Laws of Motion Test

Result

Laws of Motion Test - 63

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

A rocket is propelled by a gas which is initially at a temperature of $$4000 \ K$$. The temperature of the gas falls to $$1000 \ K$$ as it leaves the exhaust nozzle. The gas which will acquire the largest momentum
while leaving the nozzle is

A
Hydrogen

B
Helium

C
Nitrogen

D
Argon

Solution

$$\text{Momentum}=\text{Mass}\times \text{Velocity}$$
Thus, the heavier gas will acquire the largest momentum.
Argon is the heavier gas among the given options. So, it will acquire the largest momentum.
Question 2
1 / -0

What is product of force and the time for which force act called?

A
Momentum

B
Acceleration

C
Inertia

D
Impulse of force

Solution

Impulse of force,
Impulse of force $$(I) = Force \times Time =$$ Change in momentum.
Question 3
1 / -0

What is the momentum of an object of mass m, moving with a velocity $$V$$ ?

A
$$(mV)^{2}$$

B
$$mV^{2}$$

C
$$\frac{1}{2}mV^{2}$$

D
$$mV$$

Solution

Correct Answer: D
Explanation:
Momentum defined as the quantity of motion of a body. It is a vector quantity and measured by the product of mass and velocity of the body. So,
momentum = mass ×velocity
$$=m×V=mV$$
Hence Option (D) is the right answer.
Question 4
1 / -0

Two forces are acting on an object. Which of the following statements is correct?

A
The object is in equilibrium if the forces are equal in magnitude and opposite in direction.

B
The object is in equilibrium if the net torque on the object is zero.

C
The object is in equilibrium if the forces act at the same point on the abject.

D
The object is in equilibrium if the net force and the net torque on the object are both zero.

E
The object cannot be in equilibrium because more than one force acts on it.

Solution

the object is in equilibrium if the net torque on the object is zero.
hence option B is correct
Question 5
1 / -0

An ice cube has been given a push and slides without friction on a level table. Which is correct?

A
It is in stable equilibrium.

B
It is in unstable equilibrium.

C
It is in neutral equilibrium.

D
It is not in equilibrium.

Solution

Answer (C). The ice cube is in neutral equilibrium. Its zero acceleration is evidence for equilibrium.
Question 6
1 / -0

Answer the question:
The diagram shows a uniform beam being used as a balance. The beam is pivoted at its centre.
A $$1.0\,N$$ weight is attached to one end of them. An empty pan weighing $$0.2\,N$$ is attached to the other end of beam.
How many $$0.1\,N$$ weights must be placed on the pan in order to balance the beam ?

A
$$5$$

B
$$8$$

C
$$10$$

D
$$12$$

Solution

To balance the beam, net torque about pivot should be zero.

$$\tau_{net}=\tau_p+\tau_w$$

Let we need x number of weights.

$$\tau_{net}=1\times l-(0.2+0.1x)l=0$$

$$\implies 1=0.2+0.1x$$

$$x=8$$

so we need 8 weights of 0.1 N to balance beam.
Question 7
1 / -0

A uniform ladder of length $$5 m$$ is placed against the wall as shown in the figure. If coefficient of friction $$\mu $$ is the same for both the walls, what is the minimum value of $$\mu $$ for it not to slip ?

A
$$\mu =\dfrac{1}{2}$$

B
$$\mu =\dfrac{1}{4}$$

C
$$\mu =\dfrac{1}{3}$$

D
$$\mu =\dfrac{1}{5}$$

Solution

$$\textbf{Hint}$$: Draw the FBD
$$\textbf{Step 1}$$:
Given length of ladder is 5m. It is placed against the wall. Coefficient of friction is $$\mu$$.
Let mass of the ladder is $$m$$.
From the figure: for the vertical equilibrium $$N_1+\mu N_2=mg$$ $$(1)$$
For horizontal equilibrium $$N_2=\mu N_1$$ $$(2)$$
The torque about point A is given by:
$$mg(2.5)cos \theta=N_2(4)+\mu N_2(3)$$
$$mg(2.5 \times \dfrac{3}{5})=(4+3 \mu)N_2$$, since $$cos \theta=\dfrac{3}{5}$$
$$mg(\dfrac{3}{2})=(4+3 \mu)N_2$$ $$(3)$$
\textbf{Step 2}$$:
From equation $$(1)$$ and $$(2)$$
$$N_2(\dfrac{1+\mu^2}{\mu})=mg \implies N_2= \dfrac{\mu mg}{1+\mu^2}$$
So, $$mg \dfrac{3}{2}=(4+3 \mu) \dfrac{\mu mg}{1+\mu^2}$$
$$(3+3 \mu^2)=8 \mu+6 \mu^2$$
$$3 \mu^2+8 \mu-3=0$$
$$\mu=\dfrac{-8 \pm sqrt{64+36}}{6}=\dfrac{-8 \pm 10}{6}$$
$$\mu=\dfrac{1}{3}$$
Thus option C is correct.
Question 8
1 / -0

A vehicle is moving with a speed $$v$$ on a curved smooth road of width $$b$$ and radius $$R$$. For counteracting the centrifugal force on the vehicle, the difference in elevation required in between the outer and inner edges of the road is

A
$$\dfrac{v^{2}b}{Rg}$$

B
$$\dfrac{vb}{Rg}$$

C
$$\dfrac{vb^{2}}{Rg}$$

D
$$\dfrac{vb}{R^{2}g}$$

Solution

tan $$\theta $$ $$=\dfrac{v^{2}}{Rg} \Rightarrow \dfrac{h}{b}=\dfrac{v^{2}}{Rg} \Rightarrow h=\dfrac{v^{2}b}{Rg}$$.
Question 9
1 / -0

An elastic string carrying a body of mass $$m$$ at one end extends by $$1.5\ cm$$. If the body rotates in vertical circle with critical velocity, the extension in the string at the lowest position is:

A
$$3.0\ cm$$

B
$$4.5\ cm$$

C
$$1.5\ cm$$

D
$$9.0\ cm$$

Solution

When the body is at rest, the extension is $$1.5\ cm$$ while force acting on it is $$ mg $$
Force when it reaches bottom is $$F_b = \dfrac{mv^2}{r} + mg $$
Now using energy balance between topmost point and bottom most point ,
$$\dfrac{1}{2}mv^2=\dfrac{1}{2}m(\sqrt{rg})^2+mgh$$ (h is the height difference between the top and bottom point ; $$\sqrt{gr}$$ is critical velocity at topmost point )
$$ v^2 = rg + 2gh$$
or, $$ v^2 = rg + 4gr = 5 gr$$
So, $$F_b = 6mg $$
Using this in: $$\dfrac{e_2}{e_1} = \dfrac{F_2}{F_1} $$, we get:
$$e_2 = 1.5 \times 6 = 9.0\: cm $$
Question 10
1 / -0

The stick $$OA$$ rotates about a horizontal axis through $$O$$ with a constant counter clockwise velocity $$\omega =3$$ rad/sec. As it passes through the position $$\theta =0^{0},$$ a small mass $$m$$ is placed upon it at a radial distance $$r=0.5\ m$$. If the mass is observed to be slipping when $$\theta = 37^{o}$$ , then $$\mu$$ is :

A
$$\dfrac{3}{16}$$

B
$$\dfrac{9}{16}$$

C
$$\dfrac{4}{9}$$

D
$$\dfrac{5}{9}$$

Solution

As the mass is at the verge of slipping

$$\therefore $$ mg sin 37 - $$\mu$$ mg cos 37 $$=m\omega ^{2}r$$

$$6 - 8\mu = 4.5$$

$$\therefore \mu =\dfrac{3}{16}$$