Physics Test

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Physics Test - 21

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Question 1
1 / -0

In S = a + bt + ct². S is measured in metres and t in seconds. The unit of c is

A
No dimensions

B
m

C
ms−1

D
ms−2

Solution

ct2 must have dimensions of L ⇒ c must have dimensions of L/T² i.e. ,LT^-2
Question 2
1 / -0

For a body falling freely under gravity from a height H, its Potential energy will be maximum at:

A
Starting position as it starts falling

B
Lowest position as it is about to hit the ground

C
Between the start and end position of its projection

D
Data is insufficient
Solution
For a body falling freely under gravity from a height H, its Potential energy will be maximum at starting position as it starts falling.

Potential energy is defined as the energy stored due to change in position relative to others, stresses within itself, or many factors.
- The potential energy $(U)=m {~g} h$ [where $m=$ mass of body, $g=$ acceleration due to gravity, $h=$ distance from the ground].
- If the height of a body increases from the ground its energy also increases and vice versa.
Under free fall where gravity is the sole influence on the body, the total energy remains the same. Potential energy gets converted into kinetic energy.

Under free fall, the potential energy keeps on decreasing and kinetic energy keeps on increasing. This is also in confirmation with the law of conservation of energy.

From the above-given explanation, we can see that at the highest position, the kinetic energy of the particle is zero and the potential energy is maximum. Whereas at the lowest position, potential energy becomes zero and kinetic energy is maximum as the ball is about to hit the ground.
Question 3
1 / -0

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

A
Is towards east if the vehicle is accelerating

B
Is zero if the vehicle is moving with a uniform velocity

C
Both (a) and (b)

D
Must be towards east

Solution

Option (a) says friction force will be in east if car is accelerating.

Well car is going in east direction and if it is accelerating then the net force has to be in east direction. Well there can be no other force than frictional force in horizontal direction. so option (a) is correct.

Option (b) says force is 0 if constant velocity which means 0 acceleration

$\Rightarrow$ 0 net external force so no friction possible as again friction can be the only external force on the body along horizontal line.

Option (d) Must be towards east

Must

Pretty strong word

Well if the acceleration was in east direction then the option would have been correct but since nothing mentioned about it so we can't talk about Net force or its direction

Option (d) same as above both C & D are wrong
Question 4
1 / -0

The time period of revolution of geostationary satellite with respect to earth is

A
24 hrs

B
1 year

C
Infinity

D
Zero

Solution

From a non moving frame of reference, the time period of a geostationary satellite is 24 hours. But since, it is the same time for the Earth to complete one rotation, when seen from the earth the geostationary satellites look fixed at a point.

In that way, the time period of a geostationary satellite with respect to earth is infinity.
Question 5
1 / -0

A ball is projected upwards from the top of tower with a velocity 50 ms⁻¹ making angle 30^∘ with the horizontal. The height of the tower is 70 m. After how many seconds from the instant of throwing will the ball reach the ground.

A
2.33 sec

B
5.33 sec

C
7.11 sec

D
6.33 sec

Solution

Formula for calculation of time to reach the body on the ground from the tower of height 'h' (lf it is thrown vertically up with velocity u) is given by $t=\frac{u}{g}[1+\sqrt{1+\frac{2 g h}{u^{2}}}]$
So we can resolve the given velocity in vertical direction and can apply the above formula.
Initial vertical component of velocity $u \sin \theta=50 \sin 30=25 \mathrm{m} / \mathrm{s}$
$\therefore t=\frac{25}{9.8}[1+\sqrt{\left.1+\frac{2 \times 9.8 \times 70}{(25)^{2}}\right]}=7.11 \mathrm{sec}$
Question 6
1 / -0

A wagon weighing 1000 kg is moving with a velocity 50 km/h on smooth horizontal rails. A mass of 250 kg is dropped into it. The velocity with which it moves now is

A
12.5 km/hour

B
20 km/hour

C
40 km/hour

D
50 km/hour

Solution

Initially the wagon of mass $1000 \mathrm{kg}$ is moving with velocity of $50 \mathrm{km} / \mathrm{h}$ So its momentum $=1000 \times 50 \frac{\mathrm{kg} \times \mathrm{km}}{\mathrm{h}}$
When a mass $250 \mathrm{kg}$ is dropped into it. New mass of the system $=$
$1000+250=1250 \mathrm{kg}$
Let v is the velocity of the system.
By the conservation of linear momentum : Initial momentum = Final
momentum
$1000 \times 50=1250 \times v$
$v=\frac{50,000}{1250}=\frac{40 k m}{h}$
Question 7
1 / -0

The bob of a simple pendulum is imparted a velocity 5ms⁻¹when it is at its mean position. To what maximum vertical height will it rise on reaching to its extreme position if 60% of its energy is lost in overcoming friction of air?

(Take g=10ms⁻²)

A
5m

B
0.5m

C
0.25m

D
1m

Solution

When it is at mean position it will have a $K . E$ of $0.5 m v^{2}$
$K . E=0.5 \times m \times 5 \times 5=12.5 \times m J$
$60 \%$ of this $K . E$ is lost hence remaining is equal to $(40 / 100) 12.5 \times m=5 m$ $P . E=m g h=5 \times m$
\[
h=5 / g=5 / 10=0.5 m
\]
Hence it goes to a height of $0.5 \mathrm{m}$
Question 8
1 / -0

Two identical spherical masses are kept at some distance as shown. Potential energy when a mass m is taken from the surface of one sphere to the other

A
increase

B
decrease

C
increase first and then decrease

D
remain constant

Solution
Question 9
1 / -0

Two vectors of the same physical quantity are equal if

A
they have the same magnitude and the same direction.

B
they have different magnitudes but the same direction.

C
they have the same magnitude but different directions.

D
they have different magnitudes and different directions.

Solution

Two vectors are said to be equal if they have the same magnitude and direction.

In the figure below, three equal vectors have been represented.
Question 10
1 / -0

One end of a massless spring of constant 100 N/m and natural length 0.5 m is fixed and the other end is connected to a particle of mass 0.5 kg lying on a frictionless horizontal table. The spring remains horizontal. If the mass is made to rotate at angular velocity of 2 rad/s, then elongation of spring is

A
0.1 m

B
10 cm

C
1 cm

D
0.01 cm

Solution

Centrifugal force is balanced by spring force, thus,
\[
\begin{array}{l}
k x=m \omega^{2}(l+x) \\
\Rightarrow x=\frac{m \omega^{2} l}{k-m \omega^{2}}=0.01 m=1 \mathrm{cm}
\end{array}
\]