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

12th Grade > Physics

MOTION IN ONE DIMENSION MCQs

Total Questions : 30 | Page 2 of 3 pages

Question 11. 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 12. 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 13. 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 isnon uniform acceleration.

Question 14. 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}

= 0.4h
In upstream

t_{2}

\frac{2}{3 - 2}

= 2 h
Total Time = 2.4 h

Question 15. 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 16. 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

2100m/sec2 downwards
2100m/sec2 upwards
1400m/sec2
700m/sec2

Discuss Question

Answer: Option B. -> 2100m/sec2 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 17. 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?

(v1−v2):(v2−v3)=(t1−t2):(t2+t3)
(v1−v2):(v2−v3)=(t1+t2):(t2+t3)
(v1−v2):(v2−v3)=(t1−t2):(t1−t3)
(v1−v2):(v2−v3)=(t1−t2):(t2−t3)

Discuss Question

Answer: Option B. -> (v1−v2):(v2−v3)=(t1+t2):(t2+t3)
:
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 theacceleration 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 18. The relation between time and distance is

t = α x^{2} + β x

where

α

and

β

are constants. The retardation is

2αv3
2βv3
2αβv3
2β2v3

Discuss Question

Answer: Option A. -> 2αv3
:
A

\frac{d t}{d x} = 2 α x + β

\Rightarrow v = \frac{1}{2 α x + β}

∵ a = \frac{d v}{d t} = \frac{d v}{d x} . \frac{d x}{d t}

a = v \frac{d v}{d x} = \frac{- v .2 α}{(2 α x + β)^{2}} = - 2 α . v . v^{2} = - 2 α v^{3}

∴

Retardation =

2 α v^{3}

Question 19. 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 20. 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:

K1(K1−1)s(1−1/K1)
K1(K1−1)s(1−2/K1)
K1(K1−1)s(2−1/K1)
K1(K1−1)s(3−1/K1)

Discuss Question

Answer: Option B. -> K1(K1−1)s(1−2/K1)
:
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})}

← Previous
1
2
3
Next →

Latest Videos

Chapter 1 - GLOBAL STEEL SCENARIO & INDIAN STEEL INDUSTRY Part 1 : (13-04-2024) INDUSTRY AND COMPANY AWARENESS (ICA)

Direction Sense Test Part 1 Reasoning (Hindi)

Chapter 1 - RMHP / OHP / OB & BP Part 1 : (14-02-2024) GPOE

Cube & Cuboid Part 1 Reasoning (Hindi)

Data Interpretation (DI) Basic Concept Reasoning (Hindi)

Counting Figures Part 1 Counting Of Straight Lines Reasoning (Hindi)

Sail E0 free webinar Webinars

Real Numbers Part 7 Class 10 Maths

Real Numbers Part 1 Class 10 Maths

Polynomials Part 1 Class 10 Maths

Latest Test Papers

DSP Combined 1 Free Revision Test SAIL JO E-0

GPOE / GPA Combined 1 Free Revision Test SAIL JO E-0

BSP Plant Specific Review Test SAIL JO E-0

BSL Combined 1 Free Revision Test SAIL JO E-0

GFM Combined 1 Free Revision Test SAIL JO E-0