SECTION - A [PHYSICS MCQ] - JEE STD 11 Science Questions - Vidyadip

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SECTION - A [PHYSICS MCQ]

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20 questions · self-marked practice — reveal the answer and mark yourself.

MCQ 14 Marks

Consider a circular loop that is uniformly charged and has a radius $\mathrm{a} \sqrt{2}$. Find the position along the positive z - axis of the cartesian coordinate system where the electric field is maximum if the ring was assumed to be placed in xy - plane at the origin :

A
$\frac{\mathrm{a}}{\sqrt{2}}$
B
$\frac{a}{2}$
C
a
D
$0$

Answer

C. a
$E=\frac{K Q r}{\left(x^{2}+R^{2}\right)^{3 / 2}}$
$\frac{\mathrm{dE}}{\mathrm{dx}}=0$
$\therefore \mathrm{x}=\frac{\mathrm{R}}{\sqrt{2}}=\frac{\sqrt{2} \mathrm{a}}{\sqrt{2}}=\mathrm{a}$

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MCQ 24 Marks

Match List-I with List-II.

List-I	List-II
(A) Heat capacity of body	(I) $\mathrm{J} \mathrm{kg}^{-1}$
(B) Specific heat capacity of body	(II) $\mathrm{JK}^{-1}$
(C) Latent heat	(III) $\mathrm{J} \mathrm{kg}^{-1} \mathrm{~K}^{-1}$
(D) Thermal conductivity	(IV) $\mathrm{Jm}^{-1} \mathrm{~K}^{-1} \mathrm{~s}^{-1}$

Choose the correct answer from the options given below :

A
(A)-(III), (B)-(I), (C)-(II), (D)-(IV)
B
(A)-(IV), (B)-(III), (C)-(II), (D)-(I)
C
(A)-(III), (B)-(IV), (C)-(I), (D)-(II)
D
(A)-(II), (B)-(III), (C)-(I), (D)-(IV)

Answer

D. (A)-(II), (B)-(III), (C)-(I), (D)-(IV)
$\mathrm{C}^{\prime}=\frac{\Delta \mathrm{Q}}{\Delta \mathrm{T}}=\mathrm{JK}^{-1}$
$\mathrm{S}=\frac{\Delta \mathrm{Q}}{\mathrm{m} \Delta \mathrm{T}}=\mathrm{Jkg}^{-1} \mathrm{~K}^{-1}$
$\mathrm{L}=\frac{\Delta \mathrm{Q}}{\mathrm{m}}=\mathrm{Jkg}^{-1}$
$\Delta \mathrm{Q}=\frac{\mathrm{KA} \Delta \mathrm{T}}{\mathrm{L}} \Rightarrow \mathrm{K}=\frac{\Delta \mathrm{Q}(\mathrm{L})}{\mathrm{A} \Delta \mathrm{T}}=\mathrm{Jm}^{-1} \mathrm{~K}^{-1} \mathrm{~s}^{-1}$

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MCQ 34 Marks

If $\mu_{0}$ and $\varepsilon_{0}$ are the permeability and permittivity of free space, respectively, then the dimension of $\left(\frac{1}{\mu_{0} \varepsilon_{0}}\right)$ is :

A
$L / T^{2}$
B
$L^{2} / T^{2}$
C
$\mathrm{T}^{2} / \mathrm{L}$
D
$\mathrm{T}^{2} / \mathrm{L}^{2}$

Answer

B. $L^{2} / T^{2}$
$\mathrm{C}=\frac{1}{\sqrt{\mu_{0} \varepsilon_{0}}} \Rightarrow \frac{1}{\mu_{0} \varepsilon_{0}}=\mathrm{C}^{2}=\mathrm{L}^{2} \mathrm{~T}^{-2}$

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MCQ 44 Marks

A bi-convex lens has radius of curvature of both the surfaces same as $1 / 6 \mathrm{~cm}$. If this lens is required to be replaced by another convex lens having different radii of curvatures on both sides ( $\mathrm{R}_{1} \neq \mathrm{R}_{2}$ ), without any change in lens power then possible combination of $R_{1}$ and $R_{2}$ is :

A
$\frac{1}{3} \mathrm{~cm}$ and $\frac{1}{3} \mathrm{~cm}$
B
$\frac{1}{5} \mathrm{~cm}$ and $\frac{1}{7} \mathrm{~cm}$
C
$\frac{1}{3} \mathrm{~cm}$ and $\frac{1}{7} \mathrm{~cm}$
D
$\frac{1}{6} \mathrm{~cm}$ and $\frac{1}{9} \mathrm{~cm}$

Answer

B. $\frac{1}{5} \mathrm{~cm}$ and $\frac{1}{7} \mathrm{~cm}$
This will happen when
$\frac{1}{f_{1}}=\frac{1}{\mathrm{f}_{2}}$
$(\mu-1)\left(\frac{1}{\mathrm{R}_{1}}-\frac{1}{-\mathrm{R}_{2}}\right)=(\mu-1)\left(\frac{2}{\mathrm{R}}\right)$
$\frac{1}{\mathrm{R}_{1}}+\frac{1}{\mathrm{R}_{2}}=\frac{2}{\mathrm{R}}$

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MCQ 54 Marks

A body of mass 1 kg is suspended with the help of two strings making angles as shown in figure. Magnitude of tensions $T_{1}$ and $T_{2}$, respectively, are (in N ) :

A
$5,5 \sqrt{3}$
B
$5 \sqrt{3}, 5$
C
$5 \sqrt{3}, 5 \sqrt{3}$
D
5,5

Answer

B. $5 \sqrt{3}, 5$

$\mathrm{T}_{1}=\mathrm{mgcos} 30^{\circ}$
$\mathrm{T}_{2}=\mathrm{mg} \sin 30^{\circ}$

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MCQ 64 Marks

A sportsman runs around a circular track of radius $r$ such that he traverses the path $A B A B$. The distance travelled and displacement, respectively, are

A
$2 \mathrm{r}, 3 \pi \mathrm{r}$
B
$3 \pi r, \pi r$
C
$\pi r, 3 r$
D
$3 \pi r, 2 r$

Answer

D. $3 \pi r, 2 r$
Displacement $=2 \mathrm{r}$
Distance $=2 \pi r+\pi r=3 \pi r$

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MCQ 74 Marks

Two identical objects are placed in front of convex mirror and concave mirror having same radii of curvature of 12 cm , at same distance of 18 cm from the respective mirrors. The ratio of sizes of the images formed by convex mirror and by concave mirror is :

A
$1 / 2$
B
2
C
3
D
$1 / 3$

Answer

Using $m=\frac{f}{u-f}$
$m_{1}=\frac{6}{18-6}=\frac{1}{2}$

$\mathrm{m}_{2}=\frac{6}{18+6}=\frac{1}{4} \quad \therefore \frac{\mathrm{~m}_{2}}{\mathrm{~m}_{1}}=\frac{1}{2}$

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MCQ 84 Marks

In the digital circuit shown in the figure, for the given inputs the P and Q values are :

A
$P=1, Q=1$
B
$\mathrm{P}=0, \mathrm{Q}=0$
C
$\mathrm{P}=0, \mathrm{Q}=1$
D
$P=1, Q=0$

Answer

B. $\mathrm{P}=0, \mathrm{Q}=0$

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MCQ 94 Marks

Energy released when two deuterons $\left({ }_{1} \mathrm{H}^{2}\right)$ fuse to form a helium nucleus $\left({ }_{2} \mathrm{He}^{4}\right)$ is :
(Given : Binding energy per nucleon of ${ }_{1} \mathrm{H}^{2}=1.1 \mathrm{MeV}$ and binding energy per nucleon of ${ }_{2} \mathrm{He}^{4}=7.0 \mathrm{MeV}$ )

A
8.1 MeV
B
5.9 MeV
C
23.6 MeV
D
26.8 MeV

Answer

$\mathrm{E}_{\mathrm{B}}=\mathrm{BE}_{\text {reactant }}-\mathrm{BE}_{\text {product }}$
$=1.1 \times 2+1.1 \times 2-7 \times 4=-23.6 \mathrm{MeV}$
$=\mathrm{Q}=23.6 \mathrm{MeV}$

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MCQ 104 Marks

Assuming the validity of Bohr's atomic model for hydrogen like ions the radius of $\mathrm{Li}^{++}$ion in its ground state is given by $\frac{1}{X} a_{0}$, where $X=$ __________ .
(Where $\mathrm{a}_{0}$ is the first Bohr's radius.)

A
2
B
1
C
3
D
9

Answer

C. 3
$\mathrm{r}=\mathrm{r}_{0} \frac{\mathrm{n}^{2}}{\mathrm{z}} \& \mathrm{z}=3$ for $\mathrm{Li}^{+2}$ and $\mathrm{n}=1$
$\therefore \mathrm{r}=\mathrm{r}_{0} \frac{1^{2}}{3}=\frac{\mathrm{r}_{0}}{3} \quad \therefore \mathrm{x}=3$

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MCQ 114 Marks

Identify the characteristics of an adiabatic process in a monoatomic gas.
(A) Internal energy is constant.
(B) Work done in the process is equal to the charge in internal energy.
(C) The product of temperature and volume is a constant.
(D) The product of pressure and volume is a constant.
(E) The work done to change the temperature from $\mathrm{T}_{1}$ to $\mathrm{T}_{2}$ is proportional to $\left(\mathrm{T}_{2}-\mathrm{T}_{1}\right)$
Choose the correct answer from the options given below :

A
(A), (C), (D) only
B
(A), (C), (E) only
C
(B), (E) only
D
(B), (D) only

Answer

C. (B), (E) only
$\mathrm{Q}=\Delta \mathrm{U}+\mathrm{W}=0 \Rightarrow-\Delta \mathrm{U}=\mathrm{W}$
$\mathrm{WD}=-\mathrm{nC}_{\mathrm{v}} \Delta \mathrm{T} \Rightarrow|\mathrm{WD}|=\mathrm{nC}_{\mathrm{v}} \Delta \mathrm{T} \propto \mathrm{T}_{2}-\mathrm{T}_{1}$
$\therefore \mathrm{B} \& \mathrm{E}$ [Only possibility]

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MCQ 124 Marks

Two large plane parallel conducting plates are kept 10 cm apart as shown in figure. The potential difference between them is V. The potential difference between the points A and B (shown in the figure) is :

A
$\frac{1}{4} \mathrm{~V}$
B
$\frac{2}{5} \mathrm{~V}$
C
$\frac{3}{4} \mathrm{~V}$
D
1 V

Answer

B. $\frac{2}{5} \mathrm{~V}$

Using $\Delta V=E(\Delta d)$
$\mathrm{V}=\mathrm{E}$ (10)
$\mathrm{V}_{\mathrm{AB}}=\mathrm{E} .4=\frac{\mathrm{V}}{10} \times 4=\frac{2 \mathrm{~V}}{5}$

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MCQ 134 Marks

A solenoid having area A and length '$l$' is filled with a material having relative permeability 2 . The magnetic energy stored in the solenoid is :

A
$\frac{\mathrm{B}^{2} \mathrm{~A} l}{\mu_{0}}$
B
$\frac{\mathrm{B}^{2} \mathrm{~A} l}{2 \mu_{0}}$
C
$\mathrm{B}^{2} \mathrm{Al}$
D
$\frac{\mathrm{B}^{2} \mathrm{~A} l}{4 \mu_{0}}$

Answer

D. $\frac{\mathrm{B}^{2} \mathrm{~A} l}{4 \mu_{0}}$
$\frac{U}{V}=\frac{B^{2}}{2 \mu_{r} u_{0}} \Rightarrow U=\frac{B^{2}}{4 \mu_{0}} V=\frac{B^{2}}{4 \mu_{0}} A \ell$

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MCQ 144 Marks

Given a charge $q$, current I and permeability of vacuum $\mu_{0}$. Which of the following quantity has the dimension of momentum?

A
$q \mathrm{I} / \mu_{0}$
B
$q \mu_{0} I$
C
$q^{2} \mu_{0} I$
D
$q \mu_{0} / I$

Answer

B. $q \mu_{0} I$
$\mathrm{Q}=\mathrm{AT}$
$\mathrm{I}=\mathrm{A}$
$\mu_{0}=\mathrm{MLT}^{-2} \mathrm{~A}^{-2}$
$\mathrm{P}=\mathrm{Q}^{\mathrm{x}} \mu_{0}^{\mathrm{y}} \mathrm{I}^{\mathrm{z}}=[\mathrm{AT}]^{\mathrm{x}}\left[\mathrm{MLT}^{-2} \mathrm{~A}^{-2}\right]^{\mathrm{y}}[\mathrm{A}]^{\mathrm{z}}$
$\mathrm{MLT}^{-1}=\mathrm{M}^{y} \mathrm{~L}^{\mathrm{y}} \mathrm{T}^{\mathrm{x}-2 \mathrm{y}} \mathrm{A}^{-2 \mathrm{y}+\mathrm{z+x}}$
Now; $\mathrm{y}=1$
$x-2 y=-1$
$-2 y+z=0$
$\therefore \mathrm{x}=\mathrm{y}=\mathrm{z}=1$

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MCQ 154 Marks

A sinusoidal wave of wavelength 7.5 cm travels a distance of 1.2 cm along the x -direction in 0.3 sec. The crest $P$ is at $x=0$ at $t=0 \mathrm{sec}$ and maximum displacement of the wave is 2 cm. Which equation correctly represents this wave ?

A
$y=2 \cos (0.83 x-3.35 t) \mathrm{cm}$
B
$y=2 \sin (0.83 x-3.5 t) \mathrm{cm}$
C
$y=2 \cos (3.35 x-0.83 t) \mathrm{cm}$
D
$y=2 \cos (0.13 x-0.5 t) \mathrm{cm}$

Answer

A. $y=2 \cos (0.83 x-3.35 t) \mathrm{cm}$
$\mathrm{v}=\frac{\text { dis tance }}{\text { time }}$
$\mathrm{v}=\frac{12}{0.3}=4 \mathrm{~cm} / \mathrm{s}$
$\mathrm{k}=\frac{2 \pi}{\lambda}=\frac{2 \pi}{7.5}=\frac{4 \pi}{15}=0.83$
$\mathrm{v}=\frac{\omega}{\mathrm{k}} \Rightarrow \omega=\mathrm{vk}=4 \times \frac{4 \pi}{15}=3.35$
So $\mathrm{y}=\mathrm{A} \cos (\mathrm{kx}-\omega \mathrm{t})$

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MCQ 164 Marks

An electron with mass '$m$' with an initial velocity $(t=0) \overrightarrow{\mathrm{v}}=\mathrm{v}_{0} \hat{\mathrm{i}}\left(\mathrm{v}_{0}>0\right)$ enters a magnetic field $\vec{B}=B_{0} \hat{j}$. If the initial de-Broglie wavelength at $t=0$ is $\lambda_{0}$ then its value after time 't' would be :

A
$\frac{\lambda_{0}}{\sqrt{1-\frac{\mathrm{e}^{2} \mathrm{~B}_{0}^{2} \mathrm{t}^{2}}{\mathrm{~m}^{2}}}}$
B
$\frac{\lambda_{0}}{\sqrt{1+\frac{\mathrm{e}^{2} \mathrm{~B}_{0}^{2} \mathrm{t}^{2}}{\mathrm{~m}^{2}}}}$
C
$\lambda_{0} \sqrt{1+\frac{\mathrm{e}^{2} \mathrm{~B}_{0}^{2} \mathrm{t}^{2}}{\mathrm{~m}^{2}}}$
D
$\lambda_{0}$

Answer

D. $\lambda_{0}$
Magnetic field does not work
$\therefore$ Speed will not charge, so De-Broglie wavelength remains same.

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MCQ 174 Marks

Two water drops each of radius 'r' coalesce to from a bigger drop. If 'T' is the surface tension, the surface energy released in this process is :

A
$4 \pi r^{2} T\left[2-2^{\frac{2}{3}}\right]$
B
$4 \pi r^{2} T\left[2-2^{\frac{1}{3}}\right]$
C
$4 \pi r^{2} \mathrm{~T}[1+\sqrt{2}]$
D
$4 \pi r^{2} \mathrm{~T}[\sqrt{2}-1]$

Answer

A. $4 \pi r^{2} T\left[2-2^{\frac{2}{3}}\right]$
$2 \times \frac{4}{3} \pi R^{3}=\frac{4}{3} \pi r^{3} \Rightarrow r=2^{1 / 3} R$
$\mathrm{U}_{\mathrm{i}}=2 \times 4 \pi \mathrm{R}^{2} \mathrm{~T}$
$\mathrm{U}_{\mathrm{f}}=4 \pi \mathrm{r}^{2} \mathrm{~T}=4 \pi \mathrm{R}^{2} \mathrm{~T}^{2 / 3}$
$\therefore$ Heat lost $=u_{i}-u_{f}=4 \pi R^{2} T\left[2-2^{2 / 3}\right]$

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MCQ 184 Marks

The moment of inertia of a circular ring of mass M and diameter r about a tangential axis lying in the plane of the ring is :

A
$\frac{1}{2} \mathrm{Mr}^{2}$
B
$\frac{3}{8} \mathrm{Mr}^{2}$
C
$\frac{3}{2} \mathrm{Mr}^{2}$
D
$2 \mathrm{Mr}^{2}$

Answer

B. $\frac{3}{8} \mathrm{Mr}^{2}$
Diameter is given as R.
$\therefore$ Radius $=\mathrm{R} / 2$
$\mathrm{I}_{\text {tan gent }}=\frac{3}{2} \mathrm{~m}\left(\frac{\mathrm{R}}{2}\right)^{2}=\frac{3}{8} \mathrm{mR}^{2}$

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MCQ 194 Marks

In a moving coil galvanometer, two moving coils $\mathrm{M}_{1}$ and $\mathrm{M}_{2}$ have the following particulars :
$\mathrm{R}_{1}=5 \Omega, \mathrm{~N}_{1}=15, \mathrm{~A}_{1}=3.6 \times 10^{-3} \mathrm{~m}^{2}, \mathrm{~B}_{1}=0.25 \mathrm{~T}$
$\mathrm{R}_{2}=7 \Omega, \mathrm{~N}_{2}=21, \mathrm{~A}_{2}=1.8 \times 10^{-3} \mathrm{~m}^{2}, \mathrm{~B}_{2}=0.50 \mathrm{~T}$
Assuming that torsional constant of the springs are same for both coils, what will be the ratio of voltage sensitivity of $M_{1}$ and $M_{2}$ ?

A
$1: 1$
B
$1: 4$
C
$1: 3$
D
$1: 2$

Answer

A. $1: 1$
Voltage sensitivity $=\frac{\theta}{\mathrm{V}}=\frac{\mathrm{NAB}}{\mathrm{cR}}$
Ratio $=\left(\frac{\mathrm{N}_{1} \mathrm{~A}_{1} \mathrm{~B}_{1}}{\mathrm{~N}_{2} \mathrm{~A}_{2} \mathrm{~B}_{2}}\right) \frac{\mathrm{R}_{2}}{\mathrm{R}_{1}}=\frac{15 \times 3.6 \times 0.25}{21 \times 1.8 \times 0.5} \times \frac{7}{5}=\frac{1}{1}$

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MCQ 204 Marks

Given below are two statements : one is labelled as Assertion (A) and the other is labelled as Reason (R).
Assertion (A) : Net dipole moment of a polar linear isotropic dielectric substance is not zero even in the absence of an external electric field.
Reason (R) : In absence of an external electric field, the different permanent dipoles of a polar dielectric substance are oriented in random directions.
In the light of the above statements, choose the most appropriate answer from the options given below :

A
(A) is correct but $(\mathbf{R})$ is not correct
B
Both (A) and (R) are correct but (R) is not the correct explanation of (A)
C
Both (A) and (R) are correct and (R) is the correct explanation of (A)
D
(A) is not correct but ( $\mathbf{R}$ ) is correct

Answer

D. (A) is not correct but ( $\mathbf{R}$ ) is correct
A : Since polar dielectrics are randomly oriental $\overrightarrow{\mathrm{P}}_{\mathrm{net}}=\overrightarrow{0}$.
$R$ : If $\vec{E}$ is absent, polar dielectric remain polar & are randomly oriented.

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