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Physics Ncert Solutions Class 12th

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Physics Ncert Solutions Class 12th Class 12th

A common transistor ratio set requires 12V (D.C) for its operation. The D.C source is constructed by using a transformer and a rectifier circuit, which are operated at 220V (A.C) on standard domestic A.C, supply. The number of turns of secondary coil are 24, then the number of turns of primary are___________.

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Vishal Baghel

Contributor-Level 10

$\frac{N_{p}}{N_{s}} = \frac{V_{p}}{V_{s}} \Rightarrow N_{p} = \frac{V_{p}}{V_{s}} * N_{s} = \frac{220}{12} * 24 = 440$

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Physics Ncert Solutions Class 12th Class 12th

A 5 V battery is connected across the points X and Y. Assume D₁ and D₂ to the normal silicon diodes. Find the current supplies by the battery if the +ve terminal of the battery is connected to point X.

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Vishal Baghel

Contributor-Level 10

D₁ is in forward bias and D₂ is in reverse bias.

Current, I $= \frac{5 - 0.7}{10} = 0.43 A$

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Physics Ncert Solutions Class 12th Class 12th

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Physics Ncert Solutions Class 12th Class 12th

Zener breakdown occurs in a P- n junction having P and n both:

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Vishal Baghel

Contributor-Level 10

Zener break down occurs in p-n junction having p and n both : Heavily doped and have narrow depletion layer.

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Physics Ncert Solutions Class 12th Class 12th

Choose the correct statement for amplitude modulation :

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Raj Pandey

Contributor-Level 9

In amplitude modulation, the amplitude of the career signal is varied in according with the modulating signal.

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Physics Ncert Solutions Class 12th Class 12th

How many alpha and beta particles are emitted when Uranium ₉₂U²³⁸ decays to lead ₈₂Pb²⁰⁶?

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Raj Pandey

Contributor-Level 9

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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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Vishal Baghel

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 Waves

In an experiment to determine the velocity of sound in air at room temperature using a resonance tube, the first resonance is observed when the air column has a length of 20.0 cm for a tuning form of frequency 400 Hz is used. The velocity of the sound at room temperature is 336 ms^-1. The third resonance is observed when the air column has a length of ____________cm.

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Vishal Baghel

Contributor-Level 10

According to Concept of resonance tube, we can write

$\frac{λ}{4} + e = l_{1} a n d \frac{3 λ}{4} + e = l_{2} \Rightarrow \frac{λ}{2} = l_{2} - l_{1} \Rightarrow \frac{V}{2 ν} = l_{2} - l_{1}$

$\Rightarrow l_{2} = \frac{v}{2 ν} + l_{1} = \frac{336}{400} + 0.20 = 0.84 + 0.20 = 1.04 m = 104 c m$

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Physics Ncert Solutions Class 12th Class 12th

Two capacitors having C₁ and C₂ respectively are connected as shown in figure. Initially, capacitor C₁ is charged to a potential difference V volt by a battery. The battery is then removed and the charged capacitor C₁ is not connected to uncharged capacitor C₂ by closing the switch S. The amount of charge on the capacitor C₂, after equilibrium, is:

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Raj Pandey

Contributor-Level 9

After switch 'S' is closed

$Q_{1} + Q_{2} = C_{1} V$ - (1)

Using KVL

$\frac{Q_{1}}{C_{1}} - \frac{Q_{2}}{C_{2}} = 0$

$Q_{1} = Q_{2} \frac{C_{1}}{C_{2}}$ - (2)

from (1) & (2)

$\Rightarrow Q_{2} [\frac{C_{1} + C_{2}}{C_{2}}] = C_{1} V \Rightarrow Q_{2} = (\frac{C_{1} C_{2}}{C_{1} + C_{2}}) V$

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Physics Ncert Solutions Class 12th Communication Systems

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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Vishal Baghel

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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