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Physics NEET Questions — Free MCQ Practice

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The motion of a particle is described by the equation $x = a + b t^{2}$ where a = 15 cm and b = 3 cm/s². Its instantaneous velocity at time 3 sec will be

A.

36 cm/sec

B.

18 cm/sec

C.

16 cm/sec

D.

32 cm/sec

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Explanation

$x = a + b t^{2}, v = \frac{d x}{d t} = 2 b t$

Instantaneous velocity $v = 2 \times 3 \times 3 = 18 c m / \sec$

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A body travels for 15 sec starting from rest with constant acceleration. If it travels distances S₁, S₂ and S₃ in the first five seconds, second five seconds and next five seconds respectively the relation between S₁, S₂ and S₃ is

A.

$S_{1} = S_{2} = S_{3}$

B.

$5 S_{1} = 3 S_{2} = S_{3}$

C.

$S_{1} = \frac{1}{3} S_{2} = \frac{1}{5} S_{3}$

D.

$S_{1} = \frac{1}{5} S_{2} = \frac{1}{3} S_{3}$

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Explanation

Distance travelled in first 5 sec $S_{1} = \frac{1}{2} a (5)^{2}$

Distance travelled in next 5 sec $S_{2} = \frac{1}{2} a (10)^{2} - \frac{1}{2} a (5)^{2} = 3 \frac{a (5)^{2}}{2}$

Distance travelled from t =11s to t=15 s,

$S_{3} = \frac{a (15)^{2}}{2} - \frac{a (10)^{2}}{2} = \frac{5 a (5)^{2}}{2}$

If the body starts from rest and moves with constant acceleration then the ratio of distances in consecutive equal time interval $S_{1} : S_{2} : S_{3} = 1 : 3 : 5$

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A body is moving according to the equation $x = a t + b t^{2} - c t^{3}$ where x = displacement and a, b and c are constants. The acceleration of the body is

A.

$a + 2 b t$

B.

$2 b + 6 c t$

C.

$2 b - 6 c t$

D.

$3 b - 6 c t^{2}$

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Explanation

$x = a t + b t^{2} - c t^{3}, a = \frac{d^{2} x}{d t^{2}} = 2 b - 6 c t$

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A particle travels 10m in first 5 sec and 10m in next 3 sec. Assuming constant acceleration what is the distance travelled in next 2 sec ?

A.

8.3 m

B.

9.3 m

C.

10.3 m

D.

None of above

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Explanation

Let initial $(t = 0)$ velocity of particle = u

For first 5 sec motion $s_{5} = 10 m e t r e$

$s = u t + \frac{1}{2} a t^{2} \Rightarrow 10 = 5 u + \frac{1}{2} a {(5)}^{2}$

$2 u + 5 a = 4$ …(i)

For first 8 sec of motion $s_{8} = 20 m e t r e$

$20 = 8 u + \frac{1}{2} a {(8)}^{2} \Rightarrow 2 u + 8 a = 5$ …(ii)

By solving $u = \frac{7}{6} m / s and a = \frac{1}{3} m / s^{2}$

Now distance travelled by particle in Total 10 sec.

$s_{10} = u \times 10 + \frac{1}{2} a {(10)}^{2}$

By substituting the value of u and a we will get $s_{10} = 28 .3 m$

so the distance in last $2 \sec = s_{10} - s_{8}$

$= 28 .3 - 20 = 8 .3 m$

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The distance travelled by a particle is proportional to the squares of time, then the particle travels with

A.

Uniform acceleration

B.

Uniform velocity

C.

Increasing acceleration

D.

Decreasing velocity

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Explanation

$s \propto t^{2}$ (given) ∴ $s = K t^{2}$

Acceleration $a = \frac{d^{s} s}{d t^{2}} = 2 k$ (constant)

It means the particle travels with uniform acceleration.

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Velocity of a particle changes when

A.

Direction of velocity changes

B.

Magnitude of velocity changes

C.

Both of above

D.

None of the above

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Explanation

Because velocity is a vector quantity

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The motion of a particle is described by the equation u = at. The distance travelled by the particle in the first 4 seconds

A.

4 a

B.

12 a

C.

6 a

D.

8 a

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Explanation

$u = a t, x = \int_{}^{} u d t = \int_{}^{} a t d t = \frac{a t^{2}}{2}$

For $t = 4 \sec, x = 8 a$

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The relation $3 t = \sqrt{3 x} + 6$ describes the displacement of a particle in one direction where x is in metres and t in sec. The displacement, when velocity is zero, is

A.

24 metres

B.

12 metres

C.

5 metres

D.

Zero

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Explanation

$3 t = \sqrt{3 x} + 6 \Rightarrow 3 x = {(3 t - 6)}^{2}$

$\Rightarrow x = 3 t^{2} - 12 t + 12$

$v = \frac{d x}{d t} = 6 t - 12$ , for $v = 0, t = 2 \sec$

$x = 3 {(2)}^{2} - 12 \times 2 + 12 = 0$

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The average velocity of a body moving with uniform acceleration travelling a distance of 3.06 m is 0.34 ms^–1. If the change in velocity of the body is 0.18ms^–1 during this time, its uniform acceleration is

A.

0.01 ms^–2

B.

0.02 ms^–2

C.

0.03 ms^–2

D.

0.04 ms^–2

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Explanation

$Time = \frac{Distance}{Average velocity} = \frac{3 .06}{0 .34} = 9 \sec$

Acceleration $= \frac{Change in velocity}{Time} = \frac{0 .18}{9}$ =0.02 m/s²

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Equation of displacement for any particle is $s = 3 t^{3} + 7 t^{2} + 14 t + 8 m$ . Its acceleration at time t = 1 sec is

A.

10 m/s²

B.

16 m/s²

C.

25 m/s²

D.

32 m/s²

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Explanation

$s = 3 t^{3} + 7 t^{2} + 14 t + 8 m$

$a = \frac{d^{2} s}{d t^{2}} = 18 t + 14$ at $t = 1 \sec \Rightarrow a = 32 m / s^{2}$

Physics MCQs for NEET — Practice Questions with Answers

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