datavorous/entrance-exam-dataset huggingface dataset:
A uniform rectangular thin sheet ABCD of mass M has length a and breadth b, as shown in the figure. If the shaded portion HBGO is cut-off, the coordinates of the centre of mass of the remaining portion will be:

Options:
A 5a12,5b12
B 5a3,5b3
C 2a3,2b3
D 3a4,3b4

The correct answer is:
5a12,5b12

Solution:
If the total mass of object mass 'M' then the mass removed part will be M4 .

XCM=M·a2-M43a4M-M4=Ma2-3Mb163M4=512a
YCM=M·b2-M43b4M-M4=Mb2-3Ma163M4=512b

Topic: Physics, Center of Mass Momentum and Collision, JEE Main
A uniform ring is rotating about the vertical axis with angular velocity Ï initially. A point insect (S) having the same mass as that of the ring starts walking from the lowest point P1 and finally reaches the point P2 (as shown in the figure). The final angular velocity of the ring is

Options:
A ω / 4
B ω
C ω / 2
D Ï / 3

The correct answer is:
ω / 3

Solution:
From the conservation of angular momentum
12MR2ω=12MR2+MR2ω´⇒12MR2ω=32MR2ω´⇒ω´=ω3

Topic: Physics, Rotational Motion, JEE Main
A Uniform rod AB is hinged at its end A and the other end of the rod is connected to a block through a massless string as shown in the figure. The pulley is smooth and massless. Masses of the block and the rod are same and are equal to m. The acceleration of the block just after the release from this position is

Options:
A 6g13
B g4
C 3g8
D None

The correct answer is:
3g8

Solution:
For the rod: Tℓ-mgℓ2=mℓ23×aℓ ...(i)
For the: mg-T=ma ...(ii)
Solving (i) and (ii) we get: a=3g8

Topic: Physics, Rotational Motion, JEE Main
A uniform $\operatorname{rod} A B$ is suspended from a point $X$, at a variable distance from $x$ from $A$, as shown. To make the rod horizontal, a mass $m$ is suspended from its end $A$. A set of $(m, x)$ values is recorded. The appropriate variable that give a straight line, when plotted, are:

Options:
A
$m, \frac{1}{x}$

B
$m, \frac{1}{x^2}$

C
$m, x$

D
$m, x^2$

The correct answer is:
$m, \frac{1}{x}$

Solution:
Balancing torque w.r.t. point of suspension
$$
\begin{aligned}
&m g x=M g\left(\frac{\ell}{2}-x\right) \\
&\Rightarrow m x=M \frac{\ell}{2}-M x \\
&m=\left(M \frac{\ell}{2}\right) \frac{1}{x}-M
\end{aligned}
$$
$y=\alpha \frac{1}{x}-C \quad$ Straight line equation.

Topic: Physics, Rotational Motion, JEE Main, JEE Main 2018 (15 Apr Shift 1 Online)
A uniform rod AB is suspended from a point X, at a variable distance x from A, as shown. To make the rod horizontal, a mass m is suspended from its end A. A set of (m, x) value is recorded. The appropriate variables that give a straight line, when plotted, are:

Options:
A m,x2
B m,1x2
C m,1x
D m,x

The correct answer is:
m,1x

Solution:
Net torque about suspension is zero,
⇒mgx=Mgl2-x⇒mx=Ml2-Mx
⇒m=Ml21x-M⇒y= α1x-C, where,

y=m, a=Ml2 and C=M
it is the equation of straight line with variable m and 1x.

Topic: Physics, Rotational Motion, JEE Main, JEE Main 2018 (15 Apr)
A uniform rod $\mathrm{AB}$ of length $I$ and mass $m$ is free to rotate about point $A$. The rod is released from rest in the horizontal position. Given that the moment of inertia of the rod about $\mathrm{A}$ is $m l^2 / 3$, the initial angular acceleration of the rod will be

Options:
A $\frac{m g l}{2}$
B $\frac{3}{2} g l$
C $\frac{3 g}{2 l}$
D $\frac{2 g}{3 l}$

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

Solution:
Given that moment of inertia about \(A\) is
\(I=m l^3 / 3\)
Now, torque about \(A\) is given as
\(\begin{aligned}
& \tau=F \times r \\
& \tau=m g \frac{1}{2} \\
& I \alpha=m g \frac{1}{2} \\
& \frac{m l^2}{3} \alpha=m g \frac{1}{2} \\
& \alpha=\frac{3 g}{2 l}
\end{aligned}\)

Topic: Physics, Rotational Motion, JEE Main
A uniform rod AB of length l and mass m is free to rotate about point A. The rod is released from rest in the horizontal position. The initial angular acceleration of the rod will be

Options:
A 2 g 3 l
B g l 2
C 3 2 gl
D 3 g 2 l

The correct answer is:
3 g 2 l

Solution:
The moment of inertia of the uniform rod about an axis through one end and perpendicular to its length is
I = ml 2 3
Where m is mass of rod and l is length.
Torque Ï = I Î± acting on centre of gravity of rod is given by
Ï = mg l 2
or I Î± = mg l 2
or ml 2 3 Î± = mg l 2
or Î± = 3 g 2 l

Topic: Physics, Rotational Motion, JEE Main
A uniform rod AB of length L and mass M is lying on a smooth table. A small particle of mass m strikes the rod with velocity V0 at point C at a distance x from the centre O. The particle comes to rest after collision. If the value of x, so that the point A of the rod remains stationary just after the collision, is L/n, then find n.

Options:
A 6
B 4
C 12
D 8

The correct answer is:
6

Solution:
mvo = MVcm ⇒ Vcm = mvo/M
mvox = 112ML2ω⇒ω=12mvoxML2
VAG=0⇒VAC+VCG=0
⇒-ωL2+mvoM=0 ⇒ mvoM=12mvoxML2×L2
⇒ x = L/6

Topic: Physics, Rotational Motion, NEET
A uniform rod AB of mass 2 kg and length 1 m is placed on a sharp support O, such that AO=a=25 cm and OB=b=75 cm. A spring of force constant 600 N m-1 is attached to the end B, as shown. Initially, the spring is extended by 2 cm at equilibrium. As soon as the thread is burnt, find the normal force (in N) exerted by the support at O.

Options:

The correct answer is:
8

Solution:
I about O is mL212+mL216=7mL248
Spring force âKÎx=600Ã2100=12 N
Torque, just after burning the string,
Ï=2Ã10Ã14+12Ã34
â Ï= 5 + 9 = 14 N m
â´Î±=14Ã487Ã2Ã1=48 rad s-2
â´ Acceleration of COM,
a=48Ã14=12 m s-2
Now, total vertical force
12 + 20 â N = 2 Ã 12
âN = 32 â 24 = 8 N

Topic: Physics, Rotational Motion, JEE Main
A uniform rod AB of mass m and length l is at rest on a smooth horizontal surface. An impulse J is applied to the end B perpendicular to the rod in horizontal direction. Speed of particle P at a distance l6 from the centre towards A of the rod after time t=Ïml12J is

Options:
A 2 J m
B J 2 m
C J m
D 2 J m

The correct answer is:
2 J m

Solution:
Let v and Ï be the linear and angular speeds of the rod
after applying an impulse J at B. Then from :
impulse = change in momentum
we have
mv = J or v = J m ...(1)
I Ï = J Â· l 2
or m l 2 1 2 Ï = J Â· l 2 or Ï = 6 J m l ...(2)
After the given time t = Ï m l 1 2 J , the rod will rotate an angle
Î¸ = Ï t = 6 J m l Ï m l 1 2 J = Ï 2
l 6 . Ï = l 6 6 J m l = J m = v
â´ v P â = 2 v = 2 J m

Topic: Physics, Rotational Motion, JEE Main
A uniform rod is placed vertically on a smooth surface and then released. Then,

Options:
A the centre of rod follows straight line path
B the centre of mass follows circular path
C the instantaneous axis is passing through the contact point
D all the above

The correct answer is:
the centre of rod follows straight line path

Solution:
Since no external force is present on the rod in horizontal direction, the velocity of the centre of mass of rod, in horizontal direction, remains constant i.e., zero.
So, centre of mass of rod drops vertically downwards.

Topic: Physics, Rotational Motion, NEET
A uniform rod of length 2.0 m, specific gravity 0.5 and mass 2 kg is hinged at one end to the bottom of a tank of water (specific gravity = 1.0) filled up to a height of 1.0 m as shown in the figure. Taking the case Î¸â 0Â° the force exerted by the hinge on the rod is: g = 10 m s-2

Options:
A 10.2 N, upwards

B 4.2 N, downwards

C 8.3 N, downwards

D 6.2 N, upwards

The correct answer is:
8.3 N, downwards

Solution:
Length of rod inside the water =1.0 secÎ¸=secÎ¸
Upthrust F=22secÎ¸15001000(10)
Or F=20 secÎ¸
Weight of rod W=2Ã10=20N
For rotational equilibrium of rod, net torque about O should be zero.
â´FsecÎ¸2sinâ¡Î¸=W=1.0 sinÎ¸
Or 202sec2â¡Î¸=20
Or Î¸=45o
â´F=20secâ¡45o
=202 N

The hinge force is

Fhinge=F-W=8.3 N

Topic: Physics, Mechanical Properties of Fluids, JEE Main
A uniform rod of length 2.0 m is suspended through its endpoint about which it performs small angular oscillations in the vertical plane. Its time period is nearly

Options:
A 1.6 s
B 1.8 s
C 2.0 s
D 2.3 s

The correct answer is:
2.3 s

Solution:
It is a physical pendulum, and its time period T=2π I mgx
here I = moment of inertia of the pendulum about point of suspension
I= M L 2 12 + M L 2 4 = M L 2 3
and x = distance of point of suspension from its COM.
x = L/2
⇒ T=2π M L 2 /3 mgL/2 =2π 2L 3g
⇒T=2π2×23×10=2.3 sec.

Topic: Physics, Oscillations, JEE Main
A uniform rod of length \( 2.0 \mathrm{~m} \) is suspended through its endpoint about which it performs small angular oscillations in the vertical plane. Its time period is nearly

Options:
A \( 1.6 \mathrm{~s} \)
B \( 1.8 \mathrm{~s} \)
C \( 2.0 \mathrm{~s} \)
D \( 2.3 \mathrm{~s} \)

The correct answer is:
\( 2.3 \mathrm{~s} \)

Solution:
It is a physical pendulum, and its time period T=2π I mgx
Here l=moment of inertia of the pendulum about point of suspension
I= M L 2 12 + M L 2 4 = M L 2 3
and x=distance of point of suspension from its COM.
x=L2
⇒T=2πML23mgL2=2π2L3g
⇒T=2π2×23×10=2.3 sec.

Topic: Physics, Oscillations, JEE Main
A uniform rod of length ' ℓ' is pivoted at one of its ends on a vertical shaft of negligible radius. When the shaft rotates at angular speed ω the rod makes an angle θwith it (see figure). To find θ equate the rate of change of angular momentum (direction going into the paper) mℓ212ω2sinθ about the centre of mass (CM) to the torque provided by the horizontal and vertical forces FHand Fv about the CM. The value of θ is then such that:

Options:
A cosθ=2g3lω2
B cosθ=g2ℓω2
C cosθ=gℓω2
D cosθ=3g2ℓω2

The correct answer is:
cosθ=3g2ℓω2

Solution:
Torque of centrifugal force τcf=dm.x sin θω2xcos ϑ=mlω2 sin θcos θ∫0lx2dx

τef=ml2ω2sinθ cosθ3

τmg=τcf

mg.l2sin θ=ml2ω2 sinθ cos θ3

cos θ=3g2lω2

Topic: Physics, Rotational Motion, JEE Main, JEE Main 2020 (03 Sep Shift 2)
A uniform rod of length L and mass M is pivoted at the centre. Its two ends are attached to two springs of equal spring constant k. The rod is gently pushed through a small angle Î¸ in one direction and released. The frequency of oscillation is (the springs are fixed to the walls)

Options:
A 1 2 π k M
B 1 2 π 2 k M
C 1 2π 3k M
D 1 2 Ï 6 k M

The correct answer is:
1 2 π 6 k M

Solution:
Consider the situation below
From the above diagram
Î¸ = x lâ¡ / 2
â x = L 2 Î¸
In the above diagram the rod is displaced through an angle Î¸ as in above diagram
Restoring torque = - 2 kx Ã L 2 = - kxL
angular acceleration,
â a= âkÃ L 2 Î¸ÃL Î = â6k M Î¸
=â Ï 2 Î¸
Ï= 6K M
f= 1 2Ï 6K M

Topic: Physics, Oscillations, JEE Main
A uniform rod of length l is acted upon by a force F in a gravity-free region, as shown in the figure. If the area of cross-section of the rod is A and it's Young's modulus is Y, then the elastic potential energy stored in the rod due to elongation is

Options:
A U=F2l6A2Y
B U=F2l3A2Y
C U=F2l24A2Y
D U=F2l22A2Y

The correct answer is:
U=F2l6A2Y

Solution:
The acceleration of the rod is

a=Fm

So, the tension in the rod at a distance x from the free end is

Tx=mxl×Fm=Fxl

The longitudinal stress developed inside the rod at a distance x from the free end is

σ=TxA=FxlA

The strain energy density (energy per unit volume) is

dUdV=12σ2Y=12F2x2YA2l2, where dV=Adx is the volume of a small part of the rod

So the energy stored in the rod is

U=12F2YA2l2∫0lx2dx
U=F2l6A2Y

Topic: Physics, Mechanical Properties of Solids, JEE Main
A uniform rod of length l is being rotated in a horizontal plane with a constant angular speed about an axis passing through one of its ends. If the tension generated in the rod due to rotation is T(x) at a distance x from the axis, then which of the following graphs depicts it most closely?

Options:
A
B
C
D

The correct answer is:

Solution:
The mass of the rod at a distance x from the axis is dm=ML.dx
T=(dm)ω2x

Integrating the above equation from x to L
=∫xLMLω2x dx
=MLω2x22xL
M2Lω2L2-x2
The above equation depicts the relation between tension and length of rod as downward opening parabola.

Topic: Physics, Rotational Motion, JEE Main, JEE Main 2019 (12 Apr Shift 1)
A uniform rod of length \( L \) is fixed at one end. It is initially held vertically upward, then released. Find the vertical reaction
at hinge when it becomes horizontal.

Options:
A \( \frac{m g}{2} \)
B \( \frac{m g}{3} \)
C \( \frac{3 m g}{2} \)
D \( \frac{m g}{4} \)

The correct answer is:
\( \frac{m g}{4} \)

Solution:
mg-Ny=may

ay= Acceleration of the center of mass in y-direction.

=L2×α (Since it is tangential acceleration)

Now,

Ny=mg-mLα2

But,

α=τAIA=mgL2mL23=3g2L

So, Ny=mg-mL2×3g2L=mg4

Topic: Physics, Laws of Motion, JEE Main
A uniform rod of length L is free to rotate in a vertical plane about a fixed horizontal axis through B. The rod begins rotating from rest from its unstable equilibrium position. When it has turned through an angle Î¸, its angular velocity Ï is given by

Options:
A 6g L  sin θ
B 6g L sin Î¸ 2
C 6g L cos θ 2
D 6g L   cos θ

The correct answer is:
6g L sin θ 2

Solution:
When the rod rotates through angle θ, the fall h of centre of gravity is given by θ

L2-hL2=cosθ

or h=L2(1-cosθ)

Decrease in potential energy

=Mgh=MgL2(1-cosθ)

Now, KE of rotation =12Iω2

=12×ML23ω2

=12×ML23ω2

I=ML2/3 (because rod is rotating about an axis passing through its one end)

According to law of conservation of energy, decrease in potential energy will be equal to gain in kinetic energy.

MgL2(1-cosθ)=ML26ω2

ω=3gL(1-Cosθ)=6gLSinθ2

Topic: Physics, Rotational Motion, JEE Main
A uniform rod of length \( L \) is hinged at one end and initially held in horizontal position. After it is released, find its angular
speed when the rod becomes vertical.

Options:
A \( (g L)^{1 / 2} \)
B \( \left(\frac{2 g}{L}\right)^{1 / 2} \)
C \( \left(\frac{3 g}{L}\right)^{1 / 2} \)
D \( \left(\frac{4 g}{L}\right)^{1 / 2} \)

The correct answer is:
\( \left(\frac{3 g}{L}\right)^{1 / 2} \)

Solution:
The gain in kinetic energy, ∆KE=12ml23ω2

Loss in potential energy, -∆U=mghloss, where hloss=L2 is the loss of height of the center of mass.

By conservation of mechanical energy, ∆KE=-∆U

mgL2=12mL23ω2⇒ω=3gL12

Topic: Physics, Rotational Motion, JEE Main
A uniform rod of length \( l \) is pivoted at point \( A \). It is struck by a horizontal force which delivers an impulse \( J \) at a distance \( x \) from point \( A \) as shown in the figure. The impulse delivered by pivot is zero if \( x \) is equal to,

Options:
A \( \frac{l}{2} \)
B \( \frac{\frac{1}{3}}{3} \)
C \( \frac{2 l}{3} \)
D \( \frac{3 l}{4} \)

The correct answer is:
\( \frac{2 l}{3} \)

Solution:
From impulse-momentum theorem, we can write,

Impulse=change in momentum or, J=∆p.

Let velocity of the centre of mass is v, then, we can write,

J=mv-0=mv ...i

The rod starts rotating. Let the angular speed of rod about point A is ω. Then, from conservation of angular momentum, we have,

J.x=Iω=ml23ω …ii

And from the relation between linear and angular velocity of the centre of mass, we have,

v=l2ω …iii

Solving these three equations, we get,

x=23l.

Topic: Physics, Center of Mass Momentum and Collision, JEE Main
A uniform rod of length $L$ is rotated in a horizontal plane about a vertical axis through one of its ends. The angular speed of rotation is $\omega$. Find increase in length of the rod, if $\rho$ and $Y$ are the density and Young's modulus of the rod respectively,

Options:
A $\frac{\rho \omega^3 Y}{4 L^2}$
B $\frac{4 \rho \omega^2 L^3}{3 Y}$
C $\frac{\rho \omega^3 L^3}{3 Y}$
D $\frac{\rho \omega^3 L^3}{8 Y}$

The correct answer is:
$\frac{\rho \omega^3 L^3}{3 Y}$

Solution:
Given,
length of uniform $\operatorname{rod}=L$
angular speed $=\omega$
and density of rod $=\rho$
When rod is rotated in a horizontal plane, centrifugal force is responsible for increasing the length of rod, which is also equal to stress.
If $d l$ be the change in length of small element $d x$ due to rotation, by application of force $d F$.
Young's modulus,
$$
\begin{aligned}
& Y=\frac{\text { stress }}{\text { strain }}=\frac{d F / A}{d l / x} \\
\Rightarrow \quad & d l=\frac{d F \cdot x}{A Y} \\
\Rightarrow \quad & d F=\frac{A Y d l}{x}
\end{aligned}
$$
Where, $d F=$ centrifugal force due to mass of the element of length, $d x$.
$$
\begin{aligned}
d F & =d m \cdot x \omega^2 \\
d F & =\frac{m}{L} \cdot x d x \omega^2 \\
& \quad \ldots \text { (ii) } \\
& {\left[\because \text { Change in mass } d m=\frac{m}{L} d x\right] }
\end{aligned}
$$
From Eqs. (i) and (ii), we get
$$
\begin{aligned}
\frac{A Y d l}{x} & =\frac{m}{L} \cdot x d x \omega^2 \\
\Rightarrow \quad d l & =\frac{m x^2 \omega^2}{L A Y} \cdot d x
\end{aligned}
$$
$\Rightarrow$ Total change in length,
$$
\begin{aligned}
l & =\int d l=\int_0^L \frac{m x^2 \omega^2}{L A Y} d x \\
& =\frac{m \omega^2 L^2}{3 A Y}=\frac{m}{A L} \cdot \frac{\omega^2 L^3}{3 Y}=\frac{\rho \omega^2 L^3}{3 Y}
\end{aligned}
$$

Topic: Physics, Mechanical Properties of Solids, JEE Main
A uniform rod of length $\ell$, mass $\mathrm{m}$ is hung from two strings of equal length from a ceiling as shown in figure. Determine the tensions in the strings?

Options:
A $\mathrm{mg} / 3$
B $2 \mathrm{mg} / 3$
C $7 \mathrm{mg} / 6$
D $\mathrm{mg} / 8$

The correct answer is:
$\mathrm{mg} / 3$

Solution:
Let us assume that tension in left and right string is $T_A$ and $T_B$ respectively.
The rod is in equilibrium, then $\vec{F}_{\text {net }}=0$ and $\vec{\tau}_{\text {net }}=0$
$\begin{aligned}
& \text { From, } \vec{F}_{\text {net }}=0 \\
& \mathrm{mg}=\mathrm{T}_{\mathrm{A}}+\mathrm{T}_{\mathrm{B}} \ldots
\end{aligned}$
$\begin{aligned}
& \text { From } \tau_{\text {net }}=0 \text { about } A, \\
& \mathrm{mg} \frac{1}{2}-\frac{31}{4} T_B=0 \\
& \Rightarrow T_B=\frac{2 \mathrm{mg}}{3}
\end{aligned}$
So, from eq. (1),
$\begin{aligned}
& \mathrm{T}_{\mathrm{A}}+\frac{2 \mathrm{mg}}{3}=\mathrm{mg} \\
& \Rightarrow \mathrm{T}_{\mathrm{A}}=\frac{\mathrm{mg}}{3}
\end{aligned}$

Topic: Physics, Rotational Motion, JEE Main
A uniform rod of mass $20 \mathrm{~kg}$ and length $1.6 \mathrm{~m}$ is pivoted at its end and swings freely in the vertical plane. Angular acceleration of rod just after the rod is released from rest in the horizontal position is:

Options:
A $\frac{15 g}{16}$
B $\frac{17 \mathrm{~g}}{16}$
C $\frac{16 \mathrm{~g}}{15}$
D $\frac{\mathrm{g}^{15}}{15}$

The correct answer is:
$\frac{15 g}{16}$

Solution:
Just after the rod is released, Torque of weight $=\mathrm{I} \alpha$
$\begin{aligned}
& \mathrm{mg} \times \frac{1}{2}=\frac{\mathrm{ml}^2}{3} \times \alpha, \\
& \Rightarrow \alpha=\frac{3 \mathrm{~g}}{2 \mathrm{l}} \\
& \Rightarrow \alpha=\frac{3 \times \mathrm{g}}{2 \times 1.6} \\
& \Rightarrow \alpha=\frac{15 \mathrm{~g}}{16}
\end{aligned}$

Topic: Physics, Rotational Motion, JEE Main
A uniform rod of mass M1 is hinged at its upper end. A particle of mass M2 moving horizontally strikes the rod at its midpoint elastically. If the particle comes to rest after collision, then, Find the value of M1M2

Options:
A 34
B 43
C 23
D 32

The correct answer is:
34

Solution:
Angular Momentum about Hinge
L H = I H .Ï
( I H = m 1 l 2 3 )
LH= M2vL2=M1L23Ï
Ï=3M2v2M1L ââââ â¦1
for e=1, vCM=v=ÏL2 â¦2
â 2vL=3M2v2M1L
â M1M2=34
=0.75

Topic: Physics, Rotational Motion, JEE Main
A uniform rod of mass M1 is hinged at its upper end as shown in the figure. A particle of mass M2 which is moving horizontally, strikes the rod elastically at its midpoint. If the particle comes to rest after collision, then the value of M1M2 is ______. If your answer is K, mark 4K as answer after rounding off to nearest integer.

Options:

The correct answer is:
3

Solution:
Conserving angular Momentum about the hinge, just before and after the collision
LH=M2vL2=M1L23Ï
Ï=3M2v2M1L

e=1, vCM=v=ÏL2
â 2vL=3M2v2M1L
â M1M2=34
=0.75

Topic: Physics, Rotational Motion, JEE Main
A uniform rod of mass M1 is hinged at its upper end as shown in the figure. A particle of mass M2 which is moving horizontally, strikes the rod elastically at its midpoint. If the particle comes to rest after collision, then the value of M1M2 is

Options:

The correct answer is:
0.75

Solution:
Conserving angular Momentum about the hinge, just before and after the collision
LH=M2vL2=M1L23Ï
Ï=3M2v2M1L

e=1, vCM=v=ÏL2
â 2vL=3M2v2M1L
â M1M2=34
=0.75

Topic: Physics, Rotational Motion, JEE Main
A uniform rod of mass m and length L is hinged at one end and free to rotate in the horizontal plane. All the surface are smooth. A particle of same mass m collides with the rod perpendicular to the length of rod with a speed V0. The coefficient of restitution for the collision is e=12. If hinge reaction during the collision is zero then the value of x is:

Options:
A No such value of x is possible
B x=L2
C x=2L3
D x=L

The correct answer is:
x=2L3

Solution:
Angular momentum conservation about the hinge
mv0x=mv1x+mL23ω
v0x=v1x+ωL23 ...(i)
e=ωx-v1v0=12 ...(ii)
Impulse of hinge
J=mωL2+mv1-mv0
J=mωL2-mωL23x
J=0
x=2L3
It is independent of 'e'

Topic: Physics, Rotational Motion, JEE Main
A uniform rod of mass m and length L is hinged at one of its end with the ceiling and another end of the rod is attached with a thread which is attached with the horizontal ceiling at point P. If one end of the rod is slightly displaced horizontally and perpendicular to the rod and released. If the time period of small oscillation is 2π2lsin⁡θxg. Find x.

Options:

The correct answer is:
3

Solution:
Consider a small angular displacement Î²
Torque about the rotational axis
Ï=-mgl2sinâ¡Î¸ Î²
Ï=IÎ±
1=mlsinâ¡Î¸23=ml2sin2â¡Î¸3
-mgl2sinâ¡Î¸ Î²=mlsin2â¡Î¸3 Î±
Î±=-3g2lsinâ¡Î¸Î²
â´ T=2Ï2lsinâ¡Î¸3g

Topic: Physics, Oscillations, JEE Main
A uniform rod of mass \( \mathrm{m} \) and length \( 2 a \) lies at rest on a smooth horizontal table. A perfectly elastic particle of same mass \( \mathrm{m} \), moving with speed \( \mathrm{v} \) on the table in a direction perpendicular to the rod, strikes one end of the rod. The kinetic energy generated in the rod is

Options:
A \( \frac{4}{13} \mathrm{mv}^{2} \)
B \( \frac{1}{4} \mathrm{mv}^{2} \)
C \( \frac{8}{25} \mathrm{mv}^{2} \)
D None of these

The correct answer is:
\( \frac{8}{25} \mathrm{mv}^{2} \)

Solution:
From conservation of linear momentum we have,

v = v 1 + v 2 ...(1)
From conservation of angular momentum about centre
of rod we have,
mva=mv2a+ma23·ω
or v = v 2 + a ω 3 ...(2)
Further from the definition of coefficient of restitution
(e = 1) at point of impact.
Relative speed of approach = relative speed of separation
∴ v = v 1 + a ω - v 2 ...(3)
solving these three Eqs. (1), (2) and (3) we get,
v 1 = 2 5 v and ω = 6 v 5 a
∴ Kinetic energy of rod,
K = 1 2 × m × 2 5 v 2 + 1 2 × m a 2 3 × 6 v 5 a 2
= 8 2 5 m v 2

Topic: Physics, Center of Mass Momentum and Collision, JEE Main
A uniform rod PQ of length l and mass m is placed on a smooth horizontal surface. It is hinged at one of its ends and touches the conducting circular loop at another end. The system is subjected to vertical magnetic field B0. Rod is given angular velocity Ï0 about the axis passing through P. If the time after which its angular velocity becomes half is abmRln2B2l2 . Find the value of a+b, where a and b are smallest positive integers.

Options:

The correct answer is:
7

Solution:
Îµ=12B0Ïl2
I=12B0Ïl2R
Ï=Il2B04=14B0Ïl2Rl2B0
B02Ïl4R=ml23dÏdt
3B2l24mRt=-â«dÏdt
t=4mRln23B2l2
∴a=4 and b=3 ⇒a+b=4+3=7

Topic: Physics, Electromagnetic Induction, JEE Main
A uniform rope of length L and mass m1, hangs vertically from a rigid support. A block of mass m2 is attached to the free end of the rope. A transverse pulse of wavelength λ1 is produced at the lower end of the rope. The wavelength of the pulse when it reaches the top of the rope is λ2. The ratio λ2λ1 is:

Options:
A m2m1
B m1+m2m2
C m1m2
D m1+m2m1

The correct answer is:
m1+m2m2

Solution:
The situation in the rope is given below,

The velocity of the wave at the bottom of the rope is given by,

v=Tμ; where T is tension and μ is linear mass density of the string.

v1=m2gLm1

Hence, the wavelength for frequency f will be,

λ1=v1f=1fm2gLm2

Similarly, at top the velocity will be,

v1=(m1+m2)gLm1

And, the wavelength will be,

λ2=1f(m1+m2)gLm1

Hence, the ratio of both wavelengths will be,

λ 2 λ 1 = m 1 + m 2 m 2 .

Topic: Physics, Waves and Sound, JEE Main