M.C.Q (1 Marks)

MCQ 11 Mark

The force exerted by the floor of an elevator on the foot of a person standing there is more than the weight of the person if the elevator is:

A
Going up and slowing down.
✓
Going up and speeding up.
C
Going down and slowing down.
D
Going down and speeding up.

Answer

Correct option: B.

Going up and speeding up.

It means normal force exerted by the floor of the elevator on the person is greater that the weight of the person.
i.e. $N > mg$

Going up and speeding up:

$a_{eff} = g + a$
$N = ma_{eff} = mg + ma (N > mg)$

Going down and speeding up:

$a_{eff}= g - a$
$N = mg - ma (N < mg)$

Going down and slowing down:

$a_{eff} = g - (-a) = g + a$
$N = mg + ma (N > mg)$

Going up and slowing down:

$a_{eff} = g - a$
$N = mg - ma (N < mg)$

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MCQ 21 Mark

A block of mass $10\ kg$ is suspended through two light spring balances as shown in figure:

✓
Both the scales will read $10\ kg.$
B
Both the scales will read $5\ kg.$
C
The upper scale will read $10\ kg$ and the lower zero.
D
The readings may be anything but their sum will he $10\ kg.$

Answer

Correct option: A.

Both the scales will read $10\ kg.$

From the free$-$body diagram,
$\mathrm{K}_1 \mathrm{x}_1=\mathrm{mg}=10 \times 9.8=98 \mathrm{~N}$
$\mathrm{~K}_2 \mathrm{x}_2=\mathrm{K}_1 \mathrm{x}_1$
So, $\mathrm{K}_1 \mathrm{x}_1=\mathrm{K}_2 \mathrm{x}_2=98 \mathrm{~N}$

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MCQ 31 Mark

Three rigid rods are joined to form an equilateral triangle $\text{ABC}$ of side $1m.$ Three particles carrying charges $20\mu\text{C}$ each are attached to the vertices of the triangle. The whole system is at rest in an inertial frame. The resultant force on the charged particle at $A$ has the magnitude:

✓
Zero
B
$3.6N$
C
$3.6\sqrt{3}\text{N}$
D
$7.2N.$

Answer

Correct option: A.

Zero

Using, $F_{\text {net }}=\mathrm{ma}$,
$a=0 \Rightarrow F_{\text {net }}=0$
As the whole system is at rest, the resultant force on the charged particle at $A$ is zero.

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MCQ 41 Mark

A force $F_1$ acts on a particle so as to accelerate it from rest to a velocity $v$. The force $F_1$ is then replaced by $F_2$ which decelerates it to rest:

A
$F_1$ must be equal to $F_2$.
✓
$F_1$ may be equal to $F_2$.
C
$F_1$ must be unequal to $F_2$.
D
None of these.

Answer

Correct option: B.

$F_1$ may be equal to $F_2$.

Any force applied in the direction opposite the motion of the particle decelerates it to rest.

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MCQ 51 Mark

A free ${}^{238}U$ nucleus kept in a train emits an alpha particle. When the train is stationary, a nucleus decays and a passenger measures that the separation between the alpha particle and the recoiling nucleus becomes $x$ at time $t$ after the decay. If the decay takes place while the train is moving at a uniform velocity $v,$ the distance between the alpha particle and the recoiling nucleus at a time $t$ after the decay as measured by the passenger is:

A
$x + vt$
B
$x - vt$
✓
$x$
D
Depends on the direction of the train.

Answer

Correct option: C.

$x$

The moving train does not put any extra force on the alpha particle and the recoiling nucleus.
So, the distance between the alpha particle and the recoiling nucleus at a time tafter the decay, as measured by the passenger, will be same as before, i.e. $x.$

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MCQ 61 Mark

A block of mass m is placed on a smooth inclined plane of inclination 0 with the horizontal. The force exerted by the plane on the block has a magnitude:

A
$\text{mg}$
B
$\frac{\text{mg}}{\cos\theta}$
✓
$\text{mg}\cos\theta$
D
$\text{mg}\tan\theta$

Answer

Correct option: C.

$\text{mg}\cos\theta$

From the free-body diagram,
$\text{N = mg}\cos\theta$
Normal force exerted by the plane on the block is $\text{mg}\cos\theta.$

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MCQ 71 Mark

Neglect the effect of rotation of the earth. Suppose the earth suddenly stops attracting objects placed near its surface. A person standing on the surface of the earth will:

A
Fly up.
B
Slip along the surface.
C
Fly along a tangent to the earth's surface.
✓
Remain standing.

Answer

Correct option: D.

Remain standing.

If the earth suddenly stops attracting objects placed near its surface, the net force on the person will become zero and according to the first law of motion, the person will remain standing.

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MCQ 81 Mark

A smooth wedge A is fitted in a chamber hanging from a fixed ceiling near the earth's surface. A block B placed at the top of the wedge takes a time T to slide down the length of the wedge. If the block is placed at the top of the wedge and the cable supporting the chamber is broken at the same instant, the block will:

A
Take a time longer than T to slide down the wedge.
B
Take a time shorter than T to slide down the wedge.
✓
Remain at the top of the wedge.
D
Jump off the wedge.

Answer

Correct option: C.

Remain at the top of the wedge.

Downward gravitational force will be balanced by the upward pseudo force (because of the motion of the wedge in downward direction). The block will remain at its position, as both the box and the inclined plane are falling with the same acceleration (g).

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MCQ 91 Mark

In an imaginary atmosphere, the air exerts a small force $F$ on any particle in the direction of the particle's motion. A particle of mass $m$ projected upward takes a time $t_1$ in reaching the maximum height and $t_2$ in the return journey to the original point. Then:

A
$\mathrm{t}_1<\mathrm{t}_2$
✓
$t_1>t_2$
C
$t_1=t_2$
D
The relation between $\mathrm{t}_1$ and $\mathrm{t}_2$ depends on the mass of the particle.

Answer

Correct option: B.

$t_1>t_2$

Let acceleration due to air resistance force be a.
Let $H$ be maximum height attained by the particle.
Direction of air resistance force is in the direction of motion.
In the upward direction of motion, $\text{a}_{\text{eff}}=|\text{g}-\text{a}|.$
$\text{t}_{1}=\sqrt{\frac{2\text{H}}{|\text{g}-\text{a}|}} \ ...(1)$
In the downward direction of motion, $\text{a}_{\text{eff}}=\text{g}+\text{a}.$
$\text{t}_{2}=\sqrt{\frac{2\text{H}}{\text{g}+\text{a}}} \ ...(2)$
So, $t_1>t_2$

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MCQ 101 Mark

A person standing on the floor of an elevator drops a coin. The coin reaches the floor of the elevator in a time $\mathrm{t}_1$ if the elevator is stationary and in time $\mathrm{t}_2$ if it is moving uniformly. Then:

✓
$t_1=t_2$
B
$t_1<t_2$
C
$t_1>t_2$
D
$t_1t_2$ depending on whether the lift is going up or down.

Answer

Correct option: A.

$t_1=t_2$

After the coin is dropped, the only force acting on it is gravity, which is same for both the cases.
So $t_1=t_2$.

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MCQ 111 Mark

A person says that he measured the acceleration of a particle to be non-zero while no force was acting on the particle:

A
He is a liar.
B
His clock might have run slow.
C
His meter scale might have been longer than the standard.
✓
He might have used non-inertial frame.

Answer

Correct option: D.

He might have used non-inertial frame.

If no force is acting on a particle and yet, its acceleration is non-zero, it means the observer is in a non-inertial frame.

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MCQ 121 Mark

A reference frame attached to the earth:

Is an inertial frame by definition.
Cannot be an inertial frame because the earth is revolving around the sun.
Is an inertial frame because Newton's laws are applicable in this frame.
Cannot be an inertial frame because the earth is rotating about its axis.

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

Answer

Correct option: B.

$B$ and $D$

A reference frame attached to the earth cannot be an inertial frame because the earth is revolving around the sun and also rotating about its axis.

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MCQ 131 Mark

A particle is observed from two frames $S_1$ and. $S_2$. The frame $S_2$ moves with respect to $S_1$ with an acceleration $a$. Let $F_1$ and $F_2$ be the pseudo forces on the particle when seen from $S_1$ and $S_2$ respectively. Which of the following are not possible?

A
$F_1=0, F_2 \neq 0$
B
$F_1 \neq 0, F_2=0$
C
$F_1 \neq 0, F_2 \neq 0$
✓
$F_1=0, F_2=0$

Answer

Correct option: D.

$F_1=0, F_2=0$

$\text{a}_{\text{s}_1\text{s}_2}=\text{a} \ ...(1) $
Acceleration of the particle w.r.t. to $\text{S}_1=\frac{\text{F}_1}{\text{m}}$
Acceleration of the particle w.r.t. to $\text{S}_2=\frac{\text{F}_2}{\text{m}}$
If we assume $F_1 = 0$ and $F_2 = 0$,
we can conclude that $\text{a}_{\text{s}_2\text{s}_1}=0 \ ...(2)$
From equations $(1)$ and $(2)$, we can say that our assumption is wrong.
And $F_1 = 0, F_2 = 0$ is not possible.

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MCQ 141 Mark

Figure shows a heavy block kept on a frictionless surface and being pulled by two ropes of equal mass $m.$ At $t = 0$, the force on the left rope is withdrawn but the
force on the right end continues to act. Let $F_1$ and $F_2$ be the magnitudes of the forces by the right rope and the left rope on the block respectively:

✓
$F_1=F_2=F$ for $t<0$.
B
$F_1=F_2=F+m g$ for $t<0$.
C
$F_1=F, F_2=F$ for $t>0$.
D
$F_10$.

Answer

Correct option: A.

$F_1=F_2=F$ for $t<0$.

At $t < 0$, the block is in equilibrium in the horizontal direction.
So, $F_1 = F_2 = F$
At $t > 0, F_2= 0$ and $F_1= F.$

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MCQ 151 Mark

A particle stays at rest as seen in a frame. We can conclude that:

The frame is inertial.
Resultant force on the particle is zero.
The frame may be inertial but the resultant force on the particle is zero.
The frame may be non$-$inertial but there is a non$-$zero resultant force.

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

Answer

Correct option: D.

$C$ and $D$

According to Newton's second law which says that net force acting on the particle is equal to rate of change of momentum $($or mathematically $F = ma),$ so if a particle is at rest then $\text{F}_{\text{net}}=\text{ma = m}\frac{\text{dv}}{\text{dt}}=\text{m}\frac{\text{d}(0)}{\text{dt}}=\text{m}\times0=0.$
Now, if the frame is inertial, then the resultant force on the particle is zero.
If the frame is non$-$inertial, vector sum of all the forces plus a pseudo force is zero.
i.e $\text{F}_{\text{net}}\neq0.$

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MCQ 161 Mark

A car accelerates on a horizontal road due to the force exerted by:

A
The engine of the car.
B
The driver of the car.
C
The earth.
✓
The road.

Answer

Correct option: D.

The road.

The car pushes the ground in the backward direction and according to the third law of motion, reaction force of the ground in the forward direction acts on the car.

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MCQ 171 Mark

When a horse pulls a cart, the force that helps the horse to move forward is the force exerted by:

A
The cart on the horse.
✓
The ground on the horse.
C
The ground on the cart.
D
The horse on the ground.

Answer

Correct option: B.

The ground on the horse.

The horse pushes the ground in the backward direction and, in turn, the ground pushes the horse in the forward direction, according to Newton's third law of motion.

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MCQ 181 Mark

If the tension in the cable supporting an elevator is equal to the weight of the elevator, the elevator may be:

Going up with increasing speed.
Going down with increasing speed.
Going up with uniform speed.
Going down with uniform speed.

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

Answer

Correct option: D.

$C$ and $D$

Tension in the cable $=$ Weight of the elevator
Or, total upward force $=$ total downward force
That is, there's no acceleration or uniform velocity.
So, the elevator is going up/ down with uniform speed.

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MCQ 191 Mark

A block of mass m is placed on a smooth wedge of inclination 0. The whole system is accelerated horizontally so that the block does not slip on the wedge. The force exerted by the wedge on the block has a magnitude:

A
$\text{mg}$
✓
$\frac{\text{mg}}{\cos\theta}$
C
$\text{mg}\cos\theta$
D
$\text{mg}\tan\theta$

Answer

Correct option: B.

$\frac{\text{mg}}{\cos\theta}$

Free-body Diagram of the Small Block of Mass 'm'
The block is at equilibrium w.r.t. to wedge. Therefore,
$\text{mg}\sin\theta=\text{ma}\cos\theta$
$\Rightarrow\text{a}=\text{g}\tan\theta$
Normal reaction on the block is
$\text{N = mg}\cos\theta+\text{ma}\sin\theta$
Putting the value of a, we get:
$\text{N = mg}\cos\theta+\text{mg}\tan\theta\sin\theta$
$\text{N = mg}\cos\theta+\text{mg}\frac{\sin\theta}{\cos\theta}\sin\theta$
$\text{N}=\frac{\text{mg}}{\cos\theta}$

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MCQ 201 Mark

Two objects A and B are thrown upward simultaneously with the same speed. The mass of A is greater than the mass of B. Suppose the air exerts a constant and equal force of resistance on the two bodies:

A
The two bodies will reach the same height.
✓
A will go higher than B.
C
B will go higher than A.
D
Any of the above three may happen depending on the speed with which the objects are thrown.

Answer

Correct option: B.

A will go higher than B.

Let the air exert a constant resistance force = F (in downward direction).
Acceleration of particle A in downward direction due to air resistance, $\text{a}_{\text{A}}=\frac{\text{F}}{\text{m}_{\text{A}}}.$
Acceleration of particle B in downward direction due to air resistance, $\text{a}_{\text{B}}=\frac{\text{F}}{\text{m}_{\text{B}}}.$
$\text{m}_{\text{A}}>\text{m}_{\text{B}}$
$\text{a}_{\text{A}}<\text{a}_{\text{B}}$
$\text{S = us}+\frac{1}2{}\text{at}^2$
So, $\text{H}_{\text{A}}=\text{ut}-\frac{1}{2}\big(\text{a}_{\text{A}}+\text{g}\big)\text{t}^2$
$\text{H}_{\text{B}}=\text{ut}-\frac{1}{2}\big(\text{a}_{\text{B}}+\text{g}\big)\text{t}^2$
$\text{H}_{\text{A}}>\text{H}_{\text{B}}$
Therefore, A will go higher than B.

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MCQ 211 Mark

A particle is found to be at rest when seen from a frame $S_1$ and moving with constant velocity when seen from another frame $S_2$. Mark out the possible options:

A
Both the frames are inertial.
B
Both the frames are non-inertial.
C
$S_1$ is inertial and $S_2$ is non-inertial.
✓
Both A and B.

Answer

Correct option: D.

Both A and B.

d. Both A and B.
Explanation:
$S_1$ is moving with constant velocity w.r.t frame $S_2$. So, if $S_1$ is inertial, then $S_2$ will be inertial and if $S_1$ is non-inertial, then $S_2$ will be non-inertial.

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MCQ 221 Mark

Figure shows the displacement of a particle going along the $X-$axis as a function of time. The force acting on the particle is zero in the region:

$AB$
$BC$
$CD$
$DE$

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

Answer

Correct option: C.

$A$ and $C$

Slope of the $x-t$ graph gives velocity. In the regions $AB$ and $CD$, slope or velocity is constant, i.e. acceleration is zero. Hence, from the second law, force is zero in these regions.

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MCQ 231 Mark

A body of weight $w_1$ is suspended from the ceiling of a room through a chain of weight $w_2$. The ceiling pulls the chain by a force:

A
$\mathrm{w}_1$
B
$w_2$
✓
$w_1+w_2$
D
$\frac{\text{w}_1+\text{w}_2}{2}$

Answer

Correct option: C.

$w_1+w_2$

From the free$-$body diagram,
$(w_1 + w_2) - N = 0$
$N = w_1 + w_2$
The ceiling pulls the chain by a force $(w_1 + w_2)$.

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Physics M.C.Q (1 Marks)

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