Physics Test

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Physics Test - 21

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Weekly Quiz Competition

Question 1
4 / -1

A stone is dropped from a certain height. After falling for 5 sec, the stone breaks through a pane of glass and instantaneously, loses half of its velocity. If the stone takes one more second to reach the ground, what is the height of the glass above the ground? (Given g = 10 m/s²)

A
60 m

B
50 m

C
30 m

D
20 m

Solution

Velocity of the stone at \(B=0+g t=8 t\)
\(=10 \times 5=50 \mathrm{ms}^{-1}\)
From \(\mathrm{B}\) to \(\mathrm{C}\) motion \(\mathrm{u}=\frac{50}{2}=25 \mathrm{m} / \mathrm{s}\)
\(a=+g=+10 m s^{-2}\)
\(t=1 s, s=y(s a y)\)
\(y=25 \times 1+\frac{1}{2} \times 10 \times 1 \times 1\)
\(=25+5=30 \mathrm{m}\)
Question 2
4 / -1

Water is falling on the blades of a turbine at the rate of 6000 kg/min. The height of the fall is 100 m. The power given to the turbine is (Take g = 10 m/s²)

A
10 kW

B
6 kW

C
100 kW

D
150 kW

Solution

Power \(=\frac{\text { energy }}{\text { time }}\)
\(=\frac{m g h}{t}=\frac{6000 \times 10 \times 100}{60} \mathrm{W}\)
Power \(=100 \mathrm{kW}\)
Question 3
4 / -1

Three particles, each of mass 'm' g, are situated at the vertices of an equilateral triangle ABC of side l cm (as shown in the figure). The moment of inertia of the system about a line AX perpendicular to AB in the plane of ABC in gram-cm² unit will be

A
(3/4) ml²

B
2 ml²

C
(5/4) ml²

D
(3/2) ml²

Solution

The moment of inertia of

system \(=m_{A} r_{A}^{2}+m_{B} r_{B}^{2}+m_{C} r_{C}^{2}\)
\(=m_{A}(0)^{2}+m(l)^{2}+m\left(l \sin 30^{\circ}\right)^{2}\)
\(=m l^{2}+m l^{2} \times(1 / 4)=(5 / 4) m l^{2}\)
Question 4
4 / -1

If two balls each of mass 0.06 kg moving in opposite directions with speed of 4 ms^-1 collide and rebound with same speed, then the impulse imparted to each ball due to other is

A
0.48 kg ms^-1

B
0.53 kg ms^-1

C
0.81 kg ms^-1

D
0.92 kg ms^-1

Solution

Impulse = (0.06 x 4 + 0.06 x 4) x 1 = 0.48 kg m/s
Question 5
4 / -1

Which of the following magnitudes is expressed most accurately?

A
1.3 x 10^-1 m

B
1.30 x 10^-2 m

C
1.3000 x 10^-4 m

D
1.3000 x 10^-6m

Solution

The result is expressed in maximum number of significant figures in the 4th option, so this option is the answer.
Question 6
4 / -1

A body constrained to move in y-direction is subjected to a force\(\mathrm{F}=2 \hat{\mathrm{i}}+15 \hat{\mathrm{j}}+6 \hat{\mathrm{k}} \mathrm{N}\). The work done by this force in moving the body through a distance of 10 m along the y-axis is

A
100 J

B
150 J

C
120 J

D
200 J

Solution

Displacement, \(\vec{s}=10 \hat{j} \mathrm{m}\) (Along the \(\mathrm{y}\) -axis only \()\)
Force, \(\vec{F}=(2 i+15 \hat{j}+6 \hat{k}) N\)
work done, \(W=\vec{F} \cdot \vec{s}=(2 \hat{i}+15 \hat{j}+6 \hat{k}) \cdot(10 \hat{j})=150 \mathrm{J}\)
Question 7
4 / -1

A 1 kg ball moving at 12 m/s collides head-on with a 2 kg ball moving in the opposite direction at 24 m/s. What is the velocity of each ball after the impact, if the coefficient of restitution is 2/3 ?

A
v₁ = -28 m/s, v₂ = -4 m/s

B
v₁ = 20 m/s, v₂= 2 m/s

C
v₁ = -30 m/s, v₂= -18 m/s

D
v₁ = -20 m/s, v₂= -4 m/s

Solution

Using conservation of linear momentum
\(\mathrm{m}_{1} \mathrm{u}_{1}+\mathrm{m}_{2} \mathrm{u}_{2}=\mathrm{m}_{1} \mathrm{v}_{1}+\mathrm{m}_{2} \mathrm{v}_{2}\)
\(1 \times 12+2 \times(-24)=1 \times v_{1}+2 \times v_{2}\)
\(\mathrm{v}_{1}+2 \mathrm{v}_{2}=-36\)
We know,
\(e=\frac{v_{2}-v_{1}}{u_{1}-u_{2}}\)
Hence,
\(\frac{2}{3}=\frac{v_{2}-v_{1}}{12-(-24)}=\frac{v_{2}-v_{1}}{36}\)
\(\therefore v_{2}-v_{1}=24\)
Solving equations (1) and \((2),\) we get \(\mathrm{v}_{1}=-28 \mathrm{m} / \mathrm{s}, \mathrm{v}_{2}=-4 \mathrm{m} / \mathrm{s}\)
Question 8
4 / -1

A small drop of water of surface tension is squeezed between two clean glass plates, so that a thin layer of thickness d and area A is formed between them. If the angle of contact is zero, then the force required to pull the plates apart is

A
\(\frac{\sigma \mathrm{A}}{\mathrm{d}}\)

B
\(\frac{2 \sigma A}{d}\)

C
\(\frac{4 \sigma A}{d}\)

D
\(\frac{8 \sigma A}{d}\)

Solution

An extremely thin layer of a liquid can be regarded as a collection of a large number of semi-spherical drops. Hence the excess pressure across a thin layer of a liquid is \(\frac{\sigma}{r}\) instead of \(\frac{2 \sigma}{r}\) as in the case of a spherical drop, where \(r=d / 2\). Therefore, excess pressure is

\[

p=\frac{\sigma}{r}=\frac{\sigma}{d / 2}=2 \frac{\sigma}{d}

\]

\(\therefore\) Force due to surface tension pushing the two plates together is

\[

\begin{aligned}

F &=\text { excess pressure } \times \text { area of layer } \\

&=\frac{2 \sigma A}{d}

\end{aligned}

\]

Hence the correct choice is ( 2 ).
Question 9
4 / -1

The Leaning Tower of Pisa does not fall because

A
it is tapered at the top.

B
it covers a large base area.

C
the vertical line passing through the centre of gravity of the tower falls within its base.

D
its centre of gravity remains at the lowest position.

Solution

The Leaning Tower of Pisa does not fall because the vertical line through the centre of gravity of the tower falls within its base.
Question 10
4 / -1

A stone weighing 0.5 Kg tied to a rope of length 0.5 m revolves along a circular path in a vertical plane. The tension of the rope at the bottom point of the circle is 45 N. To what height will the stone rise if the rope breaks the moment the velocity is directed upwards? (g = 10m/s²)

A
2 m

B
4 m

C
6 m

D
8 m

Solution

Let v be the velocity of the stone at the bottom point. \(\mathrm{T} \cdot \mathrm{mg}=\frac{m v^{2}}{r}\)
\(45-0.5 \times 10=0.5 \times v^{2} / 0.5\)
\(v^{2}=40\)
When the rope breaks, the stone rises to vertical height h until its kinetic energy is converted into PE
\(\frac{1}{2} m v^{2}=m g h\)
\(h=40 / 20=2 \mathrm{m}\)

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Physics Test - 21

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Physics Test - 21

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SHARING IS CARING

Question 1

60 m

50 m

30 m

20 m

Solution

Question 2

10 kW

6 kW

100 kW

150 kW

Solution

Question 3

(3/4) ml2

2 ml2

(5/4) ml2

(3/2) ml2

Solution

Question 4

0.48 kg ms-1

0.53 kg ms-1

0.81 kg ms-1

0.92 kg ms-1

Solution

Question 5

1.3 x 10-1 m

1.30 x 10-2 m

1.3000 x 10-4 m

1.3000 x 10-6 m

Solution

Question 6

100 J

150 J

120 J

200 J

Solution

Question 7

v1 = -28 m/s, v2 = -4 m/s

v1 = 20 m/s, v2 = 2 m/s

v1 = -30 m/s, v2 = -18 m/s

v1 = -20 m/s, v2 = -4 m/s

Solution

Question 8

\(\frac{\sigma \mathrm{A}}{\mathrm{d}}\)

\(\frac{2 \sigma A}{d}\)

\(\frac{4 \sigma A}{d}\)

\(\frac{8 \sigma A}{d}\)

Solution

Question 9

it is tapered at the top.

it covers a large base area.

the vertical line passing through the centre of gravity of the tower falls within its base.

its centre of gravity remains at the lowest position.

Solution

Question 10

2 m

4 m

6 m

8 m

Solution

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(3/4) ml²

2 ml²

(5/4) ml²

(3/2) ml²

0.48 kg ms^-1

0.53 kg ms^-1

0.81 kg ms^-1

0.92 kg ms^-1

1.3 x 10^-1 m

1.30 x 10^-2 m

1.3000 x 10^-4 m

1.3000 x 10^-6m

v₁ = -28 m/s, v₂ = -4 m/s

v₁ = 20 m/s, v₂= 2 m/s

v₁ = -30 m/s, v₂= -18 m/s

v₁ = -20 m/s, v₂= -4 m/s