Motion In One Dimension(11th And 12th > Physics ) Questions and answers for exam Preparation

11th And 12th > Physics

MOTION IN ONE DIMENSION MCQs

Total Questions : 30 | Page 2 of 3 pages

A 150 m long train is moving with a uniform velocity of 45 km/h. The time taken by the train to cross a bridge of length 850 meters is

56 sec
68 sec
80 sec
92 sec

Discuss Question

Answer: Option C. -> 80 sec
:
C

Total distance to be covered for crossing the bridge

= length of train + length of bridge

= 150m + 850m = 1000m

Time = $\frac{D i s t a n c e}{v e l o c i t y} = \frac{1000}{45 \times \frac{5}{18}} = 80$ sec

Question 12.

A point moves with uniform acceleration and $v_{1}, v_{2}$ and $v_{3}$ denote the average velocities in the three successive intervals of time $t_{1}, t_{2}$ and $t_{3}$ . Which of the following relations is correct?

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

Discuss Question

Answer: Option B. ->

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

:
B

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

Question 13.

A stone falls from a balloon that is descending at a uniform rate of 12 m/s. The magnitude of the displacement of the stone from the point of release after 10 sec is

490 m
510 m
610 m
725 m

Discuss Question

Answer: Option C. -> 610 m
:
C

u = -12m/s, g = - 9.8 m/ $s e c^{2}$ , t = 10sec

Displacement = $u t + \frac{1}{2} g t^{2}$

= $- 12 \times 10 + \frac{1}{2} \times - 9.8 \times 100 = - 610 m$ (since downwards)

The magnitude is 610m.

Question 14.

A ball is dropped on the floor from a height of 10 m. It rebounds to a height of 2.5 m. If the ball is in contact with the floor for 0.01 sec, the average acceleration during contact is

$2100 m / s e c^{2}$ downwards
$2100 m / s e c^{2}$ upwards
$1400 m / s e c^{2}$
$700 m / s e c^{2}$

Discuss Question

Answer: Option B. ->

2100 m / s e c^{2}

upwards
:
B

Velocity at the time of striking the floor,

$u = \sqrt{2 g h} = \sqrt{2 \times 9.8 \times 10} = 14 m / s$ (-ve since downwards)

Velocity with which it rebounds.

$v = \sqrt{2 g h_{2}} = \sqrt{2 \times 9.8 \times 2.5} = 7 m / s$ (+ve since upwards)

$∴$ Change in velocity $△ v = 7 - (- 14) = 21 m / s$

$∴$ Acceleration = $\frac{△ v}{△ t} = \frac{21}{0.01} = 2100 m / s^{2}$ (upwards since positive)

Question 15.

A body A is projected upwards with a velocity of 98 m/s. The second body B is projected upwards with the same initial velocity but after 4 sec. Both the bodies will meet after

6 sec
8 sec
10 sec
12 sec

Discuss Question

Answer: Option D. -> 12 sec
:
D

Let t be the time of flight of the first body when they meet.

Then the time of flight of the second body will be (t - 4) sec.

Since the displacement of both the bodies from the ground will be the same,

$h_{1} = h_{2}$

$∴ 98 t - \frac{1}{2} g t^{2} = 98 (t - 4) - \frac{1}{2} g (t - 4)^{2}$

On solving, we get t = 12 seconds.

Question 16.

A motor car moving with a uniform speed of $20 m s^{- 1}$ comes to rest on the application of brakes after travelling a distance of 10 m. Its acceleration is

$20 m / s e c^{2}$
$- 20 m / s e c^{2}$
$- 40 m / s e c^{2}$
$+ 2 m / s e c^{2}$

Discuss Question

Answer: Option B. ->

- 20 m / s e c^{2}

:
B

Using, $v^{2} = u^{2} + 2 a S$

$\Rightarrow o = u^{2} + 2 a S$

$\Rightarrow a = - \frac{u^{2}}{2 S}$

$a = - \frac{(20)^{2}}{2 \times 10}$

$a = - 20 m s^{- 2}$

Question 17.

A body is moving in a straight line, starting its journey from origin. At any instant, its velocity is given by, $K_{1} t^{(K_{1} - 1)}$ when $K_{1}$ is a constant and t is the time. Find the acceleration of the particle, when it is at a distance s from the origin:

$K_{1} (K_{1} - 1) s^{(1 - 1 / K_{1})}$
$K_{1} (K_{1} - 1) s^{(1 - 2 / K_{1})}$
$K_{1} (K_{1} - 1) s^{(2 - 1 / K_{1})}$
$K_{1} (K_{1} - 1) s^{(3 - 1 / K_{1})}$

Discuss Question

Answer: Option B. ->

K_{1} (K_{1} - 1) s^{(1 - 2 / K_{1})}

:
B

$v = K_{1} t^{(K_{1} - 1)} = \frac{d x}{d t} \int_{o}^{s} d x = s = \int K_{1} t^{(K_{1} - 1)} . d t = t^{(K_{1})} t = (S)^{1 / K_{1}} a = \frac{d v}{d t} = K_{1} (K_{1} - 1) t^{K_{1} - 2} = K_{1} (K_{1} - 1) (S)^{(1 - 2 / K_{1})}$

Question 18.

A stone dropped from the top of the tower touches the ground in 4 sec. The height of the tower is about

80 m
40 m
20 m
160 m

Discuss Question

Answer: Option A. -> 80 m
:
A

We have,

$h = \frac{1}{2} g t^{2} = \frac{1}{2} \times 10 \times (4)^{2} = 80 m$

Question 19.

A particle is moving along the path given by y = $\frac{c}{6}$ $t^{6}$ (where c is positive constant). The relation between the acceleration (a) and the velocity (v) of the particle at t=5 sec. Is

$5 a = v$
a = $\sqrt{v}$
$a = 5 v$
$a = v$

Discuss Question

Answer: Option D. ->

a = v

:
D

Differentiate y we get $v = c (t^{5})$
Differentiate v we get $a = 5 c (t^{4})$

In expressions for v and a, substitute t = 5 s and take ratio of the two quantities.

We get, v=a

Question 20.

A particle located at x = 0, at time t = 0 starts moving along the positive x direction with a velocity v that varies as v = $α$ $\sqrt{x}$ .The displacement of the particle varies with time as