datavorous/entrance-exam-dataset huggingface dataset:
Two soap bubbles combine to form a single bubble. In this process, the change in volume and surface area are respectively $V$ and $A$. If $P$ is the atmospheric pressure, and $T$ is the surface tension of the soap solution, the following relation is true :

Options:
A $4 P V+3 T A=0$
B $3 P V-4 T A=0$
C $4 P V-3 T A=0$
D $3 P V+4 T A=0$

The correct answer is:
$3 P V+4 T A=0$

Solution:
Let radii of two soap bubbles are $a$ and $b$ respectively and radius of single larger bubble is $c$.
As excess pressure for a soap bubble is $\frac{4 T}{r}$ and external pressure $P$.
$P_i=P+\frac{4 T}{r}$
So, $\quad P_a=P+\frac{4 T}{a}, P_b=P+\frac{4 T}{b}$
and $\quad P_c=P+\frac{4 T}{c}$ ...(i)
and $\quad V_a=\frac{4}{3} \pi a^3, \quad V_b=\frac{4}{3} \pi b^3$
and $\quad V_c=\frac{4}{3} \pi c^3$ ...(ii)
Now as mass is conserved.
$\mu_a+\mu_b=\mu_c$
i.e., $\quad \frac{P_a V_a}{R T_a}+\frac{P_b V_b}{R T_b}=\frac{P_c V_c}{R T_c} \quad($ as $P V=\mu R T)$
As temperature is constant, i.e., $T_a=T_b=T_c$
So, $\quad P_a V_a+P_b V_b=P_c V_c$
which in the light of Eqs. (i) and (ii) becomes,
$\left(P+\frac{4 T}{a}\right)\left(\frac{4}{3} \pi a^3\right)+\left(P+\frac{4 T}{b}\right)\left(\frac{4}{3} \pi b^3\right)$ $=\left(P+\frac{4 T}{c}\right)\left(\frac{4}{3} \pi c^3\right)$
i.e., $4 T\left(a^2+b^2-c^2\right)=P\left(c^3-a^3-b^3\right)$ ...(iii)
Now, $\quad V=\frac{4}{3} \pi\left(a^3+b^3-c^3\right)$
and $\quad A=4 \pi\left(a^2+b^2-c^2\right)$
$\therefore \quad \frac{T A}{\pi}=-\frac{3}{4 \pi} V P$
or $\quad 4 T A+3 P V=0$

Topic: Physics, Mechanical Properties of Fluids, JEE Main
Two soap bubbles of different radii are formed at the two ends of a hollow tube. Initially, a valve in the middle of the tube separates the air in the two halves of the tube. When the valve is opened, then

Options:
A air flows from the bigger bubble to the smaller bubble till the sizes become equal
B air flows from bigger bubble to the smaller bubble till their sizes are interchanged
C air flows from the smaller bubble to the larger bubble
D there is no flow of air

The correct answer is:
air flows from the smaller bubble to the larger bubble

Solution:
P=2TR

R, for the smaller bubble is less as compared to the larger one. Thus, pressure P will be more in the smaller bubble. Since air flows from higher pressure to lower pressure, therefore, here air will flow from smaller bubble to the larger bubble.

Topic: Physics, Mechanical Properties of Fluids, JEE Main
Two soap bubbles of radii ${ }^{r_1}$ and $r_2$ equal to $4 \mathrm{~cm}$ and $5 \mathrm{~cm}$ are touching each other over a common surface $S_1 S_2$ (shown in figure). Its radius will be

Options:
A $4 \mathrm{~cm}$
B $20 \mathrm{~cm}$
C $5 \mathrm{~cm}$
D $4.5 \mathrm{~cm}$

The correct answer is:
$20 \mathrm{~cm}$

Solution:
$r=\frac{r_1 r_2}{r_1-r_2}=\frac{5 \times 4}{5-4}=20 \mathrm{~cm}$

Topic: Physics, Mechanical Properties of Fluids, JEE Main
Two soap bubbles of radii $r$ and $2 r$ are arrangement shown in the diagram. The valve is now opened. Then which one of the following will result.

Options:
A The radii of the bubbles will remain unchanged
B The bubbles will have equal radii
C The radius of the smaller bubble will decrease and that of the bigger bubble will decrease
D The radius of the smaller bubble will decrease and that of the bigger bubble will increase

The correct answer is:
The radius of the smaller bubble will decrease and that of the bigger bubble will increase

Solution:
Pressure difference $=\frac{4 T}{r}$
For smaller soap. $\mathrm{p_{atm}}-\mathrm{p}_{i1}=\frac{4 \mathrm{T}}{\mathrm{r}}$
For bigger soap, $\mathrm{p_{atm}}-\mathrm{p}_{i2}=\frac{4 \mathrm{T}}{2 \mathrm{r}}$
As the pressure inside smaller bubble is greater than pressure inside bigger bubble, so ar fown from smaller to bigger and thu's radius of the smaller bubble will decrease and that of the bigger bubble will increase.

Topic: Physics, Mechanical Properties of Fluids, WBJEE, WBJEE 2013
Two soap bubbles of radii x and y coalesce to constitute a bubble of radius z. Then z is equal to

Options:
A x2+y2
B x+y
C x+y
D x+y2

The correct answer is:
x2+y2

Solution:
n=n1+n2
pV=p1V1+p2V2
p1=p0+4Tx, p2=p0+4Ty, p=p0+4Tz
If the process takes place is vacuum then p0=0
p1=4Tx, p2=4Ty, p=4Tz
If process is isothermal
∴ p1V1+p2V2=pV
∴ z=x2+y2

Topic: Physics, Thermodynamics, JEE Main
Two solid hemispheres of radii R and R2 with centers O and O′ respectively as shown in figure. The density of bigger hemisphere is ρ and that of smaller hemisphere is 2ρ . Taking center of bigger hemisphere is at origin and the distance between centers of two hemispheres OO′ is R10 find co-ordinates of center of mass of the system.

Options:
A -R50,21R80
B -R30,21R80
C -R50,7R16
D -R30,7R16

The correct answer is:
-R50,21R80

Solution:
Let, M2=M, M1=4M

xcm=4MO-MR104M+M⇒ xcm=-R50
ycm=4M3R8-MR164M+M⇒ ycm=21R80

Topic: Physics, Center of Mass Momentum and Collision, JEE Main
Two solid rubber balls $A$ and $B$ having masses 200 and $400 \mathrm{gm}$ respectively are moving in opposite directions with velocity of $A$ equal to $0.3 \mathrm{~m} / \mathrm{s}$. After collision the two balls come to rest, then the velocity of $B$ is

Options:
A $0.15 \mathrm{~m} / \mathrm{sec}$
B $1.5 \mathrm{~m} / \mathrm{sec}$
C $-0.15 \mathrm{~m} / \mathrm{sec}$
D None of the above

The correct answer is:
$-0.15 \mathrm{~m} / \mathrm{sec}$

Solution:
Here, two solid rubber balls A and B having masses $200 \mathrm{~g}$ and $400 \mathrm{~g}$ respectively are moving in opposite directions with velocity of A equal to $0.3 \mathrm{~ms}^{-1}$ and after collision the two balls come to rest.
Final velocities $\mathrm{v}_{\mathrm{A}}$ and $\mathrm{v}_{\mathrm{B}}$ are zero because both come to rest.
Initial momentum $=$ Final momentum
$\begin{aligned}
& \mathrm{m}_{\mathrm{A}} \mathrm{u}_{\mathrm{A}}+\mathrm{m}_{\mathrm{B}} \mathrm{u}_{\mathrm{B}}=\mathrm{m}_{\mathrm{A}} \times \mathrm{o}+\mathrm{m}_{\mathrm{B}} \times \mathrm{o} \\
& \mathrm{m}_{\mathrm{A}} \mathrm{u}_{\mathrm{A}}+\mathrm{m}_{\mathrm{B}} \mathrm{u}_{\mathrm{B}}=\mathrm{o} \\
& \mathrm{m}_{\mathrm{B}} \mathrm{u}_{\mathrm{B}}=-\mathrm{m}_{\mathrm{A}} \mathrm{u}_{\mathrm{A}} \\
& \mathrm{u}_{\mathrm{B}}=-\frac{\mathrm{m}_{\mathrm{A}} \mathrm{u}_{\mathrm{A}}}{\mathrm{m}_{\mathrm{B}}} \\
& \mathrm{u}_{\mathrm{B}}=-\frac{200 \times 0.3}{400}=-0.15 \mathrm{~ms}^{-1}
\end{aligned}$

Topic: Physics, Center of Mass Momentum and Collision, JEE Main
Two solid spheres A and B made of the same material have radii rA and rBrespectively. Both the spheres are cooled from the same temperature under the conditions valid for Newton's law of cooling. The ratio of the rate of change of temperature of A and B is

Options:
A rArB
B rBrA
C rA2rB2
D rB2rA2

The correct answer is:
rBrA

Solution:
4π3r3ρc-dTdt=σ4πr2(T4-T04)
∴ -dTdt=3σρrcT4-T04=H (say)
Ratio of rates of fall of temperature
HAHB=rBrA

Topic: Physics, Thermal Properties of Matter, JEE Main
Two solids $A$ and $B$ float in water. It is observed that $A$ floats with half its volume immersed and $B$ floats with $2 / 3$ of its volume immersed. Compare the densities of $A$ and $B$

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

The correct answer is:
$3: 4$

Solution:
If two different bodies $A$ and $B$ are floating in the same liquid then $\frac{\rho_A}{\rho_B}=\frac{\left(f_in\right)_A}{\left(f_in\right)_B}=\frac{1 / 2}{2 / 3}=\frac{3}{4}$

Topic: Physics, Mechanical Properties of Fluids, JEE Main
Two solids A and B of mass 1 kg and 2 kg respectively are moving with equal linear momentum. The ratio of their kinetic energies K.E.A:K.E.B will be A1, so the value of A will be _________.

Options:

The correct answer is:
2

Solution:
Kinetic energy K=P22m,PA=PB

K∝1 m

KAKB=mBmA

=21

Topic: Physics, Work Power Energy, JEE Main, JEE Main 2021 (24 Feb Shift 2)
Two solids dissociate as follows:
A s⇌B g+C g; KP1=x  atm2
D s⇌C g+E g; KP2=y  atm2
The total pressure when both the solids dissociate simultaneously is:

Options:
A x+y atm
B x2+y2 atm
C x+y atm
D 2x+y atm

The correct answer is:
2x+y atm

Solution:
kp1=P1P1+P2
kp2=P2(P1+P2)
kP1+kP2=(P1+P2)2
x+y=(P1+P2)2
P1+P 2=x+y
2(P1+P2)=2x+y
PTotal=PB+PC+PE
=2(P1+P2)=2x+y

Topic: Chemistry, Chemical Equilibrium, JEE Main, JEE Main 2019 (12 Jan Shift 1)
Two solutions A and B are prepared by dissolving 1 g of non-volatile solutes X and Y. respectively in 1 kg of water. The ratio of depression in freezing points for A and B is found to be 1:4. The ratio of molar masses of X and Y is

Options:
A 1:4
B 1:0.25
C 1:0.20
D 1:5

The correct answer is:
1:0.25

Solution:
We know that for an non-electrolyte

∆Tf = Kfm where m = molality

ΔTfxΔTfy=kf·mxkf·my=1Mx1My

⇒14=MyMx

⇒Mx:My=1:0.25

Topic: Chemistry, Solutions, JEE Main, JEE Main 2022 (25 Jul Shift 2)
Two solutions of a substance (non electrolyte) are mixed in the following manner. 480 ml of 1.5 M first solution + 520 mL of 1.2 M second solution. What is the molarity of the final mixture?

Options:
A
1.20 M

B
1.50 M

C
1.344 M

D
2.70 M

The correct answer is:
1.344 M

Solution:
No Solution Available

Topic: Chemistry, Some Basic Concepts of Chemistry, JEE Main, JEE Main 2005
Two sonometer wires A and B are fixed on a sonometer. The ratio of their lengths, diameters, densities and tensions are given below:

lAlB=3635; dAdB=41; TATB=81; ÏAÏB=12

If the higher frequency among the two wires is 360 Hz, then what is the beat frequency (in Hz) observed when the two wires are sounded together?

Options:

The correct answer is:
10

Solution:
Frequency of sonometer wire is given by
υ=12lTm=12lTπr2ρ
υ1=12l1T1πr12ρ1
υ2=12l2T2πr22ρ2
υ1υ2=l2l1T1T2×r22r12×ρ2ρ1
υ1υ2=353681×116×21=3536
Since υ1<υ2, this means υ2=360 Hz

Therefore,

υ1=360×3536=350 Hz

fbeat=υ2-υ1=10 Hz

Topic: Physics, Waves and Sound, JEE Main
Two sound waves (expressed in CGS units) given by $y_1=0.3 \sin \frac{2 \pi}{\lambda}(v t-x)$ and $y_2=0.4 \sin \frac{2 \pi}{\lambda}(v t-x+\theta) \quad$ interfere. The resultant amplitude at a place where phase difference is $\pi / 2$ will be

Options:
A 0.7 cm
B 0.1 cm
C 0.5 cm
D $\frac{1}{10} \sqrt{7} \mathrm{~cm}$

The correct answer is:
0.5 cm

Solution:
Resultant amplitude $=\sqrt{a_1^2+a_2^2+2 a_1 a_2 \cos \phi}$
$=\sqrt{0.3^2+0.4^2+2 \times 0.3 \times 0.4 \times \cos \frac{\pi}{2}}=0.5 \mathrm{~cm}$

Topic: Physics, Waves and Sound, JEE Main
Two sound waves having the same wavelength Î» and same frequency but with intensities 9I0 and 4I0 are interfering. When the path difference between the waves is Î»4 the resulting intensity is nI0. What is the value of n?

Options:

The correct answer is:
13

Solution:
ϕ=2πλ×∆x
=2πλ×λ4=π2
I=I1+I2+2I1I2cos⁡ϕ
=9I0+4I0=13I0

Topic: Physics, Waves and Sound, JEE Main
Two sound waves of slightly different frequencies propagating in the same direction produce beats due to

Options:
A Interference
B Diffraction
C Polarization
D Refraction

The correct answer is:
Interference

Solution:
When two or more waves come together, the 'disturbances' of waves are superimposed when they come together - this is known as superposition. When crests and troughs of the two waves are precisely aligned, their superposition produces constructive interference because their disturbances add up and produce a wave having twice the amplitude but the same wavelength of the individual waves. On the other hand, when two identical waves that arrive totally out of phase, produce destructive interference because disturbances are in the opposite direction for the superposition, the resultant amplitude being zero.
Now, beats are produced by the superposition of two waves of slightly different frequencies but the same amplitude. The resulting wave is a combination of constructive and destructive interference - alternating in time. between the two, giving the wave a time-varying amplitude.
Thus, two sound waves of slightly different frequencies propagating in the same direction produce beat due to - Interference.

Topic: Physics, Waves and Sound, JEE Main
Two sound waves of wavelengths 98 cm and 100 cm arrive at the same point, from two different sources. If the speed of sound in the medium is 392 m s-1, then the number of beats heard is

Options:
A 8 Hz
B 4 Hz
C 6 Hz
D 12 Hz

The correct answer is:
8 Hz

Solution:
The beat frequency = |f1-f2|, where f=cλ

fbeat=3920.98-3921 Hz

fbeat=8 Hz

Topic: Physics, Waves and Sound, JEE Main
Two sources of equal emf are connected to an external resistance R. The internal resistance of the two sources are $R_1$ and $R_2\left(R_2>R_1\right)$. If the potential difference across the source having internal resistance $\mathrm{R}_2$ is zero, then

Options:
A
$\left.\mathrm{R}=\mathrm{R}_2 \times\left(\mathrm{R}_1+\mathrm{R}_2\right) / \mathrm{R}_2-\mathrm{R}_1\right)$

B
$\mathrm{R}=\mathrm{R}_2-\mathrm{R}_1$

C
$\mathrm{R}=\mathrm{R}_1 \mathrm{R}_2 /\left(\mathrm{R}_1+\mathrm{R}_2\right)$

D
$\mathrm{R}=\mathrm{R}_1 \mathrm{R}_2 /\left(\mathrm{R}_2-\mathrm{R}_1\right)$

The correct answer is:
$\mathrm{R}=\mathrm{R}_2-\mathrm{R}_1$

Solution:
$\mathrm{I}=\frac{2 \mathrm{E}}{\mathrm{R}_1+\mathrm{R}_2+\mathrm{R}}$
$\mathrm{E}-\mathrm{R}_2 \mathrm{I}=0$
$\Rightarrow \mathrm{R}=\mathrm{R}_2-\mathrm{R}_1$

Topic: Physics, Current Electricity, JEE Main, JEE Main 2005
Two sources of equal emfs are connected in series. This combination is connected to an external resistance R. The internal resistances of the two sources are r1 and r2r1>r2. If the potential difference across the source of internal resistance r1 is zero then the value of R will be

Options:
A r1-r2
B r1r2r1+r2
C r1+r22
D r2-r1

The correct answer is:
r1-r2

Solution:
Current through the wire will be,

I=2εR+r1+r2.

Now potential difference across cell 1 will be, ε-Ir1=0

⇒ε=2εR+r1+r2×r1⇒R+r1+r2=2r1⇒R=r1-r2

Topic: Physics, Current Electricity, JEE Main, JEE Main 2022 (27 Jul Shift 1)
Two sources of light emit with a power of 200 W. The ratio of number of photons of visible light emitted by each source having wavelengths 300 nm and 500 nm respectively, will be :

Options:
A 1:5
B 1:3
C 5:3
D 3:5

The correct answer is:
3:5

Solution:
Let n1 and n2 be the number of photons coming out from the first and the second source of light respectively per second.

So, for the first source,

n1×hcλ1=200 ...1

And, for the second source,

n2×hcλ2=200 ...2

From equations (1) and (2), it implies that

n1n2=λ1λ2=300500⇒n1n2=35

Topic: Physics, Dual Nature of Matter, JEE Main, JEE Main 2024 (29 Jan Shift 2)
Two sources of light emit X-rays of wavelength 1 nm and visible light of wavelength 500 nm, respectively. Both the sources emit light of the same power 200 W. The ratio of the number density of photons of X-rays to the number density of photons of the visible light of the given wavelengths is:

Options:
A 1500
B 250
C 1250
D 500

The correct answer is:
1500

Solution:
Ps-Power of source

Ps=nhcλ ; n = no. of photons emitted/s

⇒ n∝λ ⇒ n2n1=1/500

Topic: Physics, Electromagnetic Waves, JEE Main, JEE Main 2020 (03 Sep Shift 2)
Two sources of sound placed close to each other, are emitting progressive waves given by
$y_1=4 \sin 600 \pi t \text { and } y_2=5 \sin 608 \pi t$
An observer located near these two sources of sound will hear

Options:
A 4 beats per second with intensity ratio $25: 16$ between waxing and waning
B 8 beats per second with intensity ratio $25: 16$ between waxing and waning
C 8 beats per second with intensity ratio $81: 1$ between waxing and waning
D 4 beats per second with intensity ratio $81: 1$ between waxing and waning

The correct answer is:
4 beats per second with intensity ratio $81: 1$ between waxing and waning

Solution:
Given, $y_1=4 \sin 600 \pi t$ and $y_2=5 \sin 608 \pi t$
Comparing with general equation $y=a \sin 2 \pi f t$
we get, $f_2=300 \mathrm{~Hz}$ and $f_2=304 \mathrm{~Hz}$
Number of beats $=f_2-f_1=4 \mathrm{~s}^{-1}$
$\begin{aligned}
\frac{I_{\max }}{I_{\min }} & =\left(\frac{a_1+a_2}{a_1-a_2}\right)^2 \\
& =\left(\frac{4+5}{4-5}\right)^2=\frac{81}{1}
\end{aligned}$

Topic: Physics, Waves and Sound, JEE Main
Two sources of sound S1 and S2 produce sound waves of same frequency 600 Hz. A listener is moving from source S1 towards S2 with a constant speed u0 m/s and he hears 10 beats/s. The velocity of sound is 330 m/s. Then, u0 equals:

Options:
A 10.0 m/s
B 2.5 m/s
C 5.5 m/s
D 15.0 m/s

The correct answer is:
2.5 m/s

Solution:
Frequency observed from S1: f1=660330-u0330-0
Frequency observed from S2: f2=660330+u0330-0
As f2-f1=10
660330+u0330-0-330-u0330-0=10
22u0=10
u0=2.5 m/s

Topic: Physics, Waves and Sound, JEE Main
Two sources of sound S1 and S2 produce sound waves of same frequency 660 Hz. A listener is moving from source S1 towards S2 with a constant speed u0 m/s and he hears 10 beats/s. The velocity of sound is 330 m/s. Then, u0 equals:

Options:
A 10.0 m/s
B 2.5 m/s
C 5.5 m/s
D 15.0 m/s

The correct answer is:
2.5 m/s

Solution:
Frequency observed from S1: f1=660330-u0330-0
Frequency observed from S2: f2=660330+u0330-0
As f2-f1=10
660330+u0330-0-330-u0330-0=10
22u0=10
u0=2.5 m/s

Topic: Physics, Waves and Sound, JEE Main, JEE Main 2019 (12 Apr Shift 2)
Two sources of sound \( S_{1} \) and \( S_{2} \) produce sound waves of same frequency \( 660 \mathrm{~Hz} \). A listener is moving from source \( S_{1} \) towards \( S_{2} \) with a constant speed \( \mathrm{u}_{0} \mathrm{~m} / \mathrm{s} \) and he hears \( 10 \) beats/s. The velocity of sound is \( 330 \mathrm{~m} / \mathrm{s} . \) Then, u \( _{0} \) equals:

Options:
A \( 10.0 \mathrm{~m} / \mathrm{s} \)
B \( 2.5 \mathrm{~m} / \mathrm{s} \)
C \( 5.5 \mathrm{~m} / \mathrm{s} \)
D \( 15.0 \mathrm{~m} / \mathrm{s} \)

The correct answer is:
\( 2.5 \mathrm{~m} / \mathrm{s} \)

Solution:
Frequency observed from S1: f1=660330-u0330-0
Frequency observed from S2: f2=660330+u0330-0
As f2-f1=10
660330+u0330-0-330-u0330-0=10
22u0=10
u0=2.5 m/s

Topic: Physics, Waves and Sound, JEE Main
Two spheres $A$ and $B$ of masses $m_1$ and $m_2$ respectively collide. $A$ is at rest initially and $B$ is moving with velocity $y$ along $x$-axis. After collision $B$ has a velocity $\frac{v}{2}$ in a direction perpendicular to the original direction. The mass $A$ moves after collision in the direction

Options:
A same as that of $B$
B opposite to that of $B$
C $\theta=\tan ^{-1}\left(\frac{1}{2}\right)$ to the $x$-axis
D $\theta=\tan ^{-1}\left(\frac{-1}{2}\right)$ to the $x$-axis

The correct answer is:
$\theta=\tan ^{-1}\left(\frac{1}{2}\right)$ to the $x$-axis

Solution:
$\text {Here, } \mathbf{p}_i=m_2 v \mathbf{i}+m_2 \times 0$

$\mathbf{p}_f=m_2 \frac{v}{2} \mathbf{j}+m_1 \times \mathbf{v}_2$
Law of conservation of momentum
$\begin{gathered}
\mathbf{p}_i=\mathbf{p}_f \\
m_2 v \mathbf{i}=m_2 \frac{v}{2} \mathbf{j}+m_1 \times \mathbf{v}_1 \\
\mathbf{v}_1=\frac{m_2}{m_1} v \mathbf{i}+\frac{m_2}{m_1} \frac{v}{2} \mathbf{j}
\end{gathered}$
From this equation we can find
$\begin{aligned}
\tan \theta & =\frac{1}{2} \\
\theta & =\tan ^{-1}\left(\frac{1}{2}\right) \text { to the } x \text {-axis. }
\end{aligned}$

Topic: Physics, Center of Mass Momentum and Collision, JEE Main
Two spheres of electric charges +2 nC and -8 nC are placed at a distance d apart. If they are allowed to touch each other, what is the new distance between them to get a repulsive force of same magnitude as before?

Options:
A d
B d2
C 3d4
D 4d3

The correct answer is:
3d4

Solution:
In the first condition,
Given, q1=+2 nC=2Ã10-9
q2=-8 C
=-8Ã10-9 C
F=kq1q2r2
F=k2Ã10-9Ã8Ã10-9d2
F=k16Ã10-18d2 .....(i)
In the second condition,
F1=k q1q2dâ²2
After touching of sphere to each other the total charge =2-8=-6 nC
charge on each sphere =3 nC
=3Ã10-9C
F1=k3Ã10-93Ã10-9dâ²2
F1=k9Ã10-18dâ²2 ...(ii)
From Equations. (i) and (ii), we get
k16Ã10-18d2=k9Ã10-18dâ²2
d2dâ²2=16Ã10-189Ã10-18
d2dâ²2=169
ddâ²=169
ddâ²=43
d=43dâ²
dâ²=34d

Topic: Physics, Electrostatics, JEE Main
Two spheres of mass $\mathrm{M}$ and $7 \mathrm{M}$ are connected by a rod whose mass is negligible, and the distance between the centres of each sphere is $\mathrm{d}$. How far from the centre of the $7 \mathrm{M}$ sphere is the Centre of Mass for this object?

Options:
A $\mathrm{d} / 8$
B $\mathrm{d} / 2$
C $\mathrm{7d} / 8$
D $\mathrm{6d} / 7$

The correct answer is:
$\mathrm{d} / 8$

Solution:
Point $\mathrm{C}$ represents the centre of mass of the system.
$\begin{aligned} & \text { Given : } m 1=7 M ; m 2=M, x 2=d \\ & \text { Using, } x=m_1 x_1+m_2 x_2 / m_1+m_2 \\ & x=(7 M) \times 0+M d / 7 M+M=d / 8\end{aligned}$

Topic: Physics, Center of Mass Momentum and Collision, JEE Main
Two spheres of radii 8 cm and 2 cm are cooling. Their temperatures are 127 â and 527 â respectively. The ratio of energy radiated by them in the same amount of time is

Options:
A 0.06
B 0.5
C 1
D 2

The correct answer is:
1

Solution:
Total energy radiated from a body
Q=AÎµÏT4t
or Qt âAT4
Qt âr2T4 (âµ A=4Ïr2)
Q1Q2=r1r22 T1T24
=822273+127273+5274=1

Topic: Physics, Thermal Properties of Matter, JEE Main
Two spheres of the same material have radii 1 m and 4 m and temperature 4000 K and 2000 K respectively. The ratio of the energy radiated per second by the first sphere to that by the second is pq (where p and q are least positive integers). The value of p+q is

Options:

The correct answer is:
2

Solution:
Energy radiated
E = Ï T 4 Ã 4 Ï R 2 Ã time Ã e
E 1 E 2 = 4000 4 Ã 1 2 Ã 1 Ã 4 Ï Ï e 2000 4 Ã 4 2 Ã 1 Ã 4 Ï Ï e = 1 1 .

Topic: Physics, Thermal Properties of Matter, JEE Main
Two spheres of the same material have radii $1 \mathrm{~m}$ and $4 \mathrm{~m}$ and temperatures $4000 \mathrm{~K}$ and $2000 \mathrm{~K}$ respectively. The ratio of the energy radiated per second by the first sphere to that by the second is

Options:
A
1 : 1

B
16 : 1

C
4 : 1

D
5 : 3

The correct answer is:
1 : 1

Solution:
Energy radiated $\alpha R^2 T^4$

Topic: Physics, Thermal Properties of Matter, JEE Main, JEE Main 2002
Two spherical balls having equal masses with radius of 5 cm each are thrown upwards along the same vertical direction at an interval of 3 s with the same initial velocity of 35 m s-1, then these balls collide at a height of _____m,

(take g=10 m s-2)

Options:

The correct answer is:
50

Solution:
S1=S2

35t+12(−g)t2=35(t−3)+12(−g)(t−3)2

35t−12gt2=35t−35×3−12gt2−6t+9

35×3+45=30t

t=15030=5

height h=35×5−12×10×52

h=175−125=50 m

Topic: Physics, Motion In One Dimension, JEE Main, JEE Main 2021 (26 Aug Shift 1)
Two spherical black bodies have radii r1 and r 2. Their surface temperatures are T1 and T 2. If they radiate same power then r2r1 is

Options:
A T1T2
B T2T1
C T1T22
D T2T12

The correct answer is:
T1T22

Solution:
Qt=σAT4
i.e. power= σAT4
∴ for same power
∴ A∝1T4
∴ A2A1=T14T24
∴ 4πr224πr12= T14T24
∴ r2r1=T1T22

Topic: Physics, Thermal Properties of Matter, JEE Main
Two spherical bodies of mass M and 5 M and radii R and 2 R respectively are released in free space with initial separation between their centres equal to 12 R . If they attract each other due to gravitational force only, then distance covered by the smaller body just before collision is

Options:
A 2.5 R
B 4.5 R
C 1.5 R
D 7.5 R

The correct answer is:
7.5 R

Solution:
Let the spheres collide after time t , when the smaller sphere covered distance x 1 and bigger sphere covered distance x 2 .
The gravitational force acting between two spheres depends on the
distance which is variable quantity.
The gravitational force, F x = GM Ã 5 M 12 R - x 2
Accelration of smaller body, a 1 x = G Ã 5 M 12 R - x 2
Acceleration of bigger body, a 2 x = G M 12 R - x 2
From equation of motion,
x 1 = 1 2 a 1 x t 2 and x 2 = 1 2 a 2 x t 2
â x 1 x 2 = a 1 x a 2 x = 5 â x 1 = 5 x 2
We know that x 1 + x 2 = 9 R

x 1 + x 1 5 = 9 R â´ x 1 + 45 R 5 = 7 . 5 R
Therefore, the two spheres collide when the smaller sphere covers a distance of 7.5 R.

Topic: Physics, Gravitation, JEE Main