datavorous/entrance-exam-dataset huggingface dataset:
A sphere of mass $2 \mathrm{~kg}$ and diameter $4.5 \mathrm{~cm}$ is attached to the lower end of a steel wire of $2 \mathrm{~m}$ length and area of cross-section $0.24 \times 10^{-6} \mathrm{~m}^2$. The wire is suspended from $205 \mathrm{~cm}$ high ceiling of a room. When the system is made to oscillate as a simple pendulum, the sphere just grazes the floor at its lowest position. The velocity of the sphere at the lowest position is (Young's modulus of steel $=2 \times 10^{11} \mathrm{Nm}^{-2}$ and acceleration due to gravity $=10 \mathrm{~ms}^{-2}$ )

Options:
A $10 \mathrm{~ms}^{-1}$
B $12 \mathrm{~ms}^{-1}$
C $15 \mathrm{~ms}^{-1}$
D $18 \mathrm{~ms}^{-1}$

The correct answer is:
$10 \mathrm{~ms}^{-1}$

Solution:
Extension in wire due to load and centrifugal force
$$
=\Delta l=(205-204.5) \mathrm{cm}=0.5 \mathrm{~cm}
$$
If velocity of sphere at lowest point is $v$, then
$$
\begin{gathered}
Y=\frac{\left(M g+\frac{M v^2}{R}\right) L}{A \Delta l} \\
\Rightarrow \quad M g+\frac{M v^2}{R}=\frac{Y A \Delta l}{L} \\
\text { where, } \quad R=202.25 \mathrm{~cm} \\
\Rightarrow \quad 2 \times 10+\frac{2 \times v^2}{202.25 \times 10^{-2}} \\
=\frac{2 \times 10^{11} \times 0.24 \times 10^{-6} \times 0.5 \times 10^{-2}}{2} \\
=0.12 \times 10^3 \quad 2 \times v^2 \\
\Rightarrow \quad \frac{202.25 \times 10^{-2}}{202.00} \Rightarrow v^2=101.125 \\
\Rightarrow \quad v=10.05 \mathrm{~ms}^{-1} \text { or } v=10 \mathrm{~ms}^{-1}
\end{gathered}
$$

Topic: Physics, Mechanical Properties of Solids, JEE Main
A sphere of mass m and radius r is projected in a gravity free space with speed v. If the coefficient of viscosity of the medium in which it moves is 1/6Ï Pa s , the distance travelled by the body before it stops is :

Options:
A mv2r
B 2mvr
C mvr
D mv4r

The correct answer is:
mvr

Solution:
The only force acting on the body is the viscous force
Here, m vdv dx = - 6 Ï Î· rv = - rv
â â« v 0 mdv = â« 0 x - rdx â x = mv r

Topic: Physics, Mechanical Properties of Fluids, JEE Main
A sphere of mass $m$ moving with a constant velocity $u$ hits another stationary sphere of the same mass. If $e$ is the coefficient of restitution, then the ratio of the velocity of two spheres after collision will be

Options:
A $\frac{1-e}{1+e}$
B $\frac{1+e}{1-e}$
C $\frac{e+1}{e-1}$
D $\frac{e-1}{e+1} t^2$

The correct answer is:
$\frac{1-e}{1+e}$

Solution:
Step 1: Apply momentum conservation
Since, there is no external force acting on the system, therefore momentum is conserved.
Let $\mathrm{v}_1$ and $\mathrm{v}_2$ be the velocities after collision
$\begin{aligned}
& \mathrm{P}_{\mathrm{i}}=\mathrm{P}_{\mathrm{f}} \\
& \Rightarrow \mathrm{mu}+\mathrm{o}=\mathrm{mv}_1+\mathrm{mv}_2 \\
& \Rightarrow \mathrm{u}=\mathrm{v}_1+\mathrm{v}_2 \ldots \text { (1) }
\end{aligned}$
Step 2: Coefficient of restitution (e)
Coefficient of restitution is defined as:
$\begin{aligned}
& e=\frac{\text { Relative velocity of separation(after collision) }}{\text { Relative velocity of approach(before collision) }}=\frac{v_2-v_1}{u_1-u_2} \\
& \Rightarrow e=\frac{\mathrm{v}_2-v_1}{\mathrm{u}-\mathrm{o}} \Rightarrow \mathrm{v}_2-\mathrm{v}_1=\mathrm{eu} \ldots . \text { (2) }
\end{aligned}$
Step 3: Solving above equations
Adding eq (1) and $(2) \Rightarrow 2 \mathrm{v}_2=(1+e) \mathrm{u} \Rightarrow \mathrm{v}_2=\frac{(1+e)}{2} \mathrm{u} \ldots$
Subtracting eq (2) from (1) $\Rightarrow 2 v_1=(1-e) u \Rightarrow v_1=\frac{(1-e)}{2}$
Dividing eq (4) by (3) $\Rightarrow \frac{\mathrm{v}_1}{\mathrm{v}_2}=\frac{1-\mathrm{e}}{1+\mathrm{e}}$

Topic: Physics, Center of Mass Momentum and Collision, JEE Main
A sphere of mass m moving with velocity v collides head-on with another sphere of the same mass at rest. If the coefficient of restitution e=12, then what is the ratio of final velocity of the second sphere to the initial velocity of the first sphere?

Options:

The correct answer is:
0.75

Solution:
mv+0=mv1+mv2
v1+v2=v ....(i)
e=v2−v1v−0
ev=v2−v1 ..(ii)
Add equation (i) & equation (ii)
ev+v=2v2
v2=v(e+1)2
v2v=e+12=0.75

Topic: Physics, Center of Mass Momentum and Collision, JEE Main
A sphere of radius 0.1 m and mass 8 π kg is attached to the lower end of a steel wire of length 5.0 m & diameter 10-3 m. The wire is suspended from 5.22 m high ceiling of a room. When the sphere is made to swing like a simple pendulum, it just grazes the floor at its lowest point. Calculate the velocity of the sphere at the lowest position. Y for steel = 1.994×1011 N m-2

Options:
A 7.7 m s-1
B 4.4 m s-1
C 2.2 m s-1
D 8.8 m s-1

The correct answer is:
8.8 m s-1

Solution:
As the length of the wire is 5 m and diameter 2×0.1=0.2 m and at the lowest point, it grazes the floor which is at a distance 5.22 m from the roof, the increase in the length of the wire at lowest point
Δ L = 5 . 2 2 - 5 + 0 . 2
= 0 . 0 2 m
So the tension in the wire (due to elasticity)
T=YALΔL=1.994×1011×π5×10-42×0.025=199.4π N
and as the equation of circular - motion of a mass m tied to a string in a vertical plane is
mv2r=T-mg cosθ
So at the lowest point
mv2r=T-mg as θ = 0
But here r = 5 + 0.02 + 0.1 = 5.12 m

So 8πv25.12=199.4π-8π×9.8
i.e., v 2 = 1 2 1 × 5 . 1 2 / 8 = 7 7 . 4 4
So v = 8.8 m s-1

Topic: Physics, Mechanical Properties of Solids, JEE Main
A sphere of radius a and mass m rolls along a horizontal plane with constant speed v0. It encounters an inclined plane at angle θ and climbs upward. Assuming that it rolls without slipping, how far up the sphere will travel?

Options:
A 25v02 gsinθ

B 7v0210 gsinθ
C v025gsinθ
D v022gsinθ

The correct answer is:
7v0210 gsinθ

Solution:
From energy conservation
mgh=12mv02+12Iω2
⇒mgh=12mv02+12×25ma2×v02a2
⇒gh=12v02+15v02
⇒gh=710v02

From triangle, sin θ=hl
Then h=l sin θ
⇒l sin θ=710v02 g
Thus, the sphere will travel distance, l=710v02 g sin θ.

Topic: Physics, Rotational Motion, JEE Main, JEE Main 2021 (25 Feb Shift 2)
A sphere of radius $R$ has a uniform distribution of electric charge in its volume. At a distance $x$ from its centre, for $x \lt R$, the electric field is directly proportional to

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

The correct answer is:
$x$

Solution:
Let sphere has uniform chare density $\rho\left(=\frac{3 Q}{4 \pi R^3}\right)$ and $E$ is the electric field at distance $x$ from the centre of the sphere.
Applying Gauss law
$E .4 \pi x^2=\frac{q}{\varepsilon_0}=\frac{\rho V^{\prime}}{\varepsilon_0}=\frac{\rho}{\varepsilon_0} \times \frac{4}{3} \pi x^3$
$\left(V^{\prime}=\right.$ Volume of dotted sphere $) \therefore \quad E=\frac{\rho}{3 \varepsilon_0} x \Rightarrow$\(\mathrm{E} \propto \mathrm{X}\)

Topic: Physics, Electrostatics, JEE Main
A sphere of radius R is gently dropped into a liquid of viscosity Î· in a vertical uniform tube. It attains a terminal velocity v. Another sphere of radius 2R when dropped into the same liquid, will attain a terminal velocity

Options:
A v
B 2v
C 4v
D 9v

The correct answer is:
4v

Solution:
Terminal velocity v=29Ãr2Ï-ÏgÎ·

Hence terminal velocity vâr2
â´ v'v=2R2R2
or v'=4v

Topic: Physics, Mechanical Properties of Fluids, JEE Main
A sphere of relative density $\sigma$ and diameter $D$ has concentric cavity of diameter $d$. The ratio of $\frac{D}{d}$, if it just floats on water in a tank is :

Options:
A $\left(\frac{\sigma-2}{\sigma+2}\right)^{1 / 3}$
B $\left(\frac{\sigma}{\sigma-1}\right)^{1 / 3}$
C $\left(\frac{\sigma-1}{\sigma}\right)^{1 / 3}$
D $\left(\frac{\sigma+1}{\sigma-1}\right)^{1 / 3}$

The correct answer is:
$\left(\frac{\sigma}{\sigma-1}\right)^{1 / 3}$

Solution:
weight $(w)=\frac{4}{3} \pi\left(\frac{\mathrm{D}^3-\mathrm{d}^3}{8}\right) \sigma \mathrm{g}$
Buoyant force $\left(\mathrm{F}_{\mathrm{b}}\right)=1 \times \frac{4}{3} \pi\left(\frac{\mathrm{D}^3}{8}\right) \cdot \mathrm{g}$
For Just Float $\Rightarrow \mathrm{w}=\mathrm{F}_{\mathrm{b}}$
$\begin{aligned} & \Rightarrow\left(D^3-d^3\right) \sigma=D^3 \\ & \Rightarrow 1-\frac{d^3}{D^3}=\frac{1}{\sigma} \\ & \Rightarrow 1-\frac{1}{\sigma}=\left(\frac{d}{D}\right)^3 \\ & \Rightarrow\left(\frac{\sigma}{\sigma-1}\right)^{\frac{1}{3}}=\left(\frac{D}{d}\right)\end{aligned}$

Topic: Physics, Mechanical Properties of Fluids, JEE Main, JEE Main 2024 (09 Apr Shift 1)
A spherical ball $A$ of mass 4 kg, moving along a straight line strikes another spherical ball $B$ of mass $1 \mathrm{kg}$ at rest. After the collision, $A$ and $B$ move with velocities $y_{1} \mathrm{ms}^{-1}$ and $v_{2} \mathrm{ms}^{-1}$ respectively making angles of $30^{\circ}$ and $60^{\circ}$ with
respect to the original direction of motion of $A .$ The ratio $\frac{y_{1}}{v_{2}}$ will be

Options:
A $\frac{\sqrt{3}}{4}$
B $\frac{4}{\sqrt{3}}$
C $\frac{1}{\sqrt{3}}$
D $\sqrt{3}$

The correct answer is:
$\frac{\sqrt{3}}{4}$

Solution:
Along $y$ -axis momentum remains zero. Here
$$
\begin{aligned}
4 v_{1} \sin 30^{\circ} &=v_{2} \sin 60^{\circ} \\
\frac{y_{1}}{v_{2}} &=\frac{\sqrt{3}}{4}
\end{aligned}
$$

Topic: Physics, Center of Mass Momentum and Collision, WBJEE, WBJEE 2012
A spherical ball $A$ of mass 4 kg, moving along a straight line strikes another spherical ball $B$ of mass $1 \mathrm{kg}$ at rest. After the collision, $A$ and $B$ move with velocities $y_{1} \mathrm{ms}^{-1}$ and $v_{2} \mathrm{ms}^{-1}$ respectively making angles of $30^{\circ}$ and $60^{\circ}$ with
respect to the original direction of motion of $A .$ The ratio $\frac{y_{1}}{v_{2}}$ will be

Options:
A $\frac{\sqrt{3}}{4}$
B $\frac{4}{\sqrt{3}}$
C $\frac{1}{\sqrt{3}}$
D $\sqrt{3}$

The correct answer is:
$\frac{\sqrt{3}}{4}$

Solution:
Along $y$ -axis momentum remains zero. Here
$$
\begin{aligned}
4 v_{1} \sin 30^{\circ} &=v_{2} \sin 60^{\circ} \\
\frac{y_{1}}{v_{2}} &=\frac{\sqrt{3}}{4}
\end{aligned}
$$

Topic: Physics, Center of Mass Momentum and Collision, JEE Main
A spherical ball A of mass 4 kg, moving along a straight line strikes another spherical ball B of mass 1 kg at rest. After the collision, A and B move with velocities v1 m sâ1 and v2 m sâ1 respectively making angles of 30Â° and 60Â° with respect to the original direction of motion of A. The ratio v1v2 will be

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

The correct answer is:
3 4

Solution:
Applying the law of conservation of linear momentum along a direction perpendicular to the direction of motion (i.e. along the y-axis), we get.
0 + 0 = 4v1 sin 30° – v2 sin 60°
4v1sin 30° = v2 sin 60°
v1v2=sin60o4sin30o=34

Topic: Physics, Center of Mass Momentum and Collision, JEE Main
A spherical ball is dropped in a long column of a highly viscous liquid. The curve in the graph shown, which represents the speed of the ball v as a function of time t is

Options:
A B
B C
C D
D A

The correct answer is:
B

Solution:
Viscous force increases with increase in velocity and becomes constant after attaining terminal velocity.

Topic: Physics, Mechanical Properties of Fluids, JEE Main
A spherical ball is dropped in a long column of viscous liquid. Which of the following graphs represent the variation of

(i) the gravitational force with time

(ii) the viscous force with time

(iii) the net force acting on the ball with time

Options:
A Q, R, P
B R, Q, P
C P, Q, R
D R, P, Q

The correct answer is:
P, Q, R

Solution:
Gravitational force remains constant on the falling spherical ball. It is represented by straight line P. The viscous force (F=6πηrv) increases as the velocity increases with time. Hence, it is represented by curve Q.

Net force = gravitational force – viscous force

As viscous force increases, net force decreases and finally becomes zero. Then the body falls with a constant terminal velocity. It is thus represented by curve R.

Topic: Physics, Mechanical Properties of Fluids, NEET
A spherical ball of density Ï and radius 0.003 m is dropped into a tube containing a viscous fluid, filled up to the 0 cm mark as shown in the figure. Viscosity of the fluid =1.260 N m-2 s-1 and its density Ï L = Ï/2=1260 kg m-3. Assume the ball reaches a terminal speed by the 10 cm mark. Find the time taken by the ball to traverse the distance between the 10 cm and 20 cm mark. [g = acceleration due to gravity = 10 m s-2ââ ]

Options:
A 2 s
B 3 s
C 5 s
D 1.5 s

The correct answer is:
5 s

Solution:
Terminal Velocity:
V T = 2 9 Ã r 2 Ï - Ï L g Î·
âVT=29Ã3Ã10-32Ã1260Ã101.260
VT=2Ã10-2 m s-1
âVT=2 cm s-1
t=10 cmVT
ât=5 s

Topic: Physics, Mechanical Properties of Fluids, JEE Main
A spherical ball of diameter 1 cm and density 5Ã103 kg m-3 is dropped gently in a large tank containing viscous liquid of density 3Ã103 kg m-3 and coefficient of viscosity 0.1 Ns m-2. The distance that the ball moves in 1 s after attaining terminal velocity is g=10 ms-2

Options:
A 109 m
B 23 m
C 49 m
D 45 m

The correct answer is:
109 m

Solution:
Given, diameter of spherical ball d=1 cm=0.01 m
Radius of spherical ball r=0.012 m
Density of spherical ball ρ=5×103 kg m-3
Density of liquid σ=3×103 kg m-3
Coefficient of viscosity η=0.1 N-s m-2
We know that,
Terminal velocity is given by
v=29gr2ρ-ση
=29×10×0.0122×5×103-3×1030.1 m s-1
v=109 m s-1
Distance travelled in one second after attaining terminal velocity=109×1 m
=109 m

Topic: Physics, Mechanical Properties of Fluids, JEE Main
A spherical ball of mass $20 \mathrm{~kg}$ is stationary at the top of a hill of height $100 \mathrm{~m}$. It rolls down a smooth surface to the ground, then climbs up another hill of height $30 \mathrm{~m}$ and finally rolls down to a horizontal base at a height of $20 \mathrm{~m}$ above the ground. The velocity attained by the ball is

Options:
A
$40 \mathrm{~m} / \mathrm{s}$

B
$20 \mathrm{~m} / \mathrm{s}$

C
$10 \mathrm{~m} / \mathrm{s}$

D
$10 \sqrt{30} \mathrm{~m} / \mathrm{s}$

The correct answer is:
$40 \mathrm{~m} / \mathrm{s}$

Solution:
$\mathrm{mgh}=1 / 2 \mathrm{mv}^2$
$\mathrm{v}=\sqrt{2 \mathrm{gh}}$
$=\sqrt{2 \times 10 \times 80}=40 \mathrm{~m} / \mathrm{s}$

Topic: Physics, Work Power Energy, JEE Main, JEE Main 2005
A spherical ball of mass m is kept at the highest point in the space between two fixed, concentric spheres A and B (see fig.). The smaller sphere A has a radius R and space between the two spheres has a width d. The ball has a diameter very slightly less than d. All surfaces are frictionless. The ball is given a gentle push (towards the right in the figure). The angle made by the radius vector of the ball with the upward vertical is denoted by Î¸. What is the total normal reaction force exerted by the spheres on the ball in terms of angle Î¸?

Options:
A mg (3cosθ –2 )
B mg ( 2cosθ – 3 )
C 3mg ( 2cosθ – 1 )
D 2mg ( 3cosθ – 1 )

The correct answer is:
mg (3cosθ –2 )

Solution:
h=R+d21-cos θ

The velocity of the ball at an angle θ is
ν2=2gh=2R+d21-cos θg ...... (i)
Let N be the normal reaction (away from the centre) at angle θ. Then, mg cos θ-N=mν2R+d2
Substituting the value of ν 2 from equation (i), we get
mg cos θ-N=2mg1-cos θ
N=mg3 cos θ-2

Topic: Physics, Work Power Energy, JEE Main
A spherical ball of radius $1 \times 10^{-4} \mathrm{~m}$ and density $10^5 \mathrm{~kg} / \mathrm{m}^3$ falls freely under gravity through a distance $h$ before entering a tank of water, If after entering in water the velocity of the ball does not change, then the value of $h$ is approximately:
(The coefficient of viscosity of water is $9.8 \times 10^{-6} \mathrm{~N} \mathrm{~s} / \mathrm{m}^2$ )

Options:
A $2296 \mathrm{~m}$
B $2518 \mathrm{~m}$
C $2249 \mathrm{~m}$
D $2396 \mathrm{~m}$

The correct answer is:
$2518 \mathrm{~m}$

Solution:
$\begin{aligned}& \mathrm{V}_{\mathrm{T}}=\frac{2 \mathrm{~g}}{9} \frac{\mathrm{R}^2\left[\rho_{\mathrm{B}}-\rho_{\mathrm{L}}\right]}{\eta} \\& \Rightarrow \mathrm{V}_{\mathrm{T}}=\frac{2}{9} \times \frac{10 \times\left(10^{-4}\right)^2}{9.8 \times 10^{-6}}\left[10^5-10^3\right] \\& \Rightarrow \mathrm{V}_{\mathrm{T}}=224.5\end{aligned}$
when ball fall from height (h)
$\begin{aligned}& \mathrm{V}=\sqrt{2 \mathrm{gh}} \\
& \mathrm{h}=\left(\frac{\mathrm{V}^2}{2 \mathrm{~g}}\right)=2518 \mathrm{~m}\end{aligned}$

Topic: Physics, Mechanical Properties of Fluids, JEE Main, JEE Main 2024 (09 Apr Shift 2)
A spherical ball of radius 1 mm and density 10.5 g cc-1 is dropped in glycerine of coefficient of viscosity 9.8 poise and density 1.5 g cc-1. Viscous force on the ball when it attains constant velocity is 3696×10-x N. The value of x is

(Given, g=9.8 m s-2 and π=227)

Options:

The correct answer is:
7

Solution:
Viscous force on the ball when it attains constant velocity i.e., terminal velocity is,

Fv=mg-FB As, a=0

Here, FB is buoyant force.

⇒Fv=Vρbg-Vρgg=Vgρb-ρg=43π10-33×9.810.5-1.5×103≈3696×10-7 N

Thus, x=7.

Topic: Physics, Mechanical Properties of Fluids, JEE Main, JEE Main 2023 (24 Jan Shift 2)
A spherical ball rolls on a table without slipping. Then the fraction of its total energy associated with rotation is

Options:
A $\frac{2}{5}$
B $\frac{2}{7}$
C $\frac{3}{5}$
D $\frac{3}{7}$

The correct answer is:
$\frac{2}{7}$

Solution:
$\begin{aligned} & \mathrm{KE}_{\mathrm{Total}}=\frac{1}{2} \mathrm{IW}^2+\frac{1}{2} \mathrm{MV}^2 \\ & =\frac{1}{2} \times \frac{2}{5} \mathrm{MR}^2 \mathrm{~W}^2+\frac{1}{2} \mathrm{MV}^2 \\ & =\frac{1}{5} \mathrm{MR}^2 \mathrm{~W}^2+\frac{1}{2} \mathrm{MV}^2=\frac{1}{5} \mathrm{MV}^2+\frac{1}{2} \mathrm{MV}^2 \\ & =\frac{7}{10} \mathrm{MV}{ }^2\end{aligned}$
$\begin{aligned} & \mathrm{KE}_{\text {rot }}=\frac{1}{2} \mathrm{IW}^2=\frac{1}{5} \mathrm{MV}^2 \\ & \frac{\mathrm{KE}_{\text {rot }}}{\mathrm{KE}_{\text {Total }}}=\frac{\frac{1}{5} \mathrm{MV}^2}{\frac{7}{10} \mathrm{MV}^2} \\ & =\frac{2}{7}\end{aligned}$

Topic: Physics, Rotational Motion, JEE Main
A spherical balloon is being inflated at the rate of $35 \mathrm{cc} / \mathrm{min}$. The rate of increase in the surface area (in $\mathrm{cm}^2 / \mathrm{min}$.) of the balloon when its diameter is $14 \mathrm{~cm}$, is :

Options:
A
10

B
$\sqrt{10}$

C
100

D
$10 \sqrt{10}$

The correct answer is:
10

Solution:
Volume of sphere $\mathrm{V}=\frac{4}{3} \pi r^3$
$\frac{d \mathrm{~V}}{d t}=\frac{4}{3} \cdot \pi \cdot 3 r^2 \cdot \frac{d r}{d t}$
$35=4 \pi r^2 \cdot \frac{d r}{d t}$ or $\frac{d r}{d t}=\frac{35}{4 \pi r^2}$
Surface area of sphere $=\mathrm{S}=4 \pi r^2$
$\frac{d \mathrm{~S}}{d t}=4 \pi \times 2 r \times \frac{d r}{d t}=8 \pi r \cdot \frac{d r}{d t}$
$\frac{d \mathrm{~S}}{d t}=\frac{70}{r}$
Now, diameter $=14 \mathrm{~cm}, r=7$
$\therefore \quad \frac{d \mathrm{~S}}{d t}=10$

Topic: Mathematics, Application of Derivatives, JEE Main, JEE Main 2013 (25 Apr Online)
A spherical balloon is expanding. If the radius is increasing at the rate of \( 2 \mathrm{~cm} / \mathrm{min} \) the rate at which the volume increases
(in cubic centimeters per minute) when the radius is \( 5 \mathrm{~cm} \), is

Options:
A \( 10 \pi \)
B \( 100 \pi \)
C \( 200 \pi \)
D \( 50 \pi \)

The correct answer is:
\( 200 \pi \)

Solution:
As, volume of sphere V=43πr3
⇒ dVdt=4πr2 drdt=4πr2.(2) [∵drdt=2]

⇒dVdt=4π.25.2=200π

Topic: Mathematics, Application of Derivatives, JEE Main
A spherical balloon is filled with $4500 ~\pi$ cubic meters of helium gas. If a leak in the balloon causes the gas to escape at the rate of $72 ~\pi$ cubic meters per minute, then the rate (in meters per minute) at which the radius of the balloon decreases $49$ minutes after the leakage began is

Options:
A
$\frac{9}{7}$

B
$\frac{7}{9}$

C
$\frac{2}{9}$

D
$\frac{9}{2}$

The correct answer is:
$\frac{2}{9}$

Solution:
$v=\frac{4}{3} \pi r^2$
After $49$ minutes volume $=4500 \pi-49(72 \pi)=972 \pi$
$\frac{4}{3} \pi r^3=972 \pi \quad \Rightarrow r^3=729 \quad \Rightarrow r=9$
$v=\frac{4}{3} \pi r^3$
$\frac{d v}{d t}=\frac{4}{3} \pi 3 r^2 \frac{d r}{d t}$
$72 \pi=4 \pi r^2 \frac{d r}{d t}$
$\frac{\mathrm{dr}}{\mathrm{dt}}=\frac{72}{4 \cdot 9 \cdot 9}=\frac{2}{9}$

Topic: Mathematics, Application of Derivatives, JEE Main, JEE Main 2012 (Offline)
A spherical balloon of 21 cm diameter is to be filled up with H2 at NTP from cylinder containing the gas at 20 atm at 27oC. The cylinder can hold 2.82 litre of water a NTP. Calculate the number of balloons that can be filled up.

Options:

The correct answer is:
10

Solution:
Volume of one balloon (to be filled)
43πr3=43×227×2123
= 4851 mL = 4.851 litre
Let ‘n’ balloons are filled, then total volume of H2 used in filling balloons =4.851×nlitre
The cylinder of H2 used in filling balloons will also have H2 in it, after n balloons are filled. Also volume of cylinder = 2.82 litre
Hence Total volume of H2 at NTP = volume of 'n' balloons + volume of cylinder
=4.851×n+2.82 ….(i)
The volume of available H2 at NTP can be derived by the data
Given P = 1 atm, V = ?, T = 273 K
and P = 20 atm, V = 2.82 litre, T = 300 K
∴P1V1T1=P2V2T2
or 1×V273=20×2.82300
∴V=20×2.82×273300=51.32 ...(ii)
From equation (1) and (II)
4.851×n+2.82=57.32
n=10

Topic: Chemistry, States of Matter, JEE Main
A spherical balloon of radius 3 cm containing helium gas has a pressure of 48×10-3 bar. At the same temperature, the pressure, of a spherical balloon of radius12 cm containing the same amount of gas will be………….×10-6 bar.

Options:

The correct answer is:
750

Solution:
if volume of 3cm balloon is =V
then volume of 12cm radius balloon =64V
So pressure will become 1 / 64 times =1/64×48×10-3 bar =750×10-3 bar

Topic: Chemistry, States of Matter, JEE Main, JEE Main 2020 (06 Sep Shift 1)
A spherical body of mass 2 kg starting from rest acquires a kinetic energy of 10000 J at the end of 5th second. The force acted on the body is _____ N.

Options:

The correct answer is:
40

Solution:
Given kinetic energy, 12mv2=10000 J

⇒12×2×v2=10000

⇒ v2=10000

⇒ v=100 m s-1

Also,

⇒ v=at=a×5=100

⇒ a=20 m s-2

Therefore, force acting

F=ma=2×20=40 N

Topic: Physics, Work Power Energy, JEE Main, JEE Main 2023 (24 Jan Shift 1)
A spherical body of mass and radius is allowed to fall in a medium of viscosity. The time in which the velocity of the body increases from zero to $0.63$ times the terminal velocity is called time constant. Dimensionally can be represented by

Options:
A $\frac{m r^2}{6 \pi \eta}$
B $\sqrt{\left(\frac{6 \pi m r \eta}{g^2}\right)}$
C $\frac{m}{6 \pi \eta r v}$
D None of the above

The correct answer is:
None of the above

Solution:
Time constant $\tau=[T]$ and Viscosity $\eta=\left[M {L}^{-1} T^{-1}\right]$
For options (a), (b) and (c) dimensions are not matching with time constant.
Option A has dimensions: \(\frac{\mathrm{ML}^2}{\mathrm{ML}^{-1} \mathrm{~T}^{-2}}=\mathrm{L}^3 \mathrm{~T}^2\)
Option B has dimensions: \(\sqrt{\frac{(\mathrm{ML}) *\left(\mathrm{ML}^{-1} \mathrm{~T}^{-2}\right)}{\mathrm{L}^2 \mathrm{~T}^{-4}}}=\mathrm{ML}^{-1} \mathrm{~T}\)
Option C has dimensions: \(\frac{\mathrm{M}}{\left(\mathrm{ML}^{-1} \mathrm{~T}^{-2}\right) *(\mathrm{~L}) *\left(\mathrm{LT}^{-1}\right)}=\frac{\mathrm{T}^3}{\mathrm{~L}}\)

Topic: Physics, Units and Dimensions, JEE Main
A spherical body, of mass m and radius r, is allowed to fall in a medium of viscosity Î· . The time in which the velocity of the body increases from 0 to 0.63 times the terminal velocity (v) is called time constant Ï . Dimensionally Ï can be represented as

Options:
A mr26πη
B 6πmrηg2
C m6πηrv
D None of these

The correct answer is:
None of these

Solution:
Time constant is the time of free fall of a body under gravity in the viscous medium, during which the velocity of the body increases to 63% of the terminal velocity. Dimensionally, none of the alternatives have dimensions of time.

Topic: Physics, Units and Dimensions, JEE Main