Physics Ncert Solutions Class 12th

A thermodynamic system is taken from an original state D to an intermediate sate E by the linear pressure shown in the figure. Its volume is then reduced to the original volume from E to F by an isobaric process. The total work done by the gas from D to E to F will be

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W_total= W_D_®E + W_E->_F = Area of triangle DEF

= $\frac{1}{2} * 3 * 300 = 450 J .$

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A ball is released from a height h. If t₁ and t₂ be the time required to complete first half and second half of the distance respectively. Then, choose the correct relation between t₁ and t₂.

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$\frac{h}{2} = \frac{1}{2} g t_{1}^{2}$

h = $\frac{1}{2} g {(t_{1} + t_{2})}^{2}$

from (i) and (ii) :-

$\frac{1}{2} g t_{1}^{2} = \frac{g}{4} {(t_{1} + t_{2})}^{2}$

$2 t_{1}^{2} = {(t_{1} + t_{2})}^{2}$

$\sqrt[]{2} t_{1} = t_{1} + t_{2}$

$t_{1} = \frac{t_{2}}{\sqrt[]{2} - 1} = (\sqrt[]{2} + 1) t_{2}$

$t_{2} = (\sqrt[]{2} - 1) t_{1}$

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$\frac{h}{2} = \frac{1}{2} g t_{1}^{2}$

h = $\frac{1}{2} g {(t_{1} + t_{2})}^{2}$

from (i) and (ii) :-

$\frac{1}{2} g t_{1}^{2} = \frac{g}{4} {(t_{1} + t_{2})}^{2}$

$2 t_{1}^{2} = {(t_{1} + t_{2})}^{2}$

$\sqrt[]{2} t_{1} = t_{1} + t_{2}$

$t_{1} = \frac{t_{2}}{\sqrt[]{2} - 1} = (\sqrt[]{2} + 1) t_{2}$

$t_{2} = (\sqrt[]{2} - 1) t_{1}$

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An object of mass 1 kg is taken to a height form the surface of earth which is equal to three times that radius of earth. The gain in potential energy of the object will be

[If, g = 10ms^-2 and radius of earth = 6400km]

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Physics Ncert Solutions Class 12th Communication Systems

Contributor-Level 10

m = 1kg ; $Δ U = \frac{- G m_{e} m}{(R + h)} - (\frac{G m_{e} m}{R})$

$Δ U = \frac{G M m}{R} - \frac{G M m}{R + h}$

$= m g_{0} R - m g_{0} \frac{R}{(1 + \frac{h}{R})} = m g_{0} {1 - \frac{1}{(1 + \frac{3 R}{R})}}$

$= m g_{0} R (\frac{3}{4})$

$= \frac{3}{4} * 1 * 10 * 6400 k m$

$= 48000 * 1 0^{3} J .$

$= 48 M J .$

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In the case of amplitude modulation to avoid distortion the modulation index $(μ)$ should be:

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Based on Theoretical data (modulation index, $μ \leq 1)$

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Identify the solar cell characteristics from the following options:

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V-l characteristics graph is plotted in 4th quadrant for solar cell.

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In normal adjustment, for a refracting telescope, the distance between objective and eye piece is 30cm. The focal length of the objective, when the angular magnification of the telescope is 2, will be:

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Angular magnification, $θ = | \frac{f_{0}}{f_{e}} |$ $\frac{_{}}{_{}}$

Length of tuber L = $| f_{0} | + | f_{e} |$ $_{}_{}$

$| f_{0} | + | f_{e} | = 30 c m$

$A s, | \frac{f_{0}}{f_{e}} | = 2$

$| f_{0} | = 2 | f_{e} |$

$\Rightarrow | f_{0} | + \frac{| f_{0} |}{2} = 30 \Rightarrow | f_{0} | = \frac{2}{3} * 30 = 20 c m$

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As shown in the figure, after passing through the medium 1. The speed of light $υ_{2}$ in medium 2 will be:

(Given c = 3 * 10⁸ ms^-¹)

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In a medium, speed of light, $V = \frac{C}{\sqrt[]{μ_{r} ε_{r}}} = \frac{3 * 1 0^{8} m / s}{\sqrt[]{1 * 9}} = 1 * 1 0^{8} m / s$

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

Assertion A : Alloys such as constantan and manganin are used in making standard resistance coils.

Reason R : Constantan and maganin have very small value of temperature coefficient of resistance.

In the light of the above statements, choose the correct answer from the options given below:

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Physics Ncert Solutions Class 12th Thermodynamics

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Based on theoretical data.

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A Carnot engine has efficiency of 50%. If the temperature of sink is reduced by 40°C, its efficiency increases by 30%. The temperature of the source will be:

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Efficiency $η = 50 %$

$\Rightarrow η = (1 - \frac{T_{2}}{T_{1}}) * 100 = 50$

$\Rightarrow 1 - \frac{T_{2}}{T_{1}} = \frac{50}{100} \Rightarrow 1 - \frac{T_{2}}{T_{1}} = \frac{1}{2}$ -(i)

After the change :- $1 - \frac{T_{2}^{1}}{T_{1}} = \frac{50 * (1.3)}{100} = \frac{65}{100} = . 65$ $\frac{_{}^{}}{_{}} \frac{}{} \frac{}{}$ -(ii)

$\Rightarrow 1 - (\frac{T_{2} - 40}{T_{1}}) = . 65$ --(ii)

$\Rightarrow \frac{40}{T_{1}} = . 65 - . 50 = . 15 \Rightarrow T_{1} = \frac{4000}{15} = 266.7 k$

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The dimensions of $(\frac{B^{2}}{μ_{0}}) w i l l b e$ $\frac{^{}}{_{}}$

(if $_{}$ $μ_{0}$ : permeability of free space and B : magnetic field)

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