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

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SECTION - A [PHYSICS MCQ]

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20 questions · 16 auto-graded MCQ + 4 self-marked written.

MCQ 14 Marks

In an electromagnetic system, a quantity defined as the ratio of electric dipole moment and magnetic dipole moment has dimension of $\left[ M ^{ P } L ^{ Q } T ^{ R } A ^{ S }\right]$. The value of P and Q are :

A
$-1,0$
B
$-1,1$
C
$1,-1$
✓
$0,-1$

Answer

Correct option: D.

$0,-1$

(D) $0,-1$
Electric dipole moment $(\overrightarrow{ P })= q \times 2 \ell$
Magnetic dipole moment $(\overrightarrow{ M })= IA$
$
\left[\frac{P}{M}\right]=\left[\frac{LTA}{L^2 A}\right]=L^{-1} T=M^0 L^{-1} T^1 A^0
$
After compering values of $P$ $\&$ $Q$ are $0,-1$

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MCQ 24 Marks

A finite size object is placed normal to the principal axis at a distance of 30 cm from a convex mirror of focal length 30 cm. A plane mirror is now placed in such a way that the image produced by both the mirrors coincide with each other. The distance between the two mirrors is :

A
45 cm
✓
7.5 cm
C
22.5 cm
D
15 cm

Answer

Correct option: B.

7.5 cm

(B) 7.5 cm

For Convex mirror
$
\begin{array}{l}
\frac{1}{v}+\frac{1}{u}=\frac{1}{f} \\
\frac{1}{v}-\frac{1}{30}=\frac{1}{30} \\
\frac{1}{v}=\frac{2}{30}=\frac{1}{15} \Rightarrow v=15 cm
\end{array}
$
Image formed by convex mirror is at 45 cm from object so plane mirror should be placed midway at 22.5 cm from object so that both of their images may coinside,
Therefore distance between both mirrors
$
=30-22.5=7.5 cm
$

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MCQ 34 Marks

Displacement of a wave is expressed as $x(t)=5 \cos \left(628 t+\frac{\pi}{2}\right) m$. The wavelength of the wave when its velocity is $300 m / s$ is :

A
5 m
✓
3 m
C
0.5 m
D
0.33 m

Answer

Correct option: B.

3 m

(B) 3 m
$x ( t )=5 \cos \left[628 t +\frac{\pi}{2}\right] m $
$\text {velocity}\left( v _{\omega}\right)=300 m / s $
$v _{ w }=\frac{\omega}{ K } $
$300=\frac{628}{K} \Rightarrow K =\frac{628}{300} $
$\frac{2 \pi}{\lambda}=\frac{628}{300} \Rightarrow \lambda=\frac{2 \times 3.14 \times 300}{628} $
$\lambda=2 m$

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MCQ 44 Marks

Match List-I with List-II.

List-I	List-II
(A) Isobaric	(I) $\Delta Q =\Delta W$
(B) Isochoric	(II) $\Delta Q=\Delta U$
(C) Adiabatic	(III) $\Delta Q=$ zero
(D) Isothermal	(IV) $\Delta Q =\Delta U + P \Delta V$

$\Delta Q =$ Heat supplied
$\Delta W =$ Work done by the system
$\Delta U =$ Change in internal energy
$P =$ Pressure of the system
$\Delta V =$ Change in volume of the system
Choose the correct answer from the options given below :

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

Answer

Correct option: C.

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

(C) (A)-(IV), (B)-(II), (C)-(III), (D)-(I)
(A) Isobaric $( P = C )$
$
\Delta Q=\Delta U+P \Delta V
$
(B) Isochoric $( V = C )$
$
\Delta Q=\Delta U
$
(C) Adiabatic $(\Delta Q =0)$
$
\Delta Q=0
$
(D) Isothermal $(\Delta U =0)$
$
\Delta Q=\Delta W
$

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MCQ 54 Marks

Consider a n-type semiconductor in which $n _{ e }$ and $n _{ h }$ are number of electrons and holes, respectively.
(A) Holes are minority carriers
(B) The dopant is a pentavalent atom
(C) $n _{ c } n _{ h } \neq n _{ i }^2$
(where $n_i$ is number of electrons or holes in semiconductor when it is intrinsic form)
(D) $n _{ e } n _{ h } \geq n _{ i }^2$
(E) The holes are not generated due to the donors
Choose the correct answer from the options given below :

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

Answer

Correct option: C.

(A), (B), (E) only

(C) (A), (B), (E) only
(A) n type semiconductor holes are minority carriers and $e ^{-}$are majority carriers
(B) Dopant are pentavalent atom.
(C) $n _{ c } \cdot n _{ h }= n _{ i }^2$ for intrinsic semiconductor
(E) In $n$ type semiconductor primary source of holes generation are thermal excitation.

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MCQ 64 Marks

Two polarisers $P_1$ and $P_2$ are placed in such a way that the intensity of the transmitted light will be zero. A third polariser $P_3$ is inserted in between $P_1$ and $P_2$, at the particular angle between $P_2$ and $P_3$. The transmitted intensity of the light passing the through all three polarisers is maximum. The angle between the polarisers $P _2$ and $P _3$ is :

✓
$\frac{\pi}{4}$
B
$\frac{\pi}{6}$
C
$\frac{\pi}{8}$
D
$\frac{\pi}{3}$

Answer

Correct option: A.

$\frac{\pi}{4}$

(A) $\frac{\pi}{4}$
Through $P_2 I_1=I_0 \sin ^2\left(\frac{\pi}{2}-\theta\right)$

$
\begin{aligned}
I_1 & =I_0 \cos ^2 \theta \\
\text { Through } P_3 \quad I_{\text {net }} & =\left(I_0 \cos ^2 \theta\right) \sin ^2 \theta \\
I_{\text {net }} & =\frac{I_0}{4}[\sin (2 \theta)]^2 \text { for } \max I_{\text {net }} \theta=45^{\circ}
\end{aligned}
$
So angle between $P_2$ and $P_3=\frac{\pi}{4}$

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MCQ 74 Marks

A block of mass 25 kg is pulled along a horizontal surface by a force at an angle $45^{\circ}$ with the horizontal. The friction coefficient between the block and the surface is 0.25 . The displacement of 5 m of the block is:

A
970 J
B
735 J
C
245 J
D
490 J

Answer

Block travels with uniform velocity
So $\quad a =0 \Rightarrow F \cos 45^{\circ}=$ friction
$\frac{F}{\sqrt{2}}=\mu\left[mg-\frac{F}{\sqrt{2}}\right] $
$\frac{F}{\sqrt{2}}=0.25\left[25 \times 9.8-\frac{F}{\sqrt{2}}\right]$
$\Rightarrow 1.25 \frac{F}{\sqrt{2}}=61.25 $
$F=\frac{61.25 \times \sqrt{2}}{1.25}=49 \sqrt{2} $
$W_{ ext }= FS \cos 45^{\circ} $
$=49 \sqrt{2} \times 5 \times \frac{1}{\sqrt{2}}=245 J$

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MCQ 84 Marks

For the determination of refractive index of glass slab, a travelling microscope is used whose main scale contains 300 equal divisions equals to 15 cm . The vernier scale attached to the microscope has 25 divisions equals to 24 divisions of main scale. The least count (LC) of the travelling microscope is (in cm) :

A
0.001
✓
0.002
C
0.0005
D
0.0025

Answer

Correct option: B.

0.002

(B) 0.002
$300$ $msd =15$ $cm $
$1$ $ msd =\frac{15}{300} $ $cm=0.05$ $ cm $
$25$ $ vsd =24$ $ msd $
$1$ $ vsd =\frac{24}{25}$ $ msd $
$LC =1 $ $msd -1$ $ vsd$
$LC =1$ $ msd -\frac{24}{25}$ $ msd =\frac{1}{25} $ $msd $
$LC =\frac{1}{25} \times 0.05=0.002$ $ cm$

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MCQ 94 Marks

A wheel is rolling on a plane surface. The speed of a particle on the highest point of the rim is $8 m / s$. The speed of the particle on the rim of the wheel at the same level as the centre of wheel, will be :

✓
$4 \sqrt{2} m / s$
B
$8 m / s$
C
$4 m / s$
D
$8 \sqrt{2} m / s$

Answer

Correct option: A.

$4 \sqrt{2} m / s$

(A) $4 \sqrt{2} m / s$

If $V_B=2 V$
Point A is instantaneous center of rotation
Given $V_B=8 m / s$
$
\begin{array}{l}
V=4 m / s \\
V_{P}=\sqrt{2} V \Rightarrow V_{p}=4 \sqrt{2} m / s
\end{array}
$

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MCQ 104 Marks

The displacement $x$ versus time graph is shown below.

(A) The average velocity during 0 to 3 s is $10 m / s$
(B) The average velocity during 3 to 5 s is $0 m / s$
(C) The instantaneous velocity at $t =2 s$ is $5 m / s$
(D) The average velocity during 5 to 7 s and instantaneous velocity at $t=6.5 s$ are equal
(E) The average velocity from $t=0$ to $t=9 s$ is zero
Choose the correct answer from the options given below:

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

Answer

Correct option: D.

(B), (C), (E) only

(D) (B), (C), (E) only
$
\begin{array}{l}
\langle\overrightarrow{v}\rangle=\frac{\Delta \overrightarrow{s}}{\Delta t}=\frac{S_{f}-S_{i}}{t_{f}-t_{i}} \\
\overrightarrow{v}=\frac{ds}{dt}=\text { slope }
\end{array}
$
(A) 0 to 3 sec ; $\langle\vec{v}\rangle=\frac{5-0}{3}=5 / 3 m / s$
(B) 0 to 5 sec ; $\langle\vec{v}\rangle=\frac{5-5}{2}=0$
(C) t = 2 ; slope $=\overrightarrow{ v }=5 m / s$
(D) t = 5 to 7 sec ; $\langle\overrightarrow{ v }\rangle=\frac{0-5}{2}=-2.5 m / s$
At t = 6.5 sec ; $\overrightarrow{ v }=10$
(E) t = 0 to t = 9 ; $\langle\overrightarrow{ v }\rangle=0$

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MCQ 114 Marks

Three parallel plate capacitors $C _1, C _2$ and $C _3$ each of capacitance $5 \mu F$ are connected as shown in figure. The effective capacitance between points A and B , when the space between the parallel plates of $C _1$ capacitor is filled with a dielectric medium having dielectric constant of 4 , is :

A
$22.5 \mu F$
B
$7.5 \mu F$
✓
$9 \mu F$
D
$30 \mu F$

Answer

Correct option: C.

$9 \mu F$

(C) $9 \mu F$
After dielectric
$C_1=4 C $
$C_1=4 \times 5=20 \mu F $
$C_2=C_3=5 \mu F$
$C _1$ $\&$ $C _2$ are in series which is parallel to $C _3$ So
$C_{eq}=\frac{C_1 C_2}{C_1+C_2}+C_3 \Rightarrow \frac{20 \times 5}{20+5}+5 $
$=4+5=9 \mu F$

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MCQ 124 Marks

An object is kept at rest at a distance of 3R above the earth's surface where R is earth's radius. The minimum speed with which it must be projected so that it does not return to earth is :
(Assume $M =$ mass of earth, $G =$ Universal gravitational constant)

✓
$\sqrt{\frac{G M}{2 R}}$
B
$\sqrt{\frac{ GM }{ R }}$
C
$\sqrt{\frac{3 GM }{ R }}$
D
$\sqrt{\frac{2 G M}{R}}$

Answer

Correct option: A.

$\sqrt{\frac{G M}{2 R}}$

(A) $\sqrt{\frac{G M}{2 R}}$

$\begin{array}{l} P _{ P }+ k _{ P }= P _{ o }+ k _0 \\ -\frac{ GMm }{4 R }+\frac{1}{2} mV _{ P }^2=0 \\ V_{ P }=\sqrt{\frac{ GM }{2 R }}\end{array}$

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MCQ 134 Marks

There are two vessels filled with an ideal gas where volume of one is double the volume of other. The large vessel contains the gas at 8 kPa at 1000 K while the smaller vessel contains the gas at 7 kPa at 500 K . If the vessels are connected to each other by a thin tube allowing the gas to flow and the temperature of both vessels is maintained at 600 K , at steady state the pressure in the vessels will be (in kPa ).

A
4.4
B
6
C
24
D
18

Answer

Number of masses will remain constant
$ n _1+ n _2= n _{ f } $
$\frac{ P _1 V_1}{ RT _1}+\frac{ P _2 V_2}{ RT _2}=\frac{ P _{ f } V _{ f }}{ RT _{ f }} $
$\frac{8 \times 2 V}{ R \times 1000}+\frac{7 \times V }{ R \times 500}=\frac{ P _{ f }(3 V)}{ R \times 600} $
$\frac{16}{1000}+\frac{14}{1000}=\frac{ P _{ f }}{ R \times 600} $
$\frac{30}{1000}=\frac{ P _{ f }}{200} $
$P _{ f }=6 kPa$

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MCQ 144 Marks

A metallic ring is uniformly charged as shown in figure. $AC$ and $BD$ are two mutually perpendicular diameters. Electric field due to arc $A B$ to '$O$' is '$E$' is magnitude. What would be the magnitude of electric field at '$O$' due to arc $ABC$ ?

A
$2 E$
✓
$\sqrt{2} E$
C
$E / 2$
D
Zero

Answer

Correct option: B.

$\sqrt{2} E$

(B) $\sqrt{2} E$

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Question 154 Marks

From the combination of resistors with resistance values $R_1=R_2=R_3=5 \Omega$ and $R_4=10 \Omega$, which of the following combination is the best circuit to get an equivalent resistance of $6 \Omega$ ?

Answer

(A)
$\frac{1}{ R _{ P }}=\frac{1}{10}+\frac{1}{15}=\frac{3+2}{30}=\frac{1}{6}$

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MCQ 164 Marks

A cylindrical rod of length 1 m and radius 4 cm is mounted vertically. It is subjected to a shear force of $10^5 N$ at the top. Considering infinitesimally small displacement in the upper edge, the angular displacement $\theta$ of the rod axis from its original position would be : (shear moduli, $G =10^{10} N / m ^2$)

✓
$1 / 160 \pi$
B
$1 / 4 \pi$
C
$1 / 40 \pi$
D
$1 / 2 \pi$

Answer

Correct option: A.

$1 / 160 \pi$

(A) $1 / 160 \pi$

$\begin{array}{l}\text { Shear moduli }=\frac{\sigma_{\text {shear }}}{\theta} \\ 10^{10}=\frac{10^5}{\pi \times 16 \times 10^{-4}} \times \frac{1}{\theta} \\ \theta=\frac{1}{160 \pi} \text { Radian }\end{array}$

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MCQ 174 Marks

Given below are two statements :
Statement (I) : The dimensions of Planck's constant and angular momentum are same.
Statement (II) : In Bohr's model electron revolve around the nucleus only in those orbits for which angular momentum is integral multiple of Planck's constant.
In the light of the above statements, choose the most appropriate answer from the options given below :

A
Both Statement I and Statement II are correct
B
Statement I is incorrect but Statement II is correct
✓
Statement I is correct but Statement II is incorrect
D
Both Statement I and Statement II are incorrect

Answer

Correct option: C.

Statement I is correct but Statement II is incorrect

(C) Statement I is correct but Statement II is incorrect
$E = hf$
$ML^2 T^{-2}=[h] \times\left[T^{-1}\right] $
${[h]=\left[ML^2 T^{-1}\right]} $
$L=[MVR]=\left[ML^2 T^{-1}\right] $
$L=\frac{nh}{2 \pi}$
L is integral multiple of $\frac{ h }{2 \pi}$

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MCQ 184 Marks

Consider a rectangular sheet of solid material of length $\ell=9 cm$ and width $d =4 cm$. The coefficient of linear expansion is $\alpha=3.1 \times 10^{-5} K^{-1}$ at room temperature and one atmospheric pressure. The mass of sheet $m =0.1 kg$ and the specific heat capacity $C _{ v }=900 J kg ^{-1} K^{-1}$. If the amount of heat supplied to the material is $8.1 \times 10^2 J$ then change in area of the rectangular sheet is :-

✓
$2.0 \times 10^{-6} m^2$
B
$3.0 \times 10^{-7} m^2$
C
$6.0 \times 10^{-7} m^2$
D
$4.0 \times 10^{-7} m^2$

Answer

Correct option: A.

$2.0 \times 10^{-6} m^2$

(A) $2.0 \times 10^{-6} m^2$
$
\begin{array}{l}
\Delta Q=ms \Delta T \\
8.1 \times 10^2=0.1 \times 900 \times \Delta T \\
\Delta A=A_0 2 \alpha \Delta T=2.0 \times 10^{-6} m^2
\end{array}
$

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MCQ 194 Marks

There are '$n$' number of identical electric bulbs, each is designed to draw a power p independently from the mains supply. They are now joined in series across the main supply. The total power drawn by the combination is :

A
np
B
$\frac{p}{n^2}$
✓
$\frac{p}{n}$
D
$p$

Answer

Correct option: C.

$\frac{p}{n}$

(C) $\frac{p}{n}$
$
\begin{array}{l}
R_s=R_1+R_2+R_3+\ldots \ldots \ldots+R_n \\
\frac{V^2}{P_s}=\frac{V^2}{P}+\frac{V^2}{P}+\ldots \ldots \ldots+\frac{V^2}{P_n} \\
P_s=\frac{P}{n}
\end{array}
$

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Question 204 Marks

A radioactive material P first decays into Q and then Q decays to non-radioactive material R. Which of the following figure represents time dependent mass of P, Q and R ?

Answer

(B)
p → Q → R

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