datavorous/entrance-exam-dataset huggingface dataset:
A rod of mass M and length L is hinged at its one end and carries a block of mass m at its lower end. A spring of force constant k1 is installed at distance a from the hinge and another of force constant k2 at a distance b as shown in the figure. If the whole arrangement rests on a smooth horizontal table top, find the frequency of vibration.

Options:
A 1 2π k 1 a 2 + k 2 b 2 L 2 ( m+ M 3 )
B 1 2π k 1 a 2 + k 2 b 2 L 2 ( m− M 3 )
C 1 2π k 1 a 2 − k 2 b 2 L 2 ( m+ M 3 )
D 1 4π k 1 a 2 + k 2 b 2 L 2 ( m− M 3 )

The correct answer is:
1 2π k 1 a 2 + k 2 b 2 L 2 ( m+ M 3 )

Solution:
For small angular displacement Î¸, net torque towards mean position is
Ï = k 1 a Î¸ a + k 2 b Î¸ b or I Î± = k 1 a 2 + k 2 b 2 Î¸
or mL 2 + 1 3 ML 2 Î± = k 1 a 2 + k 2 b 2 Î¸
or Î± = k1 a 2 + k 2 b 2 Î¸ L 2 m + M 3
â´ Ï 2 = k 1 a 2 + k 2 b 2 L 2 m + M 3
Hence, frequency
f = 2 Ï = 1 2 Ï k 1 a 2 + k2 b 2 L 2 m + M 3

Topic: Physics, Oscillations, JEE Main
A rod of mass M and length L is lying on a horizontal frictionless surface. A particle of mass m travelling along the surface hits at one end of the rod with a velocity u in a direction perpendicular to the rod. The collision is completely elastic. After collision, particle comes to rest. The ratio of masses mM is 1x. The value of x will be

Options:

The correct answer is:
4

Solution:
From momentum conservation, Pi0=Pf
mu=Mv …i
From angular momentum conservation about O,
mu·L2=ML212ω
⇒ω=6muML …(ii)
From e=R.V·SR.V·A
1=V+ωL2u

V+ωL2=u
V+3muM=u
muM+3muM=u
4muM=u
mM=14
x=4

Topic: Physics, Center of Mass Momentum and Collision, JEE Main, JEE Main 2021 (20 Jul Shift 1)
A rod $P Q$ of length $l$ is moving with ends remaining in contact with frictionless wall and floor. If at the instant, shown the velocity of end $B$ is $2 \mathrm{~m} / \mathrm{s}$ towards negative direction of $x$. The speed of end $A$ will be

Options:
A $\sqrt{3} \mathrm{~ms}^{-1}$
B $\frac{2}{\sqrt{3}} \mathrm{~ms}^{-1}$
C $\sqrt{2} \mathrm{~ms}^{-1}$
D $\frac{\sqrt{5}}{2} \mathrm{~ms}^{-1}$

The correct answer is:
$\frac{2}{\sqrt{3}} \mathrm{~ms}^{-1}$

Solution:
From the diagram in question,
$\begin{array}{rlrl}x & =1 \cos \theta \text { and } y=1 \sin \theta \\ \therefore \quad & v_x & =\frac{d x}{d t}=-1 \sin \theta \cdot \frac{d \theta}{d t}\end{array}$
and $\quad v_y=\frac{d y}{d t}=1 \cos \theta \frac{d \theta}{d t}$
$\therefore \quad \frac{v_y}{v_x}=-\cot \theta$ or $v_y=-v_x \cot \theta$
Since, $\quad v_x=-2 \mathrm{~m} / \mathrm{s}$
Hence, $v_y=(-2) \cot 60^{\circ}=-2 \times \frac{1}{\sqrt{3}}$
$=-\frac{2}{\sqrt{3}} \mathrm{~ms}^{-1}$
The negative sign indicates the motion towards negative direction of $Y$.

Topic: Physics, Motion In One Dimension, JIPMER, JIPMER 2015
A rod $P Q$ of mass $M$ and length $L$ is hinged at end $P$. The rod is kept horizontal by a massless string tied to point $\mathrm{Q}$ as shown in figure. When string is cut, the initial angular acceleration of the rod is

Options:
A $\frac{3 g}{2 L}$
B $\frac{2 \mathrm{~g}}{\mathrm{~L}}$
C $\frac{\mathrm{g}}{\mathrm{L}}$
D $\frac{2 g}{3 \mathrm{~L}}$

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

Solution:
After the string is cut the initial angular acceleratipn of the rodis generated by torque at $Q$
On balancing torque about the $Q$ point
$\begin{aligned}
& M g \frac{L}{2}=\frac{M L^2}{3} \alpha \\
& \alpha=\frac{3 g}{2 L}
\end{aligned}$
Therefore the acceleration of rod is $\alpha=\frac{3 g}{2 L}$

Topic: Physics, Rotational Motion, JEE Main
A rod with circular cross-section area 2 cm2 and length 40 cm is wound uniformly with 400 turns of an insulated wire. If a current of 0.4 A flows in the wire windings, the total magnetic flux produced inside windings is 4π×10-6 Wb. The relative permeability of the rod is

(Given : Permeability of vacuum μ0=4π×10-7 N A-2)

Options:
A 12.5
B 325
C 125
D 516

The correct answer is:
125

Solution:
Magnetic flux is given by ϕ=NBA, here, N is number of turns, B is magnetic field and A is area.

Given here, ϕ=4π×10-6 Wb, A=2 cm2, N=400, I=0.4 A.

Putting the values,

ϕ=400×μrμ0nI×A

⇒4π×10-6 =400μr×4π×10-7×4000.4×0.4×2×10-4

⇒μr=125

Topic: Physics, Magnetic Effects of Current, JEE Main, JEE Main 2023 (31 Jan Shift 1)
A roller coaster is designed such that riders experience "weightlessness" as they go round the top of a hill whose radius of curvature is $20 \mathrm{~m}$. The speed of the car at the top of the hill is between

Options:
A $13 \mathrm{~m} / \mathrm{s}$ and $14 \mathrm{~m} / \mathrm{s}$
B $14 \mathrm{~m} / \mathrm{s}$ and $15 \mathrm{~m} / \mathrm{s}$
C $15 \mathrm{~ms} /$ and $16 \mathrm{~m} / \mathrm{s}$
D $16 \mathrm{~m} / \mathrm{s}$ and $17 \mathrm{~m} / \mathrm{s}$

The correct answer is:
$14 \mathrm{~m} / \mathrm{s}$ and $15 \mathrm{~m} / \mathrm{s}$

Solution:
Actually, $m g-N=\frac{m v^2}{r}$
For weightlessness, $\quad N=0$
$$
\begin{aligned}
\therefore m g & =\frac{m v^2}{r} \\
v & =\sqrt{g r} \\
& =\sqrt{10 \times 20} \mathrm{~ms}^{-1}=14.14 \mathrm{~ms}^{-1}
\end{aligned}
$$

Topic: Physics, Work Power Energy, NEET, NEET 2008 (Mains)
A roller coaster is designed such that riders experience "weightlessness" as they go round the top of a hill whose radius of curvature is $20 \mathrm{~m}$. The speed of the car at the top of the hill is between

Options:
A $14 \mathrm{~m} / \mathrm{s}$ and $15 \mathrm{~m} / \mathrm{s}$
B $15 \mathrm{~m} / \mathrm{s}$ and $16 \mathrm{~m} / \mathrm{s}$
C $16 \mathrm{~m} / \mathrm{s}$ and $17 \mathrm{~m} / \mathrm{s}$
D $13 \mathrm{~m} / \mathrm{s}$ and $14 \mathrm{~m} / \mathrm{s}$

The correct answer is:
$14 \mathrm{~m} / \mathrm{s}$ and $15 \mathrm{~m} / \mathrm{s}$

Solution:
Balancing the forces, we get
$$
M g-N=M \frac{v^{2}}{R}
$$
For weightlessness, $N=0$
$\therefore$
$$
\frac{M v^{2}}{R}=M g
$$
where $R$ is the radius of curvature and $v$ is the speed of car. Therefore,
$$
v=\sqrt{\mathrm{Rg}}
$$
Putting the values, $R=20 \mathrm{~m}, g=10.0 \mathrm{~m} / \mathrm{s}^{2}$
So, $\quad v=\sqrt{20 \times 10.0}=14.14 \mathrm{~m} / \mathrm{s}^{2}$
Thus, the speed of the car at the top of the hill is
$$
\text { between } 14 \mathrm{~m} / \mathrm{s} \text { and } 15 \mathrm{~m} / \mathrm{s} \text { . }
$$
Note : The roller coaster is a popular amusement ride developed for amusement parks and modern theme parks.

Topic: Physics, Laws of Motion, JEE Main
A roller is made by joining together two cones at their vertices O. It is kept on two rails AB and CD which are placed asymmetrically (see figure), with its axis perpendicular to CD and its centre O at the centre of line joining AB and CD (see figure). It is given a light push so that it starts rolling with its centre O moving parallel to CD in the direction shown. As it moves, the roller will tend to:

Options:
A go straight.
B turn left and right alternately.
C turn left.
D turn right.

The correct answer is:
turn left.

Solution:
As roller moves forward radius of circle of left cone on track AB decreases. Thus right cone travel more distance compared to left and roller turns left

Topic: Physics, Rotational Motion, JEE Main, JEE Main 2016 (03 Apr)
A rolling body is kept on a plank $B$. There is sufficient friction between $A$ and $B$ and no friction between $B$ and the inclined plane. Then body:

Options:
A A rolls
B A does not experience any friction
C $A$ and $B$ has equal acceleration and unequal velocities
D A rolls depending upon the angle of inclination $\theta$

The correct answer is:
A does not experience any friction

Solution:
Both the sphere and the plank will slide down with same acceleration g $\sin \theta$.

Topic: Physics, Rotational Motion, JEE Main
A rolling wheel of 12 kg is on an inclined plane at position P and connected to a mass of 3 kg through a string of fixed length and pulley as shown in figure.
Consider PR as friction free surface.
The velocity of centre of mass of the wheel when it reaches at the bottom Q of the inclined plane PQ will be 12xgh m s-1. The value of x (rounded off to the nearest integer) is _____.

Options:

The correct answer is:
3

Solution:
If we take the wheel as a ring then its moment of inertia will be, I=Mr2.

Using mechanical energy conservation,

Mgh-mgh=12mv2+12Mv2+12Iω2⇒12gh-3gh=123v2+1212v2+1212r2v2r2

⇒9gh=15v2+12v22

⇒v=23gh=1283gh

Therefore, x=83≃3

Topic: Physics, Work Power Energy, JEE Main, JEE Main 2022 (27 Jun Shift 2)
A room at 20Â°C is heated by a heater of resistance 20 ohm connected to 200 V mains. The temperature is uniform throughout the room and the heat is transmitted through a glass window of area 1 m 2 and thickness 0.2 cm . Calculate the temperature outside. Thermal conductivity of glass is 0.2 cal/ mÂ°C s and mechanical equivalent of heat is 4.2 J/cal.

Options:
A 13.69°C
B 15.24Â°C
C 17.85Â°C
D 19.96Â°C

The correct answer is:
15.24°C

Solution:
If Î¸ is the temperature of outside, heat passing per second through the glass window,
dQ dt = KA Î¸ 1 - Î¸ 2 L
= 0.2 Ã 1 Ã 2 0 - Î¸ cal 0.2 Ã 1 0 - 2 = 1 0 0 2 0 - Î¸
And heat produced per second by the heater in the room
P= V 2 R J S = V 2 R cal s
= 2 0 0 Ã 2 0 0 2 0 Ã 4.2 = 476.2 cal s
Now as the temperature of the room is constant, the heat produced per second by heater must be equal to the heat conducted through the glass window.
1 0 0 2 0 - Î¸ = 476.2 ; Î¸ = 15.24 â C

Topic: Physics, Thermal Properties of Matter, NEET
A rope of length 10 m and linear density 0.5 kg m-1 is lying length wise on a smooth horizontal floor. It is pulled by a force of 25 N. The tension in the rope at a point 6 m away from the point of application is

Options:
A 20 N
B 15 N
C 10 N
D 5 N

The correct answer is:
10 N

Solution:
Mass of rope =10×0.5=5 kg
Given force =25 N
Acceleration =Fm=255=5 m s-2
Length of remaining rope =4 m
Hence mass of remaining rope =4×0.5=2 kg
Hence tension on the rope at a point 6m away =m×a=2×5=10 N

Topic: Physics, Laws of Motion, NEET
A rope of negligible mass is wound around a hollow cylinder of mass 3 kg and radius 40 cm and then the entire arrangement is placed on a smooth horizontal surface. What is the angular acceleration (in rad s-2) of the cylinder if the rope is pulled with a force of 30 N?

Options:

The correct answer is:
10

Solution:
Given, M = 3 kg
R = 40 cm = 0.40 m
F = 30 N
Moment of inertia of the hollow cylinder about its axis of symmetry
I = MR2
= 3 Ã (0.40)2 = 0.48 kg-m2
Torque acting on the cylinder by the applied force,
Ï = F Ã R
= 30 Ã 0.40 = 12 N-m.
Angular acceleration produced in cylinder,
Î± = Ï I = 12 0.48 = 25 rad/s 2
Linear acceleration, a = R Î±
= 0.40 Ã 25 = 10.0 m/s2

Topic: Physics, Rotational Motion, JEE Main
A rough inclined plane \(B C E\) of height \(\left(\frac{25}{6}\right) \mathrm{m}\) is kept on a rectangular wooden block \(A B C D\) of height \(10 \mathrm{~m}\), as shown in the figure. A small block is allowed to slide down from the top \(E\) of the inclined plane. The coefficient of kinetic friction between the block and the inclined plane is \(\frac{1}{8}\) and the angle of inclination of the inclined plane is \(\sin ^{-1}(0.6)\). If the small block finally reaches the ground at a point \(F\), then \(D F\) will be (Acceleration due to gravity, \(g=10 \mathrm{~ms}^{-2}\))

Options:
A \(\frac{5}{3} m\)
B \(\frac{10}{3} \mathrm{~m}\)
C \(\frac{13}{3} m\)
D \(\frac{20}{3} \mathrm{~m}\)

The correct answer is:
\(\frac{20}{3} \mathrm{~m}\)

Solution:
According to question, a small block is slide down from top \(E\) of inclined plane as shown in figure,
Force equation of a block,
\(\Rightarrow \quad m g \sin \theta-f=m a\)
\(\because\) friction force applied one block, \(f=\mu_k R\) or \(f=\mu_k R(m g \cos \theta)\)
(From figure) where, \(\mu_k=\) coefficient of kinetic friction from Eq. (i), we get
\(\begin{aligned}
& \Rightarrow \quad m g \sin \theta-\mu_k m g \cos \theta=\text { ma } \\
& \Rightarrow a=10 \sin \theta-1 / 8 \times 10 \cos \theta \ldots \text { (ii) }\left(\because \mu_k=\frac{1}{8} \text {, given }\right)
\end{aligned}\)
\(\because\) Given, angle of the inclined plane, \(\theta=\sin ^{-1}(0.6)\)
or \(\sin \theta=0.6\)
\(\begin{array}{rlrl}
\because & \cos \theta =\sqrt{1-\sin ^2 \theta}=\sqrt{1-(0.6)^2} \\
\therefore & \cos \theta =0.8
\end{array}\)
Form Eq. (ii),
\(a=10(0.6)-\frac{1}{8} \times 10(0.8) \text { or } a=5 \mathrm{~ms}^{-2}\)
when, block reached to point \(C\), then from third equation of the motion,
\(v^2=u^2+2 a s\)
where, \(v=\) final velocity of the block at point \(C\) \(u=\) initial velocity of block at point \(E\)
or
\(v=\sqrt{2 a s} \quad(\because u=0)\)
From \(\Delta \mathrm{EBC}, E C=\frac{B E}{\sin \theta} \Rightarrow s=\frac{25 / 6}{0.6}=\frac{25}{6 \times 0.6}\) or
\(\begin{aligned}
& s=\frac{125}{18} \mathrm{~m} \\
& v=\sqrt{2 \times 5 \times \frac{125}{18}}=\frac{25}{3} \mathrm{~ms}^{-1}
\end{aligned}\)
From second equation of motion of the block at point C,
In \(y\)-direction,
\(\begin{array}{ccc}
\Rightarrow & h=u t+\frac{1}{2} g t^2 \\
\Rightarrow & 10=u \sin \theta t+\frac{1}{2} g t^2 \\
\Rightarrow & 10=\frac{25}{3} \times 0.6 t+\frac{1}{2} \times 10 \times t^2 \\
\Rightarrow & \frac{25}{3} \times \frac{6}{10} t+\frac{1}{2} \times 10 \times t^2=10 \\
\Rightarrow & 5 t+5 t^2=10 \\
\Rightarrow & t^2+t-2=0 \text { or } t=1 \mathrm{sec}
\end{array}\)
Now, again from second Eqs. of motion in \(x\)-direction,
\(\Rightarrow \quad D F=v \cos \theta . t+0 \text { or } D F=\frac{25}{3} \times 0.8=\frac{20}{3} \mathrm{~m}\)

Topic: Physics, Laws of Motion, JEE Main
A rough platform OA rotates in a vertical plane about a horizontal axis through the point O with a constant counterclockwise velocity ω=3 rad s-1. As it passes through the position θ=0, a small mass m is placed upon it at a radial distance. If the mass is observed to slip at θ=37°, then the coefficient of friction between the mass & the member is

Options:
A 316
B 916
C 49
D 59

The correct answer is:
316

Solution:
As the mass is at the verge of slipping
∴mgsin⁡37-μ mgcos⁡37=mω2r
6-8μ=4.5
∴μ=316

Topic: Physics, Laws of Motion, JEE Main
A round table conference is to be held amongst 20 countries. If two particular delegates wish to sit together, then such arrangements can be done in ways.

Options:
A $2 \times(18) !$
B $\frac{19 !}{2 !}$
C $18 !$
D $19 ! \times 2 !$

The correct answer is:
$2 \times(18) !$

Solution:
We consider two particular delegates as single unit now 19 units can be arranged around a round table in 18! ways and the two delegates can be arranged mutually in 2 ! ways
Hence the required number of ways is $2 \times(18$ !)

Topic: Mathematics, Permutation Combination, JEE Main
A round uniform body of radius $\mathrm{R}$, mass $\mathrm{M}$ and moment of inertia '$\mathrm{I}$', rolls down (without slipping) an inclined plane making an angle $\theta$ with the horizontal. Then its acceleration is

Options:
A
$\frac{\mathrm{g} \sin \theta}{1+\frac{\mathrm{I}}{\mathrm{M R}^2}}$

B
$\frac{\mathrm{g} \sin \theta}{1+\frac{\mathrm{M R}^2}{\mathrm{I}}}$

C
$\frac{\mathrm{g} \sin \theta}{1-\frac{\mathrm{I}}{\mathrm{MR}^2}}$

D
$\frac{\mathrm{g} \sin \theta}{1-\frac{\mathrm{M R}^2}{\mathrm{I}}}$

The correct answer is:
$\frac{\mathrm{g} \sin \theta}{1+\frac{\mathrm{I}}{\mathrm{M R}^2}}$

Solution:
$\mathrm{M g \sin \theta-f=M a}$
$\begin{aligned}
& \mathrm{f R=I \frac{a}{R}}\\
& \Rightarrow a=\frac{g \sin \theta}{\left(1+\frac{\mathrm{I}}{\mathrm{M}\mathrm{R^2}}\right)}
\end{aligned}$

Topic: Physics, Rotational Motion, JEE Main, JEE Main 2007
A rubber ball filled with water, having a small hole is used as the bob of a simple pendulum. The time period of such a pendulum

Options:
A is a constant
B decreases with time
C increases with time
D first increases and then decreases, finally having same value as at the beginning

The correct answer is:
first increases and then decreases, finally having same value as at the beginning

Solution:
$\text { From } T=2 \pi \sqrt{\frac{L}{g}}$
The effective length increases due to this flow of water. Therefore, $\mathrm{T}$ increases. As the water flows out, the length decreases and becomes equal to the original length. Hence, the time period becomes equal to the value at the beginning.

Topic: Physics, Oscillations, JEE Main
A rubber ball is dropped from a height of 5 m on a plane. It bounces back to a height of 1.8 m. Find the coefficient of the restitution for the collision.

Options:

The correct answer is:
0.6

Solution:
If the ball is dropped from height h and bounces back to height h1

h1=e2h

âe=h1h

=1.85=0.6

Topic: Physics, Work Power Energy, JEE Main
A rubber ball is dropped from a height of $5 \mathrm{~m}$ on a planet, where the acceleration due to gravity is not known. On bouncing it rises to $1.8 \mathrm{~m}$. The ball loses its velocity on bouncing by a factor of

Options:
A $\frac{16}{25}$
B $\frac{2}{5}$
C $\frac{3}{5}$
D $\frac{9}{25}$

The correct answer is:
$\frac{2}{5}$

Solution:
Key Idea According to conservation of energy, potential energy at height $h=$ kinetic energy at
ground. Potential energy = Kinetic energy
ie,
$$
m g h=\frac{1}{2} m v^{2}
$$
$$
\Rightarrow \quad v=\sqrt{2 g h}
$$
If $h_{1}$ and $h_{2}$ are initial and final heights, then
$$
v_{1}=\sqrt{2 g h_{1}}, \quad v_{2}=\sqrt{2 g h_{2}}
$$
Loss in velocity
$$
\Delta v=v_{1}-v_{2}=\sqrt{2 g h_{1}}-\sqrt{2 g h_{2}}
$$
$\therefore$ Fractional loss in velocity $=\frac{\Delta v}{v_{1}}$
$$
\begin{array}{l}
=\frac{\sqrt{2 g h_{1}}-\sqrt{2 g h_{2}}}{\sqrt{2 g h_{1}}} \\
=1-\sqrt{\frac{h_{2}}{h_{1}}}
\end{array}
$$
Substituting the values, we have
$\therefore$
$$
\frac{\Delta v}{v_{1}}=1-\sqrt{\frac{1.8}{5}}
$$
$$
\begin{array}{l}
=1-\sqrt{0.36}=1-0.6 \\
=0.4=\frac{2}{5}
\end{array}
$$

Topic: Physics, Rotational Motion, JEE Main
A rubber ball is released from a height of 5 m above the floor. It bounces back repeatedly, always rising to 81100 of the height through which it falls. Find the average speed of the ball. (Take g=10 m s-2)

Options:
A 3.0 m s-1
B 3.5 m s-1
C 2.0 m s-1
D 2.50 m s-1

The correct answer is:
2.50 m s-1

Solution:
v0=2gh

v=e2gh=2gh

⇒e=0.9

S=h+2e2h+2e4h+………

t=2h g+2e2h g+2e22h g+………

vav=st=2.5 m s-1

Topic: Physics, Center of Mass Momentum and Collision, JEE Main, JEE Main 2021 (17 Mar Shift 2)
A rubber cord catapult has cross-sectional area $25 \mathrm{~mm}^2$ and initial length of rubber cord is $10 \mathrm{~cm}$ It is stretched to ${ }^{5 \mathrm{~cm}}$ and then released to project a missile of mass $5 \mathrm{gm}$ Taking $Y_{\text {rubber }}=5 \times 10^8 \mathrm{~N} / \mathrm{m}^2$ velocity of projected missile is

Options:
A $20 \mathrm{~ms}^{-1}$
B $100 \mathrm{~ms}^{-1}$
C $250 \mathrm{~ms}^{-1}$
D $200 \mathrm{~ms}^{-1}$

The correct answer is:
$250 \mathrm{~ms}^{-1}$

Solution:
Potential energy stored in the rubber cord catapult will be converted into kinetic energy of mass.
$\frac{1}{2} m v^2=\frac{1}{2} \frac{Y A l^2}{L} \Rightarrow v=\sqrt{\frac{Y A l^2}{m L}}$
$=\sqrt{\frac{5 \times 10^8 \times 25 \times 10^{-6} \times\left(5 \times 10^{-2}\right)^2}{5 \times 10^{-3} \times 10 \times 10^{-2}}}=250 \mathrm{~m} / \mathrm{s}$

Topic: Physics, Mechanical Properties of Solids, JEE Main
A rubber cord has a cross-sectional area 10-6 m2 and total unstretched length 0.1 m. It is stretched to 0.125 m and then released to project a particle of mass 5.0 g. The velocity of projection is [Given, Young's modulus of rubber, Y=5×108 N m-2]

Options:
A 45 m s-1
B 30 m s-1
C 25 m s-1
D 15 m s-1

The correct answer is:
25 m s-1

Solution:
Force constant K of rubber is given by

K=YAl=5×108×10-60.1=5×103 N m-1

Now, from conservation of energy,

elastic potential energy of cord = kinetic energy of particle

i.e. 12K∆l2=12mv2

⇒v=Km·∆l

=5×1035×10-30.125-0.1

=103×0.025=25 m s-1

Topic: Physics, Mechanical Properties of Solids, JEE Main
A rubber pipe of density $1.5 \times 10^3 \mathrm{~N} / \mathrm{m}^2$ and Young's modulus $5 \times 10^6 \mathrm{~N} / \mathrm{m}^2$ is suspended from the roof. The length of the pipe is $8 \mathrm{~m}$. What will be the change in length due to its own weight

Options:
A $9.6 \mathrm{~m}$
B $9.6 \times 10^3 \mathrm{~m}$
C $19.2 \times 10^{-2} \mathrm{~m}$
D $9.6 \times 10^{-2} \mathrm{~m}$

The correct answer is:
$9.6 \times 10^{-2} \mathrm{~m}$

Solution:
$l=\frac{L^2 d g}{2 T}=\frac{(8)^2 \times 1.5 \times 10^3 \times 10}{2 \times 5 \times 10^6}=9.6 \times 10^{-2} \mathrm{~m}$

Topic: Physics, Mechanical Properties of Solids, JEE Main
A running man has the same kinetic energy as that of a boy of half his mass. The man speed up by 2 m s-1 and the boy changes his speed by x m s-1, so that the kinetic energies of the boy and the man are again equal. Then x in m s-1 is

Options:
A -22
B +22
C 2
D 2

The correct answer is:
+22

Solution:
Let vM be velocity of man,vBof boy, then kinetic energy according to question ,
ie K=12MvM2=12.M2.vB2
Or vM2=VB22
Or 2vM=vB
When man speeds up 2 m s-1and boy changes his speed by x m s-1.Then ,
12MvM+22=12âM2âvB+x2
Or vM+22=vB+x22
2vM+22=2vM+x2 âµvB=2vm
Or 2vM+2=2vM+x
Or +22=x

Topic: Physics, Work Power Energy, JEE Main
As2S3 and TiO2 sol are examples of

Options:
A Negativity charge sols
B Positively charged sols
C Positively and negatively charged sols respectively
D Negatively and positively charged sols respectively

The correct answer is:
Negatively and positively charged sols respectively

Solution:
As2S3 and TiO2 sol are examples of negatively and positively charged sols respectively.

Topic: Chemistry, Surface Chemistry, JEE Main
A salt is formed when a weak acid of dissociation constant 10-4 and weak base of dissociation constant 10-5 are mixed. The pH and degree of hydrolysis of salt solution are

Options:
A 5, 1%
B 7, 14%
C 6.5, 0.3%
D 0.3, 6.5%

The correct answer is:
6.5, 0.3%

Solution:
Ka=10-4;pKa=4
Kb=10-5;pKb=5
h=Kh
=KwKa×Kb
=10-1410-5×10-5=10×10-6
=3×10-3
% Hydrolysis =h×100=3×0.001×100=0.3%
pH=7+12pKa-12pKb=7+2-2.5=6.5

Topic: Chemistry, Ionic Equilibrium, JEE Main
A sample consisting of Hydrogen atoms in the ground state is excited by monochromatic radiation of energy 12.75 eV. If we were to observe the emission spectrum of this sample, then the number of spectral lines observed, will be

Options:
A 3
B 6
C 10
D 15

The correct answer is:
6

Solution:
The energy of an electron of a hydrogen atom in the ground state is

E0=-13.6 eV

The energy of incoming photon = 12.75 eV

The energy of an electron in the excited state is

-13.6n2=-13.6+12.75=-0.85 eV

⇒n2=13.60.85=16

⇒n=4

No. of spectral lines = nn-12 = 6

Topic: Physics, Atomic Physics, JEE Main
A sample contains 10-2 kg each of two substances A and B with half lives 4 s and 8 s respectively. The ratio of their atomic weights is 1:2. The ratio of the amounts of A and B after 16 s is x100. The value of x is _____ .

Options:

The correct answer is:
25

Solution:
Ratio of number of moles of substance in the beginning will be, N0AN0B=10-2 kgMA10-2 kgMB=MBMA=12.

Number of half-life in 16 s for element A will be, nA=164=4

Number of half-life in 16 s for element B will be, nB=168=2

Hence, ratio of number of nuclei after 16 s will be, =N0A×12nAN0B×12nB=12×124-2=18=25100

Thus, value of x is 25.

Topic: Physics, Nuclear Physics, JEE Main, JEE Main 2022 (24 Jun Shift 2)
A sample contains a large number of nuclei. The probability that a nucleus in the sample will decay after four half-lives is ab where a and b are least positive integers. Value of a+b will be

Options:

The correct answer is:
31

Solution:
The probability that a nucleus will not decay is-
NN0=12n=q
When n is the number of half-lives
q=124=116
The probability that a nucleus will decay is
p=1-q=1-116=1516

Topic: Physics, Nuclear Physics, JEE Main
A sample contains mixture of helium and oxygen gas. The ratio of root mean square speed of helium and oxygen in the sample, is:

Options:
A $\frac{1}{32}$
B $\frac{2 \sqrt{2}}{1}$
C $\frac{1}{2 \sqrt{2}}$
D $\frac{1}{4}$

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

Solution:
$\begin{aligned} & \mathrm{V}_{\mathrm{ms}}=\sqrt{\frac{3 \mathrm{RT}}{\mathrm{M}_{\mathrm{w}}}} \\ & \Rightarrow \frac{\mathrm{V}_{\mathrm{O}_2}}{\mathrm{~V}_{\mathrm{He}}}=\sqrt{\frac{\mathrm{M}_{\mathrm{w}, \mathrm{He}}}{\mathrm{M}_{\mathrm{w}, \mathrm{O}_2}}} \\ & =\sqrt{\frac{4}{32}}=\frac{1}{2 \sqrt{2}} \\ & \frac{\mathrm{V}_{\mathrm{He}}}{\mathrm{V}_{\mathrm{O}_2}}=\frac{2 \sqrt{2}}{1} \\ & \end{aligned}$

Topic: Physics, Kinetic Theory of Gases, JEE Main, JEE Main 2024 (06 Apr Shift 1)