Physics Test

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Physics Test - 14

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Weekly Quiz Competition

Question 1
3 / -1

The position of a particle at any time \(t\) is given by the relation \(x(t)=\frac{v}{A}\left(1-e^{-A t}\right)\) where \(v\) is the velocity. Then what will be the dimension of \(A\)?

A
\(\left[\mathrm{T}^{-1}\right]\)

B
\(\left[\mathrm{T}^{2}\right]\)

C
\(\left[\mathrm{L}^{1}\right]\)

D
\(\left[\mathrm{L}^{-2}\right]\)

Solution

Given,

\(x(t)=\frac{v}{A}\left(1-e^{-A t}\right)\)

As we know that \( (1- {e}^{\text {-At }}) \) is a constant value and will have no dimension.

Thus, the dimension of \(\frac{v}{A}\) will be equal to the dimension of \(x\).

Dimension of position, \(x=\left[M^{0} L^{1} T^{0}\right]\)

Thedimension of velocity,\(v=\left[\mathrm{M}^{0} \mathrm{~L}^{1} \mathrm{~T}^{-1}\right]\)

\(\Rightarrow x=\frac{v}{A}\)

\(\Rightarrow\left[M^{0} L^{1} T^{0}\right]=\frac{\left[M^{0} L^{1} T^{-1}\right]}{A}\)

\(\Rightarrow A=\left[T^{-1}\right]\)
Question 2
3 / -1

A piece of blue glass heated to a high temperature and a piece of red glass at room temperature, are taken inside a dimly lit room. Then _____:

A
the blue piece will look blue and red will look as usual

B
red look brighter red and blue look ordinary blue

C
blue shines like brighter red compared to the red piece

D
both the pieces will look equally red

Solution

When the blue glass is heated at high temperatures, it absorbs all the radiation of a higher wavelength except blue.

If it is taken inside a dimly-lit room, it will emit all the radiation of higher wavelength, hence it looked brighter as compared to the red piece.
Question 3
3 / -1

A sound source is moving towards a stationary observer with \(\frac{1}{10}\)of the speed of sound. The ratio of apparent to real frequency is:

A

\(\frac{10}{9}\)

B

\(\frac{11}{10}\)

C

\(\left(\frac{11}{10}\right)^{2}\)

D

\(\left(\frac{9}{10}\right)^{2}\)

Solution

Let the speed of sound in air be \({v}\)

i.e \({v}_{\text {sound }}={v}\)

Thus the velocity of source \(v_{\text {source }}=\frac{v}{10}\)

Let \(v\) and \(v^{\prime}\) be the real frequency of the source and apparent frequency heard by the observer respectively.

Using Doppler effect when the source is moving towards the stationary observer:

\(\therefore {v}^{\prime}={v}\left[\frac{{v}_{\text {sound }}}{{v}_{\text {sound }}-{v}_{\text {source }}}\right]\)

\({v}^{\prime}={v}\left[\frac{{v}}{{v}-\frac{{v}}{10}}\right]\)

\(=\frac{{1 0}}{9} {v}\)

\(=\frac{{v}^{\prime}}{{v}}\)

\(=\frac{{1 0}}{{9}}\)
Question 4
3 / -1

A solid \(X\) is in thermal equilibrium with a solid \(Y\). Solid \(Y\) has the same temperature as solid \(Z\). If the three solids have different masses and compositions, which one of the following statements is correct?

A
\(X\) and \(Z\) have the same latent heat capacity

B
\(X\) and \(Y\) have the same heat capacity

C
\(Y\) may not necessarily be in thermal equilibrium with \(Z\)

D
There will be no net transfer of energy if \(X\) is placed in thermal contact with \(Z\)

Solution

\(X\) and \(Y\) are in thermal equilibrium. So, they both have the same temperature.

Given that \(Y\) has the same temperature as \(Z\). So \(Y\) and \(Z\) are in thermal equilibrium.Therefore, \(X, Y\), and \(Z\) are in thermal equilibrium with each other.So, if \(X\) is placed in thermal contact with \(Z\), there will be no net transfer of energy.
Question 5
3 / -1

\({ }^{1} \mathrm{n}_{0}+{ }^{235} \mathrm{U}_{92} \rightarrow{ }^{140} \mathrm{Ba}_{56}+{ }^{\mathrm{y}} \mathrm{Kr}_{36}+3^{1} \mathrm{n}_{0}\). Find \(\mathrm{y}\) from the given fission equation of an uranium nucleus?

A
\(91\)

B
\(95\)

C
\(93\)

D
\(92\)

Solution

Fission equation of a uranium nucleus is given by:

\({ }^{1} \mathrm{n}_{0}+{ }^{235} \mathrm{U}_{92} \rightarrow{ }^{140} \mathrm{Ba}_{56}+{ }^{\mathrm{y}} \mathrm{Kr}_{36}+3^{1} \mathrm{n}_{0}\)

In an equation, the right-hand side of the equation is always equal to the left-hand side. The mass number and atomic number on the left side of the reaction is:

\(A=235+1=236\)

\(Z=92+0=92\)

The mass number and atomic number on the right side of the reaction is:

\(A=140+3=143\)

\(Z=54+38+0=92\)

Right now the mass number is not the same on both sides and differ by \(93\) , therefore in order to balance out equation \(93\) should come in place of \(y\).

\({ }^{1} \mathrm{n}_{0}+{ }^{235} \mathrm{U}_{92} \rightarrow{ }^{140} \mathrm{Ba}_{56}+{ }^{93} \mathrm{Kr}_{36}+3^{1} \mathrm{n}_{0}\)

The value of ' \(y\) ' from the given fission equation of a uranium nucleus is \(93\).
Question 6
3 / -1

A rectangular coil of one turn and size \(1 \mathrm{~m} \times 0.5 \mathrm{~m}\) is placed perpendicular to a magnetic field of \(1 \mathrm{~T}\). If the field drops to \(0.5 \mathrm{~T}\), find the change in magnetic flux generated in the rectangular coil.

A
\(2.5 \mathrm{~Wb}\)

B
\(0.25 \mathrm{~Wb}\)

C
\(0.5 \mathrm{~Wb}\)

D
\(0.1 \mathrm{~Wb}\)

Solution

Given:

Number of turns \((\mathrm{N})=1\) turns,

Size \((\mathrm{A})=1 \mathrm{~m} \times 0.5 \mathrm{~m}=5 \times 10^{-1} \mathrm{~m}^{2}\),

Initial magnetic field \(\left(B_{1}\right)=1 \mathrm{~T}\), and

Final magnetic field \(\left(\mathrm{B}_{2}\right)=0.5 \mathrm{~T}\)

Change in magnetic field,

\(\Delta \mathrm{B}=1-0.5=0.5 \mathrm{~T}\)

Change in Magnatic flux,

\(\Delta \phi=A \Delta B \)

\(\Rightarrow \Delta \phi=5 \times 10^{-1} \times 0.5\)

\(=2.5 \times 10^{-1}\)

\(=0.25 \mathrm{~Wb}\)
Question 7
3 / -1

Acceleration due to gravity \('\mathrm{g}'\) and the mean density of the earth \({'\mathrm{\rho}'}\) are related by which of the following relations where \(\mathrm{G}\) is the gravitational constant and \(\mathrm{R}_{\mathrm{e}}\) is the radius of the earth:

A
\(\rho=\left(\frac{g}{G}\right) \frac{4 \pi}{3} R_{e}^{3}\)

B
\(\rho=\frac{\left(\frac{\mathrm{g}}{\mathrm{G}}\right)}{\left(\frac{4 \pi}{3} \mathrm{R}_{e}\right)}\)

C
\(\rho=\frac{\mathrm{g}}{\mathrm{G}} \frac{4 \pi}{3} \mathrm{R}_{e}^{2}\)

D
\(\rho=\frac{\left(\frac{\mathrm{g}}{\mathrm{G}}\right)}{\left(\frac{4 \pi}{3} \mathrm{R}_{e}^{3}\right)}\)

Solution

The gravitational acceleration on the earth's surface is \(\mathrm{g}=\frac{\mathrm{GM}}{\mathrm{R}_{\mathrm{e}}^{2}}\)
Now, Density \(\rho=\) mass/volume \(=\frac{\mathrm{M}}{\mathrm{V}}\)
or \(\mathrm{M}=\mathrm{\rho} \mathrm{V}=\mathrm{\rho} \times(\frac{\mathrm{4}}{ \mathrm{3}}) \pi \mathrm{R}_{\mathrm{e}}^{3}\)
Thus, \(\rho=\frac{\mathrm{G}_{\mathrm{p}}(\frac{\mathrm{4} }{ \mathrm{3}}) \pi \mathrm{R}_{\mathrm{e}}^{3}}{\mathrm{R}_{\mathrm{e}}^{2}}\)
or \(\rho=\frac{(\frac{\mathrm{g} }{\mathrm{G}})}{(\frac{\mathrm{4} }{ \mathrm{3}}) \pi \mathrm{R}_{\mathrm{e}}}\)
Question 8
3 / -1

Consider a metal-semiconductor junction. The variation of the electric field inside the semiconductor is shown in the figure below. The built-in potential (in volts) is equal to:

A
0.665

B
0.875

C
1.12

D
0.934

Solution

Given:

\(A D=75 \times 10^{-6} m\)

\(A B=10^{4} v / m\)

\(D C=10^{4} v / m\)

\(D F=25 \times 10^{-6} m\)

Built in potential at the junction is equal to area under the electric field curve at the junction.

\(V_{b i} =\text { Area of } \mathrm{ABCD}+\text { Area of } \mathrm{CDF}\)

\(V_{b i} =(A D \times A B)+\left(\frac{1}{2} \times C D \times D F\right)\)

Put the given values in above formula, we get:

\(V_{b i}=\left(75 \times 10^{-6} \times 10^{4}\right)+\left(\frac{1}{2} \times 25 \times 10^{-6} \times 10^{4}\right)\)

\(=75 \times 10^{-2}+12.5 \times 10^{-2}\)

\(=87.5 \times 10^{-2}\)

\(=0.875 \mathrm{~V}\)
Question 9
3 / -1

A capacitor of \(2 \mu \mathrm{F}\) is charged as shown in the diagram. When the switch \(S\) is turned to position 2, the percentage of its stored energy dissipated is:

A
75%

B
80%

C
0%

D
20%

Solution

Initially, the energy stored in \(2 \mu \mathrm{F}\) capacitor is:
\(U_{i}=\frac{1}{2} C V^{2}=\frac{1}{2}\left(2 \times 10^{-6}\right)\)\( V^{2}=V^{2} \times 10^{-6} \mathrm{~J}\)
Initially, the charge stored in \(2 \mu F\) capacitor is
\(Q_{i}=C v=\left(2 \times 10^{-6}\right) v=2 v x\) \(10^{-6}\) coulomb
When switch \(S\) is turned to position 2 , the charge flows and both the capacitors share charges till a common potential \(V_{c}\) is reached.
\(V_{C}=\frac{\text { total charge }}{\text { total capacitance }}\)
\(=\frac{2 V \times 10^{-6}}{(2+8) \times 10^{-6}}=\frac{V}{5}\) volt
Finally, the energy stored in both the capacitors,
\(U_{f}=\frac{1}{2}\left[(2+8) \times 10^{-6}\right]\left(\frac{V}{5}\right)^{2}=\frac{V^{2}}{5} \times 10^{-6} \mathrm{~J}\)
\(\%\) loss of energy, \(\Delta U=\frac{U_{i}-U_{f}}{U_{i}} \times 100 \%\)
\(=\frac{\left(\frac{V^{2}-V^{2}}{5}\right) \times 10^{-6}}{V^{2} \times 10^{-6}} \times 100 \%\)
\(=80 \%\)
Question 10
3 / -1

If a ball was bounced from a flat floor with linear momentum p in this case what will be the linear momentum after bouncing if there is no external force acting on it?

A
0 kg m/s

B
2p kg m/s

C
p kg m/s

D
-p kg m/s

Solution

Momentum: The product of mass and velocity is called the momentum of the body.

Momentum (P) = m × V

Where,

m = mass of the body

V = velocity of the body.

From the given condition we know that a ball was bounced from a flat floor with linear momentum and there is no external force acting on it.

Thus, according to the conservation of momentum.

But since the direction of its motion will be opposite after bouncing thus it will be represented as as shown below.