datavorous/entrance-exam-dataset huggingface dataset:
Two simple pendulums of length $1 \mathrm{~m}$ and $4 \mathrm{~m}$ respectively are both given small displacement in the same direction at the same instant. They will be again in phase after the shorter pendulum has completed number of oscillations equal to:

Options:
A
2

B
7

C
5

D
3

The correct answer is:
2

Solution:
Let $T_1$ and $T_2$ be the time period of the two pendulums $\mathrm{T}_1=2 \pi \sqrt{\frac{1}{\mathrm{~g}}}$ and $\mathrm{T}_2=2 \pi \sqrt{\frac{4}{\mathrm{~g}}}$ As $\ell_1 < \ell_2$ therefore $\mathrm{T}_1 < \mathrm{T}_2$ Let at $\mathrm{t}=0$ they start swinging together. Since their time periods are different, the swinging will not be in unison always. Only when number of completed oscillations
differ by an integer, the two pendulums will again begin to swing together Let longer length pendulum complete $\mathrm{n}$ oscillation and shorter length pendulum complete $(n+1)$ oscillation. For unison swinging
$$
\begin{aligned}
& (n+1) T_1=n_2 \\
& (n+1) \times 2 \pi \sqrt{\frac{1}{g}}=(n) \times 2 \pi \sqrt{\frac{4}{g}} \\
& \Rightarrow n=1 \\
& \therefore n+1=1+1=2
\end{aligned}
$$

Topic: Physics, Oscillations, JEE Main, JEE Main 2013 (09 Apr Online)
Two simple pendulums of length 5 m and 20 m are given small displacements in the same direction at the same time. The minimum number of oscillations, the shorter pendulum has completed, when the phase difference between them becomes zero again, is

Options:
A 3
B 4
C 2
D 5

The correct answer is:
2

Solution:
NS = Number of oscillations made by shorter length pendulum with time period TS.
NL = Number of oscillations made by longer length pendulum with time period TL.

If t is the time after which the pendulums are back in sync, then
Then t=NSTS=NLTL
⇒ NS2π5g=NL×2π20g ∵ T=2πlg
⇒NS=2NL i.e. if ⇒NL=1 NS=2

Topic: Physics, Oscillations, JEE Main
Two simple pendulums of lengths 5 m and 20 m respectively are given small displacement in one direction at the same time they will again be in the same sense when the pendulum of shorter length has completed n oscillations. Then, n is

Options:
A 5
B 1
C 2
D 3

The correct answer is:
2

Solution:
According to question

n1T1=n2T2

⇒ n·2π5g=n-12π20g

nn-1=2⇒n=2

Topic: Physics, Oscillations, JEE Main
Two sitar strings, A and B playing the note 'Dha' are slightly out of tune and produce beats of frequency 5 Hz. The tension of the string B is slightly increased and the beat frequency is found to decrease by 3 Hz. If the frequency of A is 425 Hz. The original frequency of B is

Options:
A 428 Hz
B 430 Hz
C 422 Hz
D 420 Hz

The correct answer is:
420 Hz

Solution:
The frequency of B is either 420 Hz or 430 Hz. As, the tension in B is increased its frequency will increase. If the frequency is 430 Hz, the beat frequency will increase. If the frequency is 420 Hz beat frequency will decrease, hence the correct answer is 420 Hz.

Topic: Physics, Waves and Sound, JEE Main, JEE Main 2018 (16 Apr Online)
Two skaters $\mathrm{P}$ and $\mathrm{Q}$ are skating towards each other. Skater $\mathrm{P}$ throws a ball towards $\mathrm{W}$ every $5 \mathrm{~s}$ such that it always leaves her hand with speed $2 \mathrm{~ms}^{-1}$ with respect to the ground. Consider two cases:
$\mathrm{(I)}$ $\mathrm{P}$ runs with speed $1 \mathrm{~ms}^{-1}$ towards $\mathrm{Q}$ while $\mathrm{Q}$ remains stationary
$\mathrm{(II)}$ $\mathrm{Q}$ runs with speed $1 \mathrm{~ms}^{-1}$ towards $\mathrm{P}$ while $\mathrm{P}$ remains stationary.
Note that irrespective of speed of $\mathrm{P}$, ball always leaves $\mathrm{P's}$ hand with speed $2 \mathrm{~ms}^{-1}$ with respect to the ground. Ignore gravity. Balls will be received by $\mathrm{Q}$

Options:
A one every $2.5 \mathrm{~s}$ in case $\mathrm{(I)}$ and one every $3.3 \mathrm{~s}$ in case $\mathrm{(II)}$
B one every $2 \mathrm{~s}$ in case $\mathrm{(I)}$ and one every $4 \mathrm{~s}$ in case $\mathrm{(II)}$
C one every $3.3 \mathrm{~s}$ in case $\mathrm{(I)}$ and one every $2.5 \mathrm{~s}$ in case $\mathrm{(II)}$
D one every $2.5 \mathrm{~s}$ in case $\mathrm{(I)}$ and one every $2.5 \mathrm{~s}$ in case $\mathrm{(II)}$

The correct answer is:
one every $2.5 \mathrm{~s}$ in case $\mathrm{(I)}$ and one every $3.3 \mathrm{~s}$ in case $\mathrm{(II)}$

Solution:
(I) Let initial distance between $\mathrm{P} \& \mathrm{Q}$ is $\mathrm{x}$
at $\mathrm{t}_{1}=\frac{\mathrm{x}}{2}$ a receive the ball.
Next ball.
$\begin{array}{l}
t_{2}=\frac{x-5}{2}+5 \\
\Delta t=\frac{5}{2}
\end{array}$
(II) in second case
at $t=0$ $\mathrm{P}$ throws the ball.
$t_{1}=\frac{x}{3}$.
Next ball
$\begin{array}{l}
t_{2}=\frac{x-5}{3}+5 \\
\Delta t=\frac{10}{3}
\end{array}$

Topic: Physics, Motion In One Dimension, JEE Main
Two skaters $\mathrm{P}$ and $\mathrm{Q}$ are skating towards each other. Skater $\mathrm{P}$ throws a ball towards $\mathrm{W}$ every $5 \mathrm{~s}$ such that it always leaves her hand with speed $2 \mathrm{~ms}^{-1}$ with respect to the ground. Consider two cases:
$\mathrm{(I)}$ $\mathrm{P}$ runs with speed $1 \mathrm{~ms}^{-1}$ towards $\mathrm{Q}$ while $\mathrm{Q}$ remains stationary
$\mathrm{(II)}$ $\mathrm{Q}$ runs with speed $1 \mathrm{~ms}^{-1}$ towards $\mathrm{P}$ while $\mathrm{P}$ remains stationary.
Note that irrespective of speed of $\mathrm{P}$, ball always leaves $\mathrm{P's}$ hand with speed $2 \mathrm{~ms}^{-1}$ with respect to the ground. Ignore gravity. Balls will be received by $\mathrm{Q}$

Options:
A one every $2.5 \mathrm{~s}$ in case $\mathrm{(I)}$ and one every $3.3 \mathrm{~s}$ in case $\mathrm{(II)}$
B one every $2 \mathrm{~s}$ in case $\mathrm{(I)}$ and one every $4 \mathrm{~s}$ in case $\mathrm{(II)}$
C one every $3.3 \mathrm{~s}$ in case $\mathrm{(I)}$ and one every $2.5 \mathrm{~s}$ in case $\mathrm{(II)}$
D one every $2.5 \mathrm{~s}$ in case $\mathrm{(I)}$ and one every $2.5 \mathrm{~s}$ in case $\mathrm{(II)}$

The correct answer is:
one every $2.5 \mathrm{~s}$ in case $\mathrm{(I)}$ and one every $3.3 \mathrm{~s}$ in case $\mathrm{(II)}$

Solution:
(I) Let initial distance between $\mathrm{P} \& \mathrm{Q}$ is $\mathrm{x}$
at $\mathrm{t}_{1}=\frac{\mathrm{x}}{2}$ a receive the ball.
Next ball.
$\begin{array}{l}
t_{2}=\frac{x-5}{2}+5 \\
\Delta t=\frac{5}{2}
\end{array}$
(II) in second case
at $t=0$ $\mathrm{P}$ throws the ball.
$t_{1}=\frac{x}{3}$.
Next ball
$\begin{array}{l}
t_{2}=\frac{x-5}{3}+5 \\
\Delta t=\frac{10}{3}
\end{array}$

Topic: Physics, Motion In One Dimension, KVPY, KVPY 2013 (SA)
Two slabs $A$ and $B$ of different materials but of the same thickness are joined end to end to form a composite slab. The thermal conductivities of $A$ and $B$ are $k_1$ and $k_2$ respectively. A steady temperature difference of $12^{\circ} \mathrm{C}$ is maintained across the composite slab. If $k_1=\frac{k_2}{2}$, the temperature difference across slab $A$ is

Options:
A $4^{\circ} \mathrm{C}$
B $6^{\circ} \mathrm{C}$
C $8^{\circ} \mathrm{C}$
D $10^{\circ} \mathrm{C}$

The correct answer is:
$8^{\circ} \mathrm{C}$

Solution:
The given situation can be shown as
Rate of flow of heat will be equal in both the slabs
$\begin{aligned}
& \therefore & (12-x) K_1 & =K_2(x-0) \\
& & 12-x & =2 x \quad\left(\because K_1=\frac{K_2}{2}\right) \\
& & x & =4^{\circ} \mathrm{C}
\end{aligned}$
The temperature difference across slab
$\begin{aligned}
A & =(12-x)=(12-4) \\
& =8^{\circ} \mathrm{C}
\end{aligned}$

Topic: Physics, Thermal Properties of Matter, AP EAMCET, AP EAMCET 2011
Two slabs $A$ and $B$ of different materials but of the same thickness are joined end to end to form a composite slab. The thermal conductivities of $A$ and $B$ are $k_1$ and $k_2$ respectively. A steady temperature difference of $12^{\circ} \mathrm{C}$ is maintained across the composite slab. If $k_1=\frac{k_2}{2}$, the temperature difference across slab $A$ is

Options:
A $4^{\circ} \mathrm{C}$
B $6^{\circ} \mathrm{C}$
C $8^{\circ} \mathrm{C}$
D $10^{\circ} \mathrm{C}$

The correct answer is:
$8^{\circ} \mathrm{C}$

Solution:
The given situation can be shown as
Rate of flow of heat will be equal in both the slabs
$\begin{aligned}
& \therefore & (12-x) K_1 & =K_2(x-0) \\
& & 12-x & =2 x \quad\left(\because K_1=\frac{K_2}{2}\right) \\
& & x & =4^{\circ} \mathrm{C}
\end{aligned}$
The temperature difference across slab
$\begin{aligned}
A & =(12-x)=(12-4) \\
& =8^{\circ} \mathrm{C}
\end{aligned}$

Topic: Physics, Thermal Properties of Matter, TS EAMCET, TS EAMCET 2011
Two slabs $A$ and $B$ of equal surface area are placed one over the other such that their surfaces are completely in contact. The thickness of slab $A$ is twice that of $B$. The coefficient of thermal conductivity of slab $A$ is twice that of $B$. The first surface of slab $A$ is maintained at $100^{\circ} \mathrm{C}$, while the second surface of slab $B$ is maintained at $25^{\circ} \mathrm{C}$. The temperature at the contact of their surfaces is

Options:
A $62.5^{\circ} \mathrm{C}$
B $45^{\circ} \mathrm{C}$
C $55^{\circ} \mathrm{C}$
D $85^{\circ} \mathrm{C}$

The correct answer is:
$62.5^{\circ} \mathrm{C}$

Solution:
The temperature at the contact of the surface
$$
\begin{aligned}
& =\frac{K_1 d_2 \theta_1+K_2 d_1 \theta_2}{K_1 d_2+K_2 d_1} \\
& =\frac{2 K_2 d_2 \times 100+2 d_2 \times K_2 \times 25}{2 K_2 d_2+K_2 2 d_2} \\
& =\frac{200+50}{4}=62.5^{\circ} \mathrm{C}
\end{aligned}
$$

Topic: Physics, Thermal Properties of Matter, AP EAMCET, AP EAMCET 2008
Two slabs $A$ and $B$ of equal surface area are placed one over the other such that their surfaces are completely in contact. The thickness of slab $A$ is twice that of $B$. The coefficient of thermal conductivity of slab $A$ is twice that of $B$. The first surface of slab $A$ is maintained at $100^{\circ} \mathrm{C}$, while the second surface of slab $B$ is maintained at $25^{\circ} \mathrm{C}$. The temperature at the contact of their surfaces is

Options:
A $62.5^{\circ} \mathrm{C}$
B $45^{\circ} \mathrm{C}$
C $55^{\circ} \mathrm{C}$
D $85^{\circ} \mathrm{C}$

The correct answer is:
$62.5^{\circ} \mathrm{C}$

Solution:
The temperature at the contact of the surface
$$
\begin{aligned}
& =\frac{K_1 d_2 \theta_1+K_2 d_1 \theta_2}{K_1 d_2+K_2 d_1} \\
& =\frac{2 K_2 d_2 \times 100+2 d_2 \times K_2 \times 25}{2 K_2 d_2+K_2 2 d_2} \\
& =\frac{200+50}{4}=62.5^{\circ} \mathrm{C}
\end{aligned}
$$

Topic: Physics, Thermal Properties of Matter, TS EAMCET, TS EAMCET 2008
Two slits are $1 \mathrm{~mm}$ apart and the screen is located $1 \mathrm{~m}$ away from the slits. A light of wavelength $500 \mathrm{~nm}$ is used. The width of each slit to obtain 10 maxima of the double slit pattern within the central maximum of the single slit pattern is _____$\times 10^{-4} \mathrm{~m}$

Options:

The correct answer is:
2

Solution:
$\begin{aligned} & \mathrm{d}=1 \mathrm{~mm}, \mathrm{D}=1 \mathrm{~m}, \lambda=500 \mathrm{~nm} \\ & 10\left(\frac{\lambda \mathrm{D}}{\mathrm{d}}\right)=\frac{2 \lambda \mathrm{D}}{\mathrm{a}} \\ & \mathbf{a}=\frac{\mathrm{d}}{5} \\ &=\frac{10 \times 10^{-4} \mathrm{~m}}{5} \\ &=2 \times 10^{-4}\end{aligned}$

Topic: Physics, Wave Optics, JEE Main, JEE Main 2024 (08 Apr Shift 2)
Two slits are separated by a distance of 0.5 mm and illuminated with light of wavelength Î»=6000 Aâ . If the screen is placed at 2.5 m from the slits. The distance of the third bright fringe from the centre will be

Options:
A 1.5 mm
B 3 mm
C 6Â mm
D 9 mm

The correct answer is:
9 mm

Solution:
The distance of nth bright fringe from the centre,
xn=nDλd
So, x3=3 × 2.5 × 6000 × 10-100.5 × 10-3
=9×10-3 m=9 mm

Topic: Physics, Wave Optics, JEE Main
Two slits, S1 and S2, illuminated by a white light source, give a white central maximum. A transparent sheet, of refractive index 1.25 and thickness t1, is placed in front of S1. Another transparent sheet, of refractive index 1.50 and thickness t2, is placed in front of S2. If the central maxima are not affected, then the ratio t1:t2 will be

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

The correct answer is:
2 : 1

Solution:
Since there is no shift in central maxima, the path difference introduced by the two sheets are equal.
i.e. μ1-1t1=μ2-1t2
Where μ1 and μ2 are refraction index
i.e. t1t2=(μ2-1)(μ1-1)=(1.5-1)(1.25-1)=0.50.25=2

Topic: Physics, Wave Optics, JEE Main
Two small balls of the same size, having masses m1 and m2 (m1>m2) are tied by a thin weightless thread and dropped from a certain height. Taking the force of buoyancy of air into account, the tension T in the thread during the flight, after the motion of the ball becomes uniform, will be

Options:
A (m1-m2)g
B (m1-m2)g2
C (m1+m2)g
D (m1+m2)g2

The correct answer is:
(m1-m2)g2

Solution:
As both the ball are of same size, force of buoyancy on each is same. Therefore, in equilibrium
F + F = m 1 g + m 2 g or F = m 1 + m 2 g 2
Considering the equilibrium of lower ball,
T + F = m 1 g
T = m 1 g - F
T = m 1 g - m 1 + m 2 g 2
T = m 1 - m 2 g 2

Topic: Physics, Laws of Motion, JEE Main
Two small drops of mercury each of radius 'R' coalesce to form a large single drop. The ratio of the total surface energies before and after the change is

Options:
A $2\frac{2}{3}:1$
B $\sqrt{2}: 1$
C $2\frac{1}{3}: 1$
D $2: 1$

The correct answer is:
$2\frac{1}{3}: 1$

Solution:
The volume of mercury remains same.
$$
\begin{array}{l}
\therefore \frac{4 \pi}{3} R^{3}=2\left(\frac{4 \pi}{3} r^{3}\right) \\
\therefore R^{3}=2 r^{3} \quad \text { or } \quad r^{3}=\frac{R^{3}}{2} \\
\therefore r=\frac{R}{2^{\frac{1}{3}}}
\end{array}
$$
Initial surface energy $\mathrm{E}_{1}=4 \pi \mathrm{R}^{2} \cdot \mathrm{T}$
Final surface energy
$\mathrm{E}_{2}=4 \pi \mathrm{r}^{2} \mathrm{~T}=4 \pi\left(\frac{\mathrm{R}}{2^{\frac{1}{3}}}\right)^{2} \mathrm{~T}=4 \pi \frac{\mathrm{R}^{2}}{2^{\frac{2}{3}} \mathrm{~T}}$
$\therefore \frac{E_{1}}{E_{2}}=2^{\frac{2}{3}}$

Topic: Physics, Mechanical Properties of Fluids, MHT CET, MHT CET 2020 (16 Oct Shift 1)
Two small equal point charges of magnitude $q$ are suspended from a common point on the ceiling by insulating mass less strings of equal lengths. They come to equilibrium with each string making angle $\theta$ from the vertical. If the mass of each charge is $m$, then the electrostatic potential at the centre of line joining them will be
$$
\left(\frac{1}{4 \pi \in_0}=k\right) \text {. }
$$

Options:
A
$2 \sqrt{k m g \tan \theta}$

B
$\sqrt{k m g \tan \theta}$

C
$4 \sqrt{k m g / \tan \theta}$

D
$4 \sqrt{k m g / \tan \theta}$

The correct answer is:
$4 \sqrt{k m g / \tan \theta}$

Solution:
In equilibrium, $\mathrm{F}_{\mathrm{e}}=\mathrm{T} \sin \theta$
$$
\begin{aligned}
& m g=T \cos \theta \\
& \tan \theta=\frac{F_e}{m g}=\frac{q^2}{4 \pi \epsilon_0 x^2 \times m g} \\
& \therefore x=\sqrt{\frac{q^2}{4 \pi \epsilon_0 \tan \theta m g}}
\end{aligned}
$$
Electric potential at the centre of the line
$$
\mathrm{V}=\frac{\mathrm{kq}}{\mathrm{x} / 2}+\frac{\mathrm{kq}}{\mathrm{x} / 2}=4 \sqrt{\mathrm{kmg} / \tan \theta}
$$

Topic: Physics, Electrostatics, JEE Main
Two small equal point charges of magnitude $q$ are suspended from a common point on the ceiling by insulating mass less strings of equal lengths. They come to equilibrium with each string making angle $\theta$ from the vertical. If the mass of each charge is $m$, then the electrostatic potential at the centre of line joining them will be
$$
\left(\frac{1}{4 \pi \in_0}=k\right) \text {. }
$$

Options:
A
$2 \sqrt{k m g \tan \theta}$

B
$\sqrt{k m g \tan \theta}$

C
$4 \sqrt{k m g / \tan \theta}$

D
$4 \sqrt{k m g / \tan \theta}$

The correct answer is:
$4 \sqrt{k m g / \tan \theta}$

Solution:
In equilibrium, $\mathrm{F}_{\mathrm{e}}=\mathrm{T} \sin \theta$
$$
\begin{aligned}
& m g=T \cos \theta \\
& \tan \theta=\frac{F_e}{m g}=\frac{q^2}{4 \pi \epsilon_0 x^2 \times m g} \\
& \therefore x=\sqrt{\frac{q^2}{4 \pi \epsilon_0 \tan \theta m g}}
\end{aligned}
$$
Electric potential at the centre of the line
$$
\mathrm{V}=\frac{\mathrm{kq}}{\mathrm{x} / 2}+\frac{\mathrm{kq}}{\mathrm{x} / 2}=4 \sqrt{\mathrm{kmg} / \tan \theta}
$$

Topic: Physics, Electrostatics, JEE Main, JEE Main 2013 (22 Apr Online)
Two small magnets have their masses and lengths in the ratio 1:2. The maximum torques experienced by them in a uniform magnetic field are the same. For small oscillations, the ratio of their time periods is

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

The correct answer is:
$\frac{1}{2 \sqrt{2}}$

Solution:
Let the mass of the first magnet $=m$
$\therefore$ Mass of second magnet $=2 \mathrm{~m}$
The time-period of oscillation of the magnet
$T=2 \pi \sqrt{\frac{1}{M B_H}}$
where, $I=$ Ml of magnet about the axis passing through its centre
$B_H=$ Earth's magnetic field
$M=$ Magnetic moment of magnet
$\therefore \frac{T_1}{T_2}=\sqrt{\frac{I_1}{M_1} \times \frac{M_2}{I_2}}$
Let length of the magnet $=L$
$\mathrm{Ml}$ of the magnets about the axis passing through one of the ends is $\frac{m L^2}{3}$
$\begin{aligned}
& \therefore I_1=m L^2 / 3 \text { and } I_2=\frac{(2 m)(2 L)^2}{3} \\
& \frac{T_1}{T_2}=\sqrt{\frac{m L^2 / 3}{M_1} \times \frac{M_2}{\frac{2 m \times(2 L)^2}{3}}}
\end{aligned}$
It is given that maximum torques experienced by the magnets are same
$\begin{aligned}
& \therefore & M \cdot B_H & =M_2 B_H \Rightarrow M_1=M_2\left[\because M_1=M_2 \text { (given) }\right] \\
\Rightarrow & & \frac{T_1}{T_2} & =\frac{1}{\sqrt{8}}=\frac{1}{2 \sqrt{2}}
\end{aligned}$

Topic: Physics, Magnetic Properties of Matter, JEE Main
Two small particles each of mass \( m \) are attached at the two ends of a semicircular ring of same mass as shown. \( R \) is radius of ring and \( O \) is the centre of the ring. Moment of inertia about an axis passing through \( O \) and normal to its plane is

Options:
A \( 3 m R^{2} \)
B \( \frac{3 m R^{2}}{2} \)
C \( 2 m R^{2} \)
D \( \frac{m R^{2}}{2} \)

The correct answer is:
\( 3 m R^{2} \)

Solution:
Here ring is half but its mass is m so its moment of inertia will me mR2
I=mR2 + mR2 + mR2
Inet=mR2+2mR2
I=3mR2

Topic: Physics, Center of Mass Momentum and Collision, JEE Main
Two small spheres each of mass 10 mg are suspended from a point by threads 0.5 m long. They are equally charged and repel each other to a distance of 0.20 m. The charge on each of the sphere is a21×10-8 C. The value of a will be ___.[Given g=10 m s-2]

Options:

The correct answer is:
20

Solution:
.

Tcosθ=mg=10×10-6×10=10-4

Tsinθ=9×109×q20.04=F

tanθ==0.10.24=Fmg

q=210324×10-8

0.95×10-8=a21×10-8

a=20

Topic: Physics, Electrostatics, JEE Main, JEE Main 2021 (25 Feb Shift 2)
Two small spheres each of mass m and charge q are tied from the same rigid support with the help of silk threads of length L. They make angle Î¸ with the vertical as shown in the figure. If length L is decreased, then angle Î¸ with the vertical

Options:
A increases
B decreases
C unaffected
D cannot say

The correct answer is:
increases

Solution:
When length is decreased, distance between two charges decreases and so, the force of repulsion increases. So for equilibrium, Î¸ also increases.

Topic: Physics, Electrostatics, JEE Main
Two small spherical metal balls, having equal masses, are made from materials of densities $\rho_1$ and $\rho_2\left(\rho_1=8 \rho_2\right)$ and have radii of $1 \mathrm{~mm}$ and $2 \mathrm{~mm}$, respectively. They are made to fall vertically (from rest) in viscous medium whose coefficient of viscosity equals $\eta$ and whose density is $0.1 \rho_2$. The ratio of their terminal velocities would be

Options:
A $\frac{79}{72}$
B $\frac{19}{36}$
C $\frac{39}{72}$
D $\frac{79}{36}$

The correct answer is:
$\frac{79}{36}$

Solution:
The terminal velocity achieved by ball in a viscous fluid is
$$
v_{\mathrm{t}}=\frac{2(\rho-\sigma) \mathrm{r}^2 \mathrm{~g}}{9 \eta}
$$
where, $\rho=$ density of metal of ball,
$\sigma=$ density of viscous medium,
$\mathrm{r}=$ radius of ball and
$\eta=$ coefficient of viscosity of medium
Terminal velocity of first ball,
$$
\begin{aligned}
\mathrm{v}_{\mathrm{t}_1} & =\frac{2\left(\rho_1-\sigma\right) \mathrm{r}_1^2 \mathrm{~g}}{9 \eta} \\
& =\frac{2\left(8 \rho_2-\sigma\right) \mathrm{r}_1^2 \mathrm{~g}}{9 \eta}
\end{aligned}
$$
...(i) $\left[\because \rho_1=8 \rho_2\right]$
Similarly, for second ball
$$
\mathrm{v}_{\mathrm{t}_2}=\frac{2\left(\rho_2-\sigma\right) \mathrm{r}_2^2 \mathrm{~g}}{9 \eta}
$$
From Eq. (i) and (ii), we get
$$
\begin{aligned}
\frac{\mathrm{v}_{\mathrm{t}_1}}{\mathrm{v}_{\mathrm{t}_2}} & =\frac{2\left(8 \rho_2-\sigma\right) \mathrm{r}_1^2 \mathrm{~g}}{2\left(\rho_2-\sigma\right) \mathrm{r}_2^2 \mathrm{~g}} \times \frac{9 \eta}{9 \eta} \\
& =\left(\frac{8 \rho_2-0.1 \rho_2}{\rho_2-0.1 \rho_2}\right)\left(\frac{r_1}{r_2}\right)^2
\end{aligned}
$$
..(iii) $\left[\because \sigma=0.1 \rho_2\right]$
Here, $\mathrm{r}_1=1 \mathrm{~mm}$ and $\mathrm{r}_2=2 \mathrm{~mm}$
Substituting these values in Eq. (iii), we get
$$
\Rightarrow \quad \frac{\mathrm{v}_{\mathrm{t}_1}}{\mathrm{v}_{\mathrm{t}_2}}=\left(\frac{7.9 \rho_2}{0.9 \rho_2}\right)\left(\frac{1}{2}\right)^2=\frac{79}{36}
$$

Topic: Physics, Mechanical Properties of Fluids, JEE Main
Two smooth and similar right angled prisms are arranged on a smooth horizontal plane as shown in the figure. The lower prism has a mass 3 times the upper prism. The prisms are held in an initial position as shown and are then released. As the upper prism touches the horizontal plane, the distance moved by the lower prism is

Options:
A $a-b$
B $\frac{a-b}{3}$
C $\frac{b-a}{2}$
D $\frac{a-b}{4}$

The correct answer is:
$\frac{a-b}{4}$

Solution:
If lower prism moves through a horizontal distance $\mathrm{k}$ since horizontal position of centre of mass of the two prims system remains unchanged
$\begin{aligned}
& 3 m k=m[(a-b)-k] \\
& \Rightarrow 3 k=(a-b) \\
& \Rightarrow 4 k=a-b \\
& \text { or } k=\frac{a-b}{4}
\end{aligned}$

Topic: Physics, Center of Mass Momentum and Collision, AP EAMCET, AP EAMCET 2016
Two soap bubbles are combined isothermally 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 S is the surface tension of the soap solution, the correct relation, among the following, is

Options:
A 4pV+3SA=0
B 3pV-4SA=0
C 4pV-3SA=0
D 3pV+4SA=0

The correct answer is:
3pV+4SA=0

Solution:
Let the radii of the two soap bubble be a and b, respectively, and radius of a single larger bubble be c.

As excess pressure for a soap bubble is 4Sr and external pressure is p,

pi=p+4Sr

So, pa=p+4Sa, pb=p+4Sb

and pc=p+4Tc ...(i)

and Va=43πa3, Vb=43πb3

and Vc=43πc3 ...(ii)

Now, as mass is conserved,

μa+μb=μc

i.e., paVaR Ta+pbVbR Tb=pcVcR Tc as p V=μR T

As the temperature is constant,

i.e., Ta=Tb=Tc

So, paVa+pbVb=pcVc

This, in the light of eqns. (i) and (ii) becomes,
p+4Sa 43πa3+p+4Sb 43πb3=p+4Sc 43πc3

i.e., 4Sa2+b2-c2=pc3-a3-b3 ...(iii)

Now, V=43πa3+b3-c3
and A=4πa2+b2-c2
∴ SAπ=-3Vp4π

or 4SA+3pV=0

Topic: Physics, Mechanical Properties of Fluids, JEE Main
Two soap bubbles coalesce to form a single bubble. If $\mathrm{V}$ is the subsequent change in volume of contained air and $\mathrm{S}$ change in total surface area, $\mathrm{T}$ is the surface tension and $\mathrm{P}$ atmospheric pressure, then which of the following relation is correct?

Options:
A
$4 \mathrm{PV}+3 \mathrm{ST}=0$

B
$3 \mathrm{PV}+4 \mathrm{ST}=0$

C
$2 \mathrm{PV}+3 \mathrm{ST}=0$

D
$3 \mathrm{PV}+2 \mathrm{ST}=0$

The correct answer is:
$3 \mathrm{PV}+4 \mathrm{ST}=0$

Solution:
No Solution Available

Topic: Physics, Mechanical Properties of Fluids, JEE Main, JEE Main 2014 (12 Apr Online)