Physics Test

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Physics Test - 18

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Question 1
1 / -0

A man increases his speed by 2m/s, then his kinetic energy gets doubled. The original speed of the man is:

A
\(\sqrt{2}(\sqrt{2}+1) m/s\)

B
\(\sqrt{2}(\sqrt{2}-10) m/s\)

C
\(2(\sqrt{2}+1) m/s\)

D
\(2(\sqrt{2}-1) m/s\)

Solution

The energy possessed by a body due to the virtue of its motion is called kinetic energy.
\( K E=\frac{1}{2} m v^{2}\)
Where \(KE =\) kinetic energy, \(m =\) mass and \(v =\) velocity
Given,
\(\Delta v = v _{2}- v _{1}=2 m / s\)
\(KE _{2}=2 KE _{1}\)
Where \(\Delta v =\) increase in speed
Initial kinetic energy, \(K E_{1}=\frac{1}{2} m v_{1}^{2}\)...(1)
Final kinetic energy, \(K E_{2}=\frac{1}{2} m v_{2}^{2}\)...(2)
By equation (1) and equation (2),
\( K E_{2}=2 K E_{1} \)
\(\Rightarrow \frac{1}{2} m v_{2}^{2}=2 \times \frac{1}{2} m v_{1}^{2} \)
\(\Rightarrow v_{2}^{2}=2 v_{1}^{2} \)
\(\Rightarrow v_{2}=\pm \sqrt{2} v_{1}\)...(3)
Since,
\( \Delta v=v_{2}-v_{1}=2 m / s\)
By equation (3), we get
\( \sqrt{2} v_{1}-v_{1}=2 \)
\(\Rightarrow v_1(\sqrt{2}-1)=2\)
\(\Rightarrow v_{1}=\frac{2}{\sqrt{2}-1} \)...(4)
On dividing and multiplying by \(\sqrt{2}+1\) in equation (4), we get
\( v_{1}=\frac{2}{\sqrt{2}-1} \times \frac{\sqrt{2}+1}{\sqrt{2}+1} \)
\(\Rightarrow v_{1}=\frac{2(\sqrt{2}+1)}{(\sqrt{2})^{2}-1^{2}}\)
\(\Rightarrow v_{1}=2(\sqrt{2}+1) m/s\)
So, the original speed of man is\(2(\sqrt{2}+1) m/s\).
Question 2
1 / -0

What is the correct dimensional formula for energy density?

A
\(\left[\mathrm{M}^{2} \mathrm{~L}^{-2} \mathrm{~T}^{-2}\right]\)

B
\(\left[\mathrm{M}^{2} \mathrm{~L}^{-3} \mathrm{~T}^{-2}\right]\)

C
\(\left[\mathrm{M} L^{1} \mathrm{~T}^{-3}\right]\)

D
\(\left[\mathrm{M} \mathrm{L}^{-1} \mathrm{~T}^{-2}\right]\)

Solution

Energy density is referred to as the amount of energy contained per unit of volume.
The dimensional formula of energy density will be the dimensional formula of energy divided by the dimensional formula for volume.
Dimensional formula of energy \(=\left[{M} {L}^{2} {~T}^{-2}\right]\)
Dimensional formula of volume \(=\left[{L}^{3}\right]\)
Dimensional formula of energy density \(=\frac{\left[{M} {L}^{2} {~T}^{-2}\right]} {\left[{L}^{3}\right]}\)
\(=\left[M L^{-1} T^{-2}\right]\)
Question 3
1 / -0

In an electromagnetic wave, the magnitude of electric and magnetic fields are 100 V/m and 0.265 A/m. The maximum energy flow is:

A
\(26.5 \mathrm{~W} / \mathrm{m}^{2}\)

B
\(36.5 \mathrm{~W} / \mathrm{m}^{2}\)

C
\(46.7 \mathrm{~W} / \mathrm{m}^{2}\)

D
\(765 \mathrm{~W} / \mathrm{m}^{2}\)

Solution

Given:
Electrical field \(E=100 \mathrm{~V} / \mathrm{m}\)
Magnetic field \(\mathrm{B}=0.265 \mathrm{~A} / \mathrm{m}\)
The energy flow is given by the Poynting vector
\(\vec{S}=\vec{E} \times \vec{B}\)
\(\Rightarrow S=E B \sin \emptyset\)
\(\Rightarrow S=E B \quad\left[\emptyset=90^{\circ}\right.\), as \(E\) and \(B\) are perpendicular to each other]
\(S=100 \mathrm{~V} / \mathrm{m} \times 0.265 \mathrm{~A} / \mathrm{m}\)
\(\Rightarrow S=26.5 W / m^{2}\)
Question 4
1 / -0

The flux linked with a coil is \(0.8 \mathrm{W b}\) when a \(2 \mathrm A\) current is flowing through it. If this current begins to increase at the rate of \(400 \mathrm{A} / \mathrm{s}\), the induced emf in the coil will be:

A
\(20 \mathrm{V}\)

B
\(40 \mathrm{V}\)

C
\(80 \mathrm{V}\)

D
\(160 \mathrm{V}\)

Solution

Given,
Flux linked with a coil \(,\phi\)\(=0.8 \mathrm{W b}\)
Current, \(i\)\(=2 \mathrm A\)
Increase rate of current, \(\frac{d i}{d t}\)\(=400 \mathrm{~A} / \mathrm{s}\)
\(\phi=\mathrm{Li}\)
\(\Rightarrow 0.8=\mathrm{L} \times 2\)
\(\mathrm{~L}=\frac{0.8}{2}\)
\(=0.4\)
\(\mathrm{emf}=\mathrm{L} \frac{\mathrm{di}}{\mathrm{dt}}\)
\(=0.4 \times 400\)
\(=160 \mathrm{~V}\)
Question 5
1 / -0

A paramagnetic sample shows a net magnetisation of \(6 \mathrm{~A} / \mathrm{m}\) when it is placed in an external magnetic field of \(0.4 T\) at a temperature of \(4 K\). When the sample is placed in an external magnetic field of \(0.3 T\) at a temperature of \(24 K\), then the magnetisation will be:

A
\(4 A / m\)

B
\(0.75 \mathrm{~A} / \mathrm{m}\)

C
\(2.25 \mathrm{~A} / \mathrm{m}\)

D
\(1 \mathrm{~A} / \mathrm{m}\)

Solution

Given:
\(T_{1}=4 K, T_{2}=24 K\)
For paramagnetic material,
According to curies law,
\(M=\chi H\), \(\left(\chi=\frac{C}{T}\right)\)
where,
\(\chi>0\) is the (volume) magnetic susceptibility,
\(M\) is the magnitude of the resulting magnetization \((\mathrm{A} / \mathrm{m})\),
\(H\) is the magnitude of the applied magnetic field \((\mathrm{A} / \mathrm{m})\)
\(T\) is absolute temperature \((\mathrm{K})\),
\(C\) is a material-specific Curie constant \((\mathrm{K})\).
\(\chi \propto \frac{1}{T} \)
\(\Rightarrow \chi_{1} T_{1}=\chi_{2} T_{2} \)
\(\Rightarrow \frac{6}{0.4} \times 4=\frac{I}{0.3} \times 24 \)
\(I=\frac{0.3}{0.4}=0.75 A / m\)
Question 6
1 / -0

Which of the following option is correct regarding energy band of electrons?

A
In a crystal each electrons have the same energy level.

B
The energy band below the valence band is called the conduction band.

C
The energy band that includes the energy levels of the valence electrons is called the valence band.

D
The valance band holds those electrons that are responsible for conduction.

Solution

Option (C) is correct regarding the energy band of electrons.
Each electron has a unique position inside a crystal and no two electrons see exactly the same pattern of surrounding charges. That is why no two electrons will have the same energy level. Each electron will have a different energy level.
Energy Band: The different energy levels with continuous energy variation of electrons in a material are called energy bands.
Valance Band: The energy band which includes the energy levels of the valence electrons is called the valence band.
Conduction Band: The energy bands above the valence bands are called the conduction bands. The conduction band holds those electrons that are responsible for conduction.
Question 7
1 / -0

In common emitter amplifier, the \(\frac{\mathrm{I}_{\mathrm{c}}}{\mathrm{I}_{\mathrm{e}}}\) is \(0.98 .\) The current gain will be:

A
4.9

B
7.8

C
49

D
78

Solution

Given:
\( \mathrm{I}_{\mathrm{c}}=0.98 \mathrm{I}_{\mathrm{e}}\)
We know that:
\( \mathrm{I}_{\mathrm{e}}=\mathrm{I}_{\mathrm{c}}+\mathrm{I}_{\mathrm{b}}\)
Put the given values in above formula:
\( \mathrm{I}_{\mathrm{e}}=0.98 \mathrm{I}_{\mathrm{e}}+\mathrm{I}_{\mathrm{b}}\)
\( \Rightarrow \mathrm{I}_{\mathrm{b}}=0.02 \mathrm{I}_{\mathrm{e}}\)
Current gain \((\beta)=\frac{\mathrm{I}_{\mathrm{c}}}{\mathrm{I}_{\mathrm{b}}}\)
\(=\frac{0.98 \mathrm{I}_{\mathrm{e}}}{0.02 \mathrm{I}_{\mathrm{e}}}\)
\(=49\)
Question 8
1 / -0

If the de Broglie wavelengths associated with a proton and an \(\alpha-\) particle are equal, then the ratio of velocities of the proton and the \(\alpha\) - particle will be:

A
1: 4

B
1: 2

C
4: 1

D
2: 1

Solution

De Broglie Wavelength \(\lambda_{{p}}=\lambda_{{r}}\)
So, \(\frac{{h}}{{m}_{{p}} \times {v}_{{p}}}=\frac{{h}}{{m}_{\alpha} \times {v}_{{\alpha}}}\)
Because Mass of \(\alpha\) particle is 4 times mass of proton.
\(\frac{v_{p}}{v_{a}}=\frac{m_{\alpha}}{m_{p}}=\frac{4 m_{p}}{m_{p}}\) \(=4:1\)
Question 9
1 / -0

A point object is placed at the focus of the bi-convex lens. What should be the value of d, so the final image form at infinity?

A
10 cm

B
20 cm

C
15 cm

D
None of these

Solution

The final image will be at infinity only, if the foci of lens and mirror coincides. The situation could be understood on the basis of given diagram.
Question 10
1 / -0

At what height above the surface of the earth does the weight of an object reduce by \(1 \%\). Given the radius of the earth is \(6400 .\)

A
3.2 Km

B
64 Km

C
6.4 Km

D
32 Km

Solution

Given that radius of earth is \(6400 \mathrm{~km} .~1 \%\) decrease in gravity.
Acceleration due to gravity above the surface of earth at height \(\mathrm{h}\) is given
\(g^{\prime}=g\left(1-\frac{2 h}{R_{e}}\right)\).....(1)
Here, \(g^{\prime}=0.99 \mathrm{~g}\)
Put the given values in (1).
\( 0.99=1-\frac{2 h}{R_{e}}\)
\(\Rightarrow \frac{2 h}{R_{e}}=0.01\)
\(\Rightarrow h=\frac{0.01}{2} R_{e}\)
Put the value of \(R_{e}\) in above equation.
\( h=\frac{0.01}{2} \times 6400\)
\(=32 \mathrm{Km}\)
\(\mathrm{h}=32 \mathrm{Km}\)
Question 11
1 / -0

If three particles, each of mass \(\mathrm{M}\) are placed at the three corners of an equilateral triangle of side a, the force exerted by this system on another particle of mass \(M\) is placed at the midpoint of a side is:

A
\(\frac{4 G M^{2} }{ 5 a^{2}}\)

B
\(\frac{4 G M^{2} }{ 3 a^{2}}\)

C
\(\frac{2 G M^{2} }{ 3 a^{2}}\)

D
\(\frac{2 G M^{2} }{ 5 a^{2}}\)

Solution

The gravitational force at \(\mathrm{D}\) due to \(\mathrm{C}\) is:

\(\Rightarrow F_{D C}=G \frac{M M}{\left(\frac{a}{2}\right)^{2}}=G \frac{4 M^{2}}{a^{2}}\)

The gravitational force at D due to \(B\) is

\( F_{D B}=G \frac{M M}{\left(\frac{a}{2}\right)^{2}}\)

\(=G \frac{4 M^{2}}{a^{2}}\)

Since \(F_{B D}\) and \(F_{D C}\) are equal in magnitude but they are opposite in direction, therefore they will cancel out each other.

So,

\( A D^{2}=\sqrt{A C^{2}-C D^{2}} \)

\(\Rightarrow A D^{2}=\sqrt{a^{2}-\frac{a^{2}}{4}}\)

\(=\sqrt{\frac{3 a^{2}}{4}}\)

The gravitational force at \(\mathrm{D}\) due to \(\mathrm{A}\) is

\( F_{D C}=G \frac{M M}{A D^{2}}\)

\(=G \frac{4 M^{2}}{3 a^{2}}\)
Question 12
1 / -0

Which of the following electromagnetic wave is used for thermal imaging?

A
Ultraviolet

B
Gamma ray

C
Infrared

D
β-ray
Solution
Infraredelectromagnetic wave is used for thermal imaging.
- The electromagnetic spectrum is the arrangement of electromagnetic waves, in the order of their respective wavelength.
- Infrared rays are a type of electromagnetic waves which are emitted by hot bodies.
- Within the electromagnetic spectrum, infrared waves occur at frequencies above microwaves below the wavelength of visible red light, that this type of electromagnetic waves termed infrared.
- Thermal imaging is the technique in which thermal energy and infrared radiation are used to gather information about objects.
Question 13
1 / -0

In the circuit shown below, the ammeter reading is zero. Then, the value of the resistance R is:

A

\(50 \Omega\)

B
\(100 \Omega\)

C
\(200 \Omega\)

D
\(400 \Omega\)

Solution
Question 14
1 / -0

In an experiment, electrons are made to pass through a narrow slit of width d comparable to their de Broglie wavelength. They are detected on a screen at a distance D from the slit.

Which of the following graphs can be expected to represent the number of electrons \(\mathrm{N}\) detected as a function of the detector position \(\mathrm{y}(\mathrm{y}=0\) corresponds to the middle of the slit):

A

B

C

D

Solution

As diffraction pattern has to be wider than slit width.

The line of maximum Intensity for the zeroth order will exceed very much. The electron diffraction pattern from a single slit will be as shown below.
Question 15
1 / -0

A body of mass 5 kg is placed on a rough horizontal surface. What is the maximum value of friction that can act on it? (Given μ = 0.3 and g = 10 m/s²)

A
45 N

B
50 N

C
30 N

D
15 N

Solution

Given,

Mass (m) = 5 kg

Coefficient of friction (μ) = 0.3

g = 10 m/s2

Free body diagram of the given condition,

In vertical direction forces are balanced.

∴ \(N = mg\)

The friction force is calculated by,

\(F = μ N = μ mg\)

∴ \(F = 0.3 × 5 × 10 = 15\) N