Laws of Motion Test

Result

Laws of Motion Test - 57

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Question 1
1 / -0

A particle is moving in a vertical circle. The tension in the string when passing through two positions at angles $${30}^{o}$$ and $${60}^{o}$$ from vertical (lowest position) are $${T}_{1}$$ and $${T}_{2}$$ respectively, then

A
$${T}_{1}={T}_{2}$$

B
$${T}_{2}> {T}_{1}$$

C
$${T}_{1}> {T}_{2}$$

D
tension in the string always remains the same

Solution

From the Free Body Diagram we get,
$$T= \dfrac{mv^2}{r}+ mg cos \theta$$
At $$ \theta= 30 ^0 \quad T = T_1= \dfrac{mv^2}{r}+ \dfrac{\sqrt{3}mg}{2} $$
At $$\theta= 60 ^0 \quad T= T_2= \dfrac{mv^2}{r}+ \dfrac{mg}{2}$$
From Above we get
$$T_1 > T_2$$
Question 2
1 / -0

For the body moving in vertical circular motion, work done by centripetal and centrifugal forces will be:

A
0, mg(21)

B
mg(21),mg(21),

C
0,0

D
mg(21), 0

Solution
Question 3
1 / -0

Two cubes of size 1.0 m sides, one of relative density 0.60 and another of relative density=1.5 are connected by weightless wire and placed in a large tank of water.Under equilibrium the lighter cube will project above the water surface to a height of:

A
50 cm

B
25 cm

C
10 cm

D
Zero
Question 4
1 / -0

A glass marble whose mass is 20 gms falls from a height of 0.25 m and rebounds to a height of 1.6 m. If $$g = 9.8 m/s^2$$, find the impulse acting on the fall (in N - s)

A
0.2

B
0.4

C
0.252

D
0.534

Solution
Question 5
1 / -0

A particle of mass $$m$$ is moving a uniform velocity $${v}_{1}$$. It is given an impulse such that its velocity becomes $${v}_{2}$$. The impulse is equal to

A
$$m\left[ \left| { v }_{ 2 } \right| -\left| { v }_{ 1 } \right| \right] $$

B
$$1/2m[{v}_{1}^{2}-{v}_{1}^{2}]$$

C
$$m[{v}_{1}+{v}_{2}]$$

D
$$m[{v}_{2}-{v}_{1}]$$

Solution

Given that,

Mass =$$m$$

Initial velocity =$$v_ {1} $$

Final velocity=$$v_ {2} $$

We know that,

Impulse is the change in momentum.

So, the impulse is
$$I=\Delta p$$.....(I)
We know that,
$$\Delta p=m\Delta v$$
Put the value of $$\Delta p$$ in equation (I)

$$I =m\Delta v$$
$$\Delta v$$ is the change in velocity
The impulse will be

$$I=m (v_ {2}-v_ {1}) $$

The impulse will be $$I= m (v_ {2}-v_ {1}) $$

Hence, this is the required solution.
Question 6
1 / -0

A particle of mass m is made to move with uniform speed $${ v }_{ 0 }$$ along the perimeter of regular hexagon. The magnitude of impulse at each corner of the hexagon is :

A
$$2{ mv }_{ 0 }sin\dfrac { \pi }{ 6 } $$

B
$${ mv }_{ 0 }sin\dfrac { \pi }{ 6 } $$

C
$${ mv }_{ 0 }sin\dfrac { \pi }{ 3 } $$

D
$$2{ mv }_{ 0 }sin\dfrac { \pi }{ 3 } $$

Solution
Question 7
1 / -0

The speed of a motor increases from 1200 rpm to 1800 rpm in 20 s. How many revolutions does it make during these second?

A
400

B
600

C
500

D
700

Solution
Question 8
1 / -0

An object is subjected to a force in the north-east direction. To balance this force, a second force should be applied in the direction.

A
North - East

B
South

C
South - West

D
West

Solution

An object is subjected to a force in the north-east direction. To balance this force, a second force should be applied in the south west direction.
The direction of the second force should be at $$180°$$.
Question 9
1 / -0

The tension in the string revolving in a vertical circle with a mass m at the end which is at the lowest position

A
$$\dfrac{mv^2}{r}$$

B
$$\dfrac{mv^2}{r}-mg$$

C
$$\dfrac{mv^2}{r}+mg$$

D
$$mg$$

Solution

R.E.F image
$$F_{c}$$ Fore of centrifygal
$$T=F_{c}+mg$$
$$=\frac{mv^{2}}{12}+mg$$
so, option (c)
Question 10
1 / -0

When a simple pendulum is rotated in a vertical plane with constant angular velocity, centripetal force is:

A
maximum at highest point

B
maximum at lowest point

C
same at all points

D
zero

Solution