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Physics NCERT Exemplar Solutions Class 12th Chapter Six Physics Class 12th Physics NCERT Exemplar Solutions Class 12th Chapter Fourteen Semiconductor Electronics

The variation of acceleration due to gravity (g) with distance (r) from the centre of the earth is correctly represented by: (Given R = radius of earth)

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Two 10cm log, straight wires, each carrying a current of 5A are kept parallel to each other. If each wire experienced a force of 10^-5N, then separation between the wires is__________cm.

Vishal Baghel

$F = \frac{μ_{0} * 2}{4 π} \frac{i_{1} i_{2}}{d} l$

$1 0^{- 5} = \frac{1 0^{- 7} * 2 * 5 * 5}{d} = \frac{10}{100}$

$d = 5 * 1 0^{- 2} m$

d = 5 cm

The positive feedback is required by an amplifier to act an oscillator. The feedback here means:

Vishal Baghel

A portion of the output power is returned back to the input.

The variation of acceleration due to gravity (g) with distance (r) from the centre of the earth is correctly represented by: (Given R = radius of earth)

Payal Gupta

$g \propto r 0 \leq r \leq R$

$g \propto \frac{1}{r^{2}} r > R$

A transistor is used in an amplifier circuit in common emitter mode. If the best current changes by 100 $μ A,$ it brings a change of 10 mA is collector current. If the load resistance is 2 $k Ω$ and input resistance is 1 $k Ω,$ the value of power gain is x * 10⁴. The value of x is_____________.

alok kumar singh

Power gain = ${(\frac{Δ i_{c}}{Δ i_{b}})}^{2} * \frac{R_{0}}{R_{i}}$

= ${(\frac{10 * 1 0^{- 3}}{100 * 1 0^{- 6}})}^{2} * \frac{2}{1}$

= 2 * 104 = x * 104

= 2

The typical transfer characteristics of a transistor in CE configuration is shown in figure. A load resistor of $2 k Ω$ is connected in the collector branch of the circuit used. The input resistance of the transistor is $0.50 k Ω .$ The voltage gain of the transistor is________.

alok kumar singh

Voltage gain = $\frac{Δ l_{C}}{Δ l_{B}} * \frac{R_{C}}{R_{B}}$

$\frac{5 * 1 0^{- 3}}{100 * 1 0^{- 6}} * \frac{2}{0.5}$

= 200

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