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 1 of 3 pages

Question 1.

The velocity of a body depends on time according to equation V=20 + 0.1 $t^{2}$ . the body is undergoing

Uniform Acceleration
Uniform Retardation
Non-Uniform acceleration
Zero Acceleration

Discuss Question

Answer: Option C. -> Non-Uniform acceleration
:
C

a = $\frac{d v}{d t}$ = 0.2 t,

We observe that the acceleration is dependent on time and hence it is non uniform acceleration.

Question 2.

The position of a particle travelling along x - axis is given by $x_{t}$ = $t^{3}$ - 9 $t^{2}$ + 6t where $x_{t}$ is in m and t is in second . Then

1) the particle does not come to rest at all.

2) the particle comes to rest firstly at (3 - $\sqrt{7}$ ) s and then at (3 + $\sqrt{7}$ ) s

3) the speed of the particle at t= 2s is 18 m $s^{- 1}$

4) the acceleration of the particle at t= 2s is 6 m $s^{- 2}$

2,3, 4 are correct
2,3 are correct
All of them are correct
1 and 2 are correct

Discuss Question

Answer: Option A. -> 2,3, 4 are correct
:
A

$x_{t}$ = $t^{3}$ - 9 $t^{2}$ + 6t

v= $\frac{d x_{t}}{d t}$ = (3 $t^{2}$ - 18t + 6) m $s^{- 1}$

So, the particle comes to rest since this equation has two valid roots.

$\Rightarrow$ $t_{1}$ = (3 - $\sqrt{7}$ ) s and $\Rightarrow$ $t_{2}$ = (3 - $\sqrt{7}$ ) s

Substituting the value t = 2s in the equation for velocity, we get, v= = - 18 m $s^{- 1}$

The acceleration of the particle is a=6t - 18 m $s^{- 2}$

Substituting the value t = 2s in the equation for acceleration, we get, a= = - 6 m $s^{- 2}$

Question 3.

A particle moves according to the equation $\frac{d v}{d t}$ = $α$ - $β$ v , where $α$ and $β$ are constants. Find the velocity as a funtion of time. Assume body starts from rest.

v = ( $\frac{β}{α}$ ) (1 - $e^{- β t}$ )
v = ( $\frac{β}{α}$ ) ( $e^{- β t}$ )
v = ( $\frac{α}{β}$ ) (1 - $e^{- β t}$ )
v = ( $\frac{α}{β}$ ) ( $e^{- β t}$ )

Discuss Question

Answer: Option C. -> v = (

\frac{α}{β}

) (1 -

e^{- β t}

)
:
C

Take the relation given for acceleration and integrate with the given limits to obtain the desired function for velocity.

$\frac{d v}{d t}$ = $α$ - $β$ v $\Rightarrow$ $v \int 1$ $\frac{d v}{α - β v}$ = $t \int 0$ dt

v = ( $\frac{α}{β}$ ) (1 - $e^{- β t}$ )

Question 4.

A steamer moves with a velocity 3 kmph in and against the direction of river water whose velocity is 2 kmph. Total time for total journey if the boat travels 2 km in direction of stream and then back to its place will be (in hour)

Discuss Question

Answer: Option A. -> 2.4
:
A

In direction of stream , $t_{1}$ = $\frac{2}{3 + 2}$ = 0.4 h

In upstream $t_{2}$ = $\frac{2}{3 - 2}$ = 2 h

Total Time = 2.4 h

Question 5.

A police van moving on a highway with a speed of 30 km/h fires a bullet at a thief's car speeding away in the same direction with a speed of 192 km/h. If the nuzzle speed of the bullet is 150 m/s, with what speed does the bullet hit the thief's car?

105 m $s^{- 1}$
110 m $s^{- 1}$
115 m $s^{- 1}$
120 m $s^{- 1}$

Discuss Question

Answer: Option A. -> 105 m

s^{- 1}

:
A

The speed of the bullet with respect to ground will be,

$v_{b g}$ = $v_{b v}$ + $v_{v g}$

=150 + $\frac{25}{3}$ = $\frac{475}{3} m s^{- 1}$

Speed of thief's car is,

$v_{t g}$ =192 $\times$ $\frac{5}{18}$ = $\frac{160}{3} m s^{- 1}$

$v_{b t}$ = $v_{b g}$ - $v_{t g}$

$\frac{475}{3}$ - $\frac{160}{3}$ = 105 $m s^{- 1}$

Question 6.

A Body moves 6 m north. 8 m east and 10m vertically upwards, what is the magnitude of its resultant displacement from initial position?

$10 \sqrt{2} m$
$10 m$
$\frac{10}{\sqrt{2}} m$
$10 \times 2 m$

Discuss Question

Answer: Option A. ->

10 \sqrt{2} m

:
A

\to r = x ˆ i + y ˆ j + z ˆ k ∴ r = \sqrt{x^{2} + y^{2} + z^{2}}

r = \sqrt{6^{2} + 8^{2} + 10^{2}} = 10 \sqrt{2} m

Question 7.

A frictionless wire AB is fixed on a circular frame of radius R. A very small spherical ball slips on this wire. The time taken by this ball to slip from A to B is

$\frac{2 \sqrt{g R}}{g c o s θ}$
$\frac{2 \sqrt{g R} c o s θ}{g}$
$2 \sqrt{\frac{R}{g}}$
$\frac{g R}{\sqrt{g c o s θ}}$

Discuss Question

Answer: Option C. ->

2 \sqrt{\frac{R}{g}}

:
C

$∠ A B C = 90^{0} Acceleration along AB is a = g c o s θ Distance travelled is A B = 2 R C o s θ We have 2 R C o s θ = \frac{1}{2} \times g c o s θ t^{2} \Rightarrow t = 2 \sqrt{\frac{R}{g}}$

Question 8.

A body is released from the top of a tower of height h. It takes t sec to reach the ground. Where will be the ball after time $\frac{t}{2}$ s.

At $\frac{h}{2}$ from the ground
At $\frac{h}{4}$ from the ground
Depends upon mass and volume of the body
At $\frac{3 h}{4}$ from the ground

Discuss Question

Answer: Option D. -> At

\frac{3 h}{4}

from the ground
:
D

Let the body after time $\frac{t}{2}$ fall a distance x from the top. Since it starts from rest, we have

$x = \frac{1}{2} g \frac{t^{2}}{4} = \frac{g t^{2}}{8}$ .............(i)

It reaches the ground in t seconds, we have

$h = \frac{1}{2} g t^{2}$ ..............(ii)

Eliminate t from (i) and (ii), we get x = $\frac{h}{4}$

$∴$ Height of the body from the ground

= $h - \frac{h}{4} = \frac{3 h}{4}$

Question 9.

An object is projected upwards with a velocity of 100 m/s.The time after which it will strike the ground is (g = $10 m s^{- 2})$

10 sec
20 sec
15 sec
5 sec

Discuss Question

Answer: Option B. -> 20 sec
:
B

Time of flight = $\frac{2 u}{g} = \frac{2 \times 100}{10} = 20 s e c$

Question 10.

A body is released from a great height and falls freely towards the earth. Another body is released from the same height exactly one second later. The separation between the two bodies, two seconds after the release of the second body is