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Physics NEET Questions — Free MCQ Practice

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The peak value of an alternating e.m.f. E is given by $E = E_{0} \cos ω t$ is 10 volts and its frequency is 50 Hz. At time $t = \frac{1}{600} s e c$ , the instantaneous e.m.f. is

A.

10 V

B.

$5 \sqrt{3} V$

C.

5 V

D.

1 V

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Explanation

$E = E_{0} \cos ω t = E_{0} \cos \frac{2 π t}{T}$

$= 10 \cos \frac{2 π \times 50 \times 1}{600} = 10 \cos \frac{π}{6} = 5 \sqrt{3} v o l t .$

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If a current I given by $I_{0} \sin (ω t - \frac{π}{2})$ flows in an ac circuit across which an ac potential of $E = E_{0} \sin ω t$ has been applied, then the power consumption P in the circuit will be

A.

$P = \frac{E_{0} I_{0}}{\sqrt{2}}$

B.

$P = \sqrt{2} E_{0} I_{0}$

C.

$P = \frac{E_{0} I_{0}}{2}$

D.

P = 0

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Explanation

Phase angle $ϕ = 90^{o}$ , so power $P = V i \cos ϕ = 0$

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In an ac circuit, the instantaneous values of e.m.f. and current are e = 200 sin 314 t volt and $i = \sin (314 t + \frac{π}{3})$ ampere. The average power consumed in watt is

A.

200

B.

100

C.

50

D.

25

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Explanation

$V_{r m s} = \frac{200}{\sqrt{2}}, i_{r m s} = \frac{1}{\sqrt{2}}$

$∴ P = V_{r m s} i_{r m s} \cos ϕ = \frac{200}{\sqrt{2}} \frac{1}{\sqrt{2}} \cos \frac{π}{3} = 50 w a t t$

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An alternating current is given by the equation $i = i_{1} \cos ω t + i_{2} \sin ω t$ . The r.m.s. current is given by

A.

$\frac{1}{\sqrt{2}} (i_{1} + i_{2})$

B.

$\frac{1}{\sqrt{2}} {(i_{i} + i_{2})}^{2}$

C.

$\frac{1}{\sqrt{2}} {(i_{1}^{2} + i_{2}^{2})}^{1 / 2}$

D.

$\frac{1}{2} {(i_{1}^{2} + i_{2}^{2})}^{1 / 2}$

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Explanation

$i_{r m s} = \sqrt{\frac{i_{1}^{2} + i_{2}^{2}}{2}} = \frac{1}{\sqrt{2}} {(i_{1}^{2} + i_{2}^{2})}^{1 / 2}$

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In an ac circuit, the current is given by $i = 5 \sin (100 t - \frac{π}{2})$ and the ac potential is V = 200 sin(100t) volt. Then the power consumption is :

A.

20 watts

B.

40 watts

C.

1000 watts

D.

0 watt

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Explanation

$P = V i \cos ϕ$

Phase difference $ϕ = \frac{π}{2} \Rightarrow P = z e r o$

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A resistance of 20 ohms is connected to a source of an alternating potential $V = 220 \sin (100 π t)$ . The time taken by the current to change from its peak value to r.m.s value is

A.

0.2 sec

B.

0.25 sec

C.

25 × 10^–3 sec

D.

2.5 × 10^–3 sec

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Explanation

Peak value to r.m.s. value means current becomes $\frac{1}{\sqrt{2}}$ times.

$T i m e t a k e n b y t h e c u r r e n t t o r e a c h i t s p e a k v a l u e : i = i_{0} \sin 100 π t \Rightarrow i_{0} = i_{0} \sin 100 π t_{1} \Rightarrow \sin 100 π t_{1} = \sin \frac{π}{2} \Rightarrow t_{1} = \frac{1}{200} s e c T i m e t a k e n b y t h e c u r r e n t t o b e c o m e \frac{i}{\sqrt{2}} a f t e r r e a c h i n g i t s p e a k v a l u e : i = i_{0} \sin 100 π t \Rightarrow \frac{1}{\sqrt{2}} \times i_{0} = i_{0} \sin 100 π t_{2} \Rightarrow \sin 100 π t_{2} = \sin \frac{3 π}{4} \Rightarrow t_{2} = \frac{3}{400} s e c T i m e t a k e n b y t h e c u r r e n t t o c h a n g e f r o m i t s p e a k v a l u e t o r . m . s . v a l u e : t_{2} - t_{1} = \frac{3}{400} - \frac{1}{200} = \frac{1}{400} s e c = 2 .5 \times 10^{- 3} s e c .$

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Voltage and current in an ac circuit are given by $V = 5 \sin (100 π t - \frac{π}{6})$ and $I = 4 \sin (100 π t + \frac{π}{6})$

A.

Voltage leads the current by 30°

B.

Current leads the voltage by 30°

C.

Current leads the voltage by 60°

D.

Voltage leads the current by 60°

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Explanation

Phase difference $Δ ϕ = ϕ_{2} - ϕ = \frac{π}{6} - (\frac{- π}{6}) = \frac{π}{3}$

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An alternating current of frequency ‘f’ is flowing in a circuit containing a resistance R and a choke L in series. The impedance of this circuit is

A.

R + 2πfL

B.

$\sqrt{R^{2} + 4 π^{2} f^{2} L^{2}}$

C.

$\sqrt{R^{2} + L^{2}}$

D.

$\sqrt{R^{2} + 2 π f L}$

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Explanation

$Z = \sqrt{R^{2} + X_{L}^{2}}, X_{L} = ω L$ and $ω = 2 π f$

$∴ Z = \sqrt{R^{2} + 4 π^{2} f^{2} L^{2}}$

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A resistance of 300 Ω and an inductance of $\frac{1}{π}$ henry are connected in series to a ac voltage of 20 volts and 200 Hz frequency. The phase angle between the voltage and current is :

A.

$\tan^{- 1} \frac{4}{3}$

B.

$\tan^{- 1} \frac{3}{4}$

C.

$\tan^{- 1} \frac{3}{2}$

D.

$\tan^{- 1} \frac{2}{5}$

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Explanation

Phase angle $\tan ϕ = \frac{ω L}{R} = \frac{2 π \times 200}{300} \times \frac{1}{π} = \frac{4}{3}$

$∴ ϕ = \tan^{- 1} \frac{4}{3}$

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In a region of uniform magnetic induction B = 10^–2 tesla, a circular coil of radius 30 cm and resistance π² ohm is rotated about an axis that is perpendicular to the direction of B and which forms a diameter of the coil. If the coil rotates at 200 rpm the amplitude of the alternating current induced in the coil is :

A.

4π² mA

B.

30 mA

C.

6 mA

D.

200 mA

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Explanation

Amplitude of $a c = i_{0} = \frac{V_{0}}{R} = \frac{ω N B A}{R} = \frac{(2 π ν) N B (π r^{2})}{R}$

$\Rightarrow i_{0} = \frac{2 π \times \frac{200}{60} \times 1 \times 10^{- 2} \times π \times {(0 .3)}^{2}}{π^{2}} = 6 m A$

Physics MCQs for NEET — Practice Questions with Answers

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