Motion in a Plane Questions — NEET Physics

Q1

A bob is whirled in a horizontal circle by means of a string at an initial speed of

10 \mathrm{~rpm}

. If the tension in the string is quadrupled while keeping the radius constant, the new speed is:

A 20 rpm

B 40 rpm

C 5 rpm

D 10 rpm

Correct Answer

Option A

Solution

To solve this problem, we need to understand the relationship between the tension in the string and the speed of the bob whirling in a horizontal circle.

The centripetal force acting on the bob is provided by the tension in the string, and it can be given by the formula:

F = \frac{m v^2}{r}

where: F is the centripetal force (or tension in the string) m is the mass of the bob v is the tangential speed of the bob r is the radius of the circle According to the problem, the initial speed of the bob is

10 \mathrm{~rpm}

, and the radius is kept constant. Let’s denote the initial tension in the string as

T_1

and the new tension as

T_2

. Given that the tension in the string is quadrupled, we have:

T_2 = 4 T_1

Also, the centripetal force can be written in terms of tension:

T_1 = \frac{m v_1^2}{r}

T_2 = \frac{m v_2^2}{r}

By substituting

T_2 = 4 T_1

into the second equation, we get:

4 T_1 = \frac{m v_2^2}{r}

We know from the first equation that:

T_1 = \frac{m v_1^2}{r}

Substituting this into our equation for

T_2

:

4 \left(\frac{m v_1^2}{r}\right) = \frac{m v_2^2}{r}

The masses and radii cancel out, leaving us with:

4 v_1^2 = v_2^2

Taking the square root of both sides:

v_2 = 2 v_1

The initial speed

v_1

is given as

10 \mathrm{~rpm}

. Therefore, the new speed

v_2

is:

v_2 = 2 \times 10 \mathrm{~rpm} = 20 \mathrm{~rpm}

Hence, the new speed of the bob is 20 rpm . The correct answer is Option A.

Q2

Let

\omega_1, \omega_2

and

\omega_3

be the angular speed of the second hand, minute hand and hour hand of a smoothly running analog clock, respectively. If

x_1, x_2

and

x_3

are their respective angular distances in 1 minute then the factor which remains constant

(k)

is

A

\dfrac{\omega_1}{x_1}=\dfrac{\omega_2}{x_2}=\dfrac{\omega_3}{x_3}=k

B

\omega_1 x_1=\omega_2 x_2=\omega_3 x_3=k

C

\omega_1 x_1^2=\omega_2 x_2^2=\omega_3 x_3^2=k

D

\omega_1^2 x_1=\omega_2^2 x_2=\omega_3^2 x_3=k

Correct Answer

Option A

Solution

\begin{aligned} & \omega_1=\frac{2 \pi}{60} ; \quad x_1=\frac{2 \pi}{60} \times 60=2 \pi \\\\ & \omega_2=\frac{2 \pi}{3600} ; \quad x_2=\frac{2 \pi}{3600} \times 60=\frac{2 \pi}{60} \\\\ & \omega_3=\frac{2 \pi}{3600 \times 12} ; \quad x_3=\frac{2 \pi}{3600 \times 12} \times 60=\frac{2 \pi}{720} \\\\ & \frac{\omega_1}{x_1}=\frac{\omega_2}{x_2}=\frac{\omega_3}{x_3}=\frac{1}{60}=k \end{aligned}

Q3

A ball is projected from point A with velocity

20 \mathrm{~m} \mathrm{~s}^{-1}

at an angle

60^{\circ}

to the horizontal direction. At the highest point

\mathrm{B}

of the path (as shown in figure), the velocity

\mathrm{v} \mathrm{m} \mathrm{s}^{-1}

of the ball will be:

A 20

B

10 \sqrt{3}

C Zero

D 10

Correct Answer

Option D

Solution

At the top most point of its trajectory particle will have only horizontal component of velocity

\begin{aligned} & \mathrm{V} \text{ at top }=v \cos \theta \\ & =20 \times \frac{1}{2} \\ & =10 \mathrm{~m} / \mathrm{s} \end{aligned}

Q4

A particle is executing uniform circular motion with velocity

\vec{v}

and acceleration

\vec{a}

. Which of the following is true?

A

\vec{v}

is a constant;

\vec{a}

is not a constant

B

\vec{v}

is not a constant;

\vec{a}

is not a constant

C

\vec{v}

is a constant;

\vec{a}

is a constant

D

\vec{v}

is not a constant;

\vec{a}

is a constant

Correct Answer

Option B

Solution

Direction of velocity and centripetal acceleration changes continuously so

\vec{v}

is not constant and

\vec{a}

is not a constant.

Q5

A bullet is fired from a gun at the speed of

280 \mathrm{~ms}^{-1}

in the direction

30^{\circ}

above the horizontal. The maximum height attained by the bullet is

\left(g=9.8 \mathrm{~ms}^{-2}, \sin 30^{\circ}=0.5\right)

:-

A 2000 m

B 1000 m

C 3000 m

D 2800 m

Correct Answer

Option B

Solution

\mathrm{H}_{\max }=\frac{\mathrm{u}^{2} \sin ^{2} \theta}{2 g}

=\frac{(280)^{2}\left(\sin 30^{\circ}\right)^{2}}{2(9.8)}

=1000 \mathrm{~m}

Q6

A horizontal bridge is built across a river. A student standing on the bridge throws a small ball vertically upwards with a velocity

4 \mathrm{~m} \mathrm{~s}^{-1}

. The ball strikes the water surface after

4 \mathrm{~s}

. The height of bridge above water surface is (Take

g=10 \mathrm{~m} \mathrm{~s}^{-2}

)

A 60 m

B 64 m

C 68 m

D 56 m

Correct Answer

Option B

Solution

S=u t+\frac{1}{2} a t^{2}

-\mathrm{H}=4 \times 4-\frac{1}{2} \times 10 \times 4^{2}

-\mathrm{H}=16-80

-\mathrm{H}=-64

\mathrm{H}=64 \mathrm{~m}

Q7

A cricket ball is thrown by a player at a speed of 20 m/s in a direction 30

^\circ

above the horizontal. The maximum height attained by the ball during its motion is (g = 10 m/s 2 )

A 25 m

B 5 m

C 10 m

D 20 m

Correct Answer

Option B

Solution

Maximum height reached by the ball

H = {{{v^2}{{\sin }^2}\theta } \over {2g}}

= {{{{(20)}^2}{{\sin }^2}30^\circ } \over {2g}}

H = 5\,m

Q8

A ball is projected with a velocity, 10 ms

-

1 , at an angle of 60

^\circ

with the vertical direction. Its speed at the highest point of its trajectory will be

A Zero

B 5

\sqrt3

ms

-

1

C 5 ms

-

1

D 10 ms

-

1

Correct Answer

Option B

Solution

At highest point vertical component of velocity become zero. At highest point speed of object = 10cos30

^\circ

= 5

\sqrt3

m/s

Q9

A particle moving in a circle of radius R with a uniform speed takes a time T to complete one revolution. If this particle were projected with the same speed at an angle '

\theta

' to the horizontal, the maximum height attained by it equals 4R. The angle of projection,

\theta

, is then given by :

A

\theta = {\sin ^{ - 1}}{\left( {{{2g{T^2}} \over {{\pi ^2}R}}} \right)^{1/2}}

B

\theta = {\cos ^{ - 1}}{\left( {{{g{T^2}} \over {{\pi ^2}R}}} \right)^{1/2}}

C

\theta = {\cos ^{ - 1}}{\left( {{{{\pi ^2}R} \over {g{T^2}}}} \right)^{1/2}}

D

\theta = {\sin ^{ - 1}}{\left( {{{{\pi ^2}R} \over {g{T^2}}}} \right)^{1/2}}

Correct Answer

Option A

Solution

T = {{2\pi R} \over V}

V = {{2\pi R} \over T}

H = {{{u^2}{{\sin }^2}\theta } \over {2g}}

4R = {{4{\pi ^2}{R^2}{{\sin }^2}\theta } \over {{T^2}2g}}

{\sin ^2}\theta = {{8R{T^2}g} \over {4{\pi ^2}{R^2}}}

{\sin ^2}\theta = \sqrt {{{2{T^2}g} \over {{\pi ^2}R}}}

\theta = {\sin ^{ - 1}}{\left( {{{2{T^2}g} \over {{\pi ^2}R}}} \right)^{1/2}}

Q10

A car starts from rest and accelerates at 5m/s 2 . At t = 4 s, a ball is dropped out of a window by a person sitting in the car. What is the velocity and acceleration of the ball at t = 6 s? (Take g = 10 m/s 2 )

A 20

\sqrt 2

m/s, 10 m/s 2

B 20 m/s, 5 m/s 2

C 20 m/s, 0

D 20

\sqrt 2

m/s, 0

Correct Answer

Option A

Solution

u = 0 a = 5 t = 4 velocity of car at t = 4 sec is V = u + at V = 0 + 5 $\times$ 4 V = 20 m/s For ball : At t = 4 s, A ball is dropped out of a window so velocity of ball at this instant is 20 ms–1 along horizontal.

After 2 seconds of motion : Horizontal velocity of ball , V x = 20 m/sec V y = u + ut = 10 $\times$ 2 Vertical velocity of ball, V y = 20 m/sec So magnitude of velocity of ball V =

\sqrt {V_x^2 + V_y^2}

= 20

\sqrt 2

and Acceleration of ball at t = 6 s is g = 10 m/sec 2 As ball is under free fall.