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Physics Ncert Solutions Class 12th

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Physics Ncert Solutions Class 12th Thermodynamics

Starting with the same initial conditions, an ideal gas expands form volume V₁ to V₂ in the three different ways. The work done by the gas in W₁ if the process is purely isothermal, W₂, if the process is purely adiabatic and W₃ if the process is purely isobaric. Then, choose the correct option

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Vishal Baghel

Contributor-Level 10

Process-AB $\Rightarrow$ Isobaric,

Process-AC $\Rightarrow$ Isothermal, and

Process-ADÞAdiabatic

W₂ < W₁ < W₃

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Physics Ncert Solutions Class 12th Class 12th

A telegraph line of length 100km has a capacity of $0.01 μ F / k m$ and it carries an alternating current at 0.5 kilo cycle per second. If minimum impedance is required, then the value of the inductance that need to be introduced in series is _________mH.

(if p = $\sqrt[]{10}$ )

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Raj Pandey

Contributor-Level 9

C = 10^-6 f

f = 0.5 * 10³ = 500

$z = \sqrt[]{x_{R}^{2} + {(x_{L} - x_{C})}^{2}}$

for minimum impedance :

x_L = x_C

$\Rightarrow ω^{2} = \frac{1}{L C}$

$\Rightarrow L = \frac{1}{π^{2}} = \frac{1}{10} * 1000 = 100 m H$

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Physics Ncert Solutions Class 12th Class 12th

(if p = $\sqrt[]{10}$ )

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Physics Ncert Solutions Class 12th Class 12th

Match List – I with List – II

Choose the correct answer from the options given below:

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Raj Pandey

Contributor-Level 9

Using factual data.

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Physics Ncert Solutions Class 12th Class 12th

An inductor of 0.5 mH, a capacitor of 200 $μ F$ and a resistor of 2 $Ω$ are connected in series with a 220V ac source. If the current is in phase with the emf, the frequency of ac source will be__________ * 10² Hz.

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alok kumar singh

Contributor-Level 10

Since current is in phase with voltage, it means circuit is in resonance, so we can write

f = $\frac{1}{2 π \sqrt[]{L C}} = \frac{1}{2 π \sqrt[]{(0.5 * 1 0^{- 3}) * (200 * 1 0^{- 6})}}$

$\Rightarrow f = \frac{1 0^{4}}{2 π \sqrt[]{10}} \approx 5 * 1 0^{2} H z$ $[T a k i n g π = \sqrt[]{10}]$

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Physics Ncert Solutions Class 12th Class 12th

A potential barrier of 0.4V exists across a p-n junction. An electron enters the junction from the n-side with a speed of 6.0 * 10⁵. The speed with which electron enters the p side will be $\frac{x}{3} * 1 0^{5} m s^{- 1}$ the value of x is__________.

(Given mass of electron = 9 * 10^-³¹ kg, charge on electron = 1.6 * 10^-¹⁹ C.)

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alok kumar singh

Contributor-Level 10

According to Work energy theorem, we can write

$K_{f} - K_{i} = W_{E l e c t r i c F o r c e} \Rightarrow \frac{1}{2} m v^{2} - \frac{1}{2} m v_{0}^{2} = - e V \Rightarrow v^{2} = v_{0}^{2} - \frac{2 e V}{m}$

$\Rightarrow v^{2} = {(6.0 * 1 0^{5})}^{2} - \frac{2 * 1.6 * 1 0^{- 19}}{9 * 1 0^{- 31}} = \frac{324 - 128}{9} * 1 0^{10} = \frac{196}{9} * 1 0^{10} \Rightarrow V = \frac{14}{3} * 1 0^{5} m / s$

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Physics Ncert Solutions Class 12th Class 12th

The TV transmission tower at a particular station has a height of 125 m. For doubling the coverage of its range, the height of the tower should be increased by

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alok kumar singh

Contributor-Level 10

According to question, we can write

$d = \sqrt[]{2 h R} \Rightarrow \frac{d_{2}}{d_{1}} = \sqrt[]{\frac{h_{2}}{h_{1}}} \Rightarrow h_{2} = {(\frac{d_{2}}{d_{1}})}^{2} h_{1} = {(\frac{2}{1})}^{2} * 125 = 500 m$

Increment in height of tower = h₂ – h₁ = 500 – 125 = 375 m

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Physics Ncert Solutions Class 12th Class 12th

As shown in the figure an inductor of inductance 200 mH is connected to an AC source of emf 220 V and frequency 50 Hz. The instantaneous voltage of the source is 0 V when the peak value of current is $\frac{\sqrt[]{a}}{π} A .$ The value of a is __________.

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alok kumar singh

Contributor-Level 10

Let l = l₀ cos wt

Then v = v₀ sinwt

at t = 0, v = 0

but l = l₀

$l_{r m s} = \frac{l_{0}}{\sqrt[]{2}}$

$V_{r m s} = l_{r m s} z$

$\Rightarrow 220 = \frac{l_{0}}{\sqrt[]{2}} (X_{L})$

$\Rightarrow 220 = \frac{l_{0}}{\sqrt[]{2}} (2 π * 50 * 200 * 1 0^{- 3})$

$\Rightarrow 220 = \frac{l_{0}}{\sqrt[]{2}} (20 π)$

$l_{0} = \frac{220 \sqrt[]{2}}{20 π}$

$l_{0} = \frac{11 \sqrt[]{2}}{π} = \frac{\sqrt[]{a}}{π}$

a = 121 * 2

a = 242

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Physics Ncert Solutions Class 12th Class 12th

A transistor is used in common-emitter mode in an amplifier circuit. When a signal of 10 mV is added to the base-emitter voltage, the base current changes by 10µA and the collector current changes by 1.5 mA. The load resistance is 5 kW. The voltage gain of the transistor will be ________.

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alok kumar singh

Contributor-Level 10

I_b = 10 µA

I_C = 1.5 mA

R_L = 50 kW or (Rc)

Base – emitter voltage = 10 mv

$R_{B} = \frac{V_{B}}{I_{B}} = \frac{10 * 1 0^{- 3}}{10 * 1 0^{- 6}} = 1 0^{3} Ω$

$A_{v} = (\frac{Δ l_{C}}{Δ l_{B}}) * (\frac{R_{C}}{R_{B}})$

= $\frac{1.5 * 1 0^{- 3}}{10 * 1 0^{- 6}} * \frac{5 * 1 0^{3}}{1 0^{3}}$

A_v = 750

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Physics Ncert Solutions Class 12th Class 12th

Drawing power form a 12V car battery, a 9V stabilized DC voltage is required to power a car stereo system, attached to the terminals A and B, as shown in the figure. If a Zener diode with ratings, V_z = 9V and P_max = 0.27 W, is connected as shown in the figure, for the above purpose, the minimum series resistance R_s must be:-

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Vishal Baghel

Contributor-Level 10

$i_{m a x} = \frac{P_{m a x}}{V_{z}} = \frac{3}{100}$

$∴ R_{s_{m i n}} = \frac{V_{R_{S}}}{i_{m a x}}$

$= \frac{3}{3 / 100} = 100 Ω$

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