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Physics MCQs for NEET — Practice Questions with Answers

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Under constant temperature, graph between $P$ and $1 / V$ is

Parabola

Hyperbola

Straight line

Circle

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Explanation

3. At constant temperature $PV$ = constant $\Rightarrow P \propto \frac{1}{V}$

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A particle is subjected to two simple harmonic motions in the same direction having equal amplitudes and equal frequency. If the resulting amplitude is equal to the amplitude of individual motions, the phase difference between them is:

$\frac{π}{3}$

$\frac{2 π}{3}$

$\frac{π}{6}$

$\frac{π}{2}$

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Explanation

Resultant amplitude

$\begin{aligned} R = \sqrt{a_{1}^{2} + a_{2}^{2} + 2 a_{1} a_{2} \cos ϕ} \\ a^{2} = a^{2} + a^{2} + 2 a^{2} \cos ϕ \\ \cos ϕ = - \frac{1}{2} \Rightarrow ϕ = \frac{2 π}{3} \end{aligned}$

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If a particle is executing SHM, with an amplitude A, the distance moved and the displacement of the body in a time equal to its period are

2A, A

4A, 0

A, A

0, 2A

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Explanation

Distance = 4A

Displacement = 0

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$The equation of the displacement of two particles making SHM are represented by y_{1} = a \sin (ωt + ϕ) & y_{2} = a \cos (ωt) The phase difference of the velocities of the t w o particles is$

$\frac{π}{2} + ϕ$

$- ϕ$

$ϕ$

$ϕ - \frac{π}{2}$

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The displacement of a particle executing SHM is given by y = 0.25 (sin 200t) cm. The maximum speed of the particles is:

200 cm/sec

100 cm/sec

50 cm/sec

0.25 cm/sec

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Explanation

$Maximum particle velocity = aω = 0.25 x 200 = 50 cm / s$

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$A particle is executing SHM with amplitude A and time period T . If at t = 0, it is at origin (mean position) then find the time instant, when it covers a distance equal to \frac{5 A}{2}$

$\frac{T}{12}$

$\frac{5 T}{12}$

$\frac{7 T}{12}$

$\frac{2 T}{3}$

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A particle undergoes SHM with a time period of 2 seconds. In how much time will it travel from its mean position to a displacement equal to half of its amplitude?

$\frac{1}{2} s$

$\frac{1}{6} s$

$\frac{1}{4} s$

$\frac{1}{3} s$

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Explanation

$y = A \sin ω t \frac{A}{2} = A \sin ω t \sin ω t = \frac{1}{2} = \sin (\frac{π}{6}) ω t = \frac{π}{6} \Rightarrow \frac{2 π}{T} t = \frac{π}{6} t = \frac{T}{12} = \frac{2}{12} = \frac{1}{6} s e c$

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If the displacement (x) and velocity v of a particle executing simple harmonic motion are related through the expression $4 v^{2} = 25 - x^{2}$ then its time period is:

$π$

2 $π$

4 $π$

6 $π$

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Explanation

$S \tan d a r d e q u a t i o n o f S H M v^{2} = ω^{2} (A^{2} - x^{2}) 4 v^{2} = 25 - x^{2} v^{2} = \frac{1}{4} (25 - x^{2}) C o m p a r e i t w i t h v^{2} = ω^{2} (A^{2} - x^{2}) \Rightarrow ω^{2} = \frac{1}{4} \Rightarrow ω = \frac{1}{2} T i m e p e r i o d = \frac{2 π}{ω} = 4 π$

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Two simple pendulums have time periods T and $\frac{5 T}{4}$ . They start vibrating at the same instant from the mean position in the same phase. The phase difference between them when bigger pendulum completes one oscillation will be:

$\frac{π}{6}$

$\frac{π}{4}$

$\frac{π}{3}$

$\frac{π}{2}$

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Explanation

$ϕ = ω_{s} t - ω_{i} t ϕ = \frac{2 π}{T} . \frac{5 T}{4} - \frac{2 π}{\frac{5 T}{4}} . \frac{5 T}{4}$

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There is a simple pendulum hanging from the ceiling of a lift. When the lift is stand still, the time period of the pendulum is T. If the resultant acceleration becomes g/4, then the new time period of the pendulum is

0.8 T

0.25 T

2 T

4 T

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Explanation

When lift is at rest, $T = 2 π \sqrt{l / g}$

If acceleration becomes g/4 then

$T^{'} = 2 π \sqrt{\frac{l}{g / 4}} = 2 π \sqrt{\frac{4 l}{g}} = 2 \times T$

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