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Physics Test - 42

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Weekly Quiz Competition

Question 1
1 / -0

A hydrogen atom is in the excited state of the principal quantum number n. It emits a photon of wavelength' \(\lambda^{\prime}\) when returns to ground state. The value of \(n\) is:

A
\(\sqrt{\lambda R(\lambda R-1)}\)

B
\(\sqrt{\frac{\lambda R-1}{\lambda \cdot R}}\)

C
\(\sqrt{\frac{\lambda R}{\lambda R-1}}\)

D
\(\sqrt{\lambda(\mathrm{R}-1)}\)

Solution

Energy of photon, \(E=\) Rhc \(\left(1-\frac{1}{n^{2}}\right)\)

\(\Rightarrow \frac{\mathrm{hc}}{\lambda}=\mathrm{Rhc}\left(1-\frac{1}{\mathrm{n}^{2}}\right)\)

\(\Rightarrow \frac{1}{\lambda}=\mathrm{R}\left(1-\frac{1}{\mathrm{n}^{2}}\right)\)

\(\Rightarrow \frac{1}{\lambda R}=1-\frac{1}{n^{2}}\)

\(\Rightarrow \frac{1}{\mathrm{n}^{2}}=1-\frac{1}{\lambda R}=\frac{\lambda R-1}{\lambda R}\)

\(\Rightarrow \mathrm{n}=\sqrt{\frac{\lambda \mathrm{R}}{(\lambda R-1)}}\)
Question 2
1 / -0

The increase in the width of the depletion region in a \(\mathrm{p-n}\) junction diode is due to:

A
Forward bias only

B
Reverse bias only

C
Both forward bias and reverse bias

D
Increase in forward current

Solution

The increase in the width of the depletion region in a \(\mathrm{p-n}\) junction diode is due to reverse bias only.

The increase in the width of the depletion region is due to the absence of the electrons and holes in the region. This occurs only in the case of the reverse bias only in a diode.
Question 3
1 / -0

The earth’s magnetic field at the equator is approximately \(0.4 \mathrm{G}\). Estimate the earth’s dipole moment.

A
\(3 \times 10^{23} \mathrm{Am}^{2}\)

B
\(1 \times 10^{23} \mathrm{Am}^{2}\)

C
\(1.05 \times 10^{23} \mathrm{Am}^{2}\)

D
\(2.05 \times 10^{23} \mathrm{Am}^{2}\)

Solution

Given,

\(B_{E} = 0.4 \mathrm{G}=4 \times 10^{-5} \mathrm{~T}\)

The radius ofthe earth\(r=6.4 \times 10^{6} \mathrm{~m}\)

The equatorial magnetic field is,

\(B_{E}=\frac{\mu_{0} m}{4 \pi r^{3}}\)

\(m=\frac{4 \times 10^{-5} \times\left(6.4 \times 10^{6}\right)^{3}}{\frac{\mu_{0}} { 4 \pi}}\)

As we know,

\(\frac{\mu_{0}} {4 \pi}=10^{-7}\)

\(=\frac{4 \times 10^{-5} \times\left(6.4 \times 10^{6}\right)^{3}}{10^{-7}}\)

\(=4 \times 10^{2} \times\left(6.4 \times 10^{6}\right)^{3}\)

\(=1.05 \times 10^{23} \mathrm{Am}^{2}\)
Question 4
1 / -0

A telescope has an objective lens of focal length 150 cm and an eyepiece of focal length 5 cm. If a 50 m tall tower at a distance of 1 km is observed through this telescope in normal setting, the angle formed by the image of the tower is θ , then θ is close to :

A
30°

B
15°

C
60°

D
1°

Solution

Magnifying power of telescope,

\(M P=\frac{\beta(\text { angle subtended by image at eye piece })}{\alpha(\text { angle subtended by object on objective })}\)

Also, \(\mathrm{MP}=\frac{\mathrm{f}_{\mathrm{o}}}{\mathrm{f}_{\mathrm{e}}}=\frac{150}{5}=30\)

\(\alpha=\frac{50}{1000}=\frac{1}{20} \mathrm{rad} \)

\(\therefore \beta=\theta=\mathrm{MP} \times \alpha=30 \times \frac{1}{20}=\frac{3}{2}=1.5 \mathrm{rad}\)

or, \(\beta=1.5 \times \frac{180^{\circ}}{\pi} \simeq 60^{\circ}\)
Question 5
1 / -0

What is the value of instantaneous displacement current in the free space between parallel plate capacitor of capacitance \(1 \mu F\) and the rate of changing potential difference maintain between plates is \(10^{6} {~V} / {s} ?\)

A
1 A

B
5 A

C
3 A

D
None of the above

Solution

Given:

Capacitance, (C) \(=1 \mu {F}\),

Rate of change in voltage, \(\frac{d V}{d t}=10^{6} {~V} / {s}\)

The expression for displacement current \({i}_{{d},}\) in the case of capacitance is given as:

\( i_{d}=\frac{d q}{d t}\)\(\quad\).....(i)

As we also know that, the charge on the capacitor is:

\( {q}={CV}\)

Where q = charge on the capacitors

On substituting the value of q = CV in equation (i), we get,

\( i_{d}=C \frac{d V}{d t}\)\(\quad\).....(ii)

On substituting the given values in equation (ii), we get,

\( i_{d}=\left(10^{-6} \times 10^{6}\right) A\)

\( {i}_{{d}}=1 {~A}\)
Question 6
1 / -0

The kinetic energy of the fastest moving photo electron from a metal of work function 2.8 eV is 2 eV. If the frequency of light is doubled, then find the maximum kinetic energy of photo electron.

A
6.8eV

B
68eV

C
-6.8eV

D
-68eV

Solution

Given:

\(\phi=2.8 \mathrm{eV}, \mathrm{E}=2 \mathrm{eV}\)

We know that Maximum kinetic energy \((\mathrm{E})=\mathrm{h} \nu-\phi\)

Put the given values in above formula.

\(2=\mathrm{h} \nu-2.8\)

\(\Rightarrow \mathrm{h} \nu=4.8 \mathrm{eV}\)

New frequency \( v^{\prime}=2 \nu\)

So, \(E^{\prime}=h \nu^{\prime}-\phi\)

\(\Rightarrow \mathrm{E}^{\prime}=2 \mathrm{~h} \nu-\phi\)

Put the given values in above formula.

\(=2 \times 4.8-2.8\)

\(=6.8 \mathrm{eV}\)
Question 7
1 / -0

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 8
1 / -0

A body of mass \(M\) is kept on a rough horizontal surface (friction coefficient \(=\mu\) ). A person is trying to pull the body by applying a horizontal force but the body is not moving. The force by the surface on the body is \(F\), where:

A
\(F=M g\)

B
\(F=\mu M g\)

C
\({Mg} \leq {F} \leq {Mg} \sqrt{1+\mu^{2}}\)

D
\({Mg} \geq {F} \geq {Mg} \sqrt{1-\mu^{2}}\)

Solution
Question 9
1 / -0

When a uranium isotope \({ }_{92}^{235} \mathrm{U}\) is bombarded with a neutron, it generates \({ }_{36}^{89} \mathrm{Kr},\) three neutrons and

A
\({ }_{144}^{56} \mathrm{Ba}\)

B
\({ }_{91}^{40} \mathrm{Zr}\)

C
\({ }_{101}^{36} \mathrm{Kr}\)

D
\({ }_{103}^{36} \mathrm{Kr}\)

Solution

\({ }_{92} \mathrm{U}^{235}+{ }_{o} \mathrm{n}^{1} \rightarrow{ }_{\mathrm{Q}} \mathrm{X}^{\mathrm{P}}+{ }_{36} \mathrm{Kr}^{89}+3{ }_{\circ} \mathrm{n}^{1}+\mathrm{Q}(\) energy \()\)

\(\sum\) Atomic number on LHS \(=92\)

\(\sum\) Atomic number on \(\mathrm{RHS}=\mathrm{Q}+36+3 \times 0\)

\(\therefore \mathrm{LHS}=\mathrm{RHS}\)

\(Q+36=92\)

\(Q=56\)

also, \(\sum\) atomic mass number on

\(\mathrm{LHS}=235+1=236\)

\(\sum\) Atomic mass number on

\(\mathrm{RHS}=\mathrm{P}+89+3 \times 1\)

\(\therefore \mathrm{LHS}=\mathrm{RHS}\)

\(P+92=236\)

\(P=144\)

So, the element is \({ }_{56} \mathrm{Ba}^{144}\)
Question 10
1 / -0

The refractive index of the material of a prism is \(\sqrt{2}\) and the angle of the prism is \(30^{\circ}\). One of the two refracting surfaces of the prism is made a mirror inwards, by silver coating. A beam of monochromatic light entering the prism from the other face will retrace its path (after reflection from the silvered surface) if its angle of incidence on the prism is:

A
\(30^{\circ}\)

B
\(45^{\circ}\)

C
\(60^{\circ}\)

D
Zero

Solution

Given:

Refractive index of the material of the prism, \(\mu=\sqrt{2}\)

Angle of prism, \(A=30^{\circ}\)

As we know

\(\mathrm {A=r_1+r_2}\)

\(30=\mathrm r_1+0\)

So, \(\mathrm{r}_1=30\)

According to the use of Snell's law,

\(\mathrm {1 \times \sin i=\mu \times \sin r}\)

\(1 \times \sin \mathrm i=\sqrt{2} \times \sin 30 \)

\(\sin \mathrm i=\frac{1}{\sqrt{2}} \)

\(\mathrm i=45^{\circ}\)

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Physics Test - 42

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Physics Test - 42

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Question 1

\(\sqrt{\lambda R(\lambda R-1)}\)

\(\sqrt{\frac{\lambda R-1}{\lambda \cdot R}}\)

\(\sqrt{\frac{\lambda R}{\lambda R-1}}\)

\(\sqrt{\lambda(\mathrm{R}-1)}\)

Solution

Question 2

Forward bias only

Reverse bias only

Both forward bias and reverse bias

Increase in forward current

Solution

Question 3

\(3 \times 10^{23} \mathrm{Am}^{2}\)

\(1 \times 10^{23} \mathrm{Am}^{2}\)

\(1.05 \times 10^{23} \mathrm{Am}^{2}\)

\(2.05 \times 10^{23} \mathrm{Am}^{2}\)

Solution

Question 4

30°

15°

60°

1°

Solution

Question 5

1 A

5 A

3 A

None of the above

Solution

Question 6

6.8eV

68eV

-6.8eV

-68eV

Solution

Question 7

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 8

\(F=M g\)

\(F=\mu M g\)

\({Mg} \leq {F} \leq {Mg} \sqrt{1+\mu^{2}}\)

\({Mg} \geq {F} \geq {Mg} \sqrt{1-\mu^{2}}\)

Solution

Question 9

\({ }_{144}^{56} \mathrm{Ba}\)

\({ }_{91}^{40} \mathrm{Zr}\)

\({ }_{101}^{36} \mathrm{Kr}\)

\({ }_{103}^{36} \mathrm{Kr}\)

Solution

Question 10

\(30^{\circ}\)

\(45^{\circ}\)

\(60^{\circ}\)

Zero

Solution

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