Physics MCQs for NEET — Practice Questions with Answers

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For high frequency, a capacitor offers

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Explanation

The reactance of a capacitor is given by the formula: $$ X_C = \frac{1}{\omega C} $$ where \( X_C \) is the capacitive reactance, \( \omega \) is the angular frequency, and \( C \) is the capacitance. For high frequencies, the value of \( \omega \) increases, and as a result, the capacitive reactance \( X_C \) decreases. Therefore, a capacitor offers less reactance at high frequencies.

The coil of a choke in a circuit

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Explanation

A choke coil, which is an inductor, opposes changes in current. This property is due to the inductive reactance, which is given by: $$ X_L = \omega L $$ where \( X_L \) is the inductive reactance, \( \omega \) is the angular frequency, and \( L \) is the inductance. As the inductive reactance increases, it resists the flow of current, effectively decreasing the current in the circuit.

The power factor of an ac circuit having resistance (R) and inductance (L) connected in series and an angular velocity w is,

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Explanation

The power factor (PF) of an AC circuit with resistance (R) and inductance (L) connected in series is given by the cosine of the phase angle (Ï•) between the voltage and the current. This phase angle is determined by the impedance of the circuit. The formula for power factor is:

$$ \text{Power Factor} = \cos(\varphi) = \frac{R}{\sqrt{R^2 + (\omega L)^2}} $$

Here, the impedance of the circuit is \( Z = \sqrt{R^2 + (\omega L)^2} \). Thus, the correct option is:

$$ \frac{R}{( R^2 + \omega^2 L^2 )^{1/2}} $$

An inductor of inductance L and resistor of resistance R are joined in series and connected by a source of frequency $ \omega $ power dissipated in the circuit is,

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Explanation

The power dissipated in an AC circuit with a resistor (R) and inductor (L) in series is given by the formula:

$$ P = \frac{V^2 R}{Z^2} $$

where \( Z \) is the impedance of the circuit, which is given by \( Z = \sqrt{R^2 + (\omega L)^2} \). Substituting \( Z^2 \) into the formula, we get:

$$ P = \frac{V^2 R}{R^2 + (\omega L)^2} $$

Thus, the correct option is:

$$ \frac{V^2 R}{(R^2 + \omega^2 L^2)} $$

Ina LCRcircuit capacitance is changedfrom C to 2C. For the resonant fequencyto remainunchanged, the inductance should be change from L to

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Explanation

The resonant frequency (\( f_0 \)) of an LCR circuit is given by the formula:

$$ f_0 = \frac{1}{2\pi \sqrt{LC}} $$

To keep the resonant frequency unchanged when the capacitance is changed from \( C \) to \( 2C \), the inductance \( L \) must be adjusted such that the product \( LC \) remains constant. If \( C \) becomes \( 2C \), then \( L \) must become \( L/2 \) to maintain the same resonant frequency. Therefore, the correct option is:

$$ \frac{L}{2} $$

In an LCR series ac circuit the voltage across each of the components L, C and R is 50 V. The voltage across the LC combination will be

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Explanation

In an LCR series AC circuit, the voltages across the inductor (L) and capacitor (C) are 180 degrees out of phase. When the circuit is at resonance, the inductive reactance (XL) equals the capacitive reactance (XC), and their voltages cancel each other out. Therefore, the voltage across the LC combination is zero.

In a circuit L, C and R are connected in series with an alternating voltage source of frequency f. The current leads the voltage by $ 45 ^\circ $ . The value of c is,

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Explanation

In an LCR circuit where the current leads the voltage by $45^ ext{°}$, the circuit is in resonance condition where the net reactance is zero. The capacitive reactance is equal to the inductive reactance ($X_C = X_L$). The condition $ an(45^ ext{°}) = 1$ gives $X_L - X_C = R$. Solving this using the given frequency, the value of capacitance can be found as $C = rac{1}{2 ext{π}f(2 ext{π}fL + R)}$.

In a series resonant LCR circuit, the voltage across R is 100 V and $R= 1k \Omega $ with $C =2 \mu F $ . The resonant frequency $ \Omega $ is 200 rad/s.At resonance the voltage across L is.

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Explanation

$ \upsilon _L = \upsilon_C = I_{X_C} ={\upsilon \over R\omega C} $

A metallic solid sphere is placed in a uniform electric field. The lines of force follow the path(s) shown in figure as

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Explanation

When a metallic solid sphere is placed in a uniform electric field, the lines of electric force are distorted due to the conducting nature of the sphere. The field inside the sphere is zero, and the field lines are perpendicular to the surface of the sphere. The correct representation of this phenomenon is shown in option 4, where the lines of force bend around the sphere and are perpendicular at the points where they touch the sphere.

The dimensional formula of $ l_0 E_0$ is

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