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Physics MCQs for NEET — Practice Questions with Answers

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The displacement of a particle is given by $y = a + bt + {ct}^{2} - {dt}^{4}$ . The initial velocity and acceleration are respectively

b, -4d

-d, 2c

b, 2c

2c, -4d

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Explanation

$G i v e n, y = a + b t + c t^{2} - d t^{4} \Rightarrow v = \frac{d v}{d t} = b + 2 c t - 4 d t^{3} \Rightarrow a = \frac{d v}{d t} = 2 c - 12 d t^{2} a t t = 0, v = b a n d a = 2 c$

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The momentum is given by p=4t+1, the force at t=2s is

4 N

8 N

10 N

15 N

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Explanation

(A)

$F = \frac{d p}{d t} = 4 N S o f o r c e i s c o n s \tan t a l w a y s$

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A particle moves along a straight line such that its displacement at any time t is given by $s = t^{3} - 3 t^{2} + 6 t - 12$ metres. The velocity when the acceleration is zero is:

$3 {ms}^{- 1}$

$- 12 {ms}^{- 1}$

$42 {ms}^{- 1}$

$- 9 {ms}^{- 1}$

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Explanation

Given,

$x = t^{3} - 3 t^{2} + 6 t - 12 v = \frac{d x}{d t} = 3 t^{2} - 6 t + 6 a = \frac{d v}{d t} = 6 t - 6 w h e n, a = 0 \Rightarrow t = 1 s e c \Rightarrow v = 3 m / s$

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The position x of particle varies with time t as $x = {at}^{2} - {bt}^{3}$ . The acceleration of the particle will be zero at time equal to

$\frac{a}{b}$

$\frac{2 a}{b}$

$\frac{a}{3 b}$

Zero

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Explanation

Diffrentiating x wrt t ,

$v = \frac{d x}{d t}$

$v = 2 a t - 3 b t^{2}$

$a = 2 a - 6 b t$

At $t = \frac{a}{3 b} a = 0$

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

$a + 2 bt$

$2 b + 6 ct$

$2 b - 6 ct$

$3 b - 6 {ct}^{2}$

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Explanation

Diffrentiating x wrt t,

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

$a = 2 b - 6 c t$

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A particle moves along X-axis in such a way that its X coordinate varies with time t according to the equation $x = (2 - 5 t + 6 t^{2}) m$ . The initial velocity of the particle is

-5 m/s

6 m/s

3 m/s

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Explanation

$v = \frac{d x}{d t} = - 5 + 12 t A t t = 0 v = - 5 m / s$

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If the momentum of a particle is given by P=(180-8t) kg m/s, then its force will be

Zero

8 N

-8 N

4 N

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Explanation

$F = \frac{d P}{d t}$

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A particle moves along x-axis as $x = 4 (t - 2) + a {(t - 2)}^{2}$

Which of the following is true?

The initial velocity of the particle is 4.

The acceleration of the particle is 2a.

The particle is at the origin at t=0.

None of these

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Explanation

$x = 4 (t - 2) + a {(t - 2)}^{2} At t = 0, x = - 8 + 4 a v = \frac{dx}{dt} = 4 + 2 a (t - 2) At t = 0, v_{in} = 4 - 4 a a = \frac{dv}{dt} = 2 a$

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A particle is moving along x-axis. The velocity v of particle varies with its position x as $v = \frac{1}{x}$ . Find velocity of particle as a function of time t given that at t=0, x=1 .

$v = \sqrt{2 t + 1}$

$v = \frac{1}{\sqrt{2 t + 1}}$

$v = \frac{1}{t}$

None of these

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Explanation

$v = \frac{1}{x} \frac{d x}{d t} = \frac{1}{x}$

$x d x = d t \int_{1}^{x} x d x = \int_{0}^{t} d t \frac{x^{2} - 1}{2} = t x = \sqrt{2 t + 1} v = \frac{d x}{d t} = \frac{1}{\sqrt{2 t + 1}}$

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A vector $\vec{A}$ is directed along $30 °$ west of north direction and another vector $\vec{B}$ along $15 °$ south of east. Their resultant cannot be in ____________ direction.

North

East

North-East

South

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Explanation

The resultant of two vectors A and B will be in a direction between the directions of A and B. Since A is directed 30Â° west of north and B is directed 15Â° south of east, their resultant cannot be in the south direction.

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