Motion Test

Result

Motion Test - 71

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Question 1
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A smooth track of incline of length $$l$$ is joined smoothly with circular track of radius $$R$$. A mass of $$m$$kg is projected up from the bottom of the inclined plane. The minimum speed of the mass to reach the top of the track is given by, $$v=$$

A
$${ \left[ 2g(l\cos { \theta } +R)(1+\cos { \theta } ) \right] }^{ 1/2 }$$

B
$${ \left( 2gl\sin { \theta } +R \right) }^{ 1/2 }$$

C
$${ \left[ 2g\{ l\sin { \theta } +R(1-\cos { \theta } )\} \right] }^{ 1/2 }$$

D
$${ \left( 2gl\cos { \theta } +R \right) }^{ 1/2 }$$

Solution

When the mass reach to the top , the final velocity $$=0$$ and height, $$H=h+h'$$
or, $$H=l\sin\theta+R(1-\cos\theta)$$
using, $$v^2-u^2=2aS$$, we get
$$0-u^2=2(-g)H \Rightarrow u^2=2g[l\sin\theta+R(1-\cos\theta)]$$
or $$u=\left[2g\{l\sin\theta+R(1-\cos\theta)\}\right]^{1/2}$$
Question 2
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In the figure, the displacement time graph of a body is shown at different time-intervals. Calculate the velocity of the body as it moves for
$$(i)\ 0 \ to\ 5 s$$,
$$(ii)\ 5 s \ to\ 7 s$$ and
$$(iii)\ 7 s\ to\ 9 s$$.

A
$$(i) \ 1.2 m s^{-1}
(ii)\ 0 m s^{-1}
(iii) \ 1 m s^{-1}$$

B
$$(i) \ 0.6 m s^{-1}
(ii) \ 0 m s^{-1}
(iii)\ 2 m s^{-1}$$

C
$$(i) \ 0 m s^{-1}
(ii) \ 0.6 m s^{-1}
(iii) \ 4 m s^{-1}$$

D
$$(i)\ 0.3 m s^{-1}
(ii) \ 1 m s^{-1}
(iii)\ 1.2 m s^{-1}$$

Solution

$$Velocity, v = \dfrac{Displacement, d}{Time, t}$$

$$(i)$$ $$v = \dfrac{3m}{5s} = 0.6 m/s $$

$$(ii)$$ $$v = \dfrac{0m}{2s} = 0 m/s $$

$$(iii)$$ $$v = \dfrac{7m-3m}{2s} = 2 m/s$$
Question 3
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Two masses of mass $$40\ gm $$ and $$30\ gm $$ are connected across a pulley with the help of light string, then the acceleration of the system is

A
$$ 4.9 m/s^{2}$$

B
$$2.94 m/s^{2}$$

C
$$1.4 m/s^{2}$$

D
$$9.8 m/s^{2}$$

Solution

The resultant forces for the given masses are:
$$0.04g-T = 0.04a$$ ...(1)
and $$ T -0.03g = 0.03a$$...(2)
Adding (1) and (2), we have
$$ \therefore 0.04g-(0.03a + 0.03g) = 0.04a$$
$$0.01g=0.07a$$ or $$a=\frac{0.098}{.07}$$ or $$ a = 1.4$$
The acceleration of the system $$ a = 1.4 m/s^2$$
Question 4
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Figure represents the displacement-time graph of motion of two cars A and B. Find the velocities of car A and car B.

A
$$v_A = 20 km h^{-1}, v_B = 10 km h^{-1}$$

B
$$v_A = 10 km h^{-1}, v_B = 20 km h^{-1}$$

C
$$v_A = 20 km h^{-1}, v_B = 40 km h^{-1}$$

D
$$v_A = 40 km h^{-1}, v_B = 20 km h^{-1}$$

Solution

For the time interval of $$4 h$$,

Displacement of car A, $$D_A = 160 km$$
Velocity of car A, $$v_A = \dfrac{160km}{4 h} = 40 km/h $$

Displacement of car B, $$D_B = 120 km-40 km = 80 km $$
Velocity of car B, $$v_B = \dfrac{80 km}{4 h} = 20 km/h $$
Question 5
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The diagram drawn above shows the distance-time graphs of four cars A,B, C and D moving on a levelled road. Choose the correct statement.

A
Car B is moving faster than car C.

B
Car C is the fastest.

C
Car A is the slowest.

D
Car D is moving slower than car C.
Question 6
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The ratio of SI units to CGS units of retardation is.

A
$$10^2$$

B
$$10$$

C
$$10^3$$

D
$$10^{-2}$$
Question 7
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Which of the following is a measurement of velocity ?

A
$$15$$ $$m$$ $$east$$

B
$$35$$ $$km/h$$ $$north$$

C
$$18$$ $$m/s$$$$^2$$

D
$$50$$ $$km/h$$

Solution

Velocity is defined as speed plus direction so unit of velocity should be unit of speed plus direction.So option B has unit of speed $$km/h$$ and direction as $$north$$.Hence it is correct option.
Question 8
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Which of the following is not an example of acceleration?

A
A person jogging at $$3\ m/s$$ along a winding path

B
A car stopping at a stop sign

C
A cheetah running $$27\ m/s$$ east

D
A plane taking off

Solution
Question 9
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Directions For Questions

Arrange the following speeds in the increasing order

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$$a$$. A boy moving on a bicycle with a speed of $$18~ km/h$$.
$$b$$. A fast runner running with a speed of $$10 ~m/s$$.
$$c$$. A scooter moving with a speed of $$800 ~m/min$$.

A
$$c > b > a$$

B
$$b > c > a$$

C
$$a> b > c$$

D
$$c > a > b$$

Solution
Question 10
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A particle experiences a net force that is always parallel to y-axis. If it moves along the curve $$xy=a^2$$ where $$a$$ is a constant and the magnitude of the force is $$y^n$$, then $$n$$ is

A
1

B
2

C
3

D
4

Solution

$$\vec{F} = y^{n} \hat{j}$$
$$\vec{S} = \dfrac{a^2}{y} \hat{i} + y \hat{j}$$

$$v^2 - u^2 = 2aS$$
$$ => v^2 - u^2 = \dfrac {2FS} {m}$$
$$ => v^2 - u^2 = \dfrac {2y^{n+1}} {m}$$

The difference of the two square terms is expressed as a constant multiplied by (n+1)th power of third term.
So (n+1) = 2
=> n = 1