (ii) & (iii)
b)(ii) & (iii)
c) (iii) & (iv) d) (i) & (iv)
d)(i)&(iv)
Answer:
a) (i) & (ii)

Question 37.
Which is produced during digestion of starch by the enzyme a – amylase?
a) Glucose
b) Maltose
c) Lactose
d) Sucrose
Answer:
b) Maltose

Question 38.
Which carbon atoms of a -D glucose are linked to form maltose?
a) C1 and C2
b) C1 and C3
c) C1 and C4
d) C3 and C4
Answer:
c) C1 and C4

Question 39.
Which among the following is a non-sugar?
a) glucose
b) cellulose
c) sucrose
d) fructose
Answer:
b) cellulose – polysaccharide which is a non – sugar

Question 40.
Starch contains
a) 80% amylopectin and 20% arnylose
b) 80% amylose and 20% arnylopectin
c) 50% amvlopectin and 50% amylose
d) 80% glucose and 20% fructose
Answer:
a) 80% amylopectin and 20% arnylose

Question 41.
Starch is a polymer of glucose molecules linked by
a) α (1, 2) glycosidic bonds
b) α (1, 3) glycosidic bonds
c) α (1, 4) glycosidic bonds
d) α (1, 5) glycosidic bonds
Answer:
c) α (1, 4) glycosidic bonds

Question 42.
Number of α- D glucose molecules joined by α (1,4) glycosidic bonds in amylose and amylopectin are respectively
a) up to 10,000 and 4000
b) up to 4000 and 10,000
c) up to 1000 and 400
d) up to 400 and 1000
Answer:
b) up to 4000 and 10,000

Question 43.
With iodine solution amylose and amylopectin give respectively
a) blue and red colour
b) violet and red colour
c) blue and purple colour
d) violet and yellow colour
Answer:
c) blue and purple colour

Question 44.
Major constituent of plant cell walls is
a) starch
b) cellulose
c) glycogen
d) amylose
Answer:
b) cellulose

Question 45.
In cellulose, glucose molecules are linked by
a) α (1, 2) glucosidic bond
b) α (1, 4) glycosidic bond
c) β (1, 2) glycosodic bond
d) β (1,4) glycosidic bond
Answer:
d) β (1,4) glycosidic bond

Question 46.
Cotton is almost pure
a) glucose
b) starch
c) cellulose
d) amylose
Answer:
c) cellulose

Question 47.
Gun cotton is
a) cellulose acetate
b) cellulose nitrate
c) ethyl cellulose
d) cellulose bromide
Answer:
b) cellulose nitrate

Question 48.
Animal starch is
a) glucose
b) glycogen
c) amylose
d) amylopectin
Answer:
b) glycogen

Question 49.
Protein are polymers of
a) α – amino acids
b) β – amino acids
c) α – hydroxy acids
d) β – hydroxy acids
Answer:
a) α – amino acids

Question 50.
The amino acids that can be synthesised by humans are called
a) essential amino acids
b) non – essential amino acids
c) polar amino acids
d) non – polar amino acids
Answer:
b) non – essential amino acids

Question 51.
At isoelectric point an amino acid exists as
a) positive ion
b) negative ion
c) zwitter ion
d) protein
Answer:
c) Zwitter ion

Question 52.
The amino acid which contains achiral carbon is
a) alanine
b) gylcine
c) phenyl alanine
d) valine
Answer:
b) glycine

Question 53.
Which is optically inactive?
a) alanine
b) glycine
c) phenyl alanine
d) valine
Answer:
b) glycine

Question 54.
A peptide bond is

Answer:

Question 55.
Number of peptide bonds present in a dipeptide is
a) 1
b) 2
c) 3
d) 4
Answer:
a) 1

Question 56.
In proteins α – helix and β – strands are two most common substructures present in
a) Primary structure
b) Secondary structure
c) Tertiary structure
d) Quaternary structure
Answer:
b) Secondary structure

Question 57.
Enzymes are
a) antibodies
b) transporters
e) bio catalysts
d) receptors
Answer:
c) bio catalysts

Question 58.
Which among the following is water soluble vitamin?
a) Vitamin A
b) Vitamin B
c) Vitamin D
d) Vitamin
Answer:
b) Vitamin B

Question 59.
Which is essential for blood clotting?
a) Vitamin A
b) Vitamin B
c) Vitamin D
d) Vitamin K
Answer:
d) Vitamin K

Question 60.
The chemical name of vitamin B6 is
a) Niacin
b) Biotin
c) Pyridoxine
d) Folic acid
Answer:
c) Pyridoxine

Question 61.
Glucose is an aldose. Which one of the following reactions is not expected with glucose?
a) It does not form oxime
b) It does not react with Grignard reagent
c) It does not form osazones
d) It does not reduce tollens reagent
Answer:
b) It does not react with Grignard reagent

Question 62.
The pyrimdine bases present in DNA are
(i) Cytosine
(ii) Thymine
(iii) Uracil
a) (i) & (ii)
b) (i) & (iii)
c) (ii) & (iii)
d) (i), (ii) & (iii)
Answer:
a) (i) & (ii)

Question 63.
A nucleotide is derived from a nucleoside by the addition of a molecule of
a) ribose
b) deoxy ribose
c) phosphoric acid
d) purine
Answer:
c) phosphoric acid

Question 64.
Which of the following is an intercellular signalling molecule?
a) Enzymes
b) Hormone
c) Protein
d) Vitamin
Answer:
b) Hormone

Question 65.
Which among the following is a steroidal hormone?
a) insulin
b) epinephrine
c) estrogen
d) adenine
Answer:
c) estrogen

Question 66.
The sweetest of all known sugars is
a) glucose
b) sucrose
c) fructose
d) ribose
Answer:
c) fructose

II. Pick out the Correct Statement

Question 1.
i) D – glucose is so named because the H and OH on C5 carbon are in the same configuration as the H and OH on C2
carbon in D – Glycerldehyde.
ii) Dextro rotatory compounds rotate the plane of polarised light in anti clock wise direction.
iii) Leavo rotatory compounds rotate the plane of polarised light in clockwise direction.
iv) The D or L isomers can either be dextro or leavo rotatory compounds.
a) (i) & (ii)
b) (ii) & (iii)
c) (iii) & (iv)
d) (i) & (iv)
Correct statements:
Answer:
d) (i) & (iv)
(ii) Dextro rotatory compounds rotate the plane of polarised light in clockwise direction.
(iii) Laevo rotatory compounds rotate the plane of polarised light in anti clockwise direction.

Question 2.
i) During hydrolysis of sucrose the optical rotation of the reaction mixture changes from levo to dextro.
ii) Honey bees have the enzyme invertase that catalyzes the hydrolysis of sucrose to glucose and fructose.
iii) In sucrose Cl of α – D – glucose is joined to C2 of β – D fructose
iv) Sucrose is a reducing sugar.
a) (i) & (ii)
b) (ii) & (iii)
c) (iii) & (iv)
d) (i) & (iv)
Correct statements :
Answer:
b) (ii) & (iii)
(i) During hydrolysis of sucrose the optical rotation of the reaction mixture changes from dextro to levo.
(iv) Sucrose is a non-reducing sugar.

Question 3.
i) Fibrous proteins are linear molecules similar to fibres.
ii) Fibrous proteins are generally soluble in water and are held together by disulphide bridges and weak intermolecular hydrogen bonds.
iii) Globular proteins have an over all spherical shape.
iv) Globular proteins are usually insoluble in water and have many functions including catalysis.
a) (i) & (ii)
b) (i) & (iii)
c) (ii) & (iii)
d) (iii) & (iv)
Answer:
b) (i) & (iii)
Correct statements :
(ii) Fibrous proteins are together by disulphide bridges and weak intermolecular hydrogen bonds.
(iv) Globular proteins are usually soluble in water and have many functions including catalysis.

Question 4.
i) Enzymes activate a reaction by increasing the activation energy by destabilizing the transition state.
ii) Many biochemical reactions are catalyzed by special proteins called enzymes.
iii) Sucrase enzyme catalyses the hydrolysis of lactose into glucose and galactose
iv) Enzymes are specific.
a) (i) & (ii)
b) (ii) & (iii)
c) (ii) & (iv)
d) (i) & (iv)
Answer:
c) (ii) & (iv)
Correct statements :
(i) Enzymes activate a reaction by reducing the activation energy’- by stabilising the transition state.
(iii) Lactase enzyme catalyses the hydrolysis of lactose into glucose and galactose.

III. Pick out the InCorrect Statement

Question 1.
i) Mono-saccharides are carbohydrates that cannot be hydrolysed further and are called simple sugars.
ii) Glucose is a ketohexose.
iii) Glucose contain one primary alcoholic group and four secondary alcoholic group.
iv) Glucose is levulose
a) (i) & (ii)
b) (ii) & (iii)
c) (ii) & (iv)
d) (iii) & (iv)
Answer:
c) (ii) & (iv)
Correct statements:
(ii) Glucose is an Aldohexose
(iv) Glucose is dextrose.

Question 2.
i) Amino acids obtained through diet are called non-essential amino acids.
ii) Amino acids have amphoteric behaviour.
iii) Iso electric point is the pH value at which the Zwitter ion of amino acid won’t move towards an electrode.
iv) pH above the isoelectric point the amino acid is positively charged.
a) (i) & (ii)
b) (ii) & (iii)
c)(iii) & (iv)
d) (i) & (iv)
Answer:
d) (i) & (iv)
Correct statements :
(i) Amino acids obtained through diet are called essential amino acids.
(iv) pH above the isoelectric point the amino acid is negatively charged.

Question 3.
i) The relative arrangement of the amino acids in the poly peptide chain is called the primary structure of the protein,
ii) α- Helix and ( β – strands are two most common sub structures formed by proteins.
iii) In the α- helix sub structure, the amino acids are arranged in a left handed helical structure.
iv) Each turn of an α- helix contains about 6.3 residues and is about 4.5 A long.
a) (i) & (ii)
b) (ii) & (iii)
c) (iii) & (iv)
d) (i) & (iv)
Answer:
c) (iii) & (iv)
Correct statements :
(iii) In the α-helix sub structure, the amino acids are arranged in a right handed helical structure.
(iv) Each turn of an α- helix contains about 3.6 residues and is about 5.4 A long.

Question 4.
i) Each vitamin has a specific function in the living system, mostly as co-enzymes.
ii) Water soluble vitamins can be stored in fatty tissues and livers.
iii) Excess of fat soluble vitamins will be excreted through urine and not stored in our body.
iv) Vitamins are small organic compounds that can not be synthesised by our body.
a) (i) & (ii)
b) (ii) & (iii)
c) (iii) & (iv)
d) (i) & (iv)
Answer:
b) (ii) & (iii)
Correct statements :
(ii) Fat soluble vitamins can be stored in fatty tissues and livers.
(iii) Excess of water soluble vitamins will be excreted through urine and not stored in our body.

IV. Assertion & Reason

Question 1.Assertion : Thymine pairs with Adenine whereas Cytosine pairs with Guanine in DNA molecule.
Reason : The hydrogen bonding between bases of two strands is highly specific.
a) Both assertion and reason are true and reason is the correct explanation of assertion.
b) Both assertion and reason are true but reason is not the correct explanation of assertion.
c) Assertion is true and reason is false.
d) Both assertion and reason are false.
Answer:
c) Assertion is true and reason is false.

Question 2.
Assertion (A): Glycyl alanine is a dipeptide
Reason (R): In glycyl alanine the amino group of glycine react with the carboxyl group of alanine to form a peptide linkage.
a) Both A and R are correct, R explains A.
b) Both A and R are correct but R does not explain A.
c) A is correct but R is wrong.
d) A is wrong but R is correct
Answer:
c) A is correct but R is wrong.
Correct Reason (R): In glycyl alanine the carbonyl group of glycine react with the amino group of alanine to form a peptide linkage.

Question 3.
Assertion (A) : A nucleotide is derived from a nucleoside by the addition of a molecule of phosphoric acid.
Reason (R) : A nucleoside is a molecule without the phosphate group.
a) Both A and R are correct, R explains A.
b) Both A and R are correct but R does not explain A.
c) A is correct but R is wrong.
d) A is wrong but R is correct
Answer:
a) Both A and R are correct, R explains A.

Question 4.
Assertion (A) : The specific association of bases A and T, C and G of the two chains of the double helix is known as complementary base pairing.
Reason (R): There are 10.5 base pairs per turn of the helix of DNA.
a) Both A and R are correct, R explains A.
b) Both A and R are correct but R does not explain A.
c) A is correct but R is wrong.
d) A is wrong but R is correct
Answer:
b) Both A and R are correct, R does not explain A.

V. Match the following

Question 1.

Answer:
i. – Glyceraldehyde
ii – Dihydroxy acetone
iii – Erythrose
iv – Erythrulose
v – Ribose
vi – Ribulose
vii – Glucose
viii – Fructose

Question 2.

Answer:
i. – Beri – beri
ii. – Cheilosis
iii – Pernicious anaemia
iv. -Depression
v. – Scurvy

Question 3.

Answer:
i. – Thiamine
ii. – Riboflovin
iii. – Niacin
iv. – Pyridoxine
v. – Folic acid

Question 4.

Answer:
i. – glycogen
ii. – heparin
iii. – histidine
iv. – phylloquinone
v. – androgen

VI. Two Mark Questions

Question 1.
How do plants synthesise glucose?
Answer:

• Plants synthesise glucose by a complex process called photosynthesis.
• During photosynthesis green leaves of plants use sunlight to convert carbon-dioxide and water into glucose and oxygen.
• Glucose is then converted into other carbohydrates

Question 2.
What is muta rotation?
Answer:

• The specific rotation of β – and (3- D glucose are 1120 and 18.70 respectively.
• When pure form of any one of these sugars is dissolved in water, slow inter conversion of α- D Glucose and (β – D Glucose via open chain structure occurs until equilibrium is established with constant specific
• rotation + 530.
• This phenomenon is called muta rotation.

Question 3.
What are epimers? What is epimerisation?
Answer:

• Sugars differing in configuration at an asymmetric centre are known as epimers.
• The process by which one epimer is converted into other is called epimerisation.
• It requires the enzymes epimerase.
• (ex) Galactose is converted into glucose by this manner in our body.

Epimers: glucose and mannose are epimers at C₂ carbon and glucose and galactose are epimers at C₄ carbon

Question 4.
What is inversion of sucrose?
Answer:

• Fructose is obtained from sucrose by heating with dilute H2SO4 or with the enzyme invertase

• The solution having equal amount of Glucose and Fructose is termed as invert sugar and the process is known as inversion of sucrose.

Question 5.
What happens when Inulin is hydrolyzed?
Answer:
Fructose is prepared commercially by the hydrolysis of Inuiin in acidic medium

Question 6.
Why sucrose is called as invert sugar?
Answer:

• On hydrolysis sucrose yields equal amount of glucose and fructose units.

• Sucrose ( + 66.6⁰) and glucose (+ 52.5⁰) are dextrorotatory.
• Fructose is levorotatory (- 92.4⁰)
• During hydrolysis of sucrose the optical rotation of the reaction mixture changes from dextro to levo.
• Hence, sucrose is called invert sugar.

Question 7.
Write about the structure of sucrose?
Answer:

• In sucrose C – 1 of α – D – Glucose is joined to C – 2 of β- D fructose
• The glycosidic bond thus formed is called a -1, 2- glycosidic bond.
• Since both the carbonyl carbons (reducing groups) are involved in the glycosidic bonding, sucrose is a non – reducing sugar.

Question 8.
Write notes on lactose.
Answer:

• Lactose is a disaccharide which on hydrolysis yields galactose and glucose.
• It is found in the milk of mammals and hence called as milk sugar.
• The β-D- Galactose and β -D-Glucose are linked by b-1, 4 – glycosidic bond.
• The aldehyde carbon is not involved in the glycosidic bond.
• Hence it retains its reducing property and is called a reducing sugar.

(β-D-galactopyranosyl-( 1 → 4) β-D-glucopyranose)
Structure of Lactose

Question 9.
Write a note on maltose?
Answer:

• Maltose is extracted from malt and hence called as malt sugar.
• Malt from sprouting barley is the major source of maltose.
• Maltose is produced during digestion of starch by the enzyme a- amylase.
• Maltose consists of two units of α -D-glucose linked by an α -1, 4 – glycosidic bond between anomeric carbon of one unit and C – 4 of the other unit.
• Since one of the glucose has the aldehydic group intact it acts as a reducing sugar.

Maltose
(α -D-glucopymnosyl-( 1 → 4) α -D-glucopyranose)
Structure of Maltose

Question 10.
What are Polysaccharides? How are they classified?
Answer:

• Polysaccharides consist of largo number of monosaccharide units bonded together by glycosidic bonds.
• They are tire most common form of carbohs drates.
• They do not have sweet taste and are called as non – sugars.
• They form linear and branched chain molecules.
• They are of two types.

i) Homopolysaccharides :
They are composed of only one type of monosaccharides, (ex) Starch, cellulose, glycogen.
ii) Hetero polysaccharide:
They are composed of more than one type of monosacchardes. (ex) hyaluronic acid, heparin.

Question 11.
Write about starch.
Answer:

• Starch is used for energy storage in plants.
• Potatoes, corn, wheat and rice are the rich sources of starch.
• It is a polymer of glucose in which glucose molecules are linked by α-(1, 4) – glycosidic bonds.
• Starch can be separated into twro fractions namely water soluble amylose and water insoluble amylopectin.
• Starch contains 20% of amylose and about 80% of amylopectin.
• Amylose is composed of unbranched chains upto 4000 α-D-Glucose molecules joined by α -(1, 4) – glycosidic bonds.
• Amylopectin contains chains upto 10000 α- D-Glucose molecules linked by a-(1, 4) – glycosidic bonds.
• In addition, there is a branching from linear chain.
• At branch points, new chains of 24 to 30 glucose molecules are linked by α -(1, 6)- glycosidic bonds.
• With iodine solution amylose give blue colour and amylopectin gives a purple colour.

Question 12.
Write briefly about cellulose.
Answer:

• Cellulose is the major constituent of plant cell walls.
• Cotton is almost pure cellulose.
• On hydrolysis cellulose gives D- glucose molecules.
• Cellulose is a straight chain polysaccharide.
• Glucose molecules are linked by (3 (1, 4) glycosidic bond.
• Cellulose is used in the manufacture of paper., cellulose fibres, rayon, explosive gun cotton.

Question 13.
Write a short notes on glycogen.
Answer:

• Glycogen is the storage polysaccharide of animals.
• It is present in the liver and muscles of animals.
• Glycogen is also called as animal starch.
• On hydrolysis it, gives glucose molecules.
• Structurally, glycogen resembles amylopectin with more branching.
• In glycogen the branching occurs every 8 – 14 glucose units opposed to 24 – 30 units in amylopectin.
• The excessive glucose in the body is stored in the form of glycogen

Question 14.
What are essential and non – essential amino acids?
Answer:

Question 15.
What is isoelectric point?
Answer:
The p^H value at which the net charge of an amino acid is neutral is called isoelectric point.

Question 16.
What are Zwitter ions?
Answer:

• In aqueous solution the proton from carboxyl group can be transferred to the amino group of an amino acid leaving these groups with opposite charges.
• Despite having both positive and negative charges this molecule is neutral and has amphoteric behaviour.
• These ions are called Zwitter ions.

Question 17.
What are proteins? How are they classified?
Answer:

• Proteins are polymers of amino acids.
• They are classified into two major types based on their structure.

Question 18.
What is complementary base pairing in DNA?
Answer:

• In the double helix of DNA each base is hydrogen bonded to a base in opposite strand to form a planar base pair.
• Two hydrogen bonds are formed between adenine and thymine.
• Three hydrogen bonds are formed between guanine and cytosine.
• This specific association of the two chains of the double helix is known as complementary base pairing.
• Other pairing tends to destabilize the double helical structure.

Question 19.
Name the biological functions of nucleotides.
Answer:

• Energy carriers (ATP)
• Components of enzyme cofactors (co- enzyme A, NAD+, FAD)
• Chemical messengers (cyclic AMP, cAMP)

Question 20.
Name the vitamins whose deficiency cause
i) Pellagra
ii) Beri – Beri
iii) Night blindness
Answer:

• Pellagra (photo sensitive dermatitis) – Vitamin B₃ (Niacin)
• Ben – Ben – Vitamin B₁ (Thiamine)
• Night blindness – Vitamin A (Retinol)

Question 21.
What is glycosidic linkage?
Answer:

• The oxide linkage by which two monosaccharides are linked in disaccharides is called as a
  glycosidic linkage.
• A glycosidic linkage is formed by the reaction of the hydroxyl group of the anomeric carbon of one
  mono saccharide with a hydroxyl group of another mono saccharide.

VII. Three Mark Questions

Question 1.
Write about the configuration of carbohydrates.
Answer:

• Almost all carbohydrates are optically active as they have one or more chiral carbon.
• Number of optical isomers = 2n (where n = no of chiral carbons)
• Fischer has devised a projection formula to relate the structure of a carbohydrate to one
  of the two enatiomeric forms of glyceraldehyde.

• Based on the above structures, carbohydrates are named as D or L.
• For example D – Glucose is so named because the H and OH on C5 carbon are in the same configuration as H and OH on C2 carbon is D – glyceraldehyde.
• Dextro rotatory compounds rotate the plane of plane polarised light in clockwise direction and represented as (+)
• Laevo rotatory compounds rotate the plane of plane polarised light in anticlockwise direction and represented as (-).
• D or L isomers can be dextro or laevo rotatory.
• Dextro rotatory compounds are represented as D (+) or L (+) and the laevo rotatory compounds as D (-) or L ( -).

Question 2.
Explain the cyclic structure of glucose.
Answer:

• Glucose was found to crystallise in two different forms depending upon the crystallisation conditions with different melting points (419 and 423 K).
• In order to explain these it was proposed that one of the hydroxyl group reacts with the aldehyde group to form a cyclic structure (hemiacetal form) as shown in figure.

• This also results in the conversion of the achiral aldehyde carbon into a chiral one with the possibility of two isomers.
• These two isomers which differ only in the configuration of C1 carbon are called anomers.
• The two anomeric forms of glucose are called α – and β – forms.
• This cyclic structure is similar to pyran, a cyclic compound with 5 carbon and one oxygen atom, and hence is called pyranose form.
• The specific rotation of α- and (β- (D) Glucose are 112⁰ and 18.7⁰ respectively.
• When pure form of any one of these sugars is dissolved in water, slow interconversion of α- D Glucose and β- D Glucose via open chain structure occurs until equilibrium is established with constant specific rotation +53⁰
• This phenomenon is called mutarotation.

Question 3.
Write about the cyclic structure of fructose.
Answer:

• Like glucose, fructose also forms a cyclic structure.
• Unlike glucose, fructose forms a five membered ring similar to furan.
• Hence it is called furanose form.
• When fructose is a component of a saccharide as in sucrose, it usually occurs in furanose form.

Question 4.
What is a peptide bond? How is it formed?
Answer:

• The carboxyl group of the first amino acid react with the amino group of the second amino acid to give an amide linkage between these amino acids.
• This amide linkage is called peptide bond.
• The resulting compound is called a dipeptide.
• Peptide bond is

(ex) Glycine and alanine combine to form glycyl alanine a dipeptide through one peptide bond.

Question 5.
Differentiate tertiary and quaternary structure of proteins
Answer:

Question 6.
What are lipids? How are they classified?
Answer:

• The word lipids is derived from the Greek word ‘lipos’ meaning fat.
• Lipids are organic molecules that are soluble in organic solvents such as chloroform and methanol.
• Lipids are insoluble in water.
• They are the principal components of cell membranes including cell walls.
• They act as energy source for living systems.
• Fat, a lipid provide 2-3 fold higher energy compared to carbohydrates or proteins.
  Types : Simple lipids, compound lipids and derived lipids.

Question 7.
How are hormones classified?
Answer:
Hormones are classified according to the distance over which they act as endocrine, paracrine and autocrine hormones.

VIII. Five Mark Questions

Question 1.
What are carbohydrates? How are they classified?
Answer:

• Carbohydrates are defined as polyhydroxy aldehydes or Ketones.
• General formula C_n(H₂O)_nCarbohydrates

Monosaccharides : These are simple sugars that cannot be hydrolysed further.
Aldoses : Contain aldehyde functional group (ex) Glucose
ketoses : Contain Ketone functional group (ex) Fructose

Disaccharides : On hydroysis give two monosaccharides. (ex) Sucrose.
Polysaccharides : On hydrolysis give large number of monosaccharides. (ex) Starch.
Homo polysaccharides : Consist of only one type of monosaccharides. (ex) Starch.
Hetero polysaccharides: Consist of more than one type of monosaccharides. (ex) heparin.

Question 2.
Elucidate the structure of glucose.
Answer:

• The molecular formula of glucose is C₆H₁₂0₆.
• On reduction with conc.HI and red phosphorus at 373K, glucose gives a mixture of n-hexane and 2-
  iodohexane. This indicates that the six carbon atoms are bonded linearly.
  
• Glucose reacts with hydroxyl amine forms oxime, with HCN to form cyanohydrin. This shows the presence of carbonyl of group

• Glucose is oxidîzed by bromine water to form glucoriic acid. This suggests that glucose contains an aldehyde
  group. With strong oxidising agent like con.HNO₃ Saccharic acid is obtained. This shows the presence of primary alcoholic group at one end

• Glucose reduces Tollens reagent, and Fehling’s solution – Hence it contains an aldehyde group.
• Glucose forms penta acetate with acetic anhydride. This shows the presence of – OH groups in glucose.
• Glucose is a stable compound and does not undergo dehydration. This shows the presence of 5 – OH groups on different carbons.
• The glucose is referred to as D(+) glucose.
• Thus the structure of D(+) Glucose can be written as

Question 3.
Elucidate the structure of fructose.
Answer:

• The molecular formula of fructose is C₆H₁₂O₆.
• On reduction with HI and red phosphorus gives a mixture of n-hexane and 2-iodohexane. This indicates that the six carbon atoms are bonded linearly.
  
• Fructose reacts with NH₂OH and HCN to form Oxime and cyanohydrin respectively. This shows the presence of carbonyl group.
• Fructose reacts with acetic anhydride in the presence of pyridine to form pentaacetate. This shows the presence of five hydroxyl groups.
• Fructose is not oxidised by bromine water. This shows the absence of aldehyde group.
  Fructose reduced with Sodium amalgam and water.
• Partial reduction with sodium amalgam and water gives epimeric alcohols sorbitol and mannitol. New asymmetric carbon is formed at C2. This confirms the presence of Keto group at C2.

Oxidation with nitric acid gives glycollic acid and tartaric acid which contain
smaller number of carbon atoms than in fructose.

Question 4.
Explain the composition and structure of nucleic acids.
Answer:

• Particles in nucleus of the cell are responsible for the transmission of inherent characteristics of each and every species form one generation to the next.
• They are called chromosomes.
• They are made up of proteins and a type of molecules called nucleic acids.
• Nucleic acids are biopolymers of nucleotides.
• They are of two types (i) deoxyribonucleic acid (DNA) (ii) ribonucleic acid (RNA).
• They are the molecular repositories that carry genetic information in every organism.

Composition and structure :

Controlled hydrolysis of DNA and RNA yields three components namely a nitrogenous base, a pentose sugar and phosphate group.

Nitrogen base:
These are organic nitrogen compounds which are derivatives of two parent compounds, pyrimidine and purine.

Pentose Sugar:

• DNA contains 2 – deoxy – D – ribose.
• RNA contains D – ribose.
• In nucleotides both the types of pentoses are in their b- Furanose (closed five membered ring) form.

Phosphate group:

• Phosphoric acid forms phosphor diester bond between nucleotides.
• Based on the number of phosphate group present in the nucleotides, they are classified into mono nucleotide, dinucleotide and trinucleotide.

Nucleosides and nucleotides :

• The molecule without the phosphate group is called a nucleoside.
• A nucleotide is derived from a nucleoside by the addition of a molecule of phosphoric acid.
• Phosphorylation occurs generally in the 5′ OH group of the sugar.
• Nucleotides are linked in DNA and RNA by phospho diester bond between 5′ OH group of one nucleotide and 3′ OH group on another nucleotide.
  Sugar + base → Nucleoside
  Nucleoside + Phosphate → Nucleotide
  n Nucleotide → Polynucleotide (Nucleic acid)

Question 5.
Explain the double strand helix structure of DNA.

• The structure elucidation of DNA by Watson and Crick in 1953 was a momentous event in science.
• They postulated a 3 – dimensional model of DNA structure.
• This model consisted of two anti-parallel helical DNA chains wound around the same axis to form
  a right – handed double helix.
• The hydrophilic back bones of alternating deoxyribose and phosphate groups are on the outside of the double helix, facing the surrounding water.
• The purine and pyrimidine bases of both strands are stacked inside the double helix, with their hydrophobic and ring structures very close together and perpendicular to the long axis.

• This reduces the repulsions between the charged phosphate groups.
• The offset pairing of the two strands creates a major groove and minor groove on the surface of the duplex.
• This model revealed that, there are 10.5 base pairs (36A0) per turn of the helix and 3.4 AO between the stacked bases.
• Each base is hydrogen bonded to a base in opposite strand to form a planar base pair.
• Two hydrogen bonds are formed between adenine and thymine.
• Three hydrogen bonds are formed between guanine and cytosine.
• Other pairing tends to destabilise the double helical structure.
• This specific association of the two chains of the double helix is known as complementary base pairing.
• The DNA double helix or duplex is held together by two forces.
  a) Hydrogen bonding between complementary base pairs.
  b) Base – stacking interactions.

Question 6.
What are the differences between DNA and RNA?
Answer:

Question 7.
Explain DNA fingerprinting.
Answer:

• DNA fingerprinting was first invented by Sir Alec Jeffreys in 1984.
• The DNA fingerprint is unique for every person.
• It can be extracted from traces of samples from blood, saliva, hair, etc.
• By using this method the individual-specific variation in human DNA can be detected.
• The extracted DNA is cut at specific points along the strand with restriction enzymes.
• This results in the formation of DNA fragments of varying lengths which were analysed by a technique called gel electrophoresis.
• This method separates the fragments based on their size.
• The gel containing the DNA fragments are then transferred to a nylon sheet using a technique called blotting.
• Then the fragments will undergo auto-radiography in which they were exposed to DNA probes (pieces of synthetic DNA that were made radioactive and that bound to the fragments)
• A piece of X-ray film was then exposed to the fragments, and a dark mark was produced at any point where a radioactive probe had become attached.
• The resultant pattern of marks could then be compared with other samples.
• DNA fingerprinting is based on slight sequence differences (usually single base – pair changes) between individuals.
• These methods are proving decisive in court cases worldwide

Tamilnadu State Board New Syllabus Samacheer Kalvi 12th Maths Guide Pdf Chapter 10 Ordinary Differential Equations Ex 10.8 Textbook Questions and Answers, Notes.

Tamilnadu Samacheer Kalvi 12th Maths Solutions Chapter 10 Ordinary Differential Equations Ex 10.8

Question 1.
The rate of increase in the number of bacteria in a certain bacteria culture is proportional to the number present. Given that the number triples in 5 hours, find how many bacteria will be present after 10 hours?
Solution:
Let x denote the number of bacteria at time t hours.
Given = \(\frac { dx }{ dt }\) = kx hence \(\frac { dx }{ x }\) = kdt
∴ x = C e^kt
Suppose x = x₀ at time t = 0
x₀ = C e^k(0) = C e° = C
∴ C = x₀
Hence x = x₀ e^kt
At time 5, x = 3x₀
(∵ Number triple in 5 hrs)
∴ Hence 3x₀ = x₀ e^5k
∴ e^5k = 3
when t = 10, x = x₀ e^10k = x₀ (e^5k)²
= x₀ 3² = 9x₀
∴ After 10 hours, the number of bacteria as 9 times the original number of bacteria.

Question 2.
Find the population of a city at any time t, given that the rate of increase of population is proportional to the population at that instant and that in a period of 40 years, the population increased from 3,00,000 to 4,00,000.
Solution:
Let P denote the population of a city
Given \(\frac { dP }{ dt }\) = kP
This equation can be written as
\(\frac { dP }{ P }\) = kdt
Taking integration on both sides, we get
∫ \(\frac { dP }{ P }\) = k∫ dt
log P = kt + log c
log P – log c = kt
log (\(\frac { P }{ c }\)) = kt
\(\frac { P }{ c }\) = e^kt
P = ce^kt …….. (1)
Initial condition:
Given when t = 0 ; P = 3,00,000
Equation (1) becomes,
3.0. 000 = ce^k(0) = ce°
3,00,000 = c [∵ e° = 1]
∴ (1) ⇒ P = 3,00,000 e^kt ………. (2)
Again when t = 40; P = 4,00,000
Equation (2) becomes,
4,00,000 = 3,00,000 e^40k
\(\frac { 4,00,000 }{ 3,00,000 }\) = e^40k
\(\frac { 4 }{ 3 }\) = e^40k
Taking log,
log(\(\frac { 4 }{ 3 }\)) = 40k
k = \(\frac { 1 }{ 40 }\) log \(\frac { 4 }{ 3 }\)
k = log(\(\frac { 4 }{ 3 }\))^{\(\frac { 1 }{ 40 }\)} [∵ n log m = log mⁿ]
Substituting k values in equation (2), we get
P = 3,00,000 e^{log(\(\frac { 4 }{ 3 }\))\(\frac { 1 }{ 40 }\)}
P = 3,00,000 (\(\frac { 4 }{ 3 }\))^{\(\frac { 1 }{ 40 }\)} [∵ e^{log a x} = a^x]

Question 3.
The equation of electromotive force for an electric circuit containing resistance and self-inductance is E = Ri + L\(\frac { di }{ dt }\), where E is the electromotive force is given to the circuit, R the resistance and L, the coefficient of induction. Find the current i at time t when E = 0.
Solution:

This is a linear differential equation.
Integrating factor I.F = e^{∫ \(\frac { R }{ L }\)dt} = e^{\(\frac { R }{ L }\)t}
Its solution is given by

Question 4.
The engine of a motorboat moving at 10 m/s is shut off. Given that the retardation at any subsequent time (after shutting off the engine) equal to the velocity at that time. Find the velocity after 2 seconds of switching off the engine.
Solution:
Let v be the velocity
Given the engine of a motorboat moving 10 m/s.
After that the engine is shut off then the acceleration is negative.
So it be \(\frac { -dv }{ dt }\)
i.e., \(\frac { dv}{ dt}\) = -v
The equation can be written as taking integration on both sides, we get
\(\frac { dv}{ dt}\) = -dt
∫ \(\frac { dv}{ v }\) = ∫-dt
log v = -t + log c
log v – log c = -t
log (\(\frac { v }{ c }\)) = -t
\(\frac {v}{c}\) = e^-t
v = ce^-t ……….. (1)
Initial condition
Given that when t = 0, v = 10 m/sec i
substituting in equation (1), we get !
10 = ce^-0
10 = ce°
10 = c
∴ c = 10
(1) ⇒ v = 10e^-t
when t = 2 find v
v = 10 e^-2
v = \(\frac {10}{e^2}\)
The velocity after 2 seconds is \(\frac {10}{e^2}\)
i.e., v = \(\frac {10}{e^2}\)

Question 5.
Suppose a person deposits 10,000 Indian rupees in a bank account at the rate of 5% per annum compounded continuously. How much money will be in his bank account 18 months later?
Solution:
Let P be the principal amount
Given Rate of interest = 5% per annum.
∴ \(\frac {dp}{dt}\) = p(\(\frac {5}{100}\)) = 0.05P
The equation can be written as,
\(\frac {dP}{P}\) = 0.05 dt
Taking Integration on both sides, we get
∫\(\frac {dP}{P}\) = 0.051 ∫dt
log P = 0.05 t + log c
log P – log C = 0.05t
log (\(\frac {P}{C}\)) = 0.05 t
\(\frac {P}{C}\) = e^0.05t
P = C e^0.05t ………. (1)
Initial condition:
Given when t = 0; P = 10,000
Substituting these values in equation (1), we get
P = C e^0.05t
10,000 = C e^{0.05 (0)}
10,000 = C e°
C = 10,000
∴ Substituting the C value in equation (1), we get
P = 10,000 e^0.05t ……. (2)
When t= 18 months = 1\(\frac {1}{2}\)yr =3/2 years, we get
(2) ⇒ P = 10,000 e^{0.05 (3/2)}
P = 10,000 e^0.075
The amount in a bank account be
P = 10,000 e^0.075

Question 6.
Assume that the rate at which radioactive nuclei decay is proportional to the number of such nuclei that are present in a given sample. In a certain sample, 10% of the original number of radioactive nuclei have undergone disintegration in a period of 100 years. What percentage of the original radioactive nuclei will remain after 1000 years?
Solution:
Let N be the sample radioactive nuclei at any time t & N₀ be the radioactive nuclei at the initial time.
Given \(\frac {dN}{dt}\) = -kN
Where k > 0 is a constant
The equation can be written as
\(\frac {dN}{N}\) = -kdt
Taking integration on both sides, we get
∫\(\frac {dN}{N}\) = ∫-kdt
log N = -kt + log C
log N – log C = -kt
log(\(\frac {N}{C}\)) = -kt
\(\frac {N}{C}\) = e^-kt
N = Ce^-kt ……… (1)
Initial condition:
when t = 0 we have N = N₀
N₀ = Ce^-k(0)
N₀ = Ce⁰
N₀ = C
Substituting C value in equation (1), we get ;
N = N₀ e^-kt ……… (2)
Given: In a certain sample 10% of the original number of radioactive nuclei have undergone disintegration in a period of 100 years be when t = p₀

Equation (2) becomes

The percentage of the original radioactive nuclei of remain after 1000 years is
\(\frac {N}{N_0}\) × 100 = (\(\frac {9}{10}\))¹⁰ × 100 = \(\frac {9^{10}}{10^{10}}\) × 10²
= \(\frac {9^{10}}{10^{8}}\) %

Question 7.
Water at temperature 100°C cools in 10 minutes to 80° C at a room temperature of 25°C. Find
(i) The temperature of the water after 20 minutes
(ii) The time when the temperature is 40° C
[log, \(\frac {11}{15}\) = -0.3101; log_e 5 = 1.6094]
Solution:
We apply “Newton’s law of cooling” while states that the rate of decrease of the temperature of a body is proportional to the difference between the temperature of the body and that of the medium.
\(\frac {dT}{dt}\) = -k(T – T₀)
Where T is the temperature of the body at time t & T₀ the constant temperature of the medium.
Thus \(\frac {dT}{dt}\) = -k(T – 25) or
\(\frac {dT}{T-25}\) = -kdt
On Integrating equation (1), we get
∫\(\frac {dT}{T-25}\) = ∫-kdt
log (T – 25) = -kt + c₁
Now T = 100, t = 0
log (100 – 25) = -k (0) + c₁
∴ log 75 = c₁
Substituting c₁ value in equation (1), we get
log (T – 25) = -kt + log 75
∵ log m – log n = log m/n
log \(\frac {T-25}{75}\) = kt ……. (2)
Also T = 80 when t = 10 minutes

Substituting k value in equation (2), we get

T = 75 (0.5378) + 25
T = 40.33 + 25
T = 65.33°C

Question 8.
At 10.00 A.M. a woman took a cup of hot instant coffee from her microwave oven and placed it on a nearby kitchen counter to cool. At this instant, the temperature of the coffee was 180° F and 10 minutes later it was 160° F. Assume that the constant temperature of the kitchen was 70° F.
(i) What was the temperature of the coffee at 10.15 AM?
The woman likes to drink coffee when its temperature is between 130° F and 140° F. Between what times should she have drunk the coffee?
Solution:
Let T be the temperature of a coffee at time t.
T_k be the temperature of the kitchen
By Newton’s law of cooling

Initial condition:
when t = 0; T = 180°F
180 = ce^k(0) + 70
180 = ce° + 70
180 – 70 = c
∴ c = 110°
Substituting c value in equation (1), we get
T = ce^+kt + 70
T = 100 e^kt + 70 …….. (2)
Second condition:
when t = 10, T = 160
(2) ⇒ 160 = 110 e^10k + 70
160 – 70= 110 e^10k

T = 81.33 + 70
T = 151.3 F
∴ The temperature of the coffee at 10.15 A.M is 151.3° F

(ii) Woman’s like to drink a coffee between 130°F and 140°F.
(a) when T = 130°F
(2) ⇒ T = 110 e^kt + 70
130 – 70 = 110 e^kt
\(\frac { 60 }{ 110 }\)
\(\frac { 6 }{ 11 }\) = e^kt

b) When T = 140°
(2) ⇒ T = 110e^kt + 70
160 – 70 = 110 e^kt

t = 22.6 min
She drinks coffee between 10.22 & 10.30 approximately.

Question 9.
A pot of boiling water at 100° C is removed from a stove at time t = 0 and left to cool in the kitchen. After 5 minutes, the water temperature has decreased to 80° C and another 5 minutes later it has dropped to 65° C. Determine the temperature of the kitchen.
Solution:
Let T be the temperature of the boiling water.
T_m is the temperature of the kitchen.
By Newton’s law of cooling, we get
\(\frac { dT }{ dt }\) = k (T – T_m)
The equation can be written as
Taking Integration on both sides, we get


(a – b)² = a² + b² – 2ab
a = 80 b = T_m
6400 + T\(_{m}^{2}\) – 160 T_m = 6500 – 100 T_m – 65 T_m + T\(_{m}^{2}\)
6400 – 6500 = 160 T_m – 165 T_m
– 100 = – 5T_m
T_m = \(\frac { 100 }{ 5 }\)
T_m = 20°C
Hence the temperature of the kitchen be 20° C

Question 10.
A tank initially contains 50 litres of pure water. Starting at time t = 0 a brine containing 2 grams of dissolved salt per litre flows into the tank at the rate of 3 litres per minute. The mixture is kept uniform by stirring and the well-stirred mixture simultaneously flows out of the tank at the same rate. Find the amount of salt present in the tank at any time t > 0.
Solution:
Let x be the amount of salt in the tank at time t.
∴ \(\frac { dx }{ dt }\) = inflow rate – outflow rate ……… (1)
Given 2 grams of dissolved salt per litre flows into the tank at the rate of 3 litres per minute.
(i.e) inflow rate contain = 6 gram salt
[∵ for one litre = 2 gram salt]
[for 3 litre = 6 gram salt]
tank contain 50 litres of water
∴ out flow of rate of salt = \(\frac { 3x }{ 50 }\)
substitute inflow rate and outflow rate in equation (1), we get