Physics Test

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Physics Test - 24

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Question 1
4 / -1

The Young's modulus of the material of a wire is 2 x 10¹⁰ Nm^–2. If the elongation strain is 1%, then the energy stored in the wire per unit volume (in Jm^–3) is

A
10⁶

B
10⁸

C
2 x 10⁶

D
2 x 10⁸

Solution

The energy stored in the wire per unit volume is given by:
\(u=\frac{1}{2} Y(\text { strain })^{2}=\frac{1}{2} \times 2 \times 10^{10} \times\left(10^{-2}\right)^{2}\)
\(u=10^{6} \mathrm{J} / \mathrm{m}^{3}\)
Question 2
4 / -1

If Earth suddenly contracts to 1.5% of its present size, without any change in its mass, then how would the value of acceleration due to gravity change?

A
2%

B
3%

C
4%

D
5%

Solution

\(g=\frac{G M}{R^{2}}\)
\(\frac{\Delta \mathrm{R}}{\mathrm{R}} \times 100=-1.5 \%\)
Percentage error in g
\(\frac{\Delta \mathrm{g}}{\mathrm{g}} \times 100=-2 \times \frac{\Delta \mathrm{R}}{\mathrm{R}} \times 100 \Rightarrow \frac{\Delta \mathrm{g}}{\mathrm{g}} \times 100=(-2) \times(-1.5)=3 \%\)
Question 3
4 / -1

A bird flies for 4 s with a velocity of |t - 2| m/s in a straight line, where t = time in seconds. It covers a distance of

A
2 m

B
4 m

C
6 m

D
8 m

Solution

Plot its velocity \(v\) against time t. The area under the curve gives the distance.

Area under the vt graph is
\(A=\left(\frac{1}{2} \times 2 \times 2\right)+\left(\frac{1}{2} \times 2 \times 2\right)=4\) units
or distance travelled by the bird is \(4 \mathrm{m}\)
Question 4
4 / -1

A cord is wound around the circumference of a bicycle wheel (without tyre) of diameter 1 m. A mass of 2 kg is tied to the end of the cord and it is allowed to fall from rest. The weight falls 2 m in 4 s. The axle of the wheel is horizontal and the wheel rotates with its plane vertical. The angular acceleration produced is (Take g = 10 ms^–2)

A
0.5 rad s^–2

B
1.0 rad s^–2

C
2.0 rad s^–2

D
4.0 rad s^–2

Solution

We first find the linear acceleration \(a\) by using the relation
or
\[
\begin{array}{l}
s=u t+\frac{1}{2} a t^{2} \\
2=0+\frac{1}{2} \times a \times(4)^{2}
\end{array}
\]
which gives \(a=\frac{1}{4} \mathrm{ms}^{-2}\). Now \(R=0.5 \mathrm{m}\). The angular acceleration \(\alpha\) is
\[
\alpha=\frac{a}{R}=\frac{1}{4 \times 0.5}=\frac{1}{2}=0.5 \mathrm{rad} \mathrm{s}^{-2}
\]
Hence the correct choice is (1).
Question 5
4 / -1

Match the following:

A
A - 4, B - 3, C - 1, D - 2

B
A - 1, B - 4, C - 2, D - 3

C
A - 3, B - 1, C - 4, D - 2

D
A - 2, B - 4, C - 1, D - 3

Solution

Unit of magnetic field intensity is Am^-1.

Unit of magnetic flux is Wb (weber)

Unit of magnetic potential is Wbm^-1.

Unit of magnetic induction is Wnm^-2.
Question 6
4 / -1

A body of mass 1 kg is projected with velocity 40 m/s at an angle of projection 30° with the horizontal. The magnitude of change in momentum from t = 0 to t = 2 sec is

A
20 Ns

B
30 Ns

C
10 Ns

D
10√3 Ns

Solution

In projectile motion, the horizontal component of the velocity of the projectile does not change; the vertical component of the velocity changes.
So, the change in momentum is due to change in the velocity of the projectile in vertical direction only.
Vertical component of the initial velocity, \(u_{y}=u \sin 30^{\circ}=40 \times \frac{1}{2}=20 m / s\)
Vertical component of the velocity at \(t=2\) sec,
\(v_{y}=u_{y}-g t=20-10 \times 2=0\)
Change in linear momentum \(=\Delta p=m\left(v_{y}-u_{y}\right)=1(0-20)=-20 k g m / s\)
Magnitude of the change in linear momentum \(=20 \mathrm{kgm} / \mathrm{s}\) or \(20 \mathrm{Ns}\)
Question 7
4 / -1

A hollow metallic sphere, filled with water, is hung from a support by a long thread. A small hole is made at the bottom of the sphere and it is oscillated. How will the time period of oscillations be affected as water slowly flows out of the hole?

A
The time period will remain unchanged as water is flowing out.

B
The time period will keep increasing until the sphere is empty.

C
The time period will keep decreasing until the sphere is empty.

D
The time period will increase at first then decrease untill the sphere is empty; finally the period will be the same as that when the sphere was full of water.

Solution

The length of the pendulum (I) is the distance between the point of support and the centre of mass of the sphere. The centre of mass (CM) of a sphere is at its centre when it is empty and completely filled with water. The CM falls below the centre of the sphere as the water flows out until the mass of the water becomes same as the mass of the sphere after which CM will shift upward and the CM will return to its original value . Hence, the effective length of the pendulum (and hence, its time period) will increase then decreases and when sphere is empty then it becomes equal to the time period when the sphere was full of water. Thus, the correct choice is (4).
Question 8
4 / -1

A car accelerates from rest at a constant rate∝ for some time. Afterwards, it decelerates at a constant rateβ to come to rest. If the total time elapsed is t , the maximum velocity acquired by the car is given by

A
\(\frac{\alpha \beta}{\alpha+\beta} t\)

B
\(\frac{\alpha+\beta}{\alpha \beta} t\)

C
\(\frac{\alpha^{2}+\beta^{2}}{\beta \alpha} \mathrm{t}\)

D
\(\frac{\alpha^{2}-\beta^{2}}{\beta \alpha} t\)

Solution

From motion from \(A\) to \(B V=\alpha t_{1}\) or \(t_{1}=\frac{v}{\alpha}\)
From motion from \(\mathrm{B}\) to \(\mathrm{C} 0=\mathrm{v}-\beta \mathrm{t}_{2}\) or \(\mathrm{t}_{2}=\frac{\mathrm{v}}{\beta}\)

\(\therefore \quad t=t_{1}+t_{2}=\frac{v}{\alpha}+\frac{v}{\beta}=\frac{v(\alpha+\beta)}{\alpha \beta}\)
or \(\quad v=\frac{\alpha \beta}{(\alpha+\beta)} t\)
Question 9
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 magnitude and direction of the velocity with which the body is projected?

A
20 ms^–1 at an angle of 30° with the horizontal

B
20 ms^–1 at an angle of 60° with the horizontal

C
10 ms^–1at an angle of 30° with the horizontal

D
10 ms^–1 at an angle of 60° with the horizontal

Solution

We know that the position coordinates \(x\) and \(y\) are given by
\(x=\left(v_{0} \cos \theta\right) \ldots(i)\)
and \(y=\left(v_{0} \sin \theta\right) t-\frac{1}{2} g t^{2}\)
Comparing eq. (i) with \(x=10 \sqrt{3}\) t,
We have \(v_{0} \cos \theta=10 \sqrt{3} \mathrm{ms}^{-1}\)
Comparing eq. (ii) with \(y=10 t-t^{2}\)
We have \(v_{0} \sin \theta=10 \mathrm{ms}^{-1}\)
These equations give \(u_{0}^{2}=10^{2} \times 3=400\) Or \(v_{0}=20 \mathrm{ms}^{-1}\) and \(\tan \theta=\frac{1}{\sqrt{3}}\) 'which gives \(\theta=30^{\circ}\)
Question 10
4 / -1

A body of mass 0.4 kg is whirled in a vertical circle making 2 rev/s. If the radius of the circle is 1.2 m, then the tension in the string when the body is at the top of the circle is

A
71.78 N

B
89.86 N

C
109.86 N

D
115.86 N

Solution

Given: Mass \((\mathrm{m})=0.4 \mathrm{kg}\)
Frequency \((n)=2\) rev/s
Radius \((r)=1.2 \mathrm{m}\)
We know that linear velocity of the body \((v)=\omega r=(2 \pi n) r\) \(=2 \times 3.14 \times 2 \times 1.2=15.07 \mathrm{m} / \mathrm{s}\)
Therefore, tension in the string when the body is at the top of the circle (T)
\(=\frac{m v^{2}}{r}-m g=\frac{0.4 \times(15.07)^{2}}{1.2}-(0.4 \times 9.8)\)
\(=75.70-3.92=71.78 \mathrm{N}\)

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Physics Test - 24

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Physics Test - 24

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Question 1

106

108

2 x 106

2 x 108

Solution

Question 2

2%

3%

4%

5%

Solution

Question 3

2 m

4 m

6 m

8 m

Solution

Question 4

0.5 rad s–2

1.0 rad s–2

2.0 rad s–2

4.0 rad s–2

Solution

Question 5

A - 4, B - 3, C - 1, D - 2

A - 1, B - 4, C - 2, D - 3

A - 3, B - 1, C - 4, D - 2

A - 2, B - 4, C - 1, D - 3

Solution

Question 6

20 Ns

30 Ns

10 Ns

10√3 Ns

Solution

Question 7

The time period will remain unchanged as water is flowing out.

The time period will keep increasing until the sphere is empty.

The time period will keep decreasing until the sphere is empty.

The time period will increase at first then decrease untill the sphere is empty; finally the period will be the same as that when the sphere was full of water.

Solution

Question 8

\(\frac{\alpha \beta}{\alpha+\beta} t\)

\(\frac{\alpha+\beta}{\alpha \beta} t\)

\(\frac{\alpha^{2}+\beta^{2}}{\beta \alpha} \mathrm{t}\)

\(\frac{\alpha^{2}-\beta^{2}}{\beta \alpha} t\)

Solution

Question 9

20 ms–1 at an angle of 30° with the horizontal

20 ms–1 at an angle of 60° with the horizontal

10 ms–1 at an angle of 30° with the horizontal

10 ms–1 at an angle of 60° with the horizontal

Solution

Question 10

71.78 N

89.86 N

109.86 N

115.86 N

Solution

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10⁶

10⁸

2 x 10⁶

2 x 10⁸

0.5 rad s^–2

1.0 rad s^–2

2.0 rad s^–2

4.0 rad s^–2

20 ms^–1 at an angle of 30° with the horizontal

20 ms^–1 at an angle of 60° with the horizontal

10 ms^–1at an angle of 30° with the horizontal

10 ms^–1 at an angle of 60° with the horizontal