datavorous/entrance-exam-dataset huggingface dataset:
A d.c. main supply of e.m.f. 220 V is connected across a storage battery of e.m.f. 200 V through a resistance of 1 Ω . The battery terminals are connected to external resistance R. The minimum value of R, so that a current passes through the battery to charge it is:

Options:
A 9 Ω
B 7 Ω
C 11 Ω
D 0

The correct answer is:
11 Ω

Solution:
Given,
emf of cell E = 200 V
Resistance = 1Ω
D. C. main supply voltage V = 220V
External resistance R = ?

r=E-VVR
1=20220×R
∴ R=11Ω

Topic: Physics, Current Electricity, JEE Main, JEE Main 2014 (09 Apr Online)
A dc motor run by a source of emf 200 V draws a current of 10 A. If the efficiency of the motor is 40%, then the resistance of the armature coil is

Options:
A 5 Ω
B 12 Ω
C 30 Ω
D 25 Ω

The correct answer is:
12 Ω

Solution:
Let us assume that the source voltage is E, the back emf developed in the armature coil is e, the armature resistance is R and the armature current is i.

The efficiency of the motor is

η=ieiE=e200=0.4

⇒e=80 V

The armature current is

i=E-eR

∴ 10=200-80R

∴ R=12 Ω

Topic: Physics, Electromagnetic Induction, JEE Main
A dc source of emf $\mathrm{E}_1=100 \mathrm{~V}$ and internal resistance $r=0.5 \Omega$, a storage battery of emf $\mathrm{E}_2=90 \mathrm{~V}$ and an external resistance $\mathrm{R}$ are connected as shown in figure. For what value of R no current will pass through the battery?

Options:
A
$5.5 \Omega$

B
$3.5 \Omega$

C
$4.5 \Omega$

D
$2.5 \Omega$

The correct answer is:
$4.5 \Omega$

Solution:
$\frac{100}{R+r}=\frac{90}{R}$
$$
\begin{aligned}
& \Rightarrow \quad \frac{\mathrm{R}+\mathrm{r}}{\mathrm{R}}=\frac{10}{9} \\
& \Rightarrow \quad 1+\frac{0.5}{\mathrm{R}}=\frac{10}{9} \\
& \Rightarrow \quad \frac{0.5}{\mathrm{R}}=\frac{1}{9} \\
& \therefore \mathrm{R}=4.5 \Omega
\end{aligned}
$$

Topic: Physics, Current Electricity, JEE Main, JEE Main 2013 (22 Apr Online)
A deep rectangular pond of surface area $\mathrm{A}$, containing water (density $=\rho$, specific heat capacity $=s$ ), is located in a region where the outside air temperature is a steady value at the $-26^{\circ} \mathrm{C}$. The thickness of the frozen ice layer in this pond, at a certain instant is $\mathrm{x}$. Taking the thermal conductivity of ice as $\mathrm{K}$, and its specific latent heat of fusion as L, the rate of increase of the thickness of ice layer, at this instant would be given by

Options:
A $26 \mathrm{~K} / \rho r(\mathrm{~L}-4 \mathrm{~s})$
B $26 \mathrm{~K} /\left(\rho x^2-L\right)$
C $26 K /(\rho x L)$
D $26 K / \rho r(L+4 s)$

The correct answer is:
$26 K /(\rho x L)$

Solution:
Key Idea If area of cross-section of a surface is not uniform or if the steady state condition is not reached, the heat flow equation can be applied to a thin layer of material perpendicular to direction of heat flow.
The rate of heat flow by conduction for growth of ice is given by,
$$
\frac{d \theta}{d t}=\frac{\left.K A\left(\theta_0-\theta_0\right)\right]}{x}
$$
where, $d \theta=\rho A d x L, \theta_0=0$ and $\theta_1=-\theta$
Given, $\theta_0=0^{\circ} \mathrm{C}, \theta_1=-26^{\circ} \mathrm{C}$
The rate of increase of thickness can be calculated from Eq.
$$
\begin{aligned}
\frac{\mathrm{d} \theta}{\mathrm{dt}} & =\frac{\mathrm{KA}\left(\theta_0-\theta_1\right)}{\mathrm{x}} \\
\Rightarrow \quad \frac{\rho \mathrm{AdxL}}{\mathrm{dt}} & =\frac{\mathrm{KA}\left(\theta_0-\theta_1\right)}{\mathrm{x}} \\
\Rightarrow \quad \frac{\mathrm{dx}}{\mathrm{dt}} & =\frac{\mathrm{KA}\left(\theta_0-\theta_1\right)}{\rho \mathrm{AxL}} \\
& =\frac{\mathrm{K}[0-(-26)]}{\rho \times \mathrm{L}}=\frac{26 \mathrm{~K}}{\rho \times \mathrm{L}}
\end{aligned}
$$

Topic: Physics, Thermal Properties of Matter, JEE Main
A deflection magnetometer is placed with its arm along the east-west direction and a bar magnet is placed along the arm of the magnetometer. Due to the magnet, the deflection observed is Î¸ and the period of oscillation of the needle in the magnetometer is T. When the magnet is removed, the period of oscillation is T0. The relation between T and T0 is

Options:
A T2=T02cos⁡θ
B T=T0cos⁡θ
C T=T0cos⁡θ
D T2=T02cos⁡θ

The correct answer is:
T2=T02cos⁡θ

Solution:
In the usual setting of deflection magnetometer, field due to magnet (F) and horizontal component (H) of earth's field are perpendicular to each other. Therefore, the net field on the magnetic needle is F2+H2
∴ T=2πIMF2+H2 ...(i)
When the magnet is removed,
T0=2πIMH ...(ii)
Also, FH=tan⁡θ
Dividing (i) by (ii), we get
TT0=HF2+H2
=HH2tan2⁡θ+H2=HHsec2⁡θ=cos⁡θ
⇒T2T02=cos⁡θ ∴T2=T02cos⁡θ

Topic: Physics, Magnetic Properties of Matter, JEE Main
A deflection magnetometer is placed with its arm along the east-west direction (tan A position) and a short bar magnet is placed symmetrically along its axis at some distance with its north pole pointing towards east. In this position the needle of the magnetometer shows a deflection of 60°. If we double the distance of the bar magnet, then the deflection will be

Options:
A sin-1⁡38
B cos-1⁡38
C tan-1⁡38
D cot-1⁡38

The correct answer is:
tan-1⁡38

Solution:
For a short bar magnet in tan A position,
μ04π2Md3=H tanθ ...(i)
When distance is doubled, then new deflection θ′ is given by,
μ04π2M2d3=H tanθ' ....(ii)
∴ tan ⁡ θ ′ tan ⁡θ=18
⇒ tan θ′=tan θ8=tan ⁡60°8=38
⇒ θ′=tan-1⁡38

Topic: Physics, Magnetic Properties of Matter, JEE Main
A deuteron and an alpha particle having equal kinetic energy enter perpendicular into a magnetic field. Let rd and rα be their respective radii of circular path. The value of rdrα is equal to:

Options:
A 12
B 2
C 1
D 2

The correct answer is:
2

Solution:
r=mvqB=2mkqB
rdrα=mdmαqαqd=2421=2
Hence option (2).

Topic: Physics, Magnetic Effects of Current, JEE Main, JEE Main 2021 (20 Jul Shift 1)
A deviation of 2° is produced in the yellow ray when prism of crown and flint glass are achromatically combined. Taking dispersive powers of crown and flint glass are 0.02 and 0.03 respectively and refractive index for yellow light for these glasses are 1.5 and 1.6 respectively. The refracting angles for crown glass prism will be ________° (in degree)

(Round off to the Nearest Integer)

Options:

The correct answer is:
12

Solution:
ω1=0.02;μ1=1.5; ω2=0.03;μ2=1.6

Achromatic combination

∴ θnet=0

θ1-θ2=0

θ1=θ2

ω1δ1=ω2δ2

and δnet=δ1-δ2=2°

δ1-ω1δ1ω2=2°

δ11-ω1ω2=2°

δ11-23=2°

δ1=6°

δ1=μ1-1A1

6°=1.5-1A1

A1=12°

Topic: Physics, Ray Optics, JEE Main, JEE Main 2021 (16 Mar Shift 2)
A device used for investigating the vibration of a fixed string or wire is

Options:
A Sonometer
B barometer
C Hydrometer
D None of these

The correct answer is:
Sonometer

Solution:
Sonometer is used to produce resonance of sound source with stretched vibrating string.

Topic: Physics, Waves and Sound, JEE Main
A diamagnetic material in a magnetic field moves:

Options:
A from stronger to the weaker parts of the field
B from weaker to the stronger parts of the field
C perpendicular to the field
D in none of the above directions

The correct answer is:
from stronger to the weaker parts of the field

Solution:
A diamagnetic material in a magnetic field moves from stronger to the weaker part of the field.

Topic: Physics, Magnetic Properties of Matter, JEE Main
A diatomic gas (γ=1.4) does 200 J of work when it is expanded isobarically. The heat given to the gas in the process is :

Options:
A 850 J
B 800 J
C 600 J
D 700 J

The correct answer is:
700 J

Solution:
For degree of freedom, we can write

γ=1+2f=1.4⇒2f=0.4

⇒f=5

For isobaric process, W=nRΔT=200 J.

Now, we can write

Q=f+22nRΔT

=72×200=700 J

Topic: Physics, Thermodynamics, JEE Main, JEE Main 2024 (01 Feb Shift 2)
A diatomic gas γ=1.4 does 400 J of work when it is expanded isobarically. The heat given to the gas in the process is _____ J.

Options:

The correct answer is:
1400

Solution:
For isobaric process: W=nRΔT⇒400=nRΔT

Also: Q=nCpΔT

⇒Q=n7R2ΔT (Cp=7R2, for diatomic gas)

Q=72×400=1400 J

Topic: Physics, Thermodynamics, JEE Main, JEE Main 2022 (27 Jun Shift 2)
A diatomic gas $(\gamma=1.4)$ does $100 \mathrm{~J}$ of work in an isobaric expansion. The heat given to the gas is :

Options:
A $250 \mathrm{~J}$
B $150 \mathrm{~J}$
C $350 \mathrm{~J}$
D $490 \mathrm{~J}$

The correct answer is:
$350 \mathrm{~J}$

Solution:
For Isobaric process
$\begin{aligned} & \mathrm{w}=\mathrm{P} \Delta \mathrm{v}=\mathrm{nR} \Delta \mathrm{T}=100 \mathrm{~J} \\ & \mathrm{Q}=\Delta \mathrm{u}+\mathrm{w} \\ & \Delta \mathrm{Q}=\frac{\mathrm{F}}{2} \mathrm{nR} \Delta \mathrm{T}+\mathrm{nR} \Delta \mathrm{T} \\ & \left(\frac{\mathrm{f}}{2}+1\right) \mathrm{nR} \Delta \mathrm{T} \\ & \left(\frac{5}{2}+1\right) 100=350 \mathrm{~J}\end{aligned}$
SECTION - B

Topic: Physics, Thermodynamics, JEE Main, JEE Main 2024 (08 Apr Shift 2)
A diatomic gas consisting of rigid molecules is at a temperature of 87 °C. If the moment of inertia of the rotating diatomic rigid molecule is 2.76×10-39gcm2 then the rms angular speed of the molecule is (Boltzmann constant =1.38×10-23JK-1)

Options:
A 6×1012rads-1
B 3×1012rads-1
C 6×1013rads-1
D 3×1013rads-1

The correct answer is:
6×1012rads-1

Solution:
Recall the formula of the kinetic energy of one molecule in terms of Boltzmann constant and absolute temperature,

K.E.=12KT, for two rigid molecules, it will be,

K.E.=2×12KT=KT, if I is a moment of inertia of body and ω is the angular velocity then,

KT=12Iω2⇒ω=2KTI=2×1.38×10-2387+2732.76×10-39×10-7=6×1012 rad s-1

Topic: Physics, Kinetic Theory of Gases, JEE Main
A diatomic gas, having CP=72R and CV=52R, is heated at constant pressure. The ratio dU:dQ:dW

Options:
A 3:5:2
B 5:7:3
C 5:7:2
D 3:7:2

The correct answer is:
5:7:2

Solution:
dU=nCvdT⇒dQ=nCpdT⇒dW=PdV=nRdT (isobaric process)

dU:dQ:dW:Cv:Cp:R=5R2:7R2:R=5:7:2.

Topic: Physics, Kinetic Theory of Gases, JEE Main, JEE Main 2021 (25 Feb Shift 1)
A diatomic gas molecule has translational, rotational and vibrational degrees of freedom. The $C_p / C_v$ is

Options:
A $1.67$
B $1.4$
C $1.29$
D $1.33$

The correct answer is:
$1.33$

Solution:
Degree of freedom $f=3$ (Translatory) +2 (rotatory) +1 (vibratory) $=6$
$\Rightarrow \frac{C_P}{C_V}=\gamma=1+\frac{2}{f}=1+\frac{2}{6}=\frac{4}{3}=1.33$

Topic: Physics, Kinetic Theory of Gases, JEE Main
A diatomic gas with rigid molecules does 10 J of work when expanded at constant pressure. What would be the heat energy absorbed by the gas, in this process?

Options:
A 40 J
B 35 J
C 25 J
D 30 J

The correct answer is:
35 J

Solution:
Cp=72R (for diatomic gas)
ΔQ=nCpΔT ∴Isobaric process
ΔW=nRΔT=10 J
Given
∴ΔQ=n×72RΔT=72×10=35 J nRΔT=10 J

Topic: Physics, Kinetic Theory of Gases, JEE Main, JEE Main 2019 (12 Apr Shift 2)
A diatomic ideal gas is compressed adiabatically to \( \frac{1}{32} \) of its initial volume. If the initial temperature of the gas is \( T_{\mathrm{i}} \) (in kelvin) and the final temperature is \( T_{\mathrm{f}}=a T_{\mathrm{i}} \), then value of \( a \) is

Options:
A \( 4 \)
B \( 6 \)
C \( 5 \)
D \( 9 \)

The correct answer is:
\( 4 \)

Solution:
Let the initial volume of diatomic gas be Vi, then its final volume will be Vi32.

In adiabatic process,

TVγ-1= constant

∴TiViγ-1=TfVfγ-1.

As γ=1.4 for diatomic gas.

⇒TiVi0.4=aTiVi320.4

⇒a=320.4=4.

Topic: Physics, Thermodynamics, JEE Main
A diatomic ideal gas is heated at constant volume until the pressure is doubled and again heated at constant pressure until the volume is doubled. The average molar heat capacity for the whole process is

Options:
A 13R6
B 19R6
C 23R6
D 17R6

The correct answer is:
19R6

Solution:
Let initial pressure, volume and temperature be P0, V0 and T0, respectively, indicated by state A in P-V diagram. The gas is then isochorically taken to state B (2P0, V0, 2T0) and then taken from state B to state C (2P0, 2V0, 4T0) isobarically.

Total heat absorbed by 1 mol of gas
Î Q = C v 2 T 0 - T 0 + C P 4 T 0 - 2 T 0
= 5 2 RT 0 + 7 2 R Ã 2 T 0 = 1 9 2 RT 0
Total change in temperature from series A to C is Î T = 3 T 0
Therefore,
Molar heat capacity = Î Q Î T = 1 9 2 RT 0 3 T 0 = 1 9 6 R

Topic: Physics, Thermodynamics, JEE Main
A diatomic ideal gas is used in a Car engine as the working substance. If during the adiabatic expansion part of the cycle, volume of the gas increases from $\mathrm{V}$ to $32 \mathrm{~V}$ the efficiency of the engine is

Options:
A
$0.5$

B
$0.75$

C
$0.99$

D
$0.25$

The correct answer is:
$0.75$

Solution:
The efficiency of cycle is
$$
\eta=1-\frac{T_2}{T_1}
$$
for adiabatic process
$$
\text { TV }^{r^{-1}}=\text { constant }
$$
For diatomic gas $\gamma=\frac{7}{5}$
$$
\begin{aligned}
& \mathrm{T}_1 \mathrm{~V}_1^{\gamma-1}=\mathrm{T}_2 \mathrm{~V}_2^{\gamma-1} \\
& \mathrm{~T}_1=\mathrm{T}_2\left(\frac{\mathrm{V}_2}{\mathrm{~V}_1}\right)^{\gamma-1} \\
& \mathrm{~T}_1=\mathrm{T}_2(32)^{\frac{7}{5}-1} \\
&=\mathrm{T}_2\left(2^5\right)^{2 / 5} \\
&=\mathrm{T}_2 \times 4 \\
& \mathrm{~T}_1=4 \mathrm{~T}_2 . \\
& \eta=\left(1-\frac{1}{4}\right)=\frac{3}{4}=0.75
\end{aligned}
$$

Topic: Physics, Thermodynamics, JEE Main, JEE Main 2010
A diatomic ideal gas is used in a Carnot engine as the working substance. If during the adiabatic expansion part of the cycle, the volume of the gas increases from V to 32V. The efficiency of the engine is

Options:
A 0.5
B 0.75
C 0.99
D 0.25

The correct answer is:
0.75

Solution:
The efficiency of cycle is
Î·=1-T2T1
For adiabatic process
TVÎ³-1=constant
For diatomic gas Î³=75
T1V1Î³-1 =T2V2Î³-1
T1=T2V2V1Î³-1
T1=T23275-1=T2252/5=T2Ã4
T1=4T2
Î·=1-14=34=0.75

Topic: Physics, Thermodynamics, JEE Main
A diatomic molecule can be modelled as two rigid balls connected with a spring such that the balls can vibrate with respect to the centre of mass of the system (spring + balls). Consider a diatomic gas made of such diatomic molecules. If the gas performs 20 J of work under isobaric condition, then heat given to the gas (in J) is

Options:

The correct answer is:
70

Solution:
∆QW=nCp∆TnR∆T
⇒∆Q=CPR.W
=72×20
=70 J

Topic: Physics, Thermodynamics, JEE Main
A diatomic molecule has a dipole moment of 1.2 D. If the bond distance is 1.0 Å , what fraction of an electronic charge, e, exists on each atom?

Options:

The correct answer is:
0.25

Solution:
Partial charge = Dipole moment Bond distance
=1.2×10-181.0×10-8=1.2×10-10esu
The fraction of an electronic charge is
=1.2×10-104.8×10-10=0.25
or = 25%

Topic: Chemistry, Chemical Bonding and Molecular Structure, JEE Main
A diatomic molecule has how many degrees of freedom

Options:
A $3$
B $4$
C $5$
D $6$

The correct answer is:
$5$

Solution:
For diamotic molecules, it has 3 translational and 2 rotational degrees of freedom.
$\therefore$ Total D.O.F $=3+2=5$

Topic: Physics, Kinetic Theory of Gases, JEE Main
A diatomic molecule is made of two masses $m_1$ and $m_2$ which are separated by a distance $r$. If we calculate its rotational energy by applying Bohr's rule of angular momentum quantization, its energy will be given by ( $n$ is an integer)

Options:
A
$\frac{\left(m_1+m_2\right)^2 n^2 h^2}{2 m_1^2 m_2^2 r^2}$

B
$\frac{n^2 h^2}{2\left(m_1+m_2\right) r^2}$

C
$\frac{2 n^2 h^2}{\left(m_1+m_2\right) r^2}$

D
$\frac{\left(m_1+m_2\right) n^2 h^2}{2 m_1 m_2 r^2}$

The correct answer is:
$\frac{\left(m_1+m_2\right) n^2 h^2}{2 m_1 m_2 r^2}$

Solution:
$r_1=\frac{m_2 r}{m_1+m_2} ; r_2=\frac{m_1 r}{m_1+m_2}$
$\left(l_1+l_2\right) \omega=\frac{n h}{2 \pi}=nh$
$K . E=\frac{1}{2}\left(l_1+l_2\right) \omega^2=\frac{n^2 h^2\left(m_1+m_2\right)}{2 m_1 m_2 r^2}$

Topic: Physics, Atomic Physics, JEE Main
A diatomic molecule is made of two masses $m_1$ and $m_2$ which are separated by a distance $r$. If we calculate its rotational energy by applying Bohr's rule of angular momentum quantization, its energy will be given by ( $n$ is an integer)

Options:
A
$\frac{\left(m_1+m_2\right)^2 n^2 h^2}{2 m_1^2 m_2^2 r^2}$

B
$\frac{n^2 h^2}{2\left(m_1+m_2\right) r^2}$

C
$\frac{2 n^2 h^2}{\left(m_1+m_2\right) r^2}$

D
$\frac{\left(m_1+m_2\right) n^2 h^2}{2 m_1 m_2 r^2}$

The correct answer is:
$\frac{\left(m_1+m_2\right) n^2 h^2}{2 m_1 m_2 r^2}$

Solution:
$r_1=\frac{m_2 r}{m_1+m_2} ; r_2=\frac{m_1 r}{m_1+m_2}$
$\left(l_1+l_2\right) \omega=\frac{n h}{2 \pi}=nh$
$K . E=\frac{1}{2}\left(l_1+l_2\right) \omega^2=\frac{n^2 h^2\left(m_1+m_2\right)}{2 m_1 m_2 r^2}$

Topic: Physics, Atomic Physics, JEE Main, JEE Main 2012 (Offline)
A diatomic molecule X2 has a body-centred cubic (bcc) structure with a cell edge of 300 pm. The density of the molecule is 6.17 gcm-3. The number of molecules present in 200 g of X2 is :
(Avogadro constant NA=6×1023mol-1)

Options:
A 40 NA
B 8 NA
C 4 NA
D 2 NA

The correct answer is:
4 NA

Solution:
For BCC z=2

d=Z×MNA× Volume

6.17=2×M6.02×1023×3×10-83

6.17=2×M6.02×2.7

M=50

No. of mole =20050=4

No. of molecule =4 NA

Topic: Chemistry, Solid State, JEE Main, JEE Main 2020 (05 Sep Shift 1)
A dibromo derivative of an alkane reacts with sodium metal to form an alicyclic hydrocarbon. The derivative is

Options:
A 1, 1-dibromopropane
B 2, 2-dibromobutane
C 1, 2-dibromoethane
D 1 , 4-dibromobutane

The correct answer is:
1 , 4-dibromobutane

Solution:
$\omega$-dihalides when reacts with sodium metal, gives cyclic hydrocarbons. This is an example of intramolecular Wurtz reaction.

Topic: Chemistry, Haloalkanes and Haloarenes, JEE Main
A dice is thrown ten times. If getting even number is considered as a success, then the probability of four successes is

Options:
A ${ }^{10} C_4\left(\frac{1}{2}\right)^4$
B ${ }^{10} C_4\left(\frac{1}{2}\right)^6$
C ${ }^{10} C_4\left(\frac{1}{2}\right)^8$
D ${ }^{10} C_6\left(\frac{1}{2}\right)^{10}$

The correct answer is:
${ }^{10} C_6\left(\frac{1}{2}\right)^{10}$

Solution:
Required probability
$={ }^{10} C_4\left(\frac{1}{2}\right)^4\left(\frac{1}{2}\right)^6$ $={ }^{10} C_4\left(\frac{1}{2}\right)^{10}={ }^{10} C_6\left(\frac{1}{2}\right)^{10}$.

Topic: Mathematics, Probability, JEE Main
A die is thrown. Let $A$ be the event that the number obtained is greater than 3 . Let $B$ be the event that the number obtained is less than 5 . Then $P(A \cup B)$ is

Options:
A
$\frac{3}{5}$

B
0

C
1

D
$\frac{2}{5}$

The correct answer is:
1

Solution:
$$
A=\{4,5,6\}, B=\{1,2,3,4\} \text {. }
$$
Obviously $P(A \cup B)=1$.

Topic: Mathematics, Probability, JEE Main, JEE Main 2008
A die is thrown two times and the sum of the scores appearing on the die is observed to be a multiple of 4. Then the conditional probability that the score 4 has appeared at least once is

Options:
A 14
B 13
C 18
D 19

The correct answer is:
19

Solution:
A die is thrown two times.

Total number of possible outcomes=36.

Let E is the event that sum of the score appearing on the die is multiple of 4.

Possible outcomes in favor of E=1, 3, 3, 1, 2, 2, 2, 6, 6, 2, 3, 5, 5, 3, 4, 4, 6, 6.

Number of favorable outcomes=9.

PE=14.

Let F is the event that at least one 4 appear.

Possible outcomes in favor of F=1, 4, 4, 1, 2, 4, 4, 2, 3, 4, 4, 3, 4, 4, 4, 5, 5, 4, 4, 6, 6, 4

E∩F=4, 4

⇒PE∩F=136.

We have to find conditional probability of occurrence of at least one 4 given that sum of scores is multiple of 4.

PF/E=PE∩FPE=13614=19.

Topic: Mathematics, Probability, JEE Main, JEE Main 2020 (03 Sep Shift 1)
A die is tossed 5 times. Getting an odd number is considered a success. Then the variance of distribution of success is

Options:
A
8/3

B
3/8

C
4/5

D
5/4

The correct answer is:
5/4

Solution:
The event follows binomial distribution with n = 5, p = 3/6 = 1/2
$q=1-p=1 / 2 \quad \therefore$ Variance $n p q=5 / 4$

Topic: Mathematics, Probability, JEE Main, JEE Main 2002
A die is tossed thrice. If getting a four is considered a success, then the mean and variance of the probability distribution of the number of successes are

Options:
A $\frac{1}{2}, \frac{1}{12}$
B $\frac{1}{6}, \frac{5}{12}$
C $\frac{5}{6}, \frac{1}{2}$
D None of these

The correct answer is:
None of these

Solution:
We are given that $n=3, \quad p=\frac{1}{6}, \quad q=\frac{5}{6}$
Mean $=n p=3 \times \frac{1}{6}=\frac{1}{2}$
Variance $=n q p=3 \times \frac{1}{6} \times \frac{5}{6}=\frac{5}{12}$.

Topic: Mathematics, Probability, JEE Main
A die is tossed twice. Getting a number greater than 4 is considered a success. Then the variance of the probability distribution of the number of successes is

Options:
A $\frac{2}{9}$
B ${ }^{\frac{4}{9}}$
C $)^{\frac{1}{3}}$
D None of these

The correct answer is:
${ }^{\frac{4}{9}}$

Solution:
Obviously, $p=\frac{2}{6}=\frac{1}{3} \Rightarrow$ $q=1-\frac{1}{3}=\frac{2}{3}$
also $n=2$. Therefore, variance $=n p q=2 \times \frac{1}{3} \times \frac{2}{3}=\frac{4}{9}$.

Topic: Mathematics, Probability, JEE Main
A dielectric circular disc of radíus $R$ carries a unform surface charge density $\sigma$. If it rotates about its axis with angular velocity $\omega$, the magnetic fíeld at the cente of dísc is :

Options:
A $\frac{\mu_0 \sigma \omega R^2}{2 \pi}$
B $\frac{\mu_0 \sigma \omega R}{2}$
C $\frac{\mu_0 \sigma \omega R^2}{4}$
D $\frac{\mu_0 \sigma \omega R^2}{2 \sqrt{2}}$

The correct answer is:
$\frac{\mu_0 \sigma \omega R}{2}$

Solution:
Let us assumed a ring like element of the disc of radius $r$ and thickness $d r$. If $\sigma$ is the surface charge density, then the charge on the element.
$$
d q=\sigma(2 \pi r) d r
$$
Current $d i$ associated with rotating charge $d q$ is
$$
d i=\frac{d q}{T}=\frac{d q \omega}{2 \pi} \quad\left(\because T=\frac{2 \pi}{\omega}\right) \ldots
$$
From Eqs. (i) and (ii), we get
$$
\Rightarrow \quad d i=\sigma \omega r d r
$$
As, the magnetic field $d B$ at centre due to the element,
$$
\begin{aligned}
& d B=\frac{\mu_0 d i}{2 r}=\frac{\mu_0 \sigma \omega r d r}{2 r} \Rightarrow d B=\frac{\mu_0 \sigma \omega}{2} d r \\
& \Rightarrow B_{\text {net }}=\frac{\mu_0 \sigma \omega}{2} \int_0^R d r \quad \Rightarrow B_{\text {net }}=\frac{\mu_0 \sigma \omega R}{2} \\
&
\end{aligned}
$$
Hence, the correct option is (2).

Topic: Physics, Magnetic Effects of Current, JEE Main
A dielectric slab is inserted between the plates of an isolated capacitor. The force between the plates will

Options:
A increase
B decrease
C remain unchanged
D become zero

The correct answer is:
remain unchanged

Solution:
We know for a parallel plate capacitor,

F=Q22A∈0

since the capacitor is isolated charge will not change, hence force will remain unchanged

Topic: Physics, Electrostatics, NEET
A diesel engine takes in 1 mole of air at 300 K, 1 atm pressure and compresses it adiabatically to 1 3 2 th of the original volume. Considering air as a diatomic ideal gas, the change in temperature is

Options:
A 900 K
B 1200 K
C 600 K
D 2400 K

The correct answer is:
900 K

Solution:
Using T 1 V 1 γ - 1 = T 2 V 2 γ - 1
⇒T2=T1V1V2γ-1

⇒T2=T113275-1

T2=4T1=1200 K∆T = 900 K

Topic: Physics, Thermodynamics, JEE Main
A differentiable function fx satisfies f0=0 and f1=sinâ¡1, then (where f' represents derivative of f)

Options:
A f'c=cosc, ∀c∈0,1
B f'c=cosc for some c∈0,1
C f'c=−cosc, ∀c∈0,1
D f'c=2cosc, ∀c∈0,1

The correct answer is:
f'c=cosc for some c∈0,1

Solution:
Let, gx=fx-sin⁡x
Now, g0=f0-0=0 & g1=f1-sin⁡1=0
As gx is differentiable, so by Rolle’s theorem,
g ′ c =0 for some c∈0,1
i.e. f ′ c −cosc=0
⇒ f ′ c =cosc for some c∈ 0,1

Topic: Mathematics, Application of Derivatives, JEE Main
A differential equation representing the family of parabolas with axis parallel to y-axis and whose length of latus rectum is the distance of the point (2,-3) from the line 3x+4y=5, is given by:

Options:
A 11d2xdy2=10
B 11d2ydx2=10
C 10d2ydx2=11
D 10d2xdy2=11

The correct answer is:
11d2ydx2=10

Solution:
The equation of family of parabolas

(x-α)2=4a(y-β)
Distance of point (2,-3) from line 3x+4y=5 is
4a=3×2+4×-3−55

4a=6−12−55

4a=115

(x−α)2=115(y−β)
Differentiate w.r.t. x
2(x−α)=115dydx

Again diff. w. r. t. x

2=115d2ydx2

11d2ydx2=10

Topic: Mathematics, Differential Equations, JEE Main, JEE Main 2021 (27 Aug Shift 2)