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Class 12th physics ncert exemplar solution class 12th chapter one

In 1959 Lyttleton and Bondi suggested that the expansion of the Universe could be explained if matter carried a net charge. Suppose that the Universe is made up of hydrogen atoms with a number density N, which is maintained a constant. Let the charge on the proton be: ep=−(1+y)e where e is the electronic charge. Find the critical value of y such that expansion may start

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alok kumar singh

Contributor-Level 10

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

Explanation- let us consider that universe is of radius R

And we know that hydrogen is made up one electron and proton so net charge is

-(1+y)e+e = -ye

Now the number of hydrogen atom in the universe = N $* \frac{4}{3} π R^{3}$

So total charge is = -ye $*$ N $* \frac{4}{3} π R^{3}$

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

Explanation- let us consider that universe is of radius R

And we know that hydrogen is made up one electron and proton so net charge is

-(1+y)e+e = -ye

Now the number of hydrogen atom in the universe = N $* \frac{4}{3} π R^{3}$

So total charge is = -ye $*$ N $* \frac{4}{3} π R^{3}$

According to gauss theorem $? E . d s = \frac{q}{ε_{0}}$

So E $* 4 π R^{2} = \frac{- y e * N * \frac{4}{3} π R^{3}}{ε_{0}}$

E= -NRye/3 $ε_{0}$

And we know electrostatic force F=qE = -ye(-NRye/3 ) $ε_{0}$ by using the value of E and q

F= $\frac{y^{2} e^{2} N R}{3 ε_{0}}$ force is positive which shows that force is repulsive.

But the gravity is g= $\frac{G M}{R^{2}}$ and gravitational force is F= $\frac{G M m}{R^{2}}$

But M= N $* \frac{4}{3} π R^{3} m_{p}$ where $m_{p}$ is mass of proton

So using this mass in above eqn so force = $\frac{G N * \frac{4}{3} π R^{3} m_{p} m}{R^{2}}$

But her masses of proton and hydrogen are equal so take as m

F= $\frac{G N * 4 π R m^{2}}{3}$

So calculate its critical value electrostatic force and gravitational force both are equal.

So, $\frac{G N * 4 π R m^{2}}{3} = \frac{y^{2} e^{2} N R}{3 ε_{0}}$

So y²= 4 $π ε_{0} \frac{m^{2}}{e^{2}}$ after using standard values we get value of y= approx. 10^-18

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Class 12th physics ncert exemplar solution class 12th chapter one

Five charges,q each are placed at the comers of a regular pentagon of side a.

i) What will be the electric field at O, the centre of the pentagon?
(ii) What will be the electric field at O if the charge from one of the comers (say A) is removed?
(iii) What will be the electric field at O if the charge q at A is replaced by – q?
(b) How would your answer to (a) be affected if pentagon is replaced by n-sided regular polygon with charge q at each of its comers?

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alok kumar singh

Contributor-Level 10

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

Explanation- (a)

(i) The electric field at the center of pentagon is zero because the distance from the center is same.

(ii) The field through one charge is Kq/r²

(iii) When one charge is positive and other is negative then net force towards negative charge. So net force is Kq/r²+ Kq/r²= 2Kq/r²

(b) It doesn't depend upon the number of sides increasing the net electric field is zero.

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Class 12th physics ncert exemplar solution class 12th chapter one

Five charges,q each are placed at the comers of a regular pentagon of side a.

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Class 12th NEET

Can I download the NCERT Exemplar PDF for Electrostatic Potential and Capacitance for offline use?

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Pallavi Pathak

Contributor-Level 10

Yes, students can download the NCERT Exemplar PDF for Class 12 Physics Chapter 2. This offers various types of conceptual and application-based questions. Students can practice these offline at their own schedule. Practicing from PDF helps in improving problem-solving skills.

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Class 12th NEET

Are NCERT Exemplar Class 12 Physics Chapter 2 solutions sufficient for exam preparation?

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Pallavi Pathak

Contributor-Level 10

Preparing from the NCERT Exemplar Chapter 2 solutions will help in providing a strong concept foundation as it covers a wide range of question types from basic to advanced. If one practices the exemplar, they can score high in Board exams and entrance exams. However, it is advisable to first start from the NCERT textbook for the theory of the concepts and later on practice exemplars to boost the understanding and solve various types of questions based on these concepts. The exemplars are aligned with the CBSE syllabus and offer great study material for strengthening concept understanding and improving problem-solving accuracy.

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Class 12th NEET

What is the significance of solving NCERT Exemplar problems for Class 12 Physics Chapter 2?

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Pallavi Pathak

Contributor-Level 10

NCERT Exemplar problems are designed by the subject matter experts. It helps in deepening conceptual understanding and improving the application skills of the students. Preparing the NCERT exemplar Chapter 2 helps students in mastering complex topics like capacitors, electric potential, and energy storage which are important for the board exam preparations and entrance tests like JEE and NEET.

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Class 12th NEET

Are NCERT Exemplar Class 12 Physics Chapter 2 solutions sufficient for exam preparation?

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Pallavi Pathak

Contributor-Level 10

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Class 12th NCERT

What is the significance of solving NCERT Exemplar problems for Class 12 Physics Chapter 2?

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Pallavi Pathak

Contributor-Level 10

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Class 12th Maths Integrals

Kindly Consider the following

61. $\int \frac{d x}{s i n^{2} x c o s^{2} x} d x$

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Vishal Baghel

Contributor-Level 10

$\begin{array}{l} I = \int \frac{d x}{s i n^{2} x c o s^{2} x} \\ = \int \frac{1}{s i n^{2} x c o s^{2} x} d x \\ = \int \frac{s i n^{2} x + c o s^{2} x}{s i n^{2} x c o s^{2} x} d x \\ = \int \frac{s i n^{2} x}{s i n^{2} x c o s^{2} x} d x + \int \frac{c o s^{2} x}{s i n^{2} x c o s^{2} x} d x \\ = \int s e c^{2} x d x + \int c o s e c^{2} x - d x] \\ = t a n x - c o t x + c \end{array}$

Therefore, the correct answer is B.

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a year ago

Class 12th Maths Integrals

Kindly Consider the following

60. $\int \frac{10 x^{9} + 1 0^{x} l o g_{e} 10 d x}{x^{10} + 1 0^{x}}$

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Vishal Baghel

Contributor-Level 10

$\begin{array}{l} P u t x^{10} + 1 0^{x} = t \\ \Rightarrow (10 x^{9} + 1 0^{x} l o g^{e} 10) d x = d t \\ I = \int \frac{10 x^{9} + 1 0^{x} l o g_{e} 10}{x^{10} + 1 0^{x}} d x \\ = \int \frac{d t}{t} = l o g t + C \\ = l o g (1 0^{x} + x^{10}) + c \end{array}$

Therefore, the correct answer is (D)

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