﻿ Medical Entrance Exam Question and Answers | Motion in Straight Line - Zigya

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# Motion in Straight Line

#### Multiple Choice Questions

101.

A man is at a distance of  6 m from a bus. The bus begins to move with a constant acceleration of 3ms.In order to catch the bus, the minimum speed with which the man should run towards the bus is

• 2 m s-1

• 4 m s-1

• 6 m s-1

• 8 m s-1

C.

6 m s-1

If the man did not run, the bus would be at a distance s1 at time t given by

s1 = 6 + $\frac{1}{2}\mathrm{\alpha }$ t2

= 6 + $\frac{1}{2}$ × 3 × t2

s1 = 6 + $\frac{3}{2}$t2

If v is the speed of man, he would cover a distance s2 = vt in time t.

To catch the bus,

s1 = s2

6 + $\frac{3}{2}$ t2 = v t

⇒     t2 $-$ t + 4 = 0

which gives

t =

Now t will be real if   is positive or zero.

Minimum v corresponds to = 0 which gives v = 6 m s-1

102.

Two identical charged spheres suspended froma  common point by two massless strings of lengths l, are initially at a distance d (d<< l) apart because of their mutual repulsion. The charges begin to leak from both the spheres at a constant rate. As a result, the spheres approach each other with a velcoity v. Then, v varies as a function of the distance x between the sphere, as

• vx

• vx-1

• vx-1

B.

Two identical charged spheres suspended from a common point by two massless strings of length L.

In ,

The charge begins to leak from both the sphere at a constant rate. As a result, the spheres approach each other with velocity v.

Therefore, equation (i) can be rewritten as,

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103.

A body is at rest at x = 0 and t = 0, it starts moving in the positive X-direction with a constant acceleration. At the same instant, another body passes through x = O moving in the positive X-direction with a constant speed. The position of the first body is given by x1 (t) after time t and that of second body by x2 (t) after the same time interval. Which of the following graphs correctly describes (x− x2) as a function of time t ?

C.

104.

The x and y coordinates of the particle at any time are x = 5t – 2t2 and y = 10t respectively, where x and y are in meters and t in seconds. The acceleration of the particle at t = 2 s is

• 0

• 5 m/s2

• -4 m/s2

• -4 m/s2

C.

-4 m/s2

x = 5t – 2t2    y = 10t

Acceleration of particle at t = 2 s is = –4 m/s2

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105.

The displacement x of a particle varies with time t as x = ae-αt + beβt where a, b, $\mathrm{\alpha }$ and β are positive constants. The velocity of the particle will

• decrease with time

• be independent of

• drop to zero when

• increase with time

D.

increase with time

Given:-

x = a e-αt + beβt

We know that

velocity v = $\frac{\mathrm{dx}}{\mathrm{dt}}$

v  =

= -a$\mathrm{\alpha }$ e-αt + bβ eβt

v =

v = A + B

where A = - a , B = bβ e-βt

The value ot term A = -a increases and of term B = bβ e-βt increases with time. As a result velocity goes on increasing with time.

106.

A monkey of mass 30 kg climbs on a massless rope whose breaking strength is 450 N. The rope will break if the monkey (Take g = 10 m/s2 )

• climbs up with a uniform speed of 5 m/s

• climbs down with an acceleration 4 m/s2

• climbs up with an acceleration 6 m/s2

• climbs down with a uniform speed of 5 m/s2

C.

climbs up with an acceleration 6 m/s2

To move up with an acceleration 'a', the monkey will push the rope downward with a force

F = ma = 30 a

Now,  Tmax = mg + ma = 30 × 10 + 30 a

⇒        450 = 300 + 30 a

⇒            a = 5 m/s2

Hence, the rope will break if the monkey climbs up with an acceleration more than 5 m/s2.

107.

If the magnitude of sum of two vectors is equal to the magnitude of difference of the two vectors, the angle between these vectors is,

• 90o

• 45o

• 180o

• 180o

A.

90o

There are two vectors P and Q.

It is given that,

Let, angle between P and Q is ,

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108.

A body is whirled in a horizontal circle of radius 25 cm. It has an angular velocity of 13 rad/s. What is its linear velocity at any point on circular path?

• 2 m/s

• 3 m/s

• 3.25 m/s

• 4.25 m/s

C.

3.25 m/s

Given:- r = 25 cm

r = 0.25  m

Linear speed  = Radius × angular speed

V = rω

= 0.25 × 13

3.25 m/s

109.

If A + B = C and that C is perpendicular to A. What is the angle between A and B, if

• $\frac{\mathrm{\pi }}{4}$ rad

• $\frac{\mathrm{\pi }}{2}$ rad

• $\mathrm{\pi }$ rad

C.

Given:-

A + B = C

Since, B = C $-$ A

Also

B2 = 2 A2

⇒   B = $\sqrt{2}$ A

Now

A2 + B2 + 2A B cosθ = C2 = A2

∴

This gives

110.

A ball is dropped from the top of a building 100 m high. At the same instant another ball is thrown upwards with a velocity of 40 m/s from the bottom of the building. The two balls will meet after

• 3 s

• 2 s

• 2.5 s

• 5 s

C.

2.5 s

Let balls meet after t s. The distance travelled by the ball coming down is

s1$\frac{1}{2}$ gt2

Distance travelled by the other ball

s2 = 40 t $-$ $\frac{1}{2}$ gt2

s1 + s2 = 100 m

∴        $\frac{1}{2}$ gt2 + 40 t $-$ $\frac{1}{2}$gt2 = 100m

t = $\frac{100}{40}$

∴                 t = 2.5 s