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12 months ago

Physics Physics NCERT Exemplar Solutions Class 12th Chapter Eight

On the $x$ -axis and at a distance $x$ from the origin, the gravitational field due to a mass distribution is given by $\frac{A x}{{(x^{2} + a^{2})}^{3 / 2}}$ in the $x$ -direction. The magnitude of gravitational potential on the $x$ -axis at a distance $x$ , taking its value to be zero at infinity is:

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alok kumar singh

Contributor-Level 10

$g = \frac{A x}{{(x^{2} + a^{2})}^{3 / 2}}$

$\Rightarrow \int_{v}^{0} ? d V = - \int_{x}^{\infty} ? g d x$

$\Rightarrow O - V = - [\int \frac{A x}{{(a^{2} + x^{2})}^{3 / 2}}]$

Let, $a^{2} + x^{2} = t^{2}$

$\Rightarrow 2 x d x = 2 t d t$

$\Rightarrow x d x = t d t$

$\Rightarrow V = \int \frac{A t d t}{t^{3}} \Rightarrow - \frac{A}{t} \Rightarrow - {\frac{A}{\sqrt[]{a^{2} + x^{2}}}|}_{x}^{\infty}$

$\Rightarrow V = \frac{A}{\sqrt[]{a^{2} + x^{2}}}$

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12 months ago

Physics Physics NCERT Exemplar Solutions Class 12th Chapter Six

A pendulum is suspended by a string of length 250 cm. The mass of the bob of the pendulum is 200g. The bob is pulled aside until the string is at 60° with vertical as shown in the figure. After releasing the bob, the maximum velocity attained by the bob will be_________ ms^-1. (if g = 10m/s²)

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Payal Gupta

Contributor-Level 10

From conservation of Energy:

$m g l (1 - c o s 60 °) = \frac{1}{2} m v^{2}$

$\Rightarrow v^{2} = 2 g l (\frac{1}{2}) = g l$

$= 10 * \frac{25}{10} = 5 m / s$

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12 months ago

Physics Physics NCERT Exemplar Solutions Class 12th Chapter Fifteen

The velocity of sound in a gas, in which two wavelengths 4.08m and 4.16m produce 40 beats in 12s, will be

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Payal Gupta

Contributor-Level 10

$| f_{1} - f_{2} | = \frac{40}{12} = \frac{10}{3}$

$\Rightarrow v (\frac{1}{λ_{1}} - \frac{1}{λ_{2}}) = \frac{10}{3}$

$\Rightarrow$ $V$ $=$ $\frac{10}{3}$ $*$ $\frac{λ_{1} λ_{2}}{λ_{2} - λ_{1}}$ $=$ $\frac{10}{3}$ $*$ $\frac{4.08 * 4.16}{. 08}$ $=$ $\frac{10}{3}$ $*$ $\frac{408 * 4.16}{8}$ $=$ $7$ $0$ $7$ $.$ $2$ $m$ $/$

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12 months ago

Physics Communication Systems

Match List – I with List – II

	List – I		List – II
A.	Television signal	I.	03 KHz
B.	Radio signal	II.	20 KHz
C.	High Quality Music	III.	02 MHz
D.	Human speech	IV.	06 MHz

Choose the correct answer from the options given below:

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Payal Gupta

Contributor-Level 10

Using factual data.

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12 months ago

Physics Physics NCERT Exemplar Solutions Class 12th Chapter Fourteen

Given below are two statements : One is labeled as Assertion A and the other is labeled as Reason R.

Assertion A : n-p-n transistor permits more current that a p-n-p transistor.

Reason R : Electrons have greater mobility as a change carrier.

Choose the correct answer form the options given below

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Payal Gupta

Contributor-Level 10

Using factual data.

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12 months ago

Physics Physics NCERT Exemplar Solutions Class 12th Chapter Fourteen

For using a multimeter to identify diode from electrical compounds, choose the correct statement out of the following about the diode

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Payal Gupta

Contributor-Level 10

Diode conducts current only in one direction.

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12 months ago

Physics Physics NCERT Exemplar Solutions Class 12th Chapter Fourteen

In the following circuit, the correct relation between output (Y) and inputs A and B will be

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Payal Gupta

Contributor-Level 10

From Truth table $Y = \bar{A B}$ .

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12 months ago

Physics Physics NCERT Exemplar Solutions Class 12th Chapter Thirteen

The Q-value of a nuclear reaction and kinetic energy of the projectile particle, K_p are related as

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Payal Gupta

Contributor-Level 10

X + p → Y + b

$Q = K_{Y} + K_{b} - K_{P}$

$Q + K_{P} = K_{Y} + K_{b}$

$\Rightarrow Q + K_{P} > 0$

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12 months ago

Physics Physics NCERT Exemplar Solutions Class 12th Chapter Eleven

THE de Broglie wavelength for an electron and a photon are $λ_{e} a n d λ_{p}$ respectively. For the same kinetic energy of electron and photon, which of the following presents the correct relation between the de Broglie wavelengths of two

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Payal Gupta

Contributor-Level 10

$λ_{e} = \frac{h}{P_{e}} = \frac{h}{\sqrt[]{2 m_{e} K_{e}}} \Rightarrow λ_{e}^{2} = \frac{h^{2}}{2 m_{e} k_{e}} \Rightarrow K_{e} = \frac{h^{2}}{2 m_{e} λ_{e}^{2}}$ - (i)

$λ_{P} = \frac{h}{P_{P}} = \frac{h c}{K_{P}} \Rightarrow K_{P} = \frac{h c}{λ_{P}}$ - (ii)

K_e = K_P

$\Rightarrow λ_{P} α λ_{e}^{2}$

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12 months ago

Physics Physics NCERT Exemplar Solutions Class 12th Chapter Nine

The aperture of the objective is 24.4 cm. The resolving power of this telescope, if a alight of wavelength 2440 $\overset{o}{A}$ is used to see the object will be

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Payal Gupta

Contributor-Level 10

Resolving power, = $\frac{d}{1.22 λ} = \frac{24.4 * 1 0^{- 2}}{1.22 * 2440 * 1 0^{- 10}}$

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