Physics Test

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Physics Test - 32

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Weekly Quiz Competition

Question 1
1 / -0

A coil having \(100\) turns and area of \(0.001 m^{2}\) is free to rotate about an axis. The coil is placed perpendicular to magnetic field of \(1.0 wb / m^{2}\). The resistance of the coil is \(10 .\) If the coil is rotated rapidly through an angle of \(180^{\circ},\) how much charge will flow through the coil?

A
\(0.02\) coulomb

B
\(0.04\) coulomb

C
\(0.08\) coulomb

D
\(0.07\) coulomb

Solution

\(\emptyset=n A B \cos \theta=n A B \cos 0^{\circ}=n A B\)

\(\left[\theta=O^{\circ}\right] ~d \emptyset=n A B-(-n A B)=2 n A B\)

Again the charge induced is:

\(\frac{d \emptyset}{R}=\frac{2 n A B}{R 2} × 100 × 0.001 × \frac{1}{10}=0.02 \) coulomb
Question 2
1 / -0

In the given circuit, the potential difference across PQ will be nearest to

A
\(9.6 \mathrm{~V}\)

B
\(6.6 \mathrm{~V}\)

C
\(4.8 \mathrm{~V}\)

D
\(3.2 \mathrm{~V}\)

Solution

The potential difference across \(PQ\)

i.e., potential difference across the resistance of \(20 \Omega\) which is \({V}={i} \times 2 {0}\)

\(i=\frac{48}{100+100+80+20}\)

\(=0.16 \mathrm{~A}\)

\(\mathrm{V}=0.16 \times 20\)

\(=3.2 \mathrm{~V}\)
Question 3
1 / -0

According to Gauss’s law, if \(E\) is _________, the charge density in the ideal conductor is zero.

A
Positive

B
Negative

C
Zero

D
Unity

Solution

The Electric Field for a uniformly distributed spherical charge is given by

Gauss' law states that the electric flux through any closed surface is equal to the total charge inside divided by \(\varepsilon _{0}\).

Charges are the source and sinks of the electric field. Since in a conducting material the electric field is everywhere zero, the divergence of \(E\) is zero, and by Gauss law, the charge density in the interior of the conductor must be zero.
Question 4
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A geyser heats water flowing at the rate of \(3.0\) litres per minute from \(27^{\circ} C\) to \(77^{\circ} C\). If the geyser operates on a gas burner, what is the rate of consumption of the fuel if its heat of combustion is \(4.0 \times 10^{4} J / g\) ?

A
\(4.0 \times 10^{4} J / g\)

B
\(1.0 \times 10^{4} J / g\)

C
\(4.0 \times 10^{-8} J / g\)

D
\(5.0 \times 10^{5} J / g\)

Solution

It is given that,

Water flows at a rate of \(3.0\) litre \(/ min =3 \times 10^{-3} m ^{3} / min\)

Density of water, \(\rho=10^{3} kg / m ^{3}\).

Clearly, mass of water flowing per minute \(=3 \times 10^{-3} \times 10^{3} kg / min =3 kg / min\)

The geyser heats the water, raising the temperature from \(27^{\circ} C\) to \(77^{\circ} C\).

Initial temperature, \(T_{1}=27^{\circ} C\)

Final temperature, \(T_{2}=77^{\circ} C\)

Thus, rise in temperature,

\(\Delta T=T_{2}-T_{1}\)

\(\Delta T=77^{\circ} C -27^{\circ} C \)

\(\Delta T=50^{\circ} C\)

Now, heat of combustion \(=4 \times 10^{4} J / g =4 \times 10^{7} J / kg\)

Specific heat of water \(=4.2 J / g^{\circ} C\)

It is known that total heat used, \(\Delta Q=m c \Delta T\)

\(\Delta Q=3 \times 4.2 \times 10^{3} \times 50 \)

\(\Delta Q=6.3 \times 10^{5} J / min\)

Now, consider \(m kg\) of fuel to be used per minute.

Thus, the heat produced \(=m \times 4 \times 10^{7} J / min\)

However, the heat energy taken by water \(=\) heat produced by fuel

Thus, equating both the sides,

\(\Rightarrow 6.3 \times 10^{5}=m \times 4 \times 10^{7} \)

\(\Rightarrow m=\frac{6.3 \times 10^{5}}{4 \times 10^{4}} \)

\(\Rightarrow m=15.75 g / min\)

Clearly, the rate of consumption of the fuel when its heat of combustion is \(4.0 \times 10^{4} J / g\) supposing the geyser operates on a gas burner is \(15.75 g / min\).
Question 5
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The displacement of a particle executing simple harmonic motion is given by \(y=A_{0}+A \sin \omega t+B \cos \omega t\). Then the amplitude of its oscillation is given by:

A
\(A_{0}+\sqrt{A^{2}}+B^{2}\)

B
\(\sqrt{A^{2}+B^{2}}\)

C
\(\sqrt{A_{0}^{2}+(A+B)^{2}}\)

D
\((A+B)\)

Solution

Given,

\({y}={A}_{0}+{A} \sin \omega {t}+{B} \sin \omega {t}\)

Equation of SHM

\(y^{\prime}=y-A_{0}=A \sin \omega t+B \cos \omega t\)

Resultant amplitude,

\({R}=\sqrt{{A}^{2}+{B}^{2}+2 {AB} \cos 90^{\circ}}\)

\(=\sqrt{{A}^{2}+{B}^{2}}\)
Question 6
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The function \(\sin ^{2}(\omega t)\) represents :

A
A simple harmonic motion with a period \(\frac{2 \pi}{ \omega}\)

B
A simple harmonic motion with a period \(\frac{ \pi}{ \omega}\)

C
A periodic, but not a simple harmonic motion with a period \(\frac{2 \pi}{ \omega}\)

D
A periodic, but not simple harmonic motion with a period \(\frac{ \pi}{ \omega}\)

Solution

As given,

\(y=\sin ^{2} \omega t\)

\(\Rightarrow y=\frac{1-\cos 2 \omega t}{2}\)

\(\Rightarrow y=\frac{1}{2}-\frac{\cos 2 \omega t}{2}\)

It is a periodic motion but it is not SHM.

\(\therefore\) Angular speed \(=2 \omega\)

\(\therefore\) Period \(T=\frac{2 \pi}{\text { angular speed }}\)

\(\Rightarrow T=\frac{2 \pi}{2 \omega}\)

\(\Rightarrow T=\frac{\pi}{\omega}\)
Question 7
1 / -0

"Heat cannot by itself flow from a body at a lower temperature to a body at a higher temperature"- the statement is which of the following?

A
First law of thermodynamics

B
Conservation of mass

C
Conservation of momentum

D
Second law of thermodynamics

Solution

In physics, the second law of thermodynamics says that heat flows naturally from an object at a higher temperature to an object at a lower temperature, and heat doesn’t flow in the opposite direction of its own.
Question 8
1 / -0

Two spherical bodies of masses \('M'\) and \('5M'\) and radii \('R'\) and \('2R'\), respectively, are released in free space with initial separation between their centres equal to \('12R'\). If they attract each other due to gravitational force only, then what is the distance covered by the smaller body just before collision?

A
\(1.5 R\)

B
\(2.5 R\)

C
\(4.5 R\)

D
\(97.5 R\)

Solution

Distance between their surfaces
\(=12 R-R-2 R=9 R\)
Since,
\(P \propto\) \(mass\)
\(a \propto\) \(mass\)
We know that,
\(Distance\) \(\propto\) \(acceleration\)
So, we can write
\(\frac{a_{1}}{a_{2}}=\frac{m}{5 m}=\frac{s_{1}}{s_{2}}\)
\(\frac{s_{1}}{s}=51\)
\(5 s~s_{1}=s_{2}\)\(\ldots \ldots(i)\)
\(s_{1}+s_{2}=9 R\)\(\ldots \ldots(ii)\)
On solving these equations;
\(s_{1}=1.5 R\)
\(s_{2}=7.5 R\)
Since smaller ball have more acceleration in same time interval, smaller ball will cover more distance.
Question 9
1 / -0

_________ developedthe corpuscular theory of light.

A
Newton

B
Huygen

C
Foucault

D
Maxwell
Solution
The corpuscular theory was largely developed by Sir Isaac Newton.

It stated that light is made up of small discrete particles called corpuscles which travel in a straight line with a finite velocity and possess impetus.

Newton’s corpuscular theory is based on the following points:
- Light consists of very thin particles called corpuscles.
- These corpuscles after emission travel in a straight line and at a very high velocity.
- When these particles enter the eyes they cause a sensation of vision.
- Corpuscles of different colors have different sizes.
Question 10
1 / -0

A circular coil of radius \(10 \mathrm{~cm}, 500\) turns and resistance \(2 \Omega\) is placed with its plane perpendicular to the horizontal component of the earth's magnetic field. It is rotated about its vertical diameter through \(180^{\circ}\) in \(0.25 \mathrm{~s}\). Estimate the magnitudes of the emf and current induced in the coil. Horizontal component of the earth's magnetic field at the place is \(3.0 \times 10^{-5} \mathrm{~T}\):

A
\(1.5 \times 10^{-3} \mathrm{~A}\)

B
\(1.9 \times 10^{-3} \mathrm{~A}\)

C
\(1.4 \times 10^{-3} \mathrm{~A}\)

D
\(1.2 \times 10^{-3} \mathrm{~A}\)

Solution

Initial flux through the coil,

\(\Phi_{\mathrm{B} \text { (initial) }}=B A \cos \theta\)

\(=3.0 \times 10^{-5} \times\left(\pi \times 10^{-2}\right) \times \cos 0^{\circ}\)

\(=3 \pi \times 10^{-7} \mathrm{~Wb}\)

Final flux after the rotation,

\(\Phi_{\mathrm{B}(\text { final) }}=3.0 \times 10^{-5} \times\left(\pi \times 10^{-2}\right) \times \cos 180^{\circ}\)

\(=-3 \pi \times 10^{-7} \mathrm{~Wb}\)

Therefore, estimated value of the induced emf is,

\(\varepsilon=N \frac{\Delta \Phi}{\Delta t}\)

\(=500 \times \frac{\left(6 \pi \times 10^{-7}\right)}{0.25}\)

\(=3.8 \times 10^{-3} \mathrm{~V}\)

\(I=\frac{\varepsilon}{R}\)

\(=1.9 \times 10^{-3} \mathrm{~A}\)

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Physics Test - 32

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Physics Test - 32

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Question 1

\(0.02\) coulomb

\(0.04\) coulomb

\(0.08\) coulomb

\(0.07\) coulomb

Solution

Question 2

\(9.6 \mathrm{~V}\)

\(6.6 \mathrm{~V}\)

\(4.8 \mathrm{~V}\)

\(3.2 \mathrm{~V}\)

Solution

Question 3

Positive

Negative

Zero

Unity

Solution

Question 4

\(4.0 \times 10^{4} J / g\)

\(1.0 \times 10^{4} J / g\)

\(4.0 \times 10^{-8} J / g\)

\(5.0 \times 10^{5} J / g\)

Solution

Question 5

\(A_{0}+\sqrt{A^{2}}+B^{2}\)

\(\sqrt{A^{2}+B^{2}}\)

\(\sqrt{A_{0}^{2}+(A+B)^{2}}\)

\((A+B)\)

Solution

Question 6

A simple harmonic motion with a period \(\frac{2 \pi}{ \omega}\)

A simple harmonic motion with a period \(\frac{ \pi}{ \omega}\)

A periodic, but not a simple harmonic motion with a period \(\frac{2 \pi}{ \omega}\)

A periodic, but not simple harmonic motion with a period \(\frac{ \pi}{ \omega}\)

Solution

Question 7

First law of thermodynamics

Conservation of mass

Conservation of momentum

Second law of thermodynamics

Solution

Question 8

\(1.5 R\)

\(2.5 R\)

\(4.5 R\)

\(97.5 R\)

Solution

Question 9

Newton

Huygen

Foucault

Maxwell

Solution

Question 10

\(1.5 \times 10^{-3} \mathrm{~A}\)

\(1.9 \times 10^{-3} \mathrm{~A}\)

\(1.4 \times 10^{-3} \mathrm{~A}\)

\(1.2 \times 10^{-3} \mathrm{~A}\)

Solution

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