Physics Assertion (A) & Reason (B) MCQ

3. A is true but R is false.

Explanation:

When pure inductor is connected to source of an alternating emf, then instantaneous value of alternating emf is given by $\text{E}=\text{E}_0\sin\omega\text{t},$ and corresponding alternating current is given by $\text{I}=\text{I}_0\sin(\omega\text{t}-\frac{\pi}{2}).$ From these two equations, it follows that alternating current lags behind e.m.f. by a phase angle of $\frac{\pi}{2}.$ The inductive reactance, so when frequency increases correspondingly inductive reactance also increases.

2. Both A and R are true but R is not the correct explanation of A.

Explanation:

As both the inductance and resistance are joined in series, hence current through both will be same. But in case of resistance, both the current and potential vary simultaneously, hence they are in same phase. While in case of an inductance when current is zero, potential difference across it is maximum and when current reaches maximum $\big($at $\omega\text{t}=\frac{\pi}{2}\big),$ potential difference across it becomes zero i.e., potential difference leads the current by $\frac{\pi}{2}$ or current lags behind the potential difference by $\frac{\pi}{2}.$ Phase angle in case of LR circuit is given as $\phi=\tan^{-1}\Big(\frac{\omega\text{L}}{\text{R}}\Big).$

3. A is true but R is false.

Explanation:

The transmission is done at high voltage due to which current through the wire is reduced. By reduction in current corresponding dissipation of energy is also reduced $\big($as $\text{H}\propto\text{I}^2\text{R}\big).$ If transmission is done at low voltage then we have to use thick wire in order to reduce the dissipation of energy. This increase the cost of transmission lines wires. In order to reduce both energy dissipation and cost of transmission wire, transmission is done at high voltage by using step-up transformers.

2. Both A and R are true but R is not the correct explanation of A.

Explanation:

The phase angle for the LCR series circuit is given by $\tan\theta=\frac{\text{X}_\text{L}-\text{X}_\text{C}}{\text{R}}=\frac{\frac{\omega\text{L-1}}{\omega\text{C}}}{\text{R}}.$ Where X_L, X_C are inductive reactance and capacitive reactance respectively. When X_L > X_C then $\tan\theta$ is positive i.e. $\theta$ is positive $\big($between 0 and $\frac{\pi}{2}\big).$ Hence emf leads the current.

2. Both A and R are true but R is not the correct explanation of A.

Explanation:

At resonance X_L= X_C or $\omega\text{L}=\frac{1}{\omega\text{C}}.$ Because of this impedance of LCR series circuit become equal to resistance of circuit $\Big(\text{Z}=\sqrt{\text{R}^2+(\text{X}_\text{L}-\text{X}_\text{C})^2}\Big).$ Therefore from $\text{I}=\frac{\text{E}}{\text{Z}}=\frac{\text{E}}{\text{R}},$ at resonance, current in LCR series circuit is maximum. Correspondingly phase angle is also equal to zero. Therefore emf and current are in phase in LCR series circuit.

1. Both A and R are true and R is the correct explanation of A.

Explanation:

The capacitive reactance of capacitor is given by,

$\text{X}_\text{C}=\frac{1}{\omega\text{C}}=\frac{1}{2\pi\text{fC}}.$

So this is infinite for D.C. (f = 0) and has a finite value for A.C. Therefore a capacitor blocks D.C. and offers an easy path for A. C.

1. Both A and R are true and R is the correct explanation of A.

Explanation:

Capacitive reactance $\text{X}_\text{C}=\frac{1}{\omega\text{C}}.$ When roe capacitance C increases, the capacitive reactance decreases. Due to decrease in its values, the current in the circuit will increase $\Bigg(\text{I}=\frac{\text{E}}{\sqrt{\text{R}^2+\text{X}^2_\text{C}}}\Bigg)$ and hence brightness of source (or electric lamp) will also increases.

2. Both A and R are true but R is not the correct explanation of A.

Explanation:

If there is no loss of energy in transformer, then instantaneous output power is equal to instantaneous input power. From this we get $\frac{\text{e}_\text{s}}{\text{e}_\text{p}}=\frac{\text{I}_\text{p}}{\text{I}_\text{s}}.$

So in step up transformer voltage increases by decreasing the current. Similarly, step-down transformer decreases the voltage by increasing current. Therefore transformer simply transforms the voltage and current, obeying the law of conservation of energy.

1. Both A and R are true and R is the correct explanation of A.

Explanation:

The element of the heater is in the form of a coil so that it has inductance L also besides the resistance R. So, for the alternating current, the effective resistance of the heater would be $\sqrt{\text{R}^2+(\omega\text{L})^2}$ which is larger than the resistance R for the direct current. Hence on healing by alternating current, heat produced per second would be less.

$\big(\text{H}=\frac{\text{V}^2}{\text{R}}$ and V is same in both the cases$\big).$