Physics Test

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Physics Test - 23

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Weekly Quiz Competition

Question 1
1 / -0

(A) The drift velocity of electrons decreases with the increase in the temperature of conductor.

(B) The drift velocity is inversely proportional to the area of cross-section of given conductor.

(C) The drift velocity does not depend on the applied potential difference to the conductor.

(D) The drift velocity of electron is inversely proportional to the length of the conductor.

(E) The drift velocity increases with the increase in the temperature of conductor.

Choose the correct answer from the options given below :

A
(A) and (B) only

B
(A) and (D) only

C
(B) and (E) only

D
(B) and (C) only

Solution

$\text { Drift velocity }=\left(\frac{e \tau}{m}\right) E $

$ v_d=\left(\frac{e \tau}{m}\right)\left(\frac{\Delta V}{\ell}\right)$

$\Delta \mathrm{V}=$ Potential difference applied across the wire As temperature increases, relaxation time decreases, hence $\mathrm{V}_{\mathrm{d}}$ decreases.

As per formula, $\mathrm{V}_{\mathrm{d}} \propto \frac{1}{\ell}$

$\mathrm{v}_{\mathrm{d}}=\frac{\mathrm{I}}{\text { neA }}$, as it is not mentioned that current is at steady state neither it is mentioned that $\mathrm{n}$ is constant for given conductor. So it can't be said that $\mathrm{v}_{\mathrm{d}}$ is inversely proportional to $\mathrm{A}$.

$ \mathrm{I}=\text { neAv } \mathrm{V}_{\mathrm{d}}=\frac{\mathrm{V}}{\mathrm{R}}=\frac{\mathrm{V}}{\rho \ell} \mathrm{A} $

$\mathrm{v}_{\mathrm{d}}=\frac{\mathrm{V}}{\rho \ell \mathrm{ne}}\left(E=\frac{\mathrm{V}}{\ell}\right)$

$\mathrm{v}_{\mathrm{d}}=\frac{\mathrm{eE} \tau}{\mathrm{m}}$

$\tau$ decrease with temperature increase.

First and fourth statements are correct.
Question 2
1 / -0

The peak voltage of a message signal and the carrier wave are 15 volts and 30 volts respectively. Find the modulation index.

A
2

B
0.5

C
0.33

D
None of these

Solution

Given: $A_{m}=15$ volts and $A_{c}=30$ volts

Where, $A_{c}=$ amplitude of the carrier wave and $A_{m}=$ amplitude of the modulating signal

We know that if $c(t)=A_{c} \sin \left(\omega_{c} t\right)$ represent carrier wave and $m(t)=A_{m} \sin \left(\omega_{m} t\right)$ represent message signal, then the modulation index is equal to the,

$\mu=\frac{A_{m}}{A_{c}}$

$\Rightarrow \mu=\frac{15}{30}$

$\Rightarrow \mu=0.5$
Question 3
1 / -0

A conducting loop carrying a current $I$ is placed in a uniform magnetic field pointing into the plane of the paper as shown. The loop will have a tendency to

A
Contract

B
Expand

C

Move towards $+v e x-a \xi s$

D

Move towards $-v e x-a \xi s$

Solution

If we use fleming's left-hand rule. We find that a force is acting in the radially outward direction throughout the circumference of the conducting loop.
Question 4
1 / -0

A wire tension 225 N produces 6 beats per second when it tuked with a fork, when the tension changed to 256 N, it again tuned with the same tuning fork to no. of beats remained unchanged, the frequency of tuning fork will be :

A
186

B
280

C
225

D
256

Solution

The frequency of a tuning fork is directly proportional to the square root of tension applied to it.

$f \propto \sqrt{T};$

or $f = K \sqrt{T};$

The number of beats = difference between frequencies of two superimposed waves.

Let the frequency of tuning fork = f_T;

When wire tension = 225 N ⇒ f_w = $K \sqrt{225}$ = 15K;

no. of beats = 6 ⇒ f_T - f_w = 6

⇒ f_T= 6 + 15K; - (1)

When wire tension = 256 N ⇒ f_w = $K \sqrt{256}$ = 16K;

no. of beats = 6 ⇒ f_w - f_T = 6

⇒ f_T = 16K - 6; - (2)

From the equations 1 and 2,

K = 12⇒ f_T = 186;
Question 5
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Which of the following represents the binding energy of a nucleus?

A
The average energy of each nucleon

B
The energy needed to split the nucleus into its parts

C
The mass-energy difference between protons and neutrons

D
The energy necessary to overcome the neutron-neutron repulsion

Solution

It is observed that mass of a stable nucleus is always less than the total mass of constituent nucleons. This difference of mass is known as mass defect. When a nucleus is formed from the free nucleons mass defect is released in the form of energy by Einstein's mass-energy relation. This energy is used to bind the nucleons to form a nucleus therefore an equivalent amount of energy is required to split the nucleus into its parts, that is called the binding energy of the nucleus.
Question 6
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For one dimensional motion, the force F(x) and the potential energy U(x) are related as:

A
$F(x)=\frac{-d U(x)}{d x}$

B
$F(x)=\frac{d U(x)}{d x}$

C
$U(x)=\frac{d F(x)}{d x}$

D
$U(x)=\frac{-d F(x)}{d x}$

Solution

The potential energy is equal to negative work done in shifting an object from some reference point to a given position for conservative force.
In mathematical form, it is given as,
$ d U(x)=-F(x) d x $
$\therefore F(x)=\frac{-d U(x)}{d x}$
where $F$ is the force in newton and $\frac{d U }{ dx}$ is change in potential energy per unit length.
Question 7
1 / -0

A wire of length L meters carrying a current I amperes is bent in the form of a circle. The magnitude of the magnetic moment is:

A
$\frac{L I^{2}}{2 π}$

B
$\frac{L I^{2}}{4 π}$

C
$\frac{L^{2} I}{2 π}$

D
$\frac{L^{2} I}{4 π}$

Solution

Magnetic moment $m = A I = {πr}^{2} I$ where $r$ is the radius of the circular loop. Now, the circumference of the circle = length of the wire, i.e.,

$2 πr = L$

${πr}^{2} = \frac{L^{2}}{4 π^{2}}$

Therefore, $m = {πr}^{2} I = \frac{{πL}^{2} I}{4 π^{2}} = \frac{L^{2} I}{4 π}$

Thus, the magnitude of the magnetic moment is $\frac{L^{2} I}{4 π}$

Hence the correct option is (D).
Question 8
1 / -0

The transverse displacement $y(x, t)$ of a wave on a string is given by $y(x, t)=e^{-\left(a x^{2}+b t^{2}+2 \sqrt{(a b)} x t\right)}$. This represents a:

A
Wave moving in + x-direction with speed $\sqrt{\frac{a}{b}}$

B
Wave moving in - x-direction with speed $\sqrt{\frac{b}{a}}$

C
Standing wave of frequency $\sqrt{b}$

D
Standing wave of frequency $\frac{1}{\sqrt{b}}$

Solution

According to the question,

$y(x, t)=e^{-\left(a x^{2}+b t^{2}+2 \sqrt{a b} x t\right)}$

$=e^{-(\sqrt{a} x+\sqrt{b} t)^{2}}$ ....(i)

Comparing equation (i) with standard equation.

$y(x, t)=f(a x+b t)$

As there is a positive sign between $x$ and $t$ terms. So, wave travel in $-x$ direction.

Wave speed$=\frac{\text { Coefficient of } t}{\text { Coefficient of } x}=\sqrt{\frac{b}{a}}$
Question 9
1 / -0

If a liquid is heated in space under no gravity, the transfer of heat will take place by process of:

A
Conduction

B
Convection

C
Radiation

D
None of these

Solution

If a liquid is heated in space under no gravity the transfer of heat will take place by process of radiation.

In the case of conduction and convection, the presence of gravity is crucial for these processes to happen.

In the case of radiation, it does not need gravity hence heat transfer can happen only by the process of radiation.
Question 10
1 / -0

Given below are two statements: One is labelled as Assertion (A) and the other is labelled as Reason (R).

Assertion (A):

The stretching of a spring is determined by the shear modulus of the material of the spring.

Reason (R):

A coil spring of copper has more tensile strength than a steel spring of same dimensions.

In the light of the above statements, choose the most appropriate answer from the options given below:

A
(A) is true but (R) is false

B
(A) is false but (R) is true

C
Both (A) and (R) are true and (R) is the correct explanation of (A)

D
Both (A) and (R) are true and (R) is not the correct explanation of (A)

Solution

From the concept of Hook's law, we have-

Normal Stress $(\sigma)=E \epsilon$

where, $E=$ Young's Modulus of Elasticity and $\epsilon$ is strain which gives linear deformation of the object.

Also, Shear Stress $(\tau)=G \gamma$

where, $G$ = Shear Modulus of Elasticity of the material and $\gamma$ is shear strain which gives angular deformation of the object.

When we stretch a spring, the length of the wire does not change but the coil experiences an angular twist. Hence shear modulus is used to determine the stretching of a spring.

$\therefore$ The assertion is True.

Also, we know that for a given dimension, Young's Modulus of Elasticity of steel is more than the Copper hence we can say that the tensile strength of Steel is more than that of $Cu$.

$\therefore$ The reason is False.

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Physics Test - 23

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Physics Test - 23

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Question 1

(A) and (B) only

(A) and (D) only

(B) and (E) only

(B) and (C) only

Solution

Question 2

2

0.5

0.33

None of these

Solution

Question 3

Contract

Expand

Move towards \(+v e x-a \xi s\)

Move towards \(-v e x-a \xi s\)

Solution

Question 4

186

280

225

256

Solution

Question 5

The average energy of each nucleon

The energy needed to split the nucleus into its parts

The mass-energy difference between protons and neutrons

The energy necessary to overcome the neutron-neutron repulsion

Solution

Question 6

\(F(x)=\frac{-d U(x)}{d x}\)

\(F(x)=\frac{d U(x)}{d x}\)

\(U(x)=\frac{d F(x)}{d x}\)

\(U(x)=\frac{-d F(x)}{d x}\)

Solution

Question 7

LI22π

LI24π

L2I2π

L2I4π

Solution

Question 8

Wave moving in + x-direction with speed \(\sqrt{\frac{a}{b}}\)

Wave moving in - x-direction with speed \(\sqrt{\frac{b}{a}}\)

Standing wave of frequency \(\sqrt{b}\)

Standing wave of frequency \(\frac{1}{\sqrt{b}}\)

Solution

Question 9

Conduction

Convection

Radiation

None of these

Solution

Question 10

(A) is true but (R) is false

(A) is false but (R) is true

Both (A) and (R) are true and (R) is the correct explanation of (A)

Both (A) and (R) are true and (R) is not the correct explanation of (A)

Solution

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$\frac{L I^{2}}{2 π}$

$\frac{L I^{2}}{4 π}$

$\frac{L^{2} I}{2 π}$

$\frac{L^{2} I}{4 π}$