Tags Class 12th

Class 12th

Get insights from 11.9k questions on Class 12th, answered by students, alumni, and experts. You may also ask and answer any question you like about Class 12th

Follow Ask Question

11.9k

Questions

Discussions

Active Users

Followers

New answer posted

a year ago

Class 12th Polymers

Which of the following polymer can be formed by using the following monomer unit?

(A) Nylon-6,6

(B) Nylon 2–nylon 6

(C) Melamine polymer

(D) Nylon-6

0 Follower 1 View

Payal Gupta

Contributor-Level 10

This is a multiple choice answer as classified in NCERT Exemplar

Option (D)

Explanation: The given structure is caprolactam. Nylon-6 is prepared from the monomer caprolactam. It is obtained by heating caprolactam with water at a high temperature.

0 0

New question posted

a year ago

Class 12th Maths Integrals

Kindly consider the following

$\int\sqrttanx d x (Hint: Put t a n x = t^{2})$

0 Follower 1 View

New question posted

a year ago

Class 12th Integrals

Kindly consider the following

$\int\sqrttanx d x (Hint: Put t a n x = t^{2})$

0 Follower 4 Views

New answer posted

a year ago

Class 12th Polymers

Which of the following statements is not true about low density polythene?

(A) Tough

(B) Hard

(C) Poor conductor of electricity

(D) Highly branched structure

0 Follower 2 Views

Payal Gupta

Contributor-Level 10

This is a multiple choice answer as classified in NCERT Exemplar

Option (B)

Explanation: Low-density polythene is obtained from the polymerization of ethane under high pressure and temperature. Low-density polyethylene is tough, a poor conductor of electricity and has a highly branched structure.

Low-density polythene is not hard because branched structure prevents stacking and thus, reduces the intermolecular force of attraction.

So, option (B) is correct.

0 0

New question posted

a year ago

Class 12th Integrals

Kindly consider the following

$\int e^{- 3 x} c o s^{3} x d x$

0 Follower

New answer posted

a year ago

Class 12th Polymers

In which of the following polymers ethylene glycol is one of the monomer units?

0 Follower 2 Views

Payal Gupta

Contributor-Level 10

This is a multiple choice answer as classified in NCERT Exemplar

Option (A)

Explanation: Dacron or Terylene is formed by the condensation polymerization of ethylene glycol and phthalic acid with the elimination of water molecules. The structure of Terylene is,

0 0

New answer posted

a year ago

Class 12th Polymers

Which of the following polymers is biodegradable?

0 Follower 1 View

Payal Gupta

Contributor-Level 10

This is a multiple choice answer as classified in NCERT Exemplar

Option (C)

Explanation: PHBV (Poly beta-hydroxybutyrate co-beta-hydroxy valerate) is a biodegradable polymer. It undergoes bacterial degradation in the environment. The structure is shown as,

0 0

New answer posted

a year ago

Class 12th Integrals

Kindly consider the following

$\int e^{- 3 x} c o s^{3} x d x$

0 Follower 1 View

alok kumar singh

Contributor-Level 10

This is a Long Answer Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l} L e t ? ? ? ? I = ? \underset{I I}{e^{? 3 x}} \underset{I}{c o s^{3} x} d x \\ ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? = c o s^{3} x . ? e^{? 3 x} ? d x ? ? (D (c o s^{3} x) . ? e^{? 3 x} ? d x) d x \\ ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? = c o s^{3} x . \frac{e^{? 3 x}}{? 3} ? ? (3 c o s^{2} x (? s i n x) . \frac{e^{? 3 x}}{? 3}) d x \\ ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? = ? \frac{1}{3} e^{? 3 x} c o s^{3} x ? ? c o s^{2} x s i n x . e^{? 3 x} d x \\ ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? = ? \frac{1}{3} e^{? 3 x} c o s^{3} x ? ? (1 ? s i n^{2} x) s i n x . e^{? 3 x} d x \\ ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? = ? \frac{1}{3} e^{? 3 x} c o s^{3} x ? ? s i n x . e^{? 3 x} d x + ? \underset{I}{s i n^{3} x} . \underset{I I}{e^{? 3 x}} d x \\ ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? = ? \frac{1}{3} e^{? 3 x} c o s^{3} x ? ? s i n x . e^{? 3 x} d x + s i n^{3} x ? e^{? 3 x} d x ? ? (D (s i n^{3} x) . ? e^{? 3 x} ? d x) d x \\ ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? = ? \frac{1}{3} e^{? 3 x} c o s^{3} x ? ? s i n x . e^{? 3 x} d x + s i n^{3} x . \frac{e^{? 3 x}}{? 3} ? ? (3 s i n^{2} x . c o s x . \frac{e^{? 3 x}}{? 3}) d x \\ ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? = ? \frac{1}{3} e^{? 3 x} c o s^{3} x ? ? s i n x . e^{? 3 x} d x ? \frac{1}{3} e^{? 3 x} s i n^{3} x + ? s i n^{2} x c o s x . e^{? 3 x} d x \\ ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? = ? \frac{1}{3} e^{? 3 x} c o s^{3} x ? ? s i n x . e^{? 3 x} d x ? \frac{1}{3} e^{? 3 x} s i n^{3} x + ? (1 ? c o s^{2} x) c o s x . e^{? 3 x} d x \\ ? ? ? ? ? ? ? ? ? ? ? ? ? I = ? \frac{1}{3} e^{? 3 x} c o s^{3} x ? [s i n x . \frac{e^{? 3 x}}{? 3} ? ? c o s x . \frac{e^{? 3 x}}{? 3} d x] ? \frac{1}{3} e^{? 3 x} s i n^{3} x + ? c o s x . e^{? 3 x} d x ? ? c o s^{3} x . e^{? 3 x} d x \\ ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? = ? \frac{1}{3} e^{? 3 x} c o s^{3} x + s i n x . \frac{e^{? 3 x}}{3} ? ? c o s x . \frac{e^{? 3 x}}{? 3} d x ? \frac{1}{3} e^{? 3 x} s i n^{3} x + ? c o s x . e^{? 3 x} d x ? I \\ ? ? ? ? ? ? ? ? ? 2 I = \frac{e^{? 3 x}}{? 3} [c o s^{3} x + s i n^{3} x] ? [s i n x . \frac{e^{? 3 x}}{3} ? ? c o s x . \frac{e^{? 3 x}}{? 3} d x] + ? c o s x . e^{? 3 x} d x \\ ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? = \frac{e^{? 3 x}}{? 3} [c o s^{3} x + s i n^{3} x] + \frac{1}{3} s i n x . e^{? 3 x} ? \frac{1}{3} ? c o s x . e^{? 3 x} d x + ? c o s x . e^{? 3 x} d x \\ ? ? ? ? ? ? ? ? 2 I = \frac{e^{? 3 x}}{? 3} [c o s^{3} x + s i n^{3} x] + \frac{1}{3} s i n x . e^{? 3 x} + \frac{2}{3} ? c o s x . e^{? 3 x} d x \\ N o w, ? I_{1} = \frac{2}{3} ? \underset{I}{c o s x} . \underset{I I}{e^{? 3 x}} d x \\ ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? = \frac{2}{3} [c o s x . ? e^{? 3 x} d x ? ? (D (c o s x) . ? e^{? 3 x} ? d x) d x] \\ ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? = \frac{2}{3} [c o s x . \frac{e^{? 3 x}}{? 3} ? ? ? s i n x . \frac{e^{? 3 x}}{? 3} d x] \\ ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? = \frac{2}{3} [c o s x . \frac{e^{? 3 x}}{? 3} ? \frac{1}{3} ? s i n x . e^{? 3 x} d x] \end{array}$

$\begin{array}{l} = \frac{2}{? 9} c o s x . e^{? 3 x} ? \frac{2}{9} ? s i n x . e^{? 3 x} d x \\ ? ? ? ? ? ? ? ? ? ? ? I_{1} = \frac{2}{? 9} c o s x . e^{? 3 x} ? \frac{2}{9} [s i n x . \frac{e^{? 3 x}}{? 3} ? ? c o s x . \frac{e^{? 3 x}}{? 3} d x] \\ ? ? ? ? ? ? ? ? ? ? ? I_{1} = ? \frac{2}{9} c o s x . e^{? 3 x} + \frac{2}{27} s i n x . e^{? 3 x} ? \frac{2}{27} ? c o s x . e^{? 3 x} d x \\ ? ? ? ? ? ? ? ? ? ? ? I_{1} = ? \frac{2}{9} c o s x . e^{? 3 x} + \frac{2}{27} s i n x . e^{? 3 x} ? \frac{1}{9} . \frac{2}{3} ? c o s x . e^{? 3 x} d x \\ ? ? ? ? ? ? ? ? ? ? ? I_{1} = ? \frac{2}{9} c o s x . e^{? 3 x} + \frac{2}{27} s i n x . e^{? 3 x} ? \frac{1}{9} . I_{1} \\ I_{1} + \frac{1}{9} I_{1} = ? \frac{2}{9} c o s x . e^{? 3 x} + \frac{2}{27} s i n x . e^{? 3 x} \\ ? ? ? ? ? ? ? \frac{10 I_{1}}{9} = ? \frac{2}{9} c o s x . e^{? 3 x} + \frac{2}{27} s i n x . e^{? 3 x} \\ ? ? ? ? ? ? ? ? ? ? ? I_{1} = ? \frac{1}{10} c o s x . e^{? 3 x} + \frac{1}{15} s i n x . e^{? 3 x} \\ S o, ? ? ? ? ? 2 I = ? \frac{1}{3} e^{? 3 x} [s i n^{3} x + c o s^{3} x] + \frac{1}{3} s i n x . e^{? 3 x} ? \frac{1}{10} c o s x . e^{? 3 x} + \frac{1}{15} s i n x . e^{? 3 x} \\ ? ? ? ? ? ? ? ? ? ? ? ? ? I = ? \frac{1}{6} e^{? 3 x} [s i n^{3} x + c o s^{3} x] + \frac{1}{6} s i n x . e^{? 3 x} ? \frac{1}{20} c o s x . e^{? 3 x} + \frac{1}{30} s i n x . e^{? 3 x} \\ ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? = ? \frac{1}{6} e^{? 3 x} [s i n^{3} x + c o s^{3} x] + \frac{1}{5} s i n x . e^{? 3 x} ? \frac{1}{20} c o s x . e^{? 3 x} \\ ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? = \frac{e^{? 3 x}}{24} [s i n 3 x ? c o s 3 x] + \frac{3 e^{? 3 x}}{40} [s i n x \end{array}$

0 0

New answer posted

a year ago

Class 12th Polymers

The commercial name of polyacrylonitrile is ______________.

(A) Dacron

(B) Orlon (acrilan)

(C) PVC

(D) Bakelite

0 Follower 2 Views

Payal Gupta

Contributor-Level 10

This is a multiple choice answer as classified in NCERT Exemplar

Option (B)

Explanation: Polyacrylonitrile is used as a wool substitute in the manufacture of commercial fibers such as orlon or acrilan.

0 0

New answer posted

a year ago

Class 12th Polymers

Which of the following is not a semisynthetic polymer?

(A) Cis-polyisoprene

(B) Cellulose nitrate

(C) Cellulose acetate

(D) Vulcanised rubber

0 Follower 1 View

Payal Gupta

Contributor-Level 10

This is a multiple choice answer as classified in NCERT Exemplar

Option (A)

Explanation: Semi-synthetic polymers are obtained from natural polymers. Cellulose derivatives like cellulose nitrate, cellulose acetate rubber are examples of semi-synthetic polymers.

Cis-polyisoprene is not a semi-synthetic polymer while other examples are semi-synthetic polymer.

0 0

Get authentic answers from experts, students and alumni that you won't find anywhere else

On Shiksha, get access to

66k Colleges
1.2k Exams
703k Reviews
1850k Answers

Community GuidelinesUser Points System

QuestionsDiscussionsTags

Need guidance on career and education? Ask our experts

Your Question

Class 12th

Questions

Discussions

Active Users

Followers

All

Discussions

Unanswered Questions

Share Your College Life Experience