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Physics NEET Questions — Free MCQ Practice

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The relation between time and distance is $t = α x^{2} + β x$ , where α and β are constants. The retardation is

A.

$2 α v^{3}$

B.

$2 β v^{3}$

C.

$2 α β v^{3}$

D.

$2 β^{2} v^{3}$

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Explanation

$\frac{d t}{d x} = 2 α x + β \Rightarrow v = \frac{1}{2 α x + β} L e t, 2 α x + β = p \Rightarrow \frac{d p}{d x} = 2 α \Rightarrow v = \frac{1}{p} \Rightarrow \frac{d v}{d p} = - \frac{1}{p^{2}} = \frac{- 1}{{(2 α x + β)}^{2}} N o w, \frac{d v}{d x} = \frac{d v}{d p} \times \frac{d p}{d x} = \frac{- 2 α}{{(2 α x + β)}^{2}}$

$∵ a = \frac{d v}{d t} = \frac{d v}{d x} . \frac{d x}{d t}$

$a = v \frac{d v}{d x} = \frac{- v . 2 α}{{(2 α x + β)}^{2}} = - v . 2 α \times \frac{1}{{(2 α x + β)}^{2}} = - 2 α . v . v^{2} = - 2 α v^{3}$

∴ Retardation $= 2 α v^{3}$

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A point moves with uniform acceleration and v₁, v₂ and v₃ denote the average velocities in the three successive intervals of time t₁, t₂ and t₃. Which of the following relations is correct ?

A.

$(v_{1} - v_{2}) : (v_{2} - v_{3}) = (t_{1} - t_{2}) : (t_{2} + t_{3})$

B.

$(v_{1} - v_{2}) : (v_{2} - v_{3}) = (t_{1} + t_{2}) : (t_{2} + t_{3})$

C.

$(v_{1} - v_{2}) : (v_{2} - v_{3}) = (t_{1} - t_{2}) : (t_{1} - t_{3})$

D.

$(v_{1} - v_{2}) : (v_{2} - v_{3}) = (t_{1} - t_{2}) : (t_{2} - t_{3})$

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Explanation

Let $u_{1}, u_{2}, u_{3}$ and u₄ be velocities at time $t = 0, t_{1}, (t_{1} + t_{2})$ and $(t_{1} + t_{2} + t_{3})$ respectively and acceleration is a then $v_{1} = \frac{u_{1} + u_{2}}{2}, v_{2} = \frac{u_{2} + u_{3}}{2} and v_{3} = \frac{u_{3} + u_{4}}{2}$

Also $u_{2} = u_{1} + a t_{1}, u_{3} = u_{1} + a (t_{1} + t_{2})$

and $u_{4} = u_{1} + a (t_{1} + t_{2} + t_{3})$

By solving, we get $\frac{v_{1} - v_{2}}{v_{2} - v_{3}} = \frac{(t_{1} + t_{2})}{(t_{2} + t_{3})}$

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The velocity of a body moving with a uniform acceleration of 2 m/sec² is 10 m/sec. Its velocity after an interval of 4 sec is

A.

12 m/sec

B.

14 m/sec

C.

16 m/sec

D.

18 m/sec

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Explanation

$v = u + a t = 10 + 2 \times 4 = 18 m / \sec$

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A particle starting from rest moving with constant acceleration travels a distance x in first 2 seconds and a distance y in next two seconds, then

A.

y = x

B.

y = 2x

C.

y = 3x

D.

y = 4x

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Explanation

$Dist . covered in 1^{st} 2 \sec : ‐ x = o + \frac{1}{2} a \times {(2)}^{2} = 2 a Dist . covered in 4 \sec : ‐ d = o + \frac{1}{2} \times a \times {(4)}^{2} = 8 a \Rightarrow y = d - x = 8 a - 2 a = 6 a \Rightarrow \frac{x}{y} = \frac{2 a}{6 a} \Rightarrow y = 3 x$

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The initial velocity of a body moving along a straight line is 7 m/s. It has a uniform acceleration of 4 m/s². The distance covered by the body in the 5^th second of its motion is

A.

25 m

B.

35 m

C.

50 m

D.

85 m

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Explanation

$S_{n} = u + \frac{a}{2} [2 n - 1]$

$S_{5^{t h}} = 7 + \frac{4}{2} [2 \times 5 - 1] = 7 + 18 = 25 m$ .

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The velocity of a body depends on time according to the equation $v = 20 + 0 .1 t^{2}$ . The body is undergoing

A.

Uniform acceleration

B.

Uniform retardation

C.

Non-uniform acceleration

D.

Zero acceleration

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Explanation

Acceleration $a = \frac{d v}{d t} = 0 .1 \times 2 t = 0 .2 t$

Which is time dependent i.e. non-uniform acceleration.

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Which of the following four statements is false ?

A.

A body can have zero velocity and still be accelerated

B.

A body can have a constant velocity and still have a varying speed

C.

A body can have a constant speed and still have a varying velocity

D.

The direction of the velocity of a body can change when its acceleration is constant

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Explanation

Constant velocity means constant speed as well as same direction throughout.

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A particle moving with a uniform acceleration travels 24 m and 64 m in the first two consecutive intervals of 4 sec each. Its initial velocity is

A.

1 m/sec

B.

10 m/sec

C.

5 m/sec

D.

2 m/sec

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Explanation

Distance travelled in 4 sec

$24 = 4 u + \frac{1}{2} a \times 16$ …(i)

Distance travelled in total 8 sec

$88 = 8 u + \frac{1}{2} a \times 64$ …(ii)

After solving (i) and (ii), we get u = 1 m/s.

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The position of a particle moving in the xy-plane at any time t is given by $x = (3 t^{2} - 6 t)$ metres, $y = (t^{2} - 2 t)$ metres. Select the correct statement about the moving particle from the following

A.

The acceleration of the particle is zero at t = 0 second

B.

The velocity of the particle is zero at t = 0 second

C.

The velocity of the particle is zero at t = 1 second

D.

The velocity and acceleration of the particle are never zero

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Explanation

$v_{x} = \frac{d x}{d t} = \frac{d}{d t} (3 t^{2} - 6 t) = 6 t - 6$ . At $t = 1, v_{x} = 0$

$v_{y} = \frac{d y}{d t} = \frac{d}{d t} (t^{2} - 2 t) = 2 t - 2$ . At $t = 1, v_{y} = 0$

Hence $v = \sqrt{v_{x}^{2} + v_{y}^{2}} = 0$

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If body having initial velocity zero is moving with uniform acceleration 8 m/sec² , then the distance travelled by it in fifth second will be

A.

36 metres

B.

40 metres

C.

100 metres

D.

Zero

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Explanation

Distance travelled in $n^{t h}$ second $= u + \frac{a}{2} (2 n - 1)$

Distance travelled in $5^{t h}$ second $= 0 + \frac{8}{2} (2 \times 5 - 1)$ = 36m

Physics MCQs for NEET — Practice Questions with Answers

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