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Physics NEET Questions — Free MCQ Practice

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An alpha particle enters a hollow tube of 4 m length with an initial speed of 1 km/s. It is accelerated in the tube and comes out of it with a speed of 9 km/s. The time for which it remains inside the tube is

A.

$8 \times 10^{- 3} s$

B.

$80 \times 10^{- 3} s$

C.

$800 \times 10^{- 3} s$

D.

$8 \times 10^{- 4} s$

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Explanation

$v^{2} = u^{2} + 2 a s \Rightarrow {(9000)}^{2} - {(1000)}^{2} = 2 \times a \times 4$

$\Rightarrow a = 10^{7} m / s^{2}$

Now $t = \frac{v - u}{a}$

$\Rightarrow t = \frac{9000 - 1000}{10^{7}} = 8 \times 10^{- 4} s e c$

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Two cars A and B are travelling in the same direction with velocities v₁ and v₂ $(v_{1} > v_{2})$ . When the car A is at a distance d behind car B, the driver of the car A applied the brake producing a uniform retardation a. There will be no collision when

A.

$d < \frac{{(v_{1} - v_{2})}^{2}}{2 a}$

B.

$d < \frac{v_{1}^{2} - v_{2}^{2}}{2 a}$

C.

$d > \frac{{(v_{1} - v_{2})}^{2}}{2 a}$

D.

$d > \frac{v_{1}^{2} - v_{2}^{2}}{2 a}$

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Explanation

Initial relative velocity $= v_{1} - v_{2}$ , Final relative velocity = 0

From $v^{2} = u^{2} - 2 a s$ ⇒ $0 = {(v_{1} - v_{2})}^{2} - 2 \times a \times s$

⇒ $s = \frac{{(v_{1} - v_{2})}^{2}}{2 a}$

If the distance between two cars is 's' then collision will take place. To avoid collision d > s ∴ $d > \frac{{(v_{1} - v_{2})}^{2}}{2 a}$

where d = actual initial distance between two cars.

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A body starts from rest from the origin with an acceleration of $6 m / s^{2}$ along the x-axis and $8 m / s^{2}$ along the y-axis. Its distance from the origin after 4 seconds will be

A.

56 m

B.

64 m

C.

80 m

D.

128 m

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Explanation

$S_{x} = u_{x} t + \frac{1}{2} a_{x} t^{2} \Rightarrow S_{x} = \frac{1}{2} \times 6 \times 16 = 48 m$

$S_{y} = u_{y} t + \frac{1}{2} a_{y} t^{2} \Rightarrow S_{y} = \frac{1}{2} \times 8 \times 16 = 64 m$

$S = \sqrt{S_{x}^{2} + S_{y}^{2}} = 80 m$

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A car moving with a velocity of 10 m/s can be stopped by the application of a constant force F in a distance of 20 m. If the velocity of the car is 30 m/s, it can be stopped by this force in

A.

$\frac{20}{3} m$

B.

20 m

C.

60 m

D.

180 m

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Explanation

$L e t i n i t i a l v e l o c i t y o f c a r b e u a n d i t s t o p s a f t e r c o v e r i n g d i s \tan c e S B y 3 r d e q u a t i o n o f m o t i o n v^{2} = u^{2} - 2 a S W h e r e v = f i n a l v e l o c i t y T o s t o p c a r, v = 0 P u t t i n g i n a b o v e e q u a t i o n, S = \frac{u^{2}}{2 a} w h e r e a i s r e t a r d a t i o n a n d s i s t h e s t o p p i n g d i s \tan c e$

$S \propto u^{2}$ .

For constant retardation a, if u becomes 3 times then S will become 9 times i.e. $9 \times 20 = 180 m$

where S is the stopping distance.

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The displacement of a particle is given by $y = a + b t + c t^{2} - d t^{4}$ . The initial velocity and acceleration are respectively

A.

$b, - 4 d$

B.

$- b, 2 c$

C.

$b, 2 c$

D.

$2 c, - 4 d$

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Explanation

$y = a + b t + c t^{2} - d t^{4}$

$∴ v = \frac{d y}{d t} = b + 2 c t - 4 d t^{3}$ and $a = \frac{d v}{d t} = 2 c - 12 d t^{2}$

Hence, at t = 0, v_initial = b and a_initial= 2c.

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A car moving with a speed of 40 km/h can be stopped by applying brakes for atleast 2 m. If the same car is moving with a speed of 80 km/h, what is the minimum stopping distance ?

A.

8 m

B.

2 m

C.

4 m

D.

6 m

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Explanation

$S \propto u^{2}$

$\frac{S_{1}}{S_{2}} = {(\frac{u_{1}}{u_{2}})}^{2}$ ⇒ $\frac{2}{S_{2}} = \frac{1}{4} \Rightarrow S_{2} = 8 m$

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An elevator car, whose floor to ceiling distance is equal to 2.7 m, starts ascending with constant acceleration of 1.2 ms^–2. 2 sec after the start, a bolt begins falling from the ceiling of the car. The free fall time of the bolt is

A.

$\sqrt{0 .54} s$

B.

$\sqrt{6} s$

C.

0.7 s

D.

1 s

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Explanation

Using relative motion concept, ${\vec{a}}_{b o l t, c a r} = {\vec{a}}_{b o l t} - {\vec{a}}_{c a r} = 9.8 - (- 1.2) = 11 m / s^{2} A l s o i n i t i a l v e l o c i t y o f b o l t w r t c a r u = 0$

$t = \sqrt{\frac{2 h}{(g + a)}} = \sqrt{\frac{2 \times 2 .7}{(9 .8 + 1 .2)}} = \sqrt{\frac{5 .4}{11}} = \sqrt{0 .49} = 0 .7 \sec$

As u = 0 and lift is moving upward with acceleration.

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The displacement is given by $x = 2 t^{2} + t + 5$ , the acceleration at $t = 2 s$ is

A.

$4 m / s^{2}$

B.

$8 m / s^{2}$

C.

$10 m / s^{2}$

D.

$15 m / s^{2}$

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Explanation

Displacement $x = 2 t^{2} + t + 5$

Velocity $= \frac{d x}{d t} = 4 t + 1$

Acceleration $= \frac{d^{2} x}{d t^{2}} = 4$ i.e. independent of time

Hence acceleration $= 4 m / s^{2}$

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Two trains travelling on the same track are approaching each other with equal speeds of 40 m/s. The drivers of the trains begin to decelerate simultaneously when they are just 2.0 km apart. Assuming the decelerations to be uniform and equal, the value of the deceleration to barely avoid collision should be

A.

11.8 m/s²

B.

11.0 m/s²

C.

1.6 m/s²

D.

0.8 m/s²

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Explanation

Both trains will travel a distance of 1 km before to come in rest. In this case by using $v^{2} = u^{2} + 2 a s$

$\Rightarrow 0 = {(40)}^{2} + 2 a \times 1000 \Rightarrow a = - 0 .8 m / s^{2}$

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A body moves from rest with a constant acceleration of 5 m/s². Its instantaneous speed (in m/s) at the end of 10 sec is

A.

50

B.

5

C.

2

D.

0.5

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Explanation

$v = u + a t \Rightarrow v = 0 + 5 \times 10 = 50 m / s$

Physics MCQs for NEET — Practice Questions with Answers

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