Physics MCQs for NEET — Practice Questions with Answers

In a junction diode, when reverse bias is applied, the potential barrier increases. This is because the negative terminal of the battery is connected to the n-side and the positive terminal to the p-side, resulting in an increase in the depletion region and a higher potential barrier.

In a common base configuration of a transistor, the alternating current gain (α) is defined as the ratio of the change in collector current (ΔI_C) to the change in emitter current (ΔI_E), with the collector voltage (V_c) held constant. Thus, $$ ext{α} = rac{ΔI_C}{ΔI_E} igg|_{V_c = const} $$.

In an N-P-N transistor circuit, the emitter current (I_E) is the sum of the base current (I_B) and the collector current (I_C). Therefore, the correct relationship is $$ I_E = I_B + I_C $$, which implies $$ I_B < I_C < I_E $$.

Ripples refer to the residual periodic variation of the DC voltage within a power supply which has been derived from an AC source. Essentially, it is the unwanted AC component present in the DC output of a rectifier. Thus, ripples are AC mixed with DC.

$ I_c =10 mA =0.90I_E $ $ \therefore I_E \approx 11 mA $ $ \therefore I_B \approx 1mA $

When a P-type semiconductor is heated, the thermal energy supplied increases the number of electron-hole pairs generated. This means that the number of electrons and holes increases equally due to the intrinsic carrier generation.

The depletion layer in a PN junction diode is caused by the diffusion of charge carriers. When a PN junction is formed, electrons from the N-region diffuse into the P-region and recombine with holes, and holes from the P-region diffuse into the N-region and recombine with electrons. This diffusion results in the creation of a region devoid of free charge carriers, known as the depletion layer.

In a solar cell, we aim to maximize the power output. A larger active junction area allows for the absorption of more light, which in turn generates more electron-hole pairs and increases the current produced by the solar cell. Hence, the active junction area in a solar cell is large to achieve more power.

To find the value of the limiting resistor ( $R$ ) for an LED, we use Ohm's law. Given the parameters: the voltage drop across the LED ( $V_{LED} = 2V$ ), the current through the LED ( $I = 10 mA$ ), and the supply voltage ( $V_{supply} = 6V$ ). The resistor value can be calculated as follows:

$$R = \frac{V_{supply} - V_{LED}}{I}$$

Substituting the values:

$$R = \frac{6V - 2V}{10 mA}$$

$$R = \frac{4V}{10 mA}$$

$$R = 400 \Omega$$

Thus, the value of the limiting resistor is $400 \Omega$ .