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Physics NEET Questions — Free MCQ Practice

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A body A moves with a uniform acceleration a and zero initial velocity. Another body B, starts from the same point moves in the same direction with a constant velocity v. The two bodies meet after a time t. The value of t is

A.

$\frac{2 v}{a}$

B.

$\frac{v}{a}$

C.

$\frac{v}{2 a}$

D.

$\sqrt{\frac{v}{2 a}}$

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Explanation

$\frac{1}{2} a t^{2} = v t \Rightarrow t = \frac{2 v}{a}$

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

A.

–5 m/s

B.

6 m/s

C.

–3 m/s

D.

3 m/s

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Explanation

The velocity of the particle is

$\frac{d x}{d t} = \frac{d}{d t} (2 - 5 t + 6 t^{2}) = (0 - 5 + 12 t)$

For initial velocity t = 0, hence $v = - 5 m / s$ .

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A car starts from rest and moves with uniform acceleration a on a straight road from time t = 0 to t = T. After that, a constant deceleration brings it to rest. In this process the average speed of the car is

A.

$\frac{a T}{4}$

B.

$\frac{3 a T}{2}$

C.

$\frac{a T}{2}$

D.

aT

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Explanation

For First part,

u = 0, t = T and acceleration = a

$∴ v = 0 + a T = a T$ and $S_{1} = 0 + \frac{1}{2} a T^{2} = \frac{1}{2} a T^{2}$

For Second part,

$u = a T,$ retardation=a₁, $v = 0$ and time taken = T₁ (let)

∴ $0 = u - a_{1} T_{1} \Rightarrow a T = a_{1} T_{1}$

and from $v^{2} = u^{2} - 2 a S_{2} \Rightarrow S_{2} = \frac{u^{2}}{2 a_{1}} = \frac{1}{2} \frac{a^{2} T^{2}}{a_{1}}$

$S_{2} = \frac{1}{2} a T \times T_{1}$ $(A s a_{1} = \frac{a T}{T_{1}})$

∴ $v_{a v} = \frac{S_{1} + S_{2}}{T + T_{1}} = \frac{\frac{1}{2} a T^{2} + \frac{1}{2} a T \times T_{1}}{T + T_{1}}$

$= \frac{\frac{1}{2} a T (T + T_{1})}{T + T_{1}} = \frac{1}{2} a T$

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An object accelerates from rest to a velocity 27.5 m/s in 10 sec .Then find distance covered by object in next 10 sec

A.

550 m

B.

137.5 m

C.

412.5 m

D.

275 m

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Explanation

u = 0, $v = 27 .5 m / s$ and t = 10 sec

$∴ a = \frac{27 .5 - 0}{10} = 2 .75 m / s^{2}$

Now, the distance traveled in next 10 sec,

$S = u t + \frac{1}{2} a t^{2} = 27 .5 \times 10 + \frac{1}{2} \times 2 .75 \times 100$

= 275 + 137.5 = 412.5 m

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If the velocity of a particle is given by $v = {(180 - 16 x)}^{1 / 2}$ m/s, then its acceleration will be

A.

Zero

B.

8 m/s²

C.

– 8 m/s²

D.

4 m/s²

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Explanation

$v = {(180 - 16 x)}^{1 / 2}$

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

$∴ a = \frac{1}{2} {(180 - 16 x)}^{- 1 / 2} \times (- 16) (\frac{d x}{d t})$

$= - 8 {(180 - 16 x)}^{- 1 / 2} \times v$

$= - 8 {(180 - 16 x)}^{- 1 / 2} \times {(180 - 16 x)}^{1 / 2}$ $= - 8 m / s^{2}$

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The displacement of a particle is proportional to the cube of time elapsed. How does the acceleration of the particle depends on time obtained

A.

$a \propto t^{2}$

B.

$a \propto t^{4}$

C.

$a \propto t^{3}$

D.

$a \propto t$

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Explanation

$x \propto t^{3}$ ∴ $x = K t^{3}$

$\Rightarrow v = \frac{d x}{d t} = 3 K t^{2}$ and $a = \frac{d v}{d t} = 6 K t$

i.e. $a \propto t$

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Speed of two identical cars are u and 4u at a specific instant. The ratio of the respective distances in which the two cars are stopped from that instant is

A.

1 : 1

B.

1 : 4

C.

1 : 8

D.

1 : 16

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Explanation

$S \propto u^{2} \Rightarrow \frac{S_{1}}{S_{2}} = {(\frac{1}{4})}^{2} = \frac{1}{16}$

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A body is moving with uniform acceleration describes 40 m in the first 5 sec and 65 m in next 5 sec. Its initial velocity will be

A.

4 m/s

B.

2.5 m/s

C.

5.5 m/s

D.

11 m/s

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Explanation

For a body moving with uniform acceleration, the distance-time relation is given by: s = ut + (1/2)at^2. Using the given values of 40 m in 5 seconds and 65 m in the next 5 seconds, we can solve for the initial velocity (u) and acceleration (a). The solution yields an initial velocity of 5.5 m/s.

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The displacement x of a particle varies with time $t, x = a e^{- α t} + b e^{β t}$ , where $a, b, α and β$ are positive constants. The velocity of the particle will

A.

Go on decreasing with time

B.

Be independent of $α$ and $β$

C.

Drop to zero when $α$ = $β$

D.

Go on increasing with time

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Explanation

$x = a e^{- α t} + b e^{β t}$

Velocity $v = \frac{d x}{d t} = \frac{d}{d t} (a e^{- α t} + b e^{β t})$

$= a . e^{- α t} (- α) + b e^{β t} . β = - a α e^{- α t} + b β e^{β t}$

Acceleration $= - a α e^{- α t} (- α) + b β e^{b t} . β$

$= a α^{2} e^{- α t} + b β^{2} e^{β t}$

Acceleration is positive so velocity goes on increasing with time.

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A car, starting from rest, accelerates at the rate f through a distance S, then continues at constant speed for time t and then decelerates at the rate $\frac{f}{2}$ to come to rest. If the total distance traversed is 15 S, then

A.

$S = \frac{1}{2} f t^{2}$

B.

$S = \frac{1}{4} f t^{2}$

C.

$S = \frac{1}{72} f t^{2}$

D.

$S = \frac{1}{6} f t^{2}$

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Physics MCQs for NEET — Practice Questions with Answers

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