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Physics Test - 11

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Question 1
1 / -0

In a compound microscope, the objective and the eyepiece have focal lengths of \(5\) cm and \(9.5\) cm, respectively, and both are kept at a distance of \(20\) cm. If the final image is formed at the least distance of \(25\) cm from the eyepiece, find the total magnification.

A
\(19.6\)

B
\(89\)

C
\(6\)

D
\(1.2\)

Solution

In the case of eye-piece lense,

\(\frac{1}{v}-\frac{1}{u}=\frac{1}{f}\)

\(\Rightarrow \frac{1}{25}-\frac{1}{u}=\frac{1}{5}\)

\(\Rightarrow \frac{1}{u}=\frac{1}{25}-\frac{1}{5}\)

\(\Rightarrow u=\frac{-5}{4}\)

\(\Rightarrow u=-1.25\) cm

For objective the image distance,

\(v=20-1.25\)

\(=18.75\) cm

Again, by the above formula:

\(\frac{1}{18.75}-\frac{1}{u}=\frac{1}{9.5}\)

\(\Rightarrow \frac{1}{u}=\frac{4}{75}-\frac{2}{19}\)

\(\Rightarrow \frac{1}{u}=\frac{76-150}{75 \times 19}\)

\(\Rightarrow u=\frac{75 \times 19}{-74}\)

\(\Rightarrow u=-19.25 \) cm

Total magnification,

\(m=m_{1} m_{2}\)

Here, \(m = \frac{v}{u}\)

\(\Rightarrow m=\frac{18.75}{19.25} \times \frac{25}{1.25}\)

\(\Rightarrow m=19.6\)
Question 2
1 / -0

Two waves having their intensities in the ratio of \(9:1\) produce interference. In the interference pattern, the ratio of maximum to minimum intensity is equal to:

A

\(\frac{4}{1}\)

B

\(\frac{8}{4}\)

C

\(\frac{2}{3}\)

D

\(\frac{1}{2}\)

Solution

Let the intensities of the two waves be \({I}_{1}\) and \({I}_{2}\).
Given,
\({I}_{1}: {I}_{2}={9}: {1}\)
Ratio of maximum and minimum intensities \(\frac{I_{\max }}{I_{\min }}=\left(\frac{\sqrt{I_{1}}+\sqrt{I_{2}}}{\sqrt{I_{1}}-\sqrt{I_{2}}}\right)^{2}\)
or,
\(\frac{I_{\max }}{I_{\min }}=\left(\frac{\sqrt{\frac{1_{1}}{I_{2}}}+1}{\sqrt{\frac{I_{1}}{I_{2}}}-1}\right)^{2}\)
or
\(\frac{I_{\max }}{I_{\min }}=\left(\frac{\sqrt{9}+1}{\sqrt{9}-1}\right)^{2}\)
\(=\left(\frac{3+1}{3-1}\right)^{2}\)
\(\Rightarrow \frac{I_{\max }}{I_{\min }}=\frac{16}{4}\)
\(=\frac{4}{1}\)
Question 3
1 / -0

In stopping potential (V) photo current (I) graph, if \(\mathrm{V}_{2}>\mathrm{V}_{1}\), then compare the wavelengths of incident radiations:

A
\(\lambda_{1}=\sqrt{\lambda_{2}}\)

B
\(\lambda_{1}<\lambda_{2}\)

C
\(\lambda_{1}=\lambda_{2}\)

D
\(\lambda_{1}>\lambda_{2}\)

Solution

Case 1: Stopping potential for \(\lambda_1=V_1\)

\(eV_1=h\nu\Rightarrow\frac{hc}{\lambda_1}\)

\(eV_1=\frac{hc}{\lambda_1}\quad\ldots\)(1)

Case 2: Stoping potential for \(\lambda_2=V_2\)

\(eV_2=h\nu\)

\(eV_2=\frac{hc}{\lambda_2}\)

According to the question,

\(V_2>V_1\)

\(\therefore\frac{hc}{\lambda_2}>\frac{hc}{\lambda_1}\)

\(\Rightarrow \lambda_1>\lambda_2\)
Question 4
1 / -0

If simple harmonic motion is represented by\(x=A \cos (\omega t+\varphi)\) then\(' \varphi^{\prime}\) is _____________.

A
Angular frequency

B
Displacement

C
Amplitude

D
Phase constant

Solution

Simple harmonic motion is represented by,

\( x=A \cos (\omega t+\varphi)\)

Where \({A}=\) amplitude, \(\omega=\) Angular frequency, \({x}=\) Displacement and \(\varphi=\) Phase constant
Question 5
1 / -0

In two single turn circular loop of wire have radius \(20 \mathrm{~cm}\) and \(2 \mathrm{~cm}\). The loops lies in the same plane and are concentric. Find the mutual inductance of the pair is:

A
\(7.9 \times 10^{-6}\)

B
\(5.2 \times 10^{-7}\)

C
\(2.4 \times 10^{-6}\)

D
\(4.0 \times 10^{-7}\)

Solution

Let the current \(i\) in the large loop of radius \(\mathrm{R}=20 \mathrm{~cm}\) produces a magnetic field of magnitude \(B=\frac{\mu_{o} i}{2 R}\) at its centre.
Since the radius \(r=2\ \mathrm{cm}\) of the smaller loop is \(r<So, the mutual inductance of the loop is \(M=\frac{\Phi}{i} =\frac{\mu \pi r^{2}}{2 R} =\frac{4 \times 3.14 \times 10^{-7} \times 3.14 \times 2 \times 2}{2 \times 20}\)
\(M \approx 4.0 \times 10^{-7}\)
Question 6
1 / -0

Which of the following is not a fundamental unit?

A
Meter

B
Kilogram

C
Ampere

D
Volt
Solution
The fundamental units are the base units defined by International System of Units. These units are not derived from any other unit, therefore they are called fundamental units.The seven base units are:
- Meter (m) for Length
- Second (s) for Time
- Kilogram (kg) for Mass
- Ampere (A) for Electric current
- Kelvin (K) for temperature
- Mole (mol) for Amount of substance
- Candela (cd) for Luminous intensity
Voltis not a fundamental unit,it is a derived unit.
Question 7
1 / -0

Which theorem states that “If a particle under the simultaneous action of three forces is in equilibrium, then each force has a constant ratio with the sine of the angle between the other two forces”?

A
Lay’s theorem

B
Lami’s Theorem

C
Newton’s law

D
Faraday’s theorem

Solution

Lami’s Theorem states that if f a particle under the simultaneous action of three forces is in equilibrium, then each force has a constant ratio with the sine of the angle between the other two forces.
Question 8
1 / -0

In gases of monoatomic molecules, find the ratio of the two specific heat of gases \(\frac{C_{p}}{C_{v}}\).

A
1.66

B
1.33

C
1.4

D
1.00

Solution

For monoatomic:

Degree of freedom \(=3\)

For monoatomic molecules,

\(\frac{C_{p}}{C_{v}}=1+\frac{2}{f}\)

Where, \(C_{p}\) is the specific heat at constant pressure, \(C_{v}\) is the specific heat at constant volume and \(f\) is the degree of freedom.

\(=1+\frac{2}{3}\)

\(\frac{C_{p}}{C_{v}}=\frac{5}{3} \approx 1.66\)
Question 9
1 / -0

An a-particle of energy \(5 MeV\) is scattered through \(180^{\circ}\) by a fixed uranium nucleus. The distance of the closest approach is of the order of

A

\(1 \dot{A}\)

B

\(10^{-10} ~cm\)

C

\(10^{-12} ~cm\)

D

\(10^{-15} ~cm\)

Solution

At closest approach, all the kinetic energy of the a-particle will converted into the potential energy of the system, \(K.E.=P.E\)
i.e., \(\frac{1}{2} mv^{2}=\frac{1}{4 \pi \epsilon_{0}} \frac{q_{1} q_{2}}{r}\)
\(5 MeV=\frac{9 \times 10^{9} \times(2 e) \times 92 e}{r}\left(\therefore \frac{1}{2} mv^{2}=5 MeV\right)\)
\(\Rightarrow r=\frac{9 \times 10^{9} \times 2 \times 92 \times\left(1.6 \times 10^{-19}\right)^{2}}{5 \times 10^{6} \times 1.6 \times 10^{-19}}\)
\(r=5.3 \times 10^{-14} ~m=10^{-12 }\) \(cm\)
Question 10
1 / -0

A plane electromagnetic wave of frequency \(35 \mathrm{MHz}\) travels in free space along the X-direction. At a particular point (in space and time) \(\overrightarrow{\mathrm{E}}=9.6 \hat{\mathrm{j}} \mathrm{V} / \mathrm{m}\). The value of magnetic field at this point is:

A
\(3.2 \times 10^{-8} \hat{\mathrm{kT}}\)

B
\(3.2 \times 10^{-8} \hat{\mathrm{i}}\)

C
\(9.6 \hat{\mathrm{j}} \mathrm{T}\)

D
\(9.6 \times 10^{-8} \hat{\mathrm{kT}}\)

Solution

\(\frac{E}{B}=C \)

\(\frac{E}{B}=3 \times 10^8 \)

\(B=\frac{E}{3 \times 10^8}=\frac{9.6}{3 \times 10^8} \)

\( B=3.2 \times 10^{-8} \mathrm{~T} \)

\(\hat{B}=\hat{v} \times \hat{E} \)

\(=\hat{i} \times \hat{j}=\hat{k}\)

So,

\(\overrightarrow{\mathrm{B}}=3.2 \times 10^{-8 \hat{\mathrm{kT}}}\)

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Physics Test - 11

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Physics Test - 11

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Question 1

\(19.6\)

\(89\)

\(6\)

\(1.2\)

Solution

Question 2

\(\frac{4}{1}\)

\(\frac{8}{4}\)

\(\frac{2}{3}\)

\(\frac{1}{2}\)

Solution

Question 3

\(\lambda_{1}=\sqrt{\lambda_{2}}\)

\(\lambda_{1}<\lambda_{2}\)

\(\lambda_{1}=\lambda_{2}\)

\(\lambda_{1}>\lambda_{2}\)

Solution

Question 4

Angular frequency

Displacement

Amplitude

Phase constant

Solution

Question 5

\(7.9 \times 10^{-6}\)

\(5.2 \times 10^{-7}\)

\(2.4 \times 10^{-6}\)

\(4.0 \times 10^{-7}\)

Solution

Question 6

Meter

Kilogram

Ampere

Volt

Solution

Question 7

Lay’s theorem

Lami’s Theorem

Newton’s law

Faraday’s theorem

Solution

Question 8

1.66

1.33

1.4

1.00

Solution

Question 9

\(1 \dot{A}\)

\(10^{-10} ~cm\)

\(10^{-12} ~cm\)

\(10^{-15} ~cm\)

Solution

Question 10

\(3.2 \times 10^{-8} \hat{\mathrm{kT}}\)

\(3.2 \times 10^{-8} \hat{\mathrm{i}}\)

\(9.6 \hat{\mathrm{j}} \mathrm{T}\)

\(9.6 \times 10^{-8} \hat{\mathrm{kT}}\)

Solution

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