M.C.Q (1 Marks) - Physics STD 11 Science Questions - Vidyadip

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M.C.Q (1 Marks)

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15 questions · auto-graded multiple-choice test.

MCQ 11 Mark

While walking on ice, one should take small steps to avoid slipping. This is because smaller steps ensure:

A
Larger friction.
✓
Smaller friction.
C
Larger normal force.
D
Smaller normal force.

Answer

Correct option: B.

Smaller friction.

According to the first law of the limiting friction,
$\text{f}=\mu\text{N}$
where f is the frictional force
$\mu$ is the coefficient of friction
N is the normal reaction force
When we take smaller steps on ice, the normal reaction force exerted by the ice is small. Therefore, the smaller steps ensure smaller friction.

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MCQ 21 Mark

Consider a vehicle going on a horizontal road towards east. Neglect any force by the air. The frictional forces on the vehicle by the road:

Is towards east if the vehicle is accelerating.
Is zero if the vehicle is moving with a uniform velocity.
Must be towards east.
Must be towards west.

✓
$A$ and $B$
B
$B$ and $C$
C
$A$ and $C$
D
$C$ and $D$

Answer

Correct option: A.

$A$ and $B$

When the vehicle is accelerating, the force is applied $($by the tyre on the road$)$ in west direction. That causes a net resultant frictional force acting in east direction. Due to this force of friction only, the car is moving in east direction.
When the vehicle is moving with a uniform velocity, the force of friction on the wheels of the vehicle by the road is zero.

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MCQ 31 Mark

Let $F, F_N$ and f denote the magnitudes of the contact force, normal force and the friction exerted by one surface on the other kept in contact. If none of these is zero:

$F > F_N$
$F > f$
$F_N> f$
$F_N - f < F < F_N+ f$

A
Only $A$
B
$A,B$ and $C$
C
$B,C$ and $D$
✓
$A, B$ and $D$

Answer

Correct option: D.

$A, B$ and $D$

The system is in equilibrium condition when $F = f.$
Hence, the net horizontal force is zero.
$\text{f}=\mu\text{F}_{\text{N}}$
$\text{F}>\text{F}_{\text{N}}$
$\text{f}=\text{F}_{\text{N}}$ and $0\leq\mu\leq1$
Therefore, we can say that $F > f.$
So the net horizontal force is nonzero.
$F > f,$ and so the net horizontal force is zero.
$\text{F}_{\text{N}}>\text{f}\Rightarrow\text{F}_{\text{N}}>\mu\text{F}_{\text{N}}\Rightarrow\mu<1$
Here, the given relation between $F$ and $f$ i.e.
$\text{F}>\text{f}$ and $\text{f}=\mu\text{F}_{\text{N}}$ will not be satisfied So it cannot be said that the net horizontal force is zero or nonzero.
$\text{F}_{\text{N}}-\text{f}<\text{F}<\text{F}_{\text{N}}+\text{f}$
$\because\text{f}=\mu\text{F}_{\text{N}}$
$\frac{\text{f}}{\mu}-\text{}f<\text{F}<\frac{\text{f}}{\mu}+\text{f}$
$\text{f}\Big(\frac{1-\mu}{\mu}\Big)<\text{F}<\text{f}\Big(\frac{1+\mu}{\mu}\Big)$
For the above relation, we can say that $F \neq f$ and so the net horizontal force is nonzero.

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MCQ 41 Mark

A block $A$ kept on an inclined surface just begins to slide if the inclination is $30^\circ$ . The block is replaced by another block $B$ and it is found that it just begins to slide if the inclination is $40^\circ :$

A
Mass of $A >$ mass of $B.$
B
Mass of $A <$ mass of $B.$
C
Mass of $A =$ mass of $B.$
✓
All the three are possible.

Answer

Correct option: D.

All the three are possible.

We know that
$\text{N = mg}\cos\theta^{\circ}$
$\text{f}_{\text{max}}=\mu\text{N}=\mu\text{mg}\cos\theta$
where $N =$ normal reaction force
$f_{max} =$ frictional force
$\theta =$ angle of inclination
$\mu =$ coefficient of friction
When the block just begins to slide, it means
$\text{mg}\sin\theta=\text{f}_{\text{max}}$
$\text{mg}\sin\theta=\mu\text{mg}\cos\theta$
$\mu=\tan\theta$
and the coefficient of friction depends on the angle of inclination $(\theta)$ and does not depend on mass.
Now consider the block sliding condition:
$\text{mg}\sin\theta-\text{f}_{\max}=\text{ma}$
$\text{mg}\sin\theta-\mu\text{mg}\cos\theta=\text{ma}$
$\therefore\text{a = g}(\sin\theta-\mu\cos\theta)$
From the above equation it is clear that acceleration does not depend on the mass but depends on $\theta$ and $\mu.$

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MCQ 51 Mark

Mark the correct statements about the friction between two bodies:

A
Limiting friction is never less than static friction.
B
Coefficient of static friction is always greater than the coefficient of kinetic friction.
C
Limiting friction is always greater than the kinetic friction.
✓
All of the above

Answer

Correct option: D.

All of the above

All the above statements are correct. The static friction is sometimes less than the kinetic friction.

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MCQ 61 Mark

A boy of mass M is applying a horizontal force to slide a box of mass M' on a rough horizontal surface. The coefficient of friction between the shoes of the box and the floor is $\mu.$ In which of the following cases it is certainly not possible to slide the box?

✓
$\mu<\mu',\text{M < M}'$
B
$\mu>\mu',\text{M < M}'$
C
$\mu<\mu',\text{M > M}'$
D
$\mu>\mu',\text{M > M}'$

Answer

Correct option: A.

$\mu<\mu',\text{M < M}'$

Let T be the force applied by the boy on the block.

Free body diagram for the box:

The condition for preventing the slide is
$\text{f}_{\text{max}}>\text{T}$
$\mu'\text{M}'\text{g}>\text{T} \ ...(\text{i})$
Now see the free body diagram of a boy of mass M:

$\text{f}_{\text{max}}=\mu\text{mg}$
The condition for preventing the slide is
$\text{f}_{\text{max}}>\text{T}$
$\mu\text{mg}>\text{T}$
The condition for sliding the entire system (block and boy) is f' > f (block is not slide)
$\mu'\text{M}'\text{g}>\mu\text{mg}$
$\mu'\text{M}'>\mu\text{m}$
$\mu<\mu'$
$\text{m < M}'$

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MCQ 71 Mark

A body of mass $M$ is kept on a rough horizontal surface $($friction coefficient = $\mu)$. A person is trying to pull the body by applying a horizontal force but the body is not moving. The force by the surface on $A$ is $F,$ where:

A
$\text{F = Mg}$
B
$\text{F}=\mu\text{Mg}$
✓
$\text{Mg}\leq\text{F}\leq\text{Mg}\sqrt{1+\mu^2}$
D
$\text{Mg}\geq\text{F}\geq\text{Mg}\sqrt{1-\mu^2}.$

Answer

Correct option: C.

$\text{Mg}\leq\text{F}\leq\text{Mg}\sqrt{1+\mu^2}$

Let $T$ be the force applied on an object of mass $M.$
If $T = 0, \text{F}_{\text{min}} = Mg.$
If $T$ is acting in the horizontal direction, then the body is not moving.
$\therefore\text{T}=\mu(\text{mg})$
$\text{F}_{\text{max}}=\sqrt{(\text{Mg})^2+(\text{T})^2}$
$=\sqrt{(\text{Mg})^2+(\mu\text{Mg})^2}$
Thus, we have:
$\text{Mg}\leq\text{F}\leq\text{Mg}\sqrt{1+(\mu)^2}$

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MCQ 81 Mark

In a situation the contact force by a rough horizontal surface on a body placed on it has constant magnitude. If the angle between this force and the vertical is decreased, the frictional force between the surface and the body will:

A
Increase.
✓
Decrease.
C
Remain the saint.
D
May increase or decrease.

Answer

Correct option: B.

Decrease.

According to the first law of limiting friction,
$\text{f}=\mu\text{N}$
where f is the frictional force
N is the normal reaction force
$\mu$ is the coefficient of static friction
and
$\text{N = mg}-\text{F}\cos\theta$
where m is the mass of the body
F is the contact force acting on the body
If we decrease the angle between this contact force and the vertical, then $\text{F}\cos\theta$ increases and the normal reaction force (N) as well as the frictional force (f) decrease.

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MCQ 91 Mark

Two cars of unequal masses use similar tyres. If they are moving at the same initial speed, the minimum stopping distance:

A
Is smaller for the heavier car.
B
Is smaller for the lighter car.
✓
Is same for both cars.
D
Depends on the volume of the car.

Answer

Correct option: C.

Is same for both cars.

Given: both the cars have same initial speed.
Let the masses of the two cars be $\text{m}_1$ and $\text{m}_2$.
Frictional force on car with mass $\text{m}_1=\mu\text{m}_1\text{g}$
So, the deceleration due to frictional force $=\frac{\mu\text{m}_1\text{g}}{\text{m}_1}=\mu\text{g}$
Frictional force on car with mass $\text{m}_2=\mu\text{m}_2\text{g}$
So, the deceleration due to frictional force $=\frac{\mu\text{m}_2\text{g}}{\text{m}_2}=\mu\text{g}$
As both the acceleration are same, from the second equation of motion $\text{s = ut}+\frac{1}{2}\text{at}^2$
Thus, we can say that both the cars have same minimum stopping distance.

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MCQ 101 Mark

Suppose all the surfaces in the previous problem are rough. The direction of friction on $B$ due to $A:$

✓
Is upward.
B
Is downward.
C
Is zero.
D
Depends on the masses of $A$ and $B.$

Answer

Correct option: A.

Is upward.

The normal reaction force on the system $($comprising of wall and contact surface of $A$ and $B)$ is provided by $F$. As can be seen from the figure, the weight of $A$ and $B$ is in the downward direction. Therefore, the frictional force $f_A$ and $f_{B A} ($friction on $B$ due to $A)$ is in upward direction.

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MCQ 111 Mark

A block is placed on a rough floor and a horizontal force $F$ is applied on it. The force of friction $f$ by the floor on the block is measured for different values of $F$ and a graph is plotted between them.

The graph is a straight line of slope $45^\circ$ .
The graph is a straight line parallel to the $F-$axis.
The graph is a straight line of slope $45^\circ$ for small $F$ and a straight line parallel to the $F-$axis for large $F.$
There is a small kink on the graph.

A
$A$ and $B$
B
$B$ and $C$
✓
$C$ and $D$
D
All of the above

Answer

Correct option: C.

$C$ and $D$

When force $F$ is applied on the block, the force of friction $f$ comes into play. As we increase the applied F, the static friction force adjusts itself to become $($equal$)$ to the applied force $F$ and goes upto its maximum value equal to limiting friction force. After this, it is treated as a constant force $($i.e. now its value does not change until and unless the body starts moving$)$. If the applied force $F$ is greater than the limiting friction force, then the kinetic friction force comes into play at that time. The kinetic friction force is always less than the limiting friction force.

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MCQ 121 Mark

The contact force exerted by a body $A$ on another body $B$ is equal to the normal force between the bodies. We conclude that:

The surfaces must be frictionless.
The force of friction between the bodies is zero.
The magnitude of normal force equals that of friction.
The bodies may be rough but they don't slip on each other.

A
$A$ and $B$
B
$A$ and $C$
C
$B$ and $C$
✓
$B$ and $D$

Answer

Correct option: D.

$B$ and $D$

The contact force exerted by a body $A$ on another body $B$ is equal to the normal force between the bodies. Therefore, we can conclude that the force of friction between the bodies is zero or the bodies may be rough but they don't slip on each other.

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MCQ 131 Mark

A scooter starting from rest moves with a constant acceleration for a time $\triangle\text{t}_1,$ then with a constant velocity for the next $\triangle\text{t}_2,$ and finally with a constant deceleration for the next $\triangle\text{t}_3$ to come to rest. A $500N$ man sitting on the scooter behind the driver manages to stay at rest with respect to the scooter without touching any other part. The force exerted by the seat on the man is:

A
$500N$ throughout the journey.
B
Less than $500N$ throughout the journey.
C
More than $500N$ throughout the journey.
✓
$> 500N$ for time $\triangle\text{t}_1$ and $\triangle\text{t}_3$ and $500N$ for $\triangle\text{t}_2.$

Answer

Correct option: D.

$> 500N$ for time $\triangle\text{t}_1$ and $\triangle\text{t}_3$ and $500N$ for $\triangle\text{t}_2.$

During the time interval $\triangle\text{t}_2,$ the scooter is moving with a constant velocity, which implies that the force exerted by the seat on the man is $500N ($for balancing the weight of the man$).$
During the time interval $\triangle\text{t}_1$ and $\triangle\text{t}_3,$ the scooter is moving with constant acceleration and deceleration, which implies that a frictional force is also applied. Therefore, the net force exerted by the seat on the man should be $> 500N.$

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MCQ 141 Mark

In order to stop a car in shortest distance on a horizontal road, one should:

A
Apply the brakes very hard so that the wheels stop rotating.
✓
Apply the brakes hard enough to just prevent slipping.
C
Pump the brakes (press and release).
D
Shut the engine off and not apply brakes.

Answer

Correct option: B.

Apply the brakes hard enough to just prevent slipping.

When we apply hard brakes just enough to prevent slipping on wheels, it provides optimum normal reaction force, which gives the maximum friction force between tyres of the car and the road.

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MCQ 151 Mark

Consider the situation shown in figure. The wall smooth but the surface of A and B in contact are rough. The friction on B due to A in equilibrium:

A
Is upward.
B
Is downward.
C
Is zero.
✓
The system cannot remain in equilibrium.

Answer

Correct option: D.

The system cannot remain in equilibrium.

Since the wall is smooth and the surface of A and B in contact are rough, the net vertical force on the system is in the downward direction. Hence, the system cannot remain in equilibrium.

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