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

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SECTION - A [PHYSICS MCQ]

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20 questions · auto-graded multiple-choice test.

MCQ 14 Marks

A monochromatic light is incident on a metallic plate having work function $\phi$. An electron, emitted normally to the plate from a point A with maximum kinetic energy, enters a constant magnetic field, perpendicular to the initial velocity of electron. The electron passes through a curve and hits back the plate at a point B. The distance between A and B is:
(Given: The magnitude of charge of an electron is e and mass is m, h is Planck's constant and c is velocity of light. Take the magnetic field exists throughout the path of electron)

A
$\sqrt{2 m\left(\frac{ hc }{\lambda}-\phi\right)} / eB$
B
$\sqrt{ m \left(\frac{ hc }{\lambda}-\phi\right)} / eB$
✓
$\sqrt{8 m\left(\frac{ hc }{\lambda}-\phi\right)} / eB$
D
$2 \sqrt{m\left(\frac{ hc }{\lambda}-\phi\right)} / eB$

Answer

Correct option: C.

$\sqrt{8 m\left(\frac{ hc }{\lambda}-\phi\right)} / eB$

(C) $\sqrt{8 m\left(\frac{ hc }{\lambda}-\phi\right)} / eB$
Explanation: $KE _{\max }=\frac{ hc }{\lambda}-\phi$
$p =\sqrt{2 mK _{\max }}$
$p =\sqrt{2 m\left(\frac{ hc }{\lambda}-\phi\right)}$
$d_{A-B}=2 R$
$=2\left[\frac{ p }{ qB }\right]$
$d _{ AB }=\frac{2 \sqrt{2 m\left(\frac{ hc }{\lambda}-\phi\right)}}{ eB }=\frac{\sqrt{8 m\left(\frac{ hc }{\lambda}-\phi\right)}}{ eB }$

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MCQ 24 Marks

A small bob of mass 100 mg and charge $+10 \mu C$ is connected to an insulating string of length 1 m . It is brought near to an infinitely long nonconducting sheet of charge density ' $\sigma$ ' as shown in figure. If string subtends an angle of $45^{\circ}$ with the sheet at equilibrium the charge density of sheet will be :
(Given, $\varepsilon_0=8.85 \times 10^{-12} \frac{F}{ m }$ and acceleration due to gravity, $g =10 m / s ^2$ )

A
$0.885 nC / m ^2$
B
$17.7 nC / m ^2$
C
$885 nC / m ^2$
✓
$1.77 nC / m ^2$

Answer

Correct option: D.

$1.77 nC / m ^2$

(D) $1.77 nC / m ^2$
Explanation:

$\begin{array}{l} qE = mg \\ q
\left[\frac{\sigma}{2 \varepsilon_0}\right]= mg \end{array}$
$\sigma=\frac{2 \varepsilon_0 mg }{ q }$
$\sigma=\frac{2 \times 8.85 \times 10^{-12} \times 100 \times 10^{-6} \times 10}{10 \times 10^{-6}}$
$\sigma=17.7 \times 10^{-10} C / m ^2$
$\sigma=1.77 nC / m ^2$

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MCQ 34 Marks

List-I	List-II
(A) Coefficient of viscosity	(I) $\left[ ML ^0 T^{-3}\right]$
(B) Intensity of wave	(II) $\left[ ML ^{-2} T^{-2}\right]$
(C) Pressure gradient	(III) $\left[ M ^{-1} LT ^2\right]$
(D) Compressibility	(IV) $\left[ ML ^{-1} T^{-1}\right]$

Choose the correct answer from the options given below:

A
(A)-(1), (B)-(IV), (C)-(III), (D)-(II)
✓
(A)-(IV), (B)-(I), (C)-(II), (D)-(III)
C
(A)-(IV), (B)-(II), (C)-(I), (D)-(III)
D
(A)-(II), (B)-(III), (C)-(IV), (D)-(I)

Answer

Correct option: B.

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

(B) (A)-(IV), (B)-(I), (C)-(II), (D)-(III)
Explanation: (A)Coefficient of viscosity
$[\eta]=\left[ M ^1 L^{-1} T^{-1}\right]$
(B) Intensity $[ I ]=\left[ M ^1 L^0 T^{-3}\right]$
(C) Pressure gradient $=\left[ ML ^{-2} T^{-2}\right]$
(D) Compressibility $[ K ]=\left[ M ^{-1} L^1 T^2\right]$

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MCQ 44 Marks

A spherical surface separates two media of refractive indices 1 and 1.5 as shown in figure. Distance of the image of an object ' O ', is :
( C is the center of curvature of the spherical surface and $R$ is the radius of curvature)

A
0.24 m right to the spherical surface
✓
0.4 m left to the spherical surface
C
0.24 m left to the spherical surface
D
0.4 m right to the spherical surface

Answer

Correct option: B.

0.4 m left to the spherical surface

(B) 0.4 m left to the spherical surface
Explanation:
$\frac{\mu_2}{ v }-\frac{\mu_1}{ u }=\frac{\mu_2-\mu_1}{ R }$
$\frac{1.5}{ v }-\frac{1}{(-0.2)}=\frac{1.5-1}{0.4}$
$\frac{1.5}{ v }=\frac{0.5}{0.4}-\frac{1}{0.2}$
$\frac{1.5}{ v }=-\frac{1.5}{0.4}$
$v =-0.4 m$

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MCQ 54 Marks

Moment of inertia of a rod of mass ' M ' and length 'L' about an axis passing through its center and normal to its length is ' $\alpha$ '. Now the rod is cut into two equal parts and these parts are joined symmetrically to form a cross shape. Moment of inertia of cross about an axis passing through its center and normal to plane containing cross is :

A
$\alpha$
✓
$\alpha / 4$
C
$\alpha / 8$
D
$\alpha / 2$

Answer

Correct option: B.

$\alpha / 4$

(B) $\alpha / 4$
Explanation:

$\alpha=\frac{M \ell^2}{12}$ ....(i)

$\alpha^{\prime}=2\left[\frac{\frac{ M }{2}\left(\frac{\ell}{2}\right)^2}{12}\right]$
$\alpha^{\prime}=\frac{ M \ell^2}{48}=\frac{\alpha}{4}$
Correct option is (B)

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MCQ 64 Marks

Considering Bohr's atomic model for hydrogen atom :
(A) the energy of H atom in ground state is same as energy of $He ^{+}$ion in its first excited state.
(B) the energy of H atom in ground state is same as that for $Li ^{++}$ion in its second excited state.
(C) the energy of H atom in its ground state is same as that of $He ^{+}$ion for its ground state.
(D) the energy of $He ^{+}$ion in its first excited state is same as that for $Li ^{++}$ion in its ground state
Choose the correct answer from the options given below :

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

Answer

Correct option: B.

(A), (B) only

(B) (A), (B) only
Explanation:
$\begin{array}{l}E \propto \frac{Z}{n^2}
\\Z_{H}=1 \quad Z_{He^{+}}=2 \quad Z_{Li^{+2}}=3\end{array}$
$1^{\text {st }}$ excited state $\Rightarrow n =2$
$2^{\text {nd }}$ excited state $\Rightarrow n =3$
From the given statements only A & B are correct.

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MCQ 74 Marks

Let $B_1$ be the magnitude of magnetic field at center of a circular coil of radius $R$ carrying current $I$. Let $B_2$ be the magnitude of magnetic field at an axial distance ' $x$ ' from the center. For $x: R=3: 4, \frac{B_2}{B_1}$ is :

A
$4: 5$
B
$16: 25$
✓
$64: 125$
D
$25: 16$

Answer

Correct option: C.

$64: 125$

(C) $64: 125$
Explanation:

$B_1=\frac{\mu_0 i }{2 R} \quad B_2=B_1 \sin ^3 \theta$
$\therefore \frac{ B _2}{B_1}=\sin ^3 \theta=\left(\frac{4}{5}\right)^3=\frac{64}{125}$

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MCQ 84 Marks

A square Lamina OABC of length 10 cm is pivoted at ' O '. Forces act at Lamina as shown in figure. If Lamina remains stationary, then the magnitude of F is :

A
20 N
B
0 (zero)
✓
10 N
D
$10 \sqrt{2} N$

Answer

Correct option: C.

10 N

(C) 10 N
Explanation:Since the lamina is equilibrium.
$\therefore F_{net}=0 \& \tau_{net}=0$

$T _{ o }=10 \ell- F \ell \Rightarrow F =10 N$

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MCQ 94 Marks

In an adiabatic process, which of the following statements is true?

A
The molar heat capacity is infinite
B
Work done by the gas equals the increase in internal energy
✓
The molar heat capacity is zero
D
The internal energy of the gas decreases as the temperature increases

Answer

Correct option: C.

The molar heat capacity is zero

(C) The molar heat capacity is zero
Explanation: For adiabatic process, $dQ =0$
$\therefore$ Molar heat capacity $=0$
$\because dQ=0 \Rightarrow dU=-dW$
Also $dU =\frac{ f }{2} nRdT$
$\therefore$ Only option (3) is correct.

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MCQ 104 Marks

A zener diode with 5 V zener voltage is used to regulate an unregulated dc voltage input of 25 V . For a $400 \Omega$ resistor connected in series, the zener current is found to be 4 times load current. The load current $\left(I_L\right)$ and load resistance $\left(R_L\right)$ are :

A
$I _{ L }=20 mA ; R _{ L }=250 \Omega$
B
$I _{ L }=10 A ; R _{ L }=0.5 \Omega$
C
$I _{ L }=0.02 mA ; R _{ L }=250 \Omega$
✓
$I _{ L }=10 mA ; R _{ L }=500 \Omega$

Answer

Correct option: D.

$I _{ L }=10 mA ; R _{ L }=500 \Omega$

(D) $I _{ L }=10 mA ; R _{ L }=500 \Omega$
Explanation:

From the circuit diagram,
$\begin{array}{l}5 i=\frac{20}{400}=\frac{1}{20} A \\
\therefore i=\frac{1}{100} A=10 mA=\text { Load current }\end{array}$
Also, $V _{ L }=5 V$
$\therefore R_{L}=\frac{5}{10 \times 10^{-3}} \Omega=500 \Omega$

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MCQ 114 Marks

The relationship between the magnetic susceptibility $(\chi)$ and the magnetic permeability $(\mu)$ is given by :
( $\mu_0$ is the permeability of free space and $\mu_r$ is relative permeability)

✓
$\chi=\frac{\mu}{\mu_0}-1$
B
$\chi=\frac{\mu_{ r }}{\mu_0}+1$
C
$\chi=\mu_r+1$
D
$\chi=1-\frac{\mu}{\mu_0}$

Answer

Correct option: A.

$\chi=\frac{\mu}{\mu_0}-1$

(A) $\chi=\frac{\mu}{\mu_0}-1$
Explanation: We have
$\mu_{r}=(1+\chi) \Rightarrow \chi=\left(\mu_{r}-1\right)$
$\mu=\mu_0 \mu_{ r } \Rightarrow \mu_{ r }=\frac{\mu}{\mu_0}$
$\therefore \chi=\left(\frac{\mu}{\mu_0}-1\right)$

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MCQ 124 Marks

A particle is subjected two simple harmonic motions as :
$x_1=\sqrt{7} \sin 5 t cm$
and $x_2=2 \sqrt{7} \sin \left(5 t+\frac{\pi}{3}\right) cm$
where x is displacement and $t$ is time in seconds.
The maximum acceleration of the particle is $x \times 10^{-2} ms^{-2}$. The value of x is :

✓
175
B
$25 \sqrt{7}$
C
$5 \sqrt{7}$
D
125

Answer

Correct option: A.

175

(A) 175
Explanation:
$\begin{array}{l}x_1=\sqrt{7} \sin 5 t \\
x _2=2 \sqrt{7} \sin \left(5 t+\frac{\pi}{3}\right)\end{array}$
From phasor,

$\therefore$ Amplitude of resultant $SHM =7$
$\begin{array}{l}\phi=\tan ^{-1} \frac{2 \sqrt{7} \times \sqrt{3} / 2}{\sqrt{7}+2 \sqrt{7} \times \frac{1}{2}}=\tan ^{-1} \frac{\sqrt{21}}{2 \sqrt{7}}=\tan ^{-1} \frac{\sqrt{3}}{2} \\
\therefore X_R=7 \sin (5 t+\phi) \\
\quad a_{R}=-7 \times 25 \sin (5 t+\phi) \\
\therefore a_{\max }=175 cm / sec=175 \times 10^{-2} m / sec\end{array}$

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MCQ 134 Marks

The battery of a mobile phone is rated as 4.2 V , 5800 mAh . How much energy is stored in it when fully charged?

A
43.8 kJ
B
48.7 kJ
✓
87.7 kJ
D
24.4 kJ

Answer

Correct option: C.

87.7 kJ

(C) 87.7 kJ
Explanation: Given $V =4.2$ volt
$\therefore$ Energy supplied by battery
$=vq=4.2 \times 5800 \times 3600 \times 10^{-3} J=87.696 kJ$
$\therefore$ Energy stored in the battery when fully charged
$=87.696 kJ \approx 87.7 kJ$

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MCQ 144 Marks

A point charge + q is placed at the origin. A second point charge +9 q is placed at $( d , 0,0)$ in Cartesian coordinate system. The point in between them where the electric field vanishes is :

A
$(4 d / 3,0,0)$
✓
$( d / 4,0,0)$
C
$(3 d / 4,0,0)$
D
$( d / 3,0,0)$

Answer

Correct option: B.

$( d / 4,0,0)$

(B) $( d / 4,0,0)$
Explanation:

Let $E _{ p }=0$
$\begin{array}{l}\therefore \frac{kq}{x^2}=\frac{k 9 q}{(d-x)^2} \\
\Rightarrow \frac{d-x}{x}=3 \Rightarrow x=\frac{d}{4}\end{array}$
$\therefore$ co-ordinate of P is $\left(\frac{ d }{4}, 0,0\right)$

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MCQ 154 Marks

A river is flowing from west to east direction with speed of $9 km h ^{-1}$. If a boat capable of moving at a maximum speed of $27 km h ^{-1}$ in still water, crosses the river in half a minute, while moving with maximum speed at an angle of $150^{\circ}$ to direction of river flow, then the width of the river is :

A
300 m
✓
112.5 m
C
75 m
D
$112.5 \times \sqrt{3} m$

Answer

Correct option: B.

112.5 m

(B) 112.5 m
Explanation:

$\therefore V_{\perp}=\text { river flow }=27 \times \cos 60^{\circ}=\frac{27}{2} km / hr .$
Time taken $=30 sec$.
$\therefore S=Vt=\frac{27}{2} \times \frac{5}{18} \times 30 m=112.5 m$

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MCQ 164 Marks

Consider two infinitely large plane parallel conducting plates as shown below. The plates are uniformly charged with a surface charge density $+\sigma$ and $-2 \sigma$. The force experienced by a point charge + q placed at the mid point between two plates will be :

A
$\frac{\sigma q}{4 \in_0}$
✓
$\frac{3 \sigma q}{2 \in_0}$
C
$\frac{3 \sigma q}{4 \in_0}$
D
$\frac{\sigma q}{2 \in_0}$

Answer

Correct option: B.

$\frac{3 \sigma q}{2 \in_0}$

(B) $\frac{3 \sigma q}{2 \in_0}$
Explanation:

Final charge distribution will be

$\therefore F _{ net }=\frac{3 \sigma}{2 \in_0} q$

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MCQ 174 Marks

A slanted object $A B$ is placed on one side of convex lens as shown in the diagram. The image is formed on the opposite side. Angle made by the image with principal axis is :

A
$-\frac{\alpha}{2}$
✓
$-45^{\circ}$
C
$+45^{\circ}$
D
$-\alpha$

Answer

Correct option: B.

$-45^{\circ}$

(B) $-45^{\circ}$
Explanation:

Location of image of A :-
$\frac{1}{v}-\frac{1}{u}=\frac{1}{f} \Rightarrow \frac{1}{v}-\frac{1}{-30}=\frac{1}{20} \Rightarrow \frac{1}{v}=\frac{1}{60} \Rightarrow v=60 cm$
$\therefore m =2$
Since size of object is small wrt the location hence
$\begin{array}{l}d v=m^2 d u \Rightarrow d v=4 \times 1=4 cm \\
h_i=mh_0 \Rightarrow h_i(d y)=2 \times 2=4 cm\end{array}$
$\therefore$ Angle made with principle axis $=-45^{\circ}$

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MCQ 184 Marks

A cord of negligible mass is wound around the rim of a wheel supported by spokes with negligible mass. The mass of wheel is 10 kg and radius is 10 cm and it can freely rotate without any friction. Initially the wheel is at rest. If a steady pull of 20 N is applied on the cord, the angular velocity of the wheel, after the cord is unwound by 1 m , would be :

✓
$20 rad / s$
B
$30 rad / s$
C
$10 rad / s$
D
$0 rad / s$

Answer

Correct option: A.

$20 rad / s$

(A) 20 rad/s
Explanation:
$\begin{array}{l} W _{ F }=20 \times 1=20 J \\
\therefore \quad \Delta KE =20 J=\frac{1}{2} I \omega^2 \\
I = MR ^2=10 \times 0.1^2=0.1 kg m ^2 \\
\therefore \quad 20=\frac{1}{2} \times 0.1 \times \omega^2 \\
\Rightarrow \omega=20 rad / sec \end{array}$

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MCQ 194 Marks

The equation for real gas is given by $\left( P +\frac{ a }{ V ^2}\right)( V - b )= RT$, where $P , V , T$ and R are the pressure, volume, temperature and gas constant, respectively. The dimension of $ab ^{-2}$ is equivalent to that of:

A
Planck's constant
B
Compressibility
C
Strain
✓
Energy density

Answer

Correct option: D.

Energy density

(D) Energy density
Explanation: $\left[ P +\frac{ a }{ V ^2}\right]( V - b )= RT$
$\therefore[ a ]=[ P ]\left[ V ^2\right]= ML ^{-1} T^{-2} L^6= ML ^5 T^{-2}$
$[ b ]=[ V ]= L ^3$
$\left[ ab ^{-2}\right]= ML ^5 T^{-2} L^{-6}= ML ^{-1} T^{-2}$
Dimension of energy density.

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MCQ 204 Marks

A light wave is propagating with plane wave fronts of the type $x+y+z=$ constant. The angle made by the direction of wave propagation with the $x$-axis is :

✓
$\cos ^{-1}\left(\frac{1}{\sqrt{3}}\right)$
B
$\cos ^{-1}\left(\frac{2}{3}\right)$
C
$\cos ^{-1}\left(\frac{1}{3}\right)$
D
$\cos ^{-1}\left(\sqrt{\frac{2}{3}}\right)$

Answer

Correct option: A.

$\cos ^{-1}\left(\frac{1}{\sqrt{3}}\right)$

(A) $\cos ^{-1}\left(\frac{1}{\sqrt{3}}\right)$
Explanation: The direction of propagation of light is perpendicular to the wave front and is symmetric about $x , y$ and z axis.
$\therefore$ Angle made by the light with $x , y \& z$ axis is same.
$\therefore \cos \alpha=\cos \beta=\cos \gamma(\alpha, \beta \& \gamma$ are angle made by light with $x , y \& z$ axis respectively)
Also $\cos ^2 \alpha+\cos ^2 \beta+\cos ^2 \gamma=1$ [Sum of direction cosines]
$\therefore \alpha=\cos ^{-1} \frac{1}{\sqrt{3}}$

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