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6 months ago

Class 12th Physics NCERT Exemplar Solutions Class 12th Chapter Thirteen

At a certain temperature, the degrees of freedom per molecule for gas is 8. The gas performs 150 J at work when it expands under constant pressure. The amount of heat absorbed by the gas will be________J.

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Vishal Baghel

Contributor-Level 10

Degree of freedom, f = 8

$∴ C_{V} = \frac{8}{2} R = 4 R a n d C_{P} = 4 R + R = 5 R$

So, H = $Δ U + W$

at constant pressure, W = $P Δ v$

150 = $n R Δ T$

So, $H = n C_{V} Δ T = n 5 R Δ T = 5 (n R Δ T)$

= 5 * 150

= 750 J

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6 months ago

Class 12th Physics NCERT Exemplar Solutions Class 12th Chapter Nine

A string of area of cross – section 4 mm² and length 0.5 m is connected with a rigid body of mass 2 kg. The body is rotated in a vertical circular path of radius 0.5m. The body acquires a speed of 5 m/s at the bottom of the circular path. Strain produce in the string when the body is at bottom of the circle is________* 10^-5.

(use young's modulus 10¹¹ N/m² and g = 10 m/s²)

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Vishal Baghel

Contributor-Level 10

For circular motion:-

$T - m g = \frac{m v^{2}}{r}$

$\Rightarrow T = m (g + \frac{v^{2}}{r})$

$= 2 (10 + \frac{25}{1 / 2}) = 120 N$

$\Rightarrow S t r a i n = \frac{s t r e s}{Y}$

= 30 * 10^-5

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6 months ago

Class 12th Physics NCERT Exemplar Solutions Class 12th Chapter Seven

The distance of centre of mass from end A of a one dimensional rod (AB) having mass density ? = ?0 $(1 - \frac{x^{2}}{L^{2}}) k g / m$ and length L (in meter) is $\frac{3 L}{α} m .$ The value of is _________. (where x is the distance from end A)

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Vishal Baghel

Contributor-Level 10

Linear mass density, $λ = λ_{0} (1 - \frac{x^{2}}{L^{2}}) k g / m$

$M = \int λ d x = \int_{0}^{L} λ_{0} (1 - \frac{x^{2}}{L^{2}}) d x$

$= λ_{0} (L - \frac{L^{3}}{3 L^{2}}) = \frac{2 L λ_{0}}{3}$

$A s, x_{C} = \frac{3 L}{α} = \frac{9}{8} L . \Rightarrow α = \frac{3 * 8}{9} = 8 / 3$

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6 months ago

Class 12th Physics NCERT Exemplar Solutions Class 12th Chapter Thirteen

A ball is thrown vertically upwards with a velocity of 19.6 ms^-1 from the top of a tower. The ball strikes the ground after 6s. The height from the ground up to which the ball can rise will be $(\frac{k}{5}) m .$ The value of k is_______(use g = 9.8m/s²)

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Vishal Baghel

Contributor-Level 10

h = 39.2 + 19.6 = 58.8 m

Height above tower

$h' = \frac{u^{2}}{2 g} = \frac{19.6 * 19.6}{2 * 9.8}$

= 19.6

$A s, \frac{k}{5} = 78.4 \Rightarrow k = 392$

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6 months ago

Class 12th Maths NCERT Exemplar Solutions Class 12th Chapter Nine

Let y = y(x) be the solution of the differential equation $x (1 - x^{2}) \frac{d y}{d x} + (3 x^{2} y - y - 4 x^{3}) = 0, x > 1, w i t h y (2) = - 2 .$ Then y(3) is equal to

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Payal Gupta

Contributor-Level 10

$x (1 - x^{2}) \frac{d y}{d x} + (3 x^{2} y - y - 4 x^{3}) = 0$

$\Rightarrow x (1 - x^{2}) \frac{d y}{d x} + (3 x^{2} - 1) y = 4 x^{3}$

$\Rightarrow y = 2 x - 1 (x^{3} - x) ∴ y (3) = - 18$

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6 months ago

Class 12th Physics NCERT Exemplar Solutions Class 12th Chapter Fifteen

A FM Broad cast transmitter, using modulating signal of frequency 20 kHz has a deviation ratio of 10. The Bandwidth required for transmission is:

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Vishal Baghel

Contributor-Level 10

f = 20 kHz

Deviation ratio = 10

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6 months ago

Class 12th Maths NCERT Exemplar Solutions Class 12th Chapter Nine

Let the solution curve y = y(x) of the differential equation $[\frac{x}{\sqrt[]{x^{2} - y^{2}}} + e^{\frac{y}{x}}] x \frac{d y}{d x} = x + [\frac{x}{\sqrt[]{x^{2} - y^{2}}} + e^{\frac{y}{x}}] y$ pass through the points (1, 0) and (2α, α), α > 0. Then a is equal to

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Payal Gupta

Contributor-Level 10

$[\frac{x}{\sqrt[]{x^{2} - y^{2}}} + e^{y / x}] x \frac{d y}{d x} = x + [\frac{x}{\sqrt[]{x^{2} - y^{2}}} + e^{y / x}] y$

$\Rightarrow e^{y / x} [x d y - y d x] = x d x + \frac{x}{\sqrt[]{x^{2} - y^{2}}} (y d x - x d y)$

$\Rightarrow e^{y / x} d (y / x) = \frac{d x}{x} - \frac{d (y / x)}{\sqrt[]{1 - {(y / x)}^{2}}}$

Integrating

$e^{y / x} = l n x - s i n^{- 1} (\frac{y}{x}) + c$

Passes (1, 0)

1 = c

$\Rightarrow α = \frac{1}{2} e x p (\sqrt[]{e} - 1 + \frac{π}{6})$

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6 months ago

Class 12th Physics NCERT Exemplar Solutions Class 12th Chapter Twelve

Perpendicular to the plane $A B C$ is $I_{0}$ . If part $A D E$ is removed, the moment of inertia of the remaining part about the same axis is $\frac{{N I}_{0}}{16}$ where $N$ is an integer. Value of $N$ is

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alok kumar singh

Contributor-Level 10

$\frac{1}{λ_{min}} = R [\frac{1}{1} - \frac{1}{\infty}] = R [n = \infty \to n = 1]$

$\frac{1}{λ_{m a x}} = R [\frac{1}{1} - \frac{1}{4}] = \frac{3 R}{4} [n = 2 \to n = 1]$

$\Rightarrow Δ λ \Rightarrow \frac{4}{3 R} - \frac{1}{R} \Rightarrow \frac{1}{3 R} = 340$

For Paschan $\Rightarrow \frac{1}{λ_{min}} = R [\frac{1}{9}] [n = \infty \to h = 3]$

$\Rightarrow \frac{1}{λ_{m a x}} = R [\frac{1}{9} - \frac{1}{16}] = \frac{7 R}{144}$

$\Rightarrow Δ λ = \frac{81}{7 R}$

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6 months ago

Class 12th Physics NCERT Exemplar Solutions Class 12th Chapter Fourteen

An n.p.n transistor with current gain = 100in common emitter configuration is shown in figure. The output voltage of the amplifier will be

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Vishal Baghel

Contributor-Level 10

n.p.m. transistor current gain B = 100

As, B = $\frac{I_{C}}{I_{B}} V o l t a g e g a i n, \frac{V_{o u t}}{V_{i n}} = (\frac{I_{C}}{I_{B}}) \frac{R_{2}}{R_{i}}$

$V_{o u t} = \frac{B R_{2}}{R_{i n}} V i n$

$\frac{1 0^{3}}{1 0^{3}} V = 1 v$

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6 months ago

Class 12th Physics NCERT Exemplar Solutions Class 12th Chapter Thirteen

A radioactive sample decays $\frac{7}{8}$ times its original quantity in 15 minutes. The half-life of the sample is

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Vishal Baghel

Contributor-Level 10

$N_{0} \to_{}^{t = 15 m i n} (N_{0} - \frac{7}{8} N_{0}) = \frac{N_{0}}{8}$

So, $N = \frac{N_{0}}{2^{t / t_{1 / 2}}} \Rightarrow \frac{N_{0}}{8} = \frac{N_{0}}{2^{t / t_{1 / 2}}}$

$\Rightarrow 2^{3} = 2^{(\frac{15 m i n}{T})}$

$\Rightarrow T = \frac{15 m i n}{3} = 5 m i n$

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