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Physics MCQs for NEET — Practice Questions with Answers

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A simple pendulum is suspended from the roof of a trolley which moves in a horizontal direction with an acceleration a, then the time period is given by $T = 2 π \sqrt{\frac{l}{g'}}$ , where $g^{'}$ is equal to

g-a

g+a

$\sqrt{g^{2} + a^{2}}$

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If the length of second's pendulum is decreased by 2%, how many seconds it will lose per day?

3927 sec

3727 sec

3427 sec

864 sec

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Explanation

$T \propto \sqrt{l} \Rightarrow \frac{∆ T}{T} = \frac{1}{2} \frac{∆ l}{l} = \frac{0.02}{2} = 0.01 Also, time measured by pendulum t = nT where n is number of oscillations \frac{∆ t}{t} = \frac{∆ T}{T} = 0.01$

Loss of time per day = $0.01 \times 24 \times 60 \times 60 = 864 \sec$

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A body is executing Simple Harmonic Motion. At a displacement x its potential energy is $E_{1}$ and at a displacement y its potential energy is $E_{2}$ . The potential energy E at displacement $(x + y)$ is

$E = \sqrt{E_{1}} + \sqrt{E_{2}}$

$\sqrt{E} = \sqrt{E_{1}} + \sqrt{E_{2}}$

$E = E_{1} + E_{2}$

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Explanation

(b) $E_{1} = \frac{1}{2} K x^{2} \Rightarrow x = \sqrt{\frac{2 E_{1}}{K}}, E_{2} = \frac{1}{2} K y^{2} \Rightarrow y = \sqrt{\frac{2 E_{2}}{K}}$

$a n d E = \frac{1}{2} K {(x + y)}^{2} \Rightarrow x + y = \sqrt{\frac{2 E}{K}}$

$\Rightarrow \sqrt{\frac{2 E_{1}}{K}} + \sqrt{\frac{2 E_{2}}{K}} = \sqrt{\frac{2 E}{K}} \Rightarrow \sqrt{E_{1}} + \sqrt{E_{2}} = \sqrt{E}$

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The equation of motion of a particle is $\frac{d^{2} y}{d t^{2}} + K y = 0$ where K is positive constant. The time period of the motion is given by

$\frac{2 π}{K}$

$2 πK$

$\frac{2 π}{\sqrt{K}}$

$2 π \sqrt{K}$

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Explanation

(c) On comparing with standard equation $\frac{d^{2} y}{d t^{2}} + ω^{2} y = 0$ we get $ω^{2} = K \Rightarrow ω \frac{2 π}{T} = \sqrt{K} \Rightarrow T = \frac{2 π}{\sqrt{K}}$

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The kinetic energy of a particle executing S.H.M. is 16 J when it is in its mean position. If the amplitude of oscillations is 25 cm and the mass of the particle is 5.12 kg, the time period of its oscillation is -

$\frac{π}{5} s e c$

$2 π \sec$

$20 π \sec$

$5 π \sec$

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Explanation

a) At mean position, the kinetic energy is maximum.

Hence $\frac{1}{2} m a^{2} ω^{2} = 16$

On putting the values we get $ω = 10 \Rightarrow T = \frac{2 π}{w} = \frac{π}{5} s e c$

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A pendulum has time period T. If it is taken on to another planet having acceleration due to gravity half and mass 9 times that of the earth then its time period on the other planet will be

$\sqrt{T}$

$T^{1 / 3}$

$\sqrt{2}$ T

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Explanation

(d) $T = 2 π \sqrt{\frac{l}{g}} \Rightarrow T \propto \frac{1}{\sqrt{g}} \Rightarrow \frac{T_{p}}{T_{e}} = \sqrt{\frac{g_{e}}{g_{p}}} = \sqrt{\frac{2}{1}} \Rightarrow T_{p} = \sqrt{2} T$

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A particle in SHM is described by the displacement equation $x (t) = A \cos (ω t + θ) . I f t h e i n i t i a l$ position of the particle is 1 cm and its initial velocity is $π$ cm/s, what is its amplitude? (The angular frequency of the particle is $π s^{- 1}$ )

1 cm

$\sqrt{2}$ cm

2 cm

2.5 cm

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Explanation

(b) Given, $v = πcm / \sec, x = 1 cm and ω = {πs}^{2} using v = ω \sqrt{a^{2} - x^{2}} \Rightarrow π = π \sqrt{a^{2} - 1} \Rightarrow 1 = a^{2} - 1 \Rightarrow a = \sqrt{2} cm .$

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A simple pendulum hanging from the ceiling of a stationary lift has a time period T₁. When the lift moves downward with constant velocity, the time period is T₂, then

$T_{2}$ is infinity

$T_{2} > T_{1}$

$T_{2} < T_{1}$

$T_{2} = T_{1}$

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Explanation

(d) $T \propto \frac{1}{\sqrt{g}}$ and g is same in both cases so time period remains same.

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If the length of a pendulum is made 9 times and mass of the bob is made 4 times , then the value of time period becomes

3/2T

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Explanation

(a) $T = 2 π \sqrt{\frac{l}{g}} \Rightarrow T \propto \sqrt{l}$ , hence if l made 9 times, T becomes 3 times.

Also time period of simple pendulum does not depends on the mass of the bob.

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A simple harmonic wave having an amplitude a and time period T is represented by the equation $y = 5 sinπ (t + 4)$ m Then the value of amplitude (a) in (m) and time period (T) in second are

$a = 10, T = 2$

$a = 5, T = 1$

$a = 10, T = 1$

$a = 5, T = 2$

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Explanation

(d) $y = 5 \sin (πt + 4 π),$ comparing it with standard equation

$y = a \sin (ω t + ϕ) = a \sin (\frac{2 πt}{T} + ϕ)$

$a = 5 m a n d \frac{2 πt}{T} = πt \Rightarrow T = 2 \sec$

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Yes. You can attempt every Physics question on this page for free without logging in, and check the correct answer with a detailed explanation instantly.

No account is required to attempt questions and view answers. A free account adds bookmarks, personal notes, and progress tracking.

The bank mixes NEET previous year questions (PYQs) with practice questions, each tagged with its exam appearances where applicable.