Motion Test

Result

Motion Test - 74

Score
-
out of -
Rank
-
out of -

TIME Taken - -

SHARING IS CARING

If our Website helped you a little, then kindly spread our voice using Social Networks. Spread our word to your readers, friends, teachers, students & all those close ones who deserve to know what you know now.

Weekly Quiz Competition

Question 1
1 / -0

For the graph given below the ratio of average speed to the average velocity in $$( ms^{-1})$$ of the particle is.

A
$$\dfrac{5}{6}$$

B
$$\dfrac{5}{8}$$

C
3

D
4
Question 2
1 / -0

Water drops fall at regular intervals from a roof at of $$125\ m$$ from the ground. At an instant when $$11th$$ drop is about leave the roof and $$1st$$ drop is at a height of $$45\ m$$ from the ground at that instant the dist of $$10th$$ drop from the roof is $$(g=10\ m/s^{2})$$

A
$$0.40\ m$$

B
$$0.20\ m$$

C
$$0.80\ m$$

D
$$1.2\ m$$

Solution
Question 3
1 / -0

The splash is heard $$2.05 \ s$$ after the stone is dropped into a well of depth $$19.6 \ m$$. The velocity of sound is?

A
$$342$$ $$ms^{-1}$$

B
$$372$$ $$ms^{-1}$$

C
$$392$$ $$ms^{-1}$$

D
$$352$$ $$ms^{-1}$$

Solution

If we use $$g=9.8m/{ s }^{ 2 }$$ for the acceleration of gravity, and assume that there are no frictional forces to slow the stone down as it falls, then using the second equation of motion $$h=\frac { 1 }{ 2 } { gt }^{ 2 }$$ gives $$t=2 \ s$$ as the time at which the stone hits the water (we’re also ignoring precisely what point the stone is at $$19.6 \ m $$ above the water).

Assuming that the sound waves are emitted precisely at the moment of contact between the stone and the water, and that there is no physiological time lag in “hearing” the sound (e.g., time for auditory nerve conduction), then this means that the sound waves travel $$19.6 \ m$$ in $$t = (2.05 - 2) \ s$$, so that their average speed is $$\dfrac { 19.6 }{ 0.05 } =392 \ m /s $$
Question 4
1 / -0

A particle starts from the origin at $$t=0s$$ with a velocity of $$10.0j\ m/s$$ and moves in the $$xy-$$plane with a constant acceleration of the $$(8\hat {i}+2j)ms^{-2}$$. Then $$y-$$coordinate of the particle in $$2\ sec$$ is

A
$$24\ m$$

B
$$16\ m$$

C
$$8\ m$$

D
$$12\ m$$

Solution
Question 5
1 / -0

A balloon is rising vertically up at constant speed $$10\ m/s$$. A stone is dropped from it when the balloon is at a height of $$40\ m$$. Total distance covered by the stone before reaching the ground is $$(take\ g=10\ m/s^{2})$$

A
$$40\ m$$

B
$$45\ m$$

C
$$50\ m$$

D
$$60\ m$$

Solution

undefined
Question 6
1 / -0

Two cars X and Y accelerate at the rate of 2 $$m/{s^2}$$ respectively from rest.The ratio of time taken by the cars X and Y is 4:5.In that given ratio of time interval if the distance travelled by car x is 100 km then the distance travelled by car y is:

A
$$\frac{{1875}}{8}\;km$$

B
$$\frac{{1875}}{2}\;km$$

C
$$\frac{{1875}}{4}\;km$$

D
$$\frac{{375}}{4}\;km$$

Solution
Question 7
1 / -0

A particle experienced a constant acceleration for $$6$$ seconds after starting from rest. If it travels a distance $$ { d }_{ 1 }$$ in the first two seconds, $$ { d }_{ 2 }$$ in the next $$2$$ seconds, and a distance $$ { d }_{ 3 }$$ in the last $$2$$ seconds, then:

A
$$ { d }_{ 1 }:{ d }_{ 2 }:{ d }_{ 3 }=1:1:2$$

B
$$ { d }_{ 1 }:{ d }_{ 2 }:{ d }_{ 3 }=1:2:3$$

C
$$ { d }_{ 1 }:{ d }_{ 2 }:{ d }_{ 3 }=1:3:5$$

D
$$ { d }_{ 1 }:{ d }_{ 2 }:{ d }_{ 3 }=1:5:9$$

Solution
Question 8
1 / -0

A particle is moving in positive x-direction with initial velocity of $$10\ m/s$$ and uniform retardation such that it reaches the initial position after $$10\ s$$. The distance traversed by the particle in $$6\ s$$ is

A
$$24 m$$

B
$$25 m$$

C
$$26 m$$

D
$$27 m$$

Solution
Question 9
1 / -0

A water drop falls at regular intervals from a tap $$9$$ m above the ground. The fourth drop is leaving the tap at the instant, the first drop touches the ground. How high is the third drop at that instant?

A
$$2$$ m

B
$$4$$ m

C
$$6$$ m

D
$$8$$ m

Solution
Question 10
1 / -0

A particle is moving on a circular path of radius R with constant speed $$v_0$$. What is the change in magnitude of velocity when it completes $$\frac{{{3^{th}}}}{4}$$ of the complete circular path ?

A
$$\sqrt 2 {v_0}$$

B
$${v_0}$$

C
Zero

D
$$2 {v_0}$$

Solution

The magnitude of velocity is the speed.
As the particle is moving in the circular path with the constant speed. Thus the magnitude of velocity will not change whatsoever. Only the direction will be different at different positions on the track.