Physics Test

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Physics Test - 25

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Weekly Quiz Competition

Question 1
4 / -1

A rubber ball is dropped from a height of 5 m on a planet where the acceleration due to gravity is not known. On bouncing, it rises to 1.8 m. The ball loses its velocity on bouncing by a factor of

A
16/25

B
2/5

C
3/5

D
9/2

Solution

A ball dropped from a height \(h_{1}\) on reaching the planet's surface will have a velocity given by
\[
v_{1}=\sqrt{2 g h_{1}}
\]
Let \(v_{2}\) be the velocity with which the ball bounces. It will attain a height \(h_{2}\) given by
\[
v_{2}=\sqrt{2 g h_{2}}
\]
\(\therefore \quad \frac{v_{2}}{v_{1}}=\sqrt{\frac{h_{2}}{h_{1}}}=\sqrt{\frac{1.8}{5}}=0.6\)
or \(\quad 1-\frac{v_{2}}{v_{1}}=1-0.6\) or \(\frac{v_{1}-v_{2}}{v_{1}}=0.4=\frac{2}{5}\)
Hence the correct choice is (2) .
Question 2
4 / -1

Time period 'T' of a simple pendulum may depend on 'm', the mass of the bob, 'l', the length of the string, and 'g', the acceleration due to gravity. This means that 'T' ∝ m^al^bg^c. What are the respective values of 'a', 'b' and 'c'?

A
1/2, 0 and 1/2

B
1/2, 0 and - 1/2

C
0, 1/2 and - 1/2

D
0, 1/2 and 1/2

Solution

We use the dimension analysis to derive the formula.
As given \(T \alpha m^{x} L^{y} g^{z}\)
\(\left[T^{1}\right] \alpha\left[M^{x}\right]\left[L^{y}\right]\left[L^{z} T^{-2 z}\right]\)
\(=\left[M^{x} L^{y+z} T^{-2 z}\right]\)
equating the powers of dimensions on both sides
\(x=0, \quad y+z=0, y=-z, \quad-2 z=1 \quad z=-\frac{1}{2}\)
\(T \alpha \quad M^{0} L^{\frac{1}{2}} T^{-\frac{1}{2}}\)
Question 3
4 / -1

A small particle is released inside the depth of a liquid in the vessel. The volume of the particle is V. Density and viscosity of the liquid areρand η,respectively. Buoyant force on the particle is greater than the weight of the particle. The variation of velocity of the particle with time is

A

B

C

D

Solution

Initially the up thrust is more than its weight, so it will accelerate, i.e. its velocity increases. As the viscosity increases, the viscous force also increases. The acceleration decreases at a time when the viscous force and the weight acting downward balance the up thrust. Finally, the acceleration will become zero and it will move with a constant velocity.
Question 4
4 / -1

Which of the following quantities is unitless as well as dimensionless?

A
Angle

B
Solid angle

C
Mechanical equivalent of heat

D
Refractive index

Solution

In optics, the refractive index or index of refraction of a substance or medium is a measure of the speed of light in that medium. It is expressed as the ratio of the speed of light in vacuum relative to that in the considered medium. This can be written mathematically as:

n = speed of light in a vacuum/speed of light in a medium

Since refractive index is the ratio of same quantities, so it has no unit and is a dimensionless quantity.
Question 5
4 / -1

In the rectangular lamina ABCD shown in the figure below, a = AB = BC/2. The moment of inertia of the lamina is minimum along the axis passing through

A
BC

B
AB

C
HF

D
EG

Solution

\(I_{\mathrm{BC}}=\frac{m(A B)^{2}}{3}=\frac{m a^{2}}{3}\)
\(I_{\mathrm{AB}}=\frac{m(B C)^{2}}{3}=\frac{4}{3} m a^{2}\)
\(I_{\mathrm{HF}}=\frac{m(A B)^{2}}{12}=\frac{m a^{2}}{12}\)
\(I_{\mathrm{EG}}=\frac{m(B C)^{2}}{12}=\frac{m a^{2}}{3}\)
Thus, the moment of inertia about \(H F\) is the minimum, which is choice ( 3 ).
Question 6
4 / -1

A bomber plane is moving horizontally with a speed of 600 m/s and a bomb released from it strikes the ground in 10 sec. What is the measure of angle at which the bomb strikes the ground? (g = 10 m/s²)

A
sin ^-1(1/5)

B
tan^-1 (6)

C
tan^-1(1/5)

D
tan^-1 (1/6)

Solution

Horizontal component of velocity will remain constant. \(\mathrm{u}_{\mathrm{X}}=600 \mathrm{m} / \mathrm{s}\)
Let h be the height of the plane from the ground.
For the motion in vertical direction,
\(\mathrm{u}_{y}=0\)
then, \(h=\frac{1}{2} g t^{2}\)
\(h=\frac{1}{2} \times 10 \times(10)^{2}=500 m\)
\(\cup \sin g v_{y}^{2}-u_{y}^{2}=2 g h\)
\(v_{y}^{2}=2 \times 10 \times 500\)
\(\mathrm{v}_{\mathrm{y}}=100 \mathrm{m} / \mathrm{s}\)
If the angle made by the resultant velocity with horizontal at the time of striking the ground is \(\theta\).
\(\tan \theta=\frac{v_{y}}{v_{x}}=\frac{100}{600}\)
\(\tan \theta=\frac{1}{6}\)
Question 7
4 / -1

A spring of force constant k is cut into two equal halves. The force constant of each half is

A
k/√2

B
k/2

C
k

D
2k

Solution

For a force \(F,\) the extension \(x\) in a spring of force constant \(k\) is given by
\[
F=k x
\]
Now extension is proportional to the length of the spring. When the spring is cut into two equal halves, the extension \(x^{\prime}\) of each half, for a force \(F\) will be \(x^{\prime}=x / 2 .\) If \(k^{\prime}\) is the force constant of each half, we have
\[
F=k^{\prime} x^{\prime}=k^{\prime} \frac{x}{2}
\]
Comparing (i) and (ii) we get \(K=2 k\). Hence the correct choice is (4) .
Question 8
4 / -1

A body is projected at time t = 0 from a certain point on a planet’s surface with a certain velocity at a certain angle with the planet’s surface (assumed horizontal). The horizontal and vertical displacements x and y (in metres), respectively vary with time t (in seconds) as x = 10 √3 t and y = 10 t – t².

What is the value of acceleration due to the gravity on the planet’s surface?

A
1.0 ms^–2

B
2 ms^–2

C
4 ms^–2

D
9.8 ms^–2

Solution

We know that the position coordinates \(x\) and \(y\) are given by
\(x=\left(\nu_{0} \cos \theta\right) \cdots(i)\)
and \(y=\left(v_{0} \sin \theta\right) t-\frac{1}{2} g t^{2} \ldots \ldots(i i)\)
Comparing \(y=10 t-t^{2}\) with Eq. (ii) above, we get \(-1=-\frac{9}{2}\) or \(g=2 m s^{-2}\)
Question 9
4 / -1

A ball, which is at rest, is dropped from a height of 'h' metres. As it bounces off the floor, its speed is 80% of what it was just before touching the ground. The ball will rise to a height of

A
0.94 h

B
0.80 h

C
0.75 h

D
0.64 h

Solution

\(v=\sqrt{2 g h} \quad \ldots(\mathrm{i})\)
Let the velocity of ball after rebounce is u ball reaches the height \(h\). \(\therefore u^{2}=2 g h^{\prime}\)
\(\therefore \frac{v^{2}}{u^{2}}=\frac{2 g h}{2 g h^{*}}\)
\(\Rightarrow h^{*}=h \times \frac{u^{2}}{v^{2}}=h \times\left(\frac{80}{100}\right)^{2}=0.64 \mathrm{h}\)
Question 10
4 / -1

The track shown in the figure ends in a circular track of radius r with centre at O. A small solid sphere of mass m rolls from rest without slipping from a point A at a height h = 6r from the level ground. What is the speed of the sphere when it reaches a point B at height r above the level ground?

A
√(10gr)

B
√(50/7gr)

C
√(22/7gr)

D
Zero

Solution

Let \(v\) be the speed of the sphere when it reaches \(B\). Then, loss in \(\mathrm{PE}=\) gain in translation \(\mathrm{KE}+\) gain in rotational KE, i.e.
or
\[
\begin{aligned}
m g(6 r-r) &=\frac{1}{2} m v^{2}+\frac{1}{2} I \omega^{2} \\
5 m g r &=\frac{1}{2} m v^{2}+\frac{1}{2} \times \frac{2}{5} m r^{2} \times \frac{v^{2}}{r^{2}}
\end{aligned}
\]
\[
=\frac{1}{2} m v^{2}+\frac{1}{5} m v^{2}=\frac{7}{10} m v^{2}
\]
or \(v=\sqrt{\frac{50 g r}{7}},\) which is choice (2)