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Physics Test - 35

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Weekly Quiz Competition

Question 1
1 / -0

The rms value of an ac current of \(50 \mathrm{~Hz}\) is \(10 \mathrm{amp} .\) The time taken by the alternating current in reaching from zero to maximum value and the peak value of current will be:

A
\(2 \times 10^{-2} \mathrm{sec}\) and \(14.14 \mathrm{amp}\)

B
\(1 \times 10^{-2} \mathrm{sec}\) and \(7.07\) amp

C
\(5 \times 10^{-3} \mathrm{sec}\) and \(7.07 \mathrm{amp}\)

D
\(5 \times 10^{-3} \mathrm{sec}\) and \(14.14 \mathrm{amp}\)

Solution

\(\mathrm{I}=10 \mathrm{~A}\)

\(\mathrm{f}=50 \mathrm{~Hz}\)

\(\mathrm{I}_{\mathrm{rms}}=\left(\frac{\mathrm{I}_{\mathrm{m}}}{\sqrt{2}}\right)\)

where \(I_{m}\) is peak value \(I_{m}=\sqrt{2} \cdot I_{\text {rms }}\)

\(\therefore I_{m}=10 \sqrt{2} A=10 \times 1.41=14.1 \mathrm{~A}\)

Time taken to reach form \(\mathrm{0}\) to \(\mathrm{I}_{\mathrm{m}}\) is \([(\frac{\mathrm{T}} { 4})-0]=(\frac{\mathrm{T}}{4})\)

\(\mathrm{t=(\frac{T}{ 4})=(\frac{1} { 4 f})}\)

\(\therefore \mathrm{t}=[\frac{1}{ \{4 \times 50}\}]=[\frac{1} { 200}] \mathrm{sec}\)

\(\therefore \mathrm{t}=0.5 \times 10^{-2} \mathrm{sec}\)

\(\mathrm{t}=5 \times 10^{-3} \mathrm{sec}\)
Question 2
1 / -0

In a p-n-p transistor, working as a common base amplifier, the current gain is \(0.96\) and the emitter current is \(7.2\) mA. The base current is:

A
\(0.20\) mA

B
\(0.36\) mA

C
\(0.29\) mA

D
\(0.45\) mA

Solution

Current gain \(=\frac{{I}_{C}}{{I}_{B}}\)

\({I}_{{C}}=\) collector current

\(I_{E}=\) emitter current

So, \(I_{C}=\) \(96\times7.2 = 6.912\)

As \(I_{E}=I_{B}+I_{C}\)

So, \(I_{B}=I_{E}-I_{C}\)

\(=(7.2-6.912)\) mA

\(=288\) mA

\(\approx 0.29\) mA
Question 3
1 / -0

What is the direction of force of friction acting on a body moving on a fixed surface?

A
Along the different of motion

B
Opposite to the direction of motion

C
Independent of Motion

D
Both (A) and (B)

Solution

When we try to slide a body on a surface, the motion of the body is opposed by a force called the force of friction. The frictional force arises due to intermolecular interaction.

When an external force \((F)\) is applied to move the body and the body does not move, then the frictional force acts opposite to applied force \(F\) and is equal to the applied force i.e., \(F-f=0\). When the body remains at rest, the frictional force is called the static friction. Static friction is a self-adjusting force.

Friction force, \({f}_{\mathrm{s}}=\mu \mathrm{mg}\)

Where, \(\mu_{s}=\) coefficient of friction.

Thus, friction force act opposite to the direction of motion.
Question 4
1 / -0

A point charge \(+10 \mu C\) is at a distance of \(5 cm\) directly above the centre of a square of side \(10 cm\), as shown in Figure. What is the magnitude of the electric flux through the square? (Hint: Think of the square as one face of a cube with edge \(10 cm\) ).

A
\(1.88 \times 10^{5} Nm ^{2} C ^{-1}\)

B
\(2.90 \times 10^{5} N m^{2} C^{-1}\)

C
\(1.52 \times 10^{5} Nm ^{2} C ^{-1}\)

D
\(3.2 \times 10^{5} Nm ^{2} C ^{-3}\)

Solution

Let us assume that the charge \(q=\pm 10 \mu C=10^{-5} C\) is placed at a distance of \(5 cm\) from the square \(A B C D\) of each side \(10 cm\). The square \(A B C D\) can be considered as one of the six faces of a cubic Gaussian surface of each side \(10 cm\).

Now, the total electric flux through the faces of the cube as per the Gaussian theorem:

\(\phi=\frac{q}{\epsilon_{0}}\)

Therefore, the total electric flux through the square \(ABCD\) will be:

\(\phi_{E}=\frac{1}{6} \times \phi\)

\(=\frac{1}{6} \times \frac{q}{\epsilon_{0}}\)

\(=\frac{1}{6} \times \frac{10^{-5}}{8.854 \times 10^{-12}} \quad\left(\because \epsilon_{0}=8.854 \times 10^{-12}\right)\)

\(=1.88 \times 10^{5} Nm ^{2} C ^{-1}\)
Question 5
1 / -0

Which of the following does not affect the apparent frequency in doppler effect?

A
Speed of source

B
Speed of observer

C
Frequency of source

D
Distance between source and observer

Solution

When sound waves are emitted by a moving source, or when the observer of the sound is moving, the apparent frequency of the sound can change. The shift in frequency due to motion of the sound wave source or of the observer is called the Doppler effect. The distance between the source and observer does not affect the apparent frequency in Doppler effect.
Question 6
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Two masses \(m_{1}\) and \(m_{2}\) are suspended together by a mass-less spring of spring constant \(K\). When the masses are in equilibrium, \(m_{1}\) is removed without disturbing the system. The angular frequency and amplitude of oscillation of \(m_{2}\) are

A
\(\sqrt{\frac{K}{m_{2}}}, \frac{m_{1} g}{K}\)

B
\(\sqrt{\frac{K}{m_{1}}}, \frac{m_{1} g}{K}\)

C
\(\sqrt{\frac{K}{m_{1}}}, \frac{m_{2} g}{K}\)

D
\(\sqrt{\frac{K}{m_{2}}}, \frac{m_{2} g}{K}\)

Solution

Angular Frequency: \(\omega=\sqrt{\frac{k}{m_{2}}}\)
Amplitude \(: y=A \sin \sqrt{\frac{k}{m_{2}}}\)
Step-by-step explanation:
Mass \(1=m_{1}\)
Mass \(2=m_{2}\)
Spring constant = \(k\)
Angular frequency \(=?\)
Amplitude \(=?\)
Solution:
As it is given that masses are in equilibrium so initially there is no motionwhich means that there will be no angular velocity in the start
when \(m_{1}\) is removed equilibrium will be disturbed and an angular velocity is produced and it will revolve in a circle
We know the formula for angular frequency of an object
\(\omega=\sqrt{\frac{k}{m}}\)
Where \(k\) is spring constant and \(m\) is the mass of object so for the given
Angular frequency will be
\(\omega=\sqrt{\frac{k}{m_{2}}}\)
Now for amplitude we know the formula
\(y=A \sin \omega t\)
Here \(t\) is the time period, \(A\) is amplitude and \(\omega\) is angular frequency
Putting the value of \(\omega\) here,
\(y=A \sin \left(\sqrt{\frac{k}{m_{2}}}\right) t\)
Question 7
1 / -0

A man of height \('h'\) walks along a straight path towards a lamp post of height \(H\) with a uniform velocity \('u'\). The velocity of the edge of the shadow on the ground will be:

A
\(\frac{ hu }{( H - h )}\)

B
\(\frac{ hu }{( H + h )}\)

C
\(\frac{( H + h )}{ hu }\)

D
\(\frac{(H-h)}{h u}\)

Solution

Let in same time \(t,\) distances covered by man and edge of image be respectively \(x\) and \(y\).

\( \tan \theta=\frac{H-h}{x}=\frac{H}{x+y}\)

\(\Rightarrow H x=H x+H y-h x+h y\)

\(\Rightarrow \frac{x}{y}=\frac{(H-h)}{h}\) ..........(i)

Now \(x=u t\)

And \(y=v t\)

\(\Rightarrow \frac{x}{y}=\frac{u}{v}\) ............(ii)

So, \(\frac{u}{v}=\frac{(H-h)}{h}\)

\(\Rightarrow v=\frac{h u}{(H-h)}\)
Question 8
1 / -0

What is the emissive power of sun when its temperature is doubled?

A
2 times

B
4 times

C
8 times

D
16 times

Solution

Stefan's Law: According to it the radiant energy emitted by a perfectly black body per unit area per sec (i.e. emissive power of black body) is directly proportional to the fourth power of its absolute temperature i.e.

\(E \propto T^{4}\)

\(\Rightarrow E=\sigma T^{4}\)

Where, is a constant called Stefan’s constant.

The value of Stefan's constant is \(5.67 \times 10^{-8} {~W} / {m}^{2} {~K}^{4}\).

Stefan's Law is used to accurately find the temperature Sun, Stars, and the earth. A black body is an ideal body that absorbs or emits all types of electromagnetic radiation.

Mathematically, Stefan's law can be written as:

\(\Rightarrow E=\sigma T^{4}\)

When the temperature is doubled i.e., \(T_{1}=2 T\)

\(\Rightarrow {E}_{1}=\sigma\left({T}_{1}\right)^{4}\)

\(\Rightarrow {E}_{1}=\sigma(2 {~T})^{4}=16 \sigma {T}^{4} \quad\left[\because {E}=\sigma {T}^{4}\right]\)

\(\Rightarrow E_{1}=16 E\)

The emissive power of the sun will increase by 16 times when its temperature is doubled.
Question 9
1 / -0

In an n-type silicon, which of the following statement is true:

A
Electrons are majority carriers and trivalent atoms are the dopants.

B
Electrons are minority carriers and pentavalent atoms are the dopants.

C
Holes are minority carriers and pentavalent atoms are the dopants.

D
Holes are majority carriers and trivalent atoms are the dopants.

Solution

An n-type semiconductor is obtained by doping a semiconductor with a pentavalent impurity. The impurity so added produces free electrons. Therefore in an n-type semiconductor, the electrons are the majority carrier and holes are the minority carriers and pentavalent atoms are the dopants.
Question 10
1 / -0

The instantaneous voltage across a pure resistor is ______ with current.

A
Leads

B
Out of phase

C
Lags

D
In- phase

Solution

In purely resistive circuits, the current and applied voltage are in phase with each other.

In purely inductive circuits, the current lags the applied voltage by 90°.

In purely capacitive circuits, the current leads the applied voltage by 90°.

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Physics Test - 35

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Physics Test - 35

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Question 1

\(2 \times 10^{-2} \mathrm{sec}\) and \(14.14 \mathrm{amp}\)

\(1 \times 10^{-2} \mathrm{sec}\) and \(7.07\) amp

\(5 \times 10^{-3} \mathrm{sec}\) and \(7.07 \mathrm{amp}\)

\(5 \times 10^{-3} \mathrm{sec}\) and \(14.14 \mathrm{amp}\)

Solution

Question 2

\(0.20\) mA

\(0.36\) mA

\(0.29\) mA

\(0.45\) mA

Solution

Question 3

Along the different of motion

Opposite to the direction of motion

Independent of Motion

Both (A) and (B)

Solution

Question 4

\(1.88 \times 10^{5} Nm ^{2} C ^{-1}\)

\(2.90 \times 10^{5} N m^{2} C^{-1}\)

\(1.52 \times 10^{5} Nm ^{2} C ^{-1}\)

\(3.2 \times 10^{5} Nm ^{2} C ^{-3}\)

Solution

Question 5

Speed of source

Speed of observer

Frequency of source

Distance between source and observer

Solution

Question 6

\(\sqrt{\frac{K}{m_{2}}}, \frac{m_{1} g}{K}\)

\(\sqrt{\frac{K}{m_{1}}}, \frac{m_{1} g}{K}\)

\(\sqrt{\frac{K}{m_{1}}}, \frac{m_{2} g}{K}\)

\(\sqrt{\frac{K}{m_{2}}}, \frac{m_{2} g}{K}\)

Solution

Question 7

\(\frac{ hu }{( H - h )}\)

\(\frac{ hu }{( H + h )}\)

\(\frac{( H + h )}{ hu }\)

\(\frac{(H-h)}{h u}\)

Solution

Question 8

2 times

4 times

8 times

16 times

Solution

Question 9

Electrons are majority carriers and trivalent atoms are the dopants.

Electrons are minority carriers and pentavalent atoms are the dopants.

Holes are minority carriers and pentavalent atoms are the dopants.

Holes are majority carriers and trivalent atoms are the dopants.

Solution

Question 10

Leads

Out of phase

Lags

In- phase

Solution

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