datavorous/entrance-exam-dataset huggingface dataset:
Two drops of the same radius are falling through air with a steady velocity of $5 \mathrm{~cm}$ per $s e c$. If the two drops coalesce, the terminal velocity would be

Options:
A $10 \mathrm{~cm}$ per $\mathrm{sec}$
B $2.5 \mathrm{~cm}$ per $\mathrm{sec}$
C $5 \times(4)^{1 / 3} \mathrm{~cm}$ per sec
D $5 \times \sqrt{2} \mathrm{~cm}$ per sec

The correct answer is:
$5 \times(4)^{1 / 3} \mathrm{~cm}$ per sec

Solution:
If two drops of same radius $r$ coalesce then radius of new drop is given by $R$
$\frac{4}{3} \pi R^3=\frac{4}{3} \pi r^3+\frac{4}{3} \pi r^3 \Rightarrow R^3=2 r^3 \Rightarrow R=2^{1 / 3} r$
If drop of radius $r$ is falling in viscous medium then it acquire a critical velocity $v$ and $v \propto r^2$
$\frac{v_2}{v_1}=\left(\frac{R}{r}\right)^2=\left(\frac{2^{1 / 3} r}{r}\right)^2$
$\Rightarrow v_2=2^{2 / 3} \times v_1=2^{2 / 3} \times(5)=5 \times(4)^{1 / 3} \mathrm{~m} / \mathrm{s}$

Topic: Physics, Mechanical Properties of Fluids, JEE Main
Two electric bulbs \( A \) and \( B \) are rated as \( 60 \mathrm{~W} \) and \( 100 \mathrm{~W} \). They are connected in parallel to the same source. Then,

Options:
A \( B \) draws more current than \( A \).
B currents drawn are in the ratio of their resistances.
C both draw the same current.
D \( A \) draws more current than \( B \).

The correct answer is:
\( B \) draws more current than \( A \).

Solution:
Power of the bulbs, when specified without any mentioned testing voltage, voltage can be assumed to be the same.

As P = V2R and V= constant

⇒ P α 1R or R α 1P

B (100 W) has less resistance compared to that of A (60 W)

When resistances are connected in parallel (V is same)

As V=IR , V=constant (parallel)

IR= constant ⇒ I α 1R

Smaller the resistance, more is the current drawn. Hence, B (smaller resistance) draws more current than A (larger resistance).

Topic: Physics, Current Electricity, JEE Main
Two electric bulbs marked $25 \mathrm{~W}-220 \mathrm{~V}$ and $100 \mathrm{~W}-220 \mathrm{~V}$ are connected in series to a $440 \mathrm{~V}$ supply.
Which of the bulbs will fuse?

Options:
A
both

B
$100 \mathrm{~W}$

C
$25 \mathrm{~W}$

D
neither

The correct answer is:
$25 \mathrm{~W}$

Solution:
Resistances of both the bulbs are
$\mathrm{R}_1=\frac{\mathrm{V}^2}{\mathrm{P}_1}=\frac{220^2}{25}$
$\mathrm{R}_2=\frac{\mathrm{V}^2}{\mathrm{P}_2}=\frac{220^2}{100}$
Hence $R_1>R_2$
When connected in series, the voltages divide in them in the ratio of their resistances. The voltage of $440\mathrm{~V}$ devides in such a way that voltage across $25 \mathrm{~w}$ bulb will be more than $220 \mathrm{~V}$.
Hence $25 \mathrm{~w}$ bulb will fuse.

Topic: Physics, Current Electricity, JEE Main, JEE Main 2012 (Offline)
Two electric dipoles, A, B with respective dipole moments dA→=-4qai^ and dB→=-2qai^ are placed on the x -axis with a separation R, as shown in the figure
The distance from A at which both of them produce the same potential is:

Options:
A R2 -1
B 2R2-1
C 2 R2+1
D R2+1

The correct answer is:
2R2-1

Solution:
V1+V2=0
K4qax2-K2qax-R2=0
x2=2x-R2
x=2R2-1

Topic: Physics, Electrostatics, JEE Main, JEE Main 2019 (10 Jan Shift 1)
Two electric dipoles of dipole moments 1.2×10-30 cm and 2.4×10-30 cm are placed in two difference uniform electric fields of strengths 5×104 N C-1 and 15×104 N C-1 respectively. The ratio of maximum torque experienced by the electric dipoles will be 1x. The value of x is _____ .

Options:

The correct answer is:
6

Solution:
Maximum torque experienced by an electric dipole is τmax=PEsin90°.

So, τmax=PE

The ratio is τ1τ2=P1E1P2E2=1.2×10-30×5×1042·4×10-30×15×104=16

Hence, the value of x=6.

Topic: Physics, Electrostatics, JEE Main, JEE Main 2022 (28 Jul Shift 1)
Two electric lamps A and B radiate the same power. Their filaments have the same dimensions and have emissivities e A and eB respectively. Their surface temperatures are T A and TB. The ratio T A T B will be equal to

Options:
A ( e B e A ) 1/4
B ( e B e A ) 1/2
C ( e A e B ) 1/2
D ( e A e B ) 1/4

The correct answer is:
( e B e A ) 1/4

Solution:
The power radiated by a filament is P=eAσT4 (where e= emissivity, σ = Stefan's constant and T= surface temperature)
Here, e T 4 = constant or e A T A 4 = e B T B 4

Topic: Physics, Thermal Properties of Matter, JEE Main
Two electrons are moving with non-relativistic speeds perpendicular to each other. If corresponding de brogile wavelengths are λ1 and λ2, their de brogile wavelength in the frame of reference attached to their centre of mass is:

Options:
A 1λCM=1λ1+1λ2
B λCM=2λ1λ2λ12+λ22
C λCM=λ1=λ2
D λCM=λ1+λ22

The correct answer is:
λCM=2λ1λ2λ12+λ22

Solution:
Momentum of each electron hλ1 i^ and hλ2j^
Velocity of centre of mass
Vcm=h2mλ1 i^+h2mλ2j^
Velocity of 1st particle about centre of mass
Vcm=h2mλ1 i^-h2mλ2j^
λcm=hh24λ12+h24λ22=2λ1 λ2λ12+λ22

Topic: Physics, Dual Nature of Matter, JEE Main, JEE Main 2018 (15 Apr)
Two electrons are moving with non-relativistic speeds perpendicular to each other. If corresponding de Broglie wavelengths are $\lambda_1$ and $\lambda_2$, their de Broglie wavelength in the frame of reference attached to their centre of mass is:

Options:
A
$\lambda_{\mathrm{CM}}=\lambda_1=\lambda_2$

B
$\frac{1}{\lambda_1}=\frac{1}{\lambda_1}+\frac{1}{\lambda_2}$

C
$\lambda_{C M}=\frac{2 \lambda_1 \lambda_2}{\sqrt{\lambda_1^2+\lambda_2^2}}$

D
$\lambda_{C M}=\left(\frac{\lambda_1+\lambda_2}{2}\right)$

The correct answer is:
$\lambda_{C M}=\frac{2 \lambda_1 \lambda_2}{\sqrt{\lambda_1^2+\lambda_2^2}}$

Solution:
Momentum (p) of each electron $\frac{\mathrm{h}}{\lambda_1} \hat{\mathrm{i}}$ and $\frac{\mathrm{h}}{\lambda_2} \hat{\mathrm{j}}$
Velocity of centre of mass
$$
\mathrm{V}_{\mathrm{cm}}=\frac{\mathrm{h}}{2 \mathrm{~m} \lambda_1} \hat{\mathrm{i}}+\frac{\mathrm{h}}{2 \mathrm{~m} \lambda_2} \hat{\mathrm{j}} \quad(\because \mathrm{p}=\mathrm{mv})
$$
Velocity of 1 st particle about centre of mass
$$
V_{1 \mathrm{~cm}}=\frac{\mathrm{h}}{2 \mathrm{~m} \lambda_1} \hat{\mathrm{i}}-\frac{\mathrm{h}}{2 \mathrm{~m} \lambda_2} \hat{\mathrm{j}}
$$
$\lambda_{\mathrm{cm}}=\frac{\mathrm{h}}{\sqrt{\frac{\mathrm{h}^2}{4 \lambda_1^2}+\frac{\mathrm{h}^2}{4 \lambda_2^2}}}=\frac{2 \lambda_1 \lambda_2}{\sqrt{\lambda_1^2+\lambda_2^2}} \quad\left(\because \lambda=\frac{\mathrm{h}}{\mathrm{p}}\right)$

Topic: Physics, Atomic Physics, JEE Main
Two electrons are moving with non-relativistic speeds perpendicular to each other. If corresponding de Broglie wavelengths are $\lambda_1$ and $\lambda_2$, their de Broglie wavelength in the frame of reference attached to their centre of mass is:

Options:
A
$\lambda_{\mathrm{CM}}=\lambda_1=\lambda_2$

B
$\frac{1}{\lambda_1}=\frac{1}{\lambda_1}+\frac{1}{\lambda_2}$

C
$\lambda_{C M}=\frac{2 \lambda_1 \lambda_2}{\sqrt{\lambda_1^2+\lambda_2^2}}$

D
$\lambda_{C M}=\left(\frac{\lambda_1+\lambda_2}{2}\right)$

The correct answer is:
$\lambda_{C M}=\frac{2 \lambda_1 \lambda_2}{\sqrt{\lambda_1^2+\lambda_2^2}}$

Solution:
Momentum (p) of each electron $\frac{\mathrm{h}}{\lambda_1} \hat{\mathrm{i}}$ and $\frac{\mathrm{h}}{\lambda_2} \hat{\mathrm{j}}$
Velocity of centre of mass
$$
\mathrm{V}_{\mathrm{cm}}=\frac{\mathrm{h}}{2 \mathrm{~m} \lambda_1} \hat{\mathrm{i}}+\frac{\mathrm{h}}{2 \mathrm{~m} \lambda_2} \hat{\mathrm{j}} \quad(\because \mathrm{p}=\mathrm{mv})
$$
Velocity of 1 st particle about centre of mass
$$
V_{1 \mathrm{~cm}}=\frac{\mathrm{h}}{2 \mathrm{~m} \lambda_1} \hat{\mathrm{i}}-\frac{\mathrm{h}}{2 \mathrm{~m} \lambda_2} \hat{\mathrm{j}}
$$
$\lambda_{\mathrm{cm}}=\frac{\mathrm{h}}{\sqrt{\frac{\mathrm{h}^2}{4 \lambda_1^2}+\frac{\mathrm{h}^2}{4 \lambda_2^2}}}=\frac{2 \lambda_1 \lambda_2}{\sqrt{\lambda_1^2+\lambda_2^2}} \quad\left(\because \lambda=\frac{\mathrm{h}}{\mathrm{p}}\right)$

Topic: Physics, Atomic Physics, JEE Main, JEE Main 2018 (15 Apr Shift 1 Online)
Two electrons each are fixed at a distance 2d. A third charge proton placed at the midpoint is displaced slightly by a distance xx≪d perpendicular to the line joining the two fixed charges. Proton will execute simple harmonic motion having angular frequency:
(m= mass of charged particle)

Options:
A πε0md32q212
B 2πε0md3q212
C q22πε0md312
D 2q2πε0md312

The correct answer is:
q22πε0md312

Solution:
From the given condition, we have

Fnet q=-2Fq/qcosθ

Fnet q=-2·14πε0·q2d2+x22·xd2+x2

=-q22πε0xd2+x23/2

For x<<d,

Fnet q=-q22πε0 d3x

∴ a=-q22πε0·md3x

Comparing with equation of SHM a=-ω2x

∴ ω=q22πε0md3

Topic: Physics, Electrostatics, JEE Main, JEE Main 2021 (24 Feb Shift 2)
Two electrons in two hydrogen-like atoms A and B have their total energies EA and EB in the ratio EA:EB=1 :2 . Their potential energies UA and UB are in the ratio UA :UB=1 :2 . If Î»A and Î»B are their de-Broglie wavelengths, then Î»A :Î»B is

Options:
A 1 :2
B 2 :1
C 1 :2
D 2 :1

The correct answer is:
2 :1

Solution:
Given,
EAEB=12,UAUB=12
So EA=x, EB=2x
And UA=y, UB=2y
∵ EA=UA+KA
And EB=UB+KB
here KA and KB are kinetic energy of particles A and B
So KA=EA-UA=x-y .....(i)
KB=EB-UB=2x-y ....(ii)
∵ de-Broglie wavelength,
λ=h2mk
So λA=h2mKA, λB=h2mKB
∴ λAλB=KBKA .....(iii)
From Equation (i), (ii) and (iii),
λAλB=2x-yx-y=21

Topic: Physics, Dual Nature of Matter, JEE Main
Two elements A and B have similar chemical properties. They don't form solid hydrogencarbonates, but react with nitrogen to form nitrides. A and B, respectively, are :

Options:
A Na and Rb
B Na and Ca
C Cs and Ba
D Li and Mg

The correct answer is:
Li and Mg

Solution:
Li and Mg do not form solid bicarbonate. But react with N2 to give nitrides. Both of them have similar properties due to diagonal relationship.
6 Li+N2⟶Δ 2 Li3N
3 Mg+N2⟶ΔMg3N2

Topic: Chemistry, s Block Elements, JEE Main, JEE Main 2020 (02 Sep Shift 2)
Two elements $A$ and $B$ with atomic numbers $Z_{A}$ and $Z_{B}$ are used to produce characteristic X-rays with frequencies $v_{A}$ and $v_{B}$ respectively. If $Z_{A}: Z_{B}=1: 2,$ then $v_{A}: v_{B}$ will be

Options:
A $1: \sqrt{2}$
B $1: 8$
C $4: 1$
D $1: 4$

The correct answer is:
$1: 4$

Solution:
Given $Z_{A}: Z_{B}=1: 2$ We know
$v \propto Z^{2}$
Hence,
$\begin{array}{l}
\frac{v_{A}}{v_{B}}=\frac{(1)^{2}}{(2)^{2}} \\
v_{A}: v_{B}=1: 4
\end{array}$

Topic: Physics, Atomic Physics, JEE Main
Two ends of a conducting rod of varying cross-sections are maintained at 200°C and 0°C respectively. In steady-state

Options:
A temperature difference across AB and CD are equal
B temperature difference across AB is greater than that of across CD
C

temperature difference across AB is less than that of across CD

D temperature difference may be equal or different depending on the thermal conductivity of the rod

The correct answer is:

temperature difference across AB is less than that of across CD

Solution:
Rate of flow of heat dQ dt or H is equal throughout the rod.
Temperature difference = (H) X (thermal resistance)
or Temperature difference Î¼ââ thermal resistance R
where R = lâ¡ KA or R â 1 A
Area across CD is less. Therefore, temperature difference across CD will be more.

Topic: Physics, Thermal Properties of Matter, NEET
Two engines pass each other moving in opposite directions with uniform speed of 30 m/s. one of them is blowing a whistle of frequency 540 Hz. Calculate the frequency heard by driver of second engine before they pass each other. speed of sound is 330 m/sec:

Options:
A 450 Hz
B 540 Hz
C 270 Hz
D 648 Hz

The correct answer is:
648 Hz

Solution:
From Doppler's effect of sound, apparent frequency is given by,
f′=v+v0v-vsf, where v, vo, vs&f represents velocity of sound, velocity of observer, velocity of source and frequency of source repectively.
=330+30330-30×540
=360300×540
=648 Hz

Topic: Physics, Waves and Sound, JEE Main, JEE Main 2016 (09 Apr Online)
Two equal capacitors are first connected in series and then in parallel. The ratio of the equivalent capacities in the two cases will be:

Options:
A 1:2
B 1:4
C 4:1
D 2:1

The correct answer is:
1:4

Solution:
For series combination,

1Ceq1=1C+1C⇒Ceq1=C2

For parallel combination,

Ceq2=C+C⇒Ceq2=2C

⇒Ceq1Ceq2=C/22C=14=1:4

Topic: Physics, Capacitance, JEE Main, JEE Main 2021 (24 Feb Shift 1)
Two equal charges $q$ are placed at a distance of $2 a$ and a third charge $-2 q$ is placed at the midpoint. The potential energy of the system is

Options:
A $\frac{q^2}{8 \pi \varepsilon_0 a}$
B $\frac{6 q^2}{8 \pi \varepsilon_0 a}$
C $-\frac{7 q^2}{8 \pi \varepsilon_0 a}$
D $\frac{9 q^2}{8 \pi \varepsilon_0 a}$

The correct answer is:
$-\frac{7 q^2}{8 \pi \varepsilon_0 a}$

Solution:
$U_{\text {system }}=\frac{1}{4 \pi \varepsilon_0} \frac{(q)(-2 q)}{a}+$ $\frac{1}{4 \pi \varepsilon_0} \frac{(-2 q)(q)}{a}+\frac{1}{4 \pi \varepsilon_0} \frac{(q)(q)}{2 a}$
$U_{s y s t e m}=-\frac{7 q^2}{8 \pi \varepsilon_0 a}$

Topic: Physics, Electrostatics, JEE Main
Two equal masses are attached to two each of a spring of spring constant \( k \). The masses are pulled out symmetrically to stretch the spring by a length \( x \) over is natural length. The work done by the spring on each mass is

Options:
A \( \frac{1}{2} k x^{2} \)
B \( -\frac{1}{2} k x^{2} \)
C \( \frac{1}{4} k x^{2} \)
D \( -\frac{1}{4} k x^{2} \)

The correct answer is:
\( -\frac{1}{4} k x^{2} \)

Solution:
Work done by spring on both mass is given by, W=U1-U2, where U1, U2 represents initial potential energy and final potential energy.
⇒W=0-12kx2=-12kx2.
So work done by spring on each mass is given by,W'=-14kx2.

Topic: Physics, Work Power Energy, JEE Main
Two equal negative charges $-q$ are fixed at the point $(0, a)$ and $(0,-a)$ on the $y$-axis. A positive charge $Q$ is released from rest at the point $(2 a, 0)$ on the $x$-axis. The charge will

Options:
A execute SHM about the origin
B move to the origin and remain at rest
C move to infinity
D execute oscillatory but not SHM

The correct answer is:
execute oscillatory but not SHM

Solution:
Component of force on charge of $+Q$ at $P$, along $x$-axis,
$\begin{aligned} F_x & =\frac{2 Q q}{4 \pi \varepsilon_0\left(a^2+x^2\right)} \times \frac{x}{\sqrt{a^2+x^2}} \\ & =\frac{2 Q q x}{4 \pi \varepsilon_0\left(a^2+x^2\right)} \times \frac{x}{\sqrt{a^2+x^2}} \\ & =\frac{2 Q q x}{4 \pi \varepsilon_0\left(a^2+x^2\right)^{3 / 2}}\end{aligned}$
Which is not directly proportional to $x$. So, motion is oscillatory but not SHM.

Topic: Physics, Oscillations, JIPMER, JIPMER 2013
Two equal positive point charges are separated by a distance 2a. The distance of a point from the centre of the line joining two charges on the equatorial line (perpendicular bisector) at which force experienced by a test charge q0 becomes maximum is ax. The value of x is ______.

Options:

The correct answer is:
2

Solution:
Let us assume at distance y, force is maximum.

Now, force at a distance y is given by, Fnet=2Fcosθ

So, Fnet=2Kqq0yy2+a232

For Fnet to be maximum, dFnetdy=0

Or Kqq0y2+a232-y32×2yy2+a212y2+a23=0

Simplifying, we get y=a2.

Hence, the value of x=2.

Topic: Physics, Electrostatics, JEE Main, JEE Main 2023 (01 Feb Shift 1)
Two equal resistances when connected in series to a battery, consume electric power of $60 \mathrm{~W}$. If these resistance are now connected in parallel combination to the same battery, the electric power consumed will be :

Options:
A $60 \mathrm{~W}$
B $240 \mathrm{~W}$
C $120 \mathrm{~W}$
D $30 \mathrm{~W}$

The correct answer is:
$240 \mathrm{~W}$

Solution:
When two resistances are connected in series, $R_{e q}=2 R$

Power consumed, $\mathrm{P}=\frac{\varepsilon^{2}}{\mathrm{R}_{\mathrm{ea}}}=\frac{\varepsilon^{2}}{2 \mathrm{R}}$

In parallel condition, $\mathrm{R}_{\mathrm{eq}}=\mathrm{R} / 2$ New power, $\mathrm{P}^{\prime}=\frac{\varepsilon^{2}}{(\mathrm{R} / 2)}$

or $\mathrm{P}^{\prime}=4 \mathrm{P}=240 \mathrm{~W}(\because \mathrm{P}=60 \mathrm{~W})$

Topic: Physics, Current Electricity, JEE Main, JEE Main 2019 (11 Jan Shift 1)
Two equal sides of an isosceles triangle are 7x-y+3=0 and x+y-3=0 and its third side passes through the point (1,-10). The equation of the third side is

Options:
A x-3y=-31
B x-3y=31
C x+3y=31
D x+3y=-31

The correct answer is:
x-3y=31

Solution:
The third side is parallel to a bisector of the angle between equal sides.

The bisectors are 7x-y+372+-12=±(x+y-3)12+12

⇒7x-y+352=±(x+y-3)2

⇒7x-y+3=±5(x+y-3)

⇒2x-6y+18=0 or 12x+4y-12=0

⇒x-3y+9=0 or 3x+y-3=0

Let the third side be x-3y=k or 3x+y=L

It passes through (1,-10).

⇒k=31, L=-7

Hence, required lines are x-3y=31 or 3x+y=-7.

Topic: Mathematics, Straight Lines, JEE Main
Two equal -ve charges $-q$ are fixed at the point $(0, a)$ and $(0, -a)$ on the $y$-axis. A positive charge $Q$ is released from rest at the poin $(2 a, 0)$ on the $x$-axis. The charge will :

Options:
A execute SHM about the origin
B move to the origin and remain at rest
C move to infinity
D execute oscillatory but not SHM

The correct answer is:
execute oscillatory but not SHM

Solution:
Component of force on charge $+Q$ at $P$, along $x$-axis,
$\begin{aligned} F_x & =\frac{2 Q q}{4 \pi \varepsilon_0\left(a^2+x^2\right)} \cos \theta \\ & =\frac{2 Q q}{4 \pi \varepsilon_0\left(a^2+x^2\right)} \times \frac{x}{\sqrt{a^2+x^2}} \\ & =\frac{2 Q q x}{4 \pi \varepsilon_0\left(a^2+x^2\right)^{3 / 2}}\end{aligned}$
Which is not directly proportional to $x$. So, motion is oscillatory but not SHM.

Topic: Physics, Electrostatics, JIPMER, JIPMER 2006
Two factories are sounding their sirens at 800 $\mathrm{Hz}$. A man goes from one factory to other at a speed of $2 \mathrm{~m} / \mathrm{s}$. The velocity of sound is $320 \mathrm{~m} / \mathrm{s}$. The number of beats heard by the person in one second will be:

Options:
A
2

B
4

C
8

D
10

The correct answer is:
10

Solution:
Given: Frequency of sound produced by siren, $f=800 \mathrm{~Hz}$
Speed of observer, $u=2 \mathrm{~m} / \mathrm{s}$
Velocity of sound, $v=320 \mathrm{~m} / \mathrm{s}$
No. of beats heard per second $=$ ?
No. of extra waves received by the observer per second $=\pm 4 \lambda$
$\therefore \quad$ No. of beats/ sec
$$
\begin{aligned}
&=\frac{2}{\lambda}-\left(-\frac{2}{\lambda}\right)=\frac{4}{\lambda} \\
&=\frac{2 \times 2}{\frac{320}{800}} \quad\left(\because \lambda=\frac{V}{f}\right) \\
&=\frac{2 \times 2 \times 800}{320}=10
\end{aligned}
$$

Topic: Physics, Waves and Sound, JEE Main, JEE Main 2014 (11 Apr Online)
Two fair dice are thrown. The numbers on them are taken as λ and μ, and a system of linear equations
x+y+z=5

x+2y+3z=μ

x+3y+λz=1
is constructed. If p is the probability that the system has a unique solution and q is the probability that the system has no solution, then:

Options:
A p=16 and q=536
B p=56 and q=136
C p=16 and q=136
D p=56 and q=536

The correct answer is:
p=56 and q=536

Solution:
11112313λ=λ-5

Δ1=11523μ3λ1=10λ+3μ-λμ-44

Δ2=11513μ1λ1=5λ+μ-λμ-13

Δ3=11512μ131=-2μ+6

For unique solution
λ≠5,μ∈{1,2,3,4,5,6}

⇒p=56×1=56
For No Solution
λ=5, μ≠3

⇒ q=16×56=536

Topic: Mathematics, Determinants, JEE Main, JEE Main 2021 (26 Aug Shift 2)
Two families with three members each and one family with four members are to be seated in a row. In how many ways can they be seated so that the same family members are not separated ?

Options:
A 2! 3! 4!
B (3!)3·(4!)
C (3!)2.4!
D 3! 4!3

The correct answer is:
(3!)3·(4!)

Solution:
We will use the concept 'objects are together'

3! ⏟Total 3 families×3!2⏟families with all 3 members together× 4! ⏟family with all 4 together

=(3!)3×4!

Topic: Mathematics, Permutation Combination, JEE Main, JEE Main 2020 (06 Sep Shift 1)
Two Faraday of electricity is passed through a solution of CuSO4 . The mass of copper deposited at the cathode is: (Atomic mass of Cu = 63.5 amu)

Options:
A 127Â g
B 0 g
C 63.5 g
D 2Â g

The correct answer is:
63.5 g

Solution:
Cu+2 + 2e- → Cu

2 Faraday charge deposits 1 mol of Cu, or

2 F≡2 Equivalents of Cu
Mass of Cu deposited=2×63.52=63.5g

Topic: Chemistry, Electrochemistry, JEE Main, JEE Main 2015 (04 Apr)
Two finite sets have $m$ and $n$ elements. The number of subsets of the first set is 112 more than that of the second set. The values of $m$ and $n$ respectively are,

Options:
A 4,7
B 7,4
C 4,4
D 7,7

The correct answer is:
7,4

Solution:
$2^{m}-2^{n}=112 \Rightarrow 2^{n}\left(2^{m-n}-1\right)=16.7$

$\quad \therefore \quad 2^{n}\left(2^{m-n}-1\right)=2^{4}\left(2^{3}-1\right)$

Comparing we get $n=4$ and $m-n=3$ $\Rightarrow n=4$ and $m=7$

Topic: Mathematics, Sets and Relations, JEE Main
Two first order reactions proceed at 25 o C at the same rate temperature. Coefficient of the rate of the first reaction is 2 and that of second reaction is 3 . Find the ratio of the rate of these reaction at 75 o C .

Options:
A 7.0
B 7.59
C 6.52
D 8.12

The correct answer is:
7.59

Solution:
Temp coefficient =RT+10Â°CRT
R T + ΔT  =  R T (TC) ΔT 10
Rate of reaction
=7.59

Topic: Chemistry, Chemical Kinetics, JEE Main
Two fixed frictionless inclined plane making an angle $30^{\circ}$ and $60^{\circ}$ with the vertical are shown in the figure. Two block $A$ and $B$ are placed on the two planes. What is the relative vertical acceleration of $A$ with respect to $B$ ?

Options:
A
$4.9 \mathrm{~ms}^{-2}$ in horizontal direction

B
$9.8 \mathrm{~ms}^{-2}$ in vertical direction

C
zero

D
$4.9 \mathrm{~ms}^{-2}$ in vertical direction

The correct answer is:
$4.9 \mathrm{~ms}^{-2}$ in vertical direction

Solution:
$m g \sin \theta=m a$
$\therefore \quad a=g \sin \theta$
where $\mathrm{a}$ is along the inclined plane
$\therefore \quad$ vertical component of acceleration is $\mathrm{g} \sin ^2 \theta$
$\therefore \quad$ relative vertical acceleration of $A$ with respect to $B$ is

Topic: Physics, Laws of Motion, JEE Main, JEE Main 2010
Two flasks A and B have equal volumes. A is maintained at 300 K and B at 600 K. Equal masses of H2 and CO2 are taken in flasks A and B respectively. Find the ratio of total K.E of gases in flask A to that of B.

Options:
A 1:2
B 11:1
C 33:2
D 55:7

The correct answer is:
11:1

Solution:
We know that the average kinetic energy depends on the temperature and number of moles.

Now, K.E. expression for flasks A and B is given below:
K.E.A​=3nART A​​2
K.E.B=3nBRTB2
Assume common mass=m
nA=m/2nB​=m/44

K.E.AK.E.B=nATAnBTB=m/2×300m/44×600=11:1

Topic: Chemistry, Thermodynamics (C), AP EAMCET, AP EAMCET 2021 (19 Aug Shift 2)
Two forces A and B whose magnitude are in the ratio 3:5 give a resultant of 28 N. If the angle of their inclination is 60Â°, then the magnitude of force B will be ----- N.

Options:

The correct answer is:
20.00

Solution:
Let A and B be the two forces.
Then A=3x; B=5x, R=28N and Î¸=60Â° .
We know that R=A2+B2+2ABcosÎ¸
â28=3x2+5x2+23x5xcos60Â°
=9x2+25x2+15x2=7x
âx=287=4
Hence, force B=5Ã4=20 N

Topic: Physics, Mathematics in Physics, JEE Main
Two forces are such that the sum of their magnitudes is $18 \mathrm{~N}$ and their resultant is $12 \mathrm{~N}$ which is perpendicular to the smaller force. Then the magnitudes of the forces are

Options:
A
$12 \mathrm{~N}, 6 \mathrm{~N}$

B
$13 \mathrm{~N}, 5 \mathrm{~N}$

C
$10 \mathrm{~N}, 8 \mathrm{~N}$

D
$16 \mathrm{~N}, 2 \mathrm{~N}$

The correct answer is:
$13 \mathrm{~N}, 5 \mathrm{~N}$

Solution:
Use $\tan \alpha=\frac{P \sin \theta}{Q+P \cos \theta} \Rightarrow \tan 90^{\circ}=\frac{P \sin \theta}{Q+P \cos \theta}=\infty$
$\therefore Q+P \cos \theta=0 \Rightarrow P \cos \theta=-Q$
$\mathrm{R}=\sqrt{\mathrm{P}^2+\mathrm{Q}^2+2 \mathrm{PQ} \cos \theta} \mathrm{R}=\sqrt{\mathrm{P}^2+\mathrm{Q}^2-2 \mathrm{Q}^2}$ or $\mathrm{R}=\sqrt{\mathrm{P}^2-\mathrm{Q}^2}=12$
$144=(\mathrm{P}+\mathrm{Q})(\mathrm{P}-\mathrm{Q})$ or $\mathrm{P}-\mathrm{Q}=144 / 18=8 \quad \therefore \mathrm{P}=13 \mathrm{~N}$ and $\mathrm{Q}=5 \mathrm{~N}$

Topic: Physics, Mathematics in Physics, JEE Main