datavorous/entrance-exam-dataset huggingface dataset:
A uniform rope of mass 6 kg and length 6 m hangs vertically from a rigid support. A block of mass 2 kg is attached to the free end of the rope. A transverse pulse of wavelength 0.075 m is produced at the lower end of the rope. The wavelength of the pulse when it reaches the top of the rope is K cm where K is

Options:

The correct answer is:
15

Solution:
T1= 2 Ã 10 = 20 N
T2 = (2 + 6) 10 = 80 N
V2V1=8020=2
Frequency remains unchanged
So Î»2Î»1=2
Î»2=2Ã0.075
= 0.15 m
K=15 cm

Topic: Physics, Waves and Sound, JEE Main
A uniform smooth rod (mass m and length l ) placed on a smooth horizontal floor is hit by a particle (mass m) moving on the floor, at a distance l4 from one end elastically (e=1). The distance travelled by the centre of the rod after the collision, when it has completed three revolutions, will be

Options:
A 2πl
B 3πl
C πl
D 4πl

The correct answer is:
2πl

Solution:
Applying conservation of linear momentum,

mv=mv′+mV
⇒v=v′+V ........(i)
Applying conservation of angular momentum about the point of collision
0=ml212ω-mVl4

⇒lω=3V ..........(ii)
for a perfectly elastic collision e=1

v-0=V+ωl4-v′ ..........(iii)

v=V+3V4-v-V

2v=11V4

⇒V=8v11

ω=24v11l

Angle rotated by the rod θ=6π

t=θω=6πω

t=6π×11l24v=11l4v

Hence, distance travelled by the centre of the rod is

S=Vt

S=8v1111l4v=2πl

Topic: Physics, Rotational Motion, JEE Main
A uniform solid cylinder can roll without sliding on a horizontal surface as shown in the figure. The mass of cylinder is 10 kg and the spring constant for both the ideal springs is 300 N m-1. What is the angular frequency (in rad s-1) of small oscillations of the centre of the cylinder?

Options:

The correct answer is:
10

Solution:
The total energy of the system is

E=34mv2+52kx2

dEdx=32ma+5kx=0

a=-10k3mx

⇒ω=10k3m=10 rad s-1

Topic: Physics, Oscillations, JEE Main
A uniform solid cylinder of density 0.8 g cm-3 floats in equilibrium in a combination of two non-mixing liquids A and B with its axis vertical. The densities of the liquids A and B are 0.7 g cm-3 and 1.2 g cm-3, respectively. If the height of liquid A is hA=1.2 cm and the length of the part of the cylinder immersed in liquid B is 0.8 cm, then the force exerted by the liquid A on the cylinder is

Options:
A zero
B 84 N
C 96 N
D 40 N

The correct answer is:
zero

Solution:
Liquid A is applying the hydrostatic force on cylinder from all the sides. So, net force is zero.

Topic: Physics, Mechanical Properties of Fluids, JEE Main
A uniform solid cylindrical roller of mass m is being pulled on a horizontal surface with force F parallel to the surface and applied at its centre. If the acceleration of the cylinder is a and it is rolling without slipping then the value of F is:

Options:
A 32 ma
B 2Â ma
C 53Â ma
D ma

The correct answer is:
32 ma

Solution:
From free body diagram of cylinder
F-fs=ma .....(1)
∵ ∑ fext=macm
also ∑τext=Icmα
⟹fs R=Icmα
⟹fs R=12 mR2α ..... (2)
For rolling without slipping
a=Rα ...... (3)
⟹ α=aR
∴ fs R=12mR2aR
⟹ fs=12ma
Put in (1)
F-12ma=ma
⟹F=32ma

Topic: Physics, Rotational Motion, JEE Main, JEE Main 2015 (10 Apr Online)
A uniform solid cylindrical roller of mass m is being pulled on a horizontal surface with force F parallel to the surface and applied at its centre. If the acceleration of the cylinder is a and it is rolling without slipping then the value of F is:

Options:
A 32 ma
B 2Â ma
C 53Â ma
D ma

The correct answer is:
32 ma

Solution:
From free body diagram of cylinder
F-fs=ma .....(1)
∵ ∑ fext=macm
also ∑τext=Icmα
⟹fs R=Icmα
⟹fs R=12 mR2α ..... (2)
For rolling without slipping
a=Rα ...... (3)
⟹ α=aR
∴ fs R=12mR2aR
⟹ fs=12ma
Put in (1)
F-12ma=ma
⟹F=32ma

Topic: Physics, Rotational Motion, JEE Main
A uniform solid right circular cone of base radius r is joined to a uniform solid hemisphere of radius r and of the same density, so as to have a common face. The centre of mass of the composite solid lies on the common face. The height of the cone is

Options:
A 2r
B 3r
C 3r
D 6r

The correct answer is:
3r

Solution:
Volume of cone=13Ïr2h
Mass of cone, m1=ÏÃ13Ïr2h
Mass of hemisphere, m2=ÏÃ12Ã43Ïr3
=ÏÃ23Ïr3
Now, Y=m1y1+m2y2m1+m2
â0=ÏÃ13Ïr2hÃh4+23Ïr3(â3r8)ÏÃ13Ïr2h+ÏÃ23Ïr2
âÏÃ13Ïr3h24â2rÃ3r8=0
h24â3r24=0â or h=3âr

Topic: Physics, Center of Mass Momentum and Collision, JEE Main
A uniform solid sphere A of mass m is rolling without sliding on a smooth horizontal surface. It collides elastically and head-on with another stationary hollow sphere B of the same mass and radius. Assuming friction to be absent everywhere, the ratio of the kinetic energy of B to that of A just after the collision is

Options:
A 5:2
B 1:1
C 2:3
D 3:2

The correct answer is:
5:2

Solution:
Since the two bodies have the same mass and collide head-on elastically, the linear velocities get exchanged.
Hence, just after the collision B will move with velocity v0 and A becomes stationary but continues to rotate at the same initial angular velocity v0R.

So after the collision, the kinetic energies of A and B are

KB=12mv02

KA=12Iω2=1225mR2·v0R2

⇒KBKA=52

Topic: Physics, Rotational Motion, JEE Main
A uniform solid sphere of mass M and radius R is lying on a rough horizontal plane. A constant force F=4 Mg acts vertically downwards at point P such that the line OP makes an angle of 60Â° with the horizontal as shown in the figure. The minimum value of the coefficient of friction Î¼ so that sphere performs pure rolling, is

Options:
A 37
B 47
C 27
D 25

The correct answer is:
27

Solution:
Writing torque about the bottommost point,

4Mg×R2=75MR2aR

⇒a=10g7

f=μ5Mg=Ma

5μg=10g7⇒μ=27

Topic: Physics, Rotational Motion, JEE Main
A uniform solid sphere of mass M, radius R and having moment of inertia about an axis passing through center of mass is I, is recast into a uniform circular disc of thickness t, whose moment of inertia about an axis passing through its edge and perpendicular to its plane remains I. Then radius of the disc will be

Options:
A R215
B 2R15
C 4R15
D R4

The correct answer is:
2R15

Solution:
I=25MR2 and Ï=3M4ÏR3
For circular disc Ïr2ÃtÏ=MâÏr2âtâ3M4ÏR3=M
âÂ Â t=4ÏR33Ïr2=4R33r2Â âÂ r2=4R33t
25MR2=32MR2
âÂ 25R2=32â4R33tâÂ t=5R
âÂ r2⋅3⋅5RÂ =Â 4R3Â âÂ r2=4R215
âÂ rÂ =2R15

Topic: Physics, Rotational Motion, JEE Main
A uniform solid spherical ball is rolling down a smooth inclined plane from a height $h$. The velocity attained by the ball when it reaches the bottom of the inclined plane is v. If the ball is now thrown vertically upwards with the same velocity $v$, the maximum height to which the ball will rise is

Options:
A $\frac{5 h}{8}$
B $\frac{3 h}{5}$
C $\frac{5 h}{7}$
D $\frac{7 h}{9}$

The correct answer is:
$\frac{5 h}{7}$

Solution:
We know that total kinetic energy of a body rolling without slipping
$K_{\text {total }}=K_{\text {rot }}+K_{\text {trans }}$ For solid spherical ball, $I=\frac{2}{5} m R^{2}$ (along to diameter) and $v=R \omega$ where $R$ is radius of spherical ball
So.
$$
K_{\text {total }}=\frac{1}{2}\left(\frac{2}{5} m R^{2}\right) \omega^{2}+\frac{1}{2} m R^{2} \omega^{2}
$$
$$
\begin{array}{l}
=\frac{7}{10} m R^{2} \omega^{2} \\
K=\frac{7}{10} m v^{2}
\end{array}
$$
Potential energy = Kinetic energy
$$
\begin{aligned}
m g h &=\frac{7}{10} m v^{2} \\
v^{2} &=\frac{10}{7} g h
\end{aligned}
$$
For vertical projection,
$$
\begin{array}{c}
v^{2}=u^{2}+2 g h^{\prime} \\
\frac{10}{7} g h=0+2 g h^{\prime} \Rightarrow h^{\prime}=\frac{5}{7} h
\end{array}
$$

Topic: Physics, Rotational Motion, JEE Main
A uniform sphere A with radius R exerts a force F on a
small particle B situated at a distance 2R from the centre
of the sphere. A spherical portion of diameter R is cut
from the sphere A as shown in the figure. If F′ is the
new gravitational force between the remaining part of
the sphere A and the particle B then the correct relation
between F and F′.

Options:
A $F^{\prime}=\frac{9}{14} F$
B $F^{\prime}=\frac{14}{9} F$
C $F^{\prime}=\frac{7}{9} F$
D $F^{\prime}=\frac{9}{7} F$

The correct answer is:
$F^{\prime}=\frac{7}{9} F$

Solution:
No solution. Refer to answer key.

Topic: Physics, Gravitation, TS EAMCET, TS EAMCET 2022 (19 Jul Shift 1)
A uniform sphere of mass 500 g rolls without slipping on a plane horizontal surface with its centre moving at a speed of 5.00 cm s-1. Its kinetic energy is:

Options:
A 8.75×10-4 J
B 8.75×10-3 J
C 6.25×10-4 J
D 1.13×10-3 J

The correct answer is:
8.75×10-4 J

Solution:
K.E of the sphere = Translational K.E+ Rotational K.E
=12mv21+K2R2 ,K= Radius of gyration
= 12×12×210021+25
= 354×10-4 J
= 8.75×10-4 J

Topic: Physics, Rotational Motion, JEE Main, JEE Main 2020 (08 Jan Shift 2)
A uniform sphere of radius \( R \) is placed on a rough horizontal surface and given a linear velocity \( v_{0} \) and angular velocity \( \omega_{0} \) as shown. The sphere comes to rest after moving some distance to the right. It follows that

Options:
A \( \mathrm{v}_{0}=\omega_{0} \mathrm{R} \)
B \( 2 \mathrm{v}_{0}=5 \omega_{0} \mathrm{R} \)
C \( 5 \mathrm{v}_{0}=2 \omega_{0} \mathrm{R} \)
D \( 2 \mathrm{v}_{0}=\omega_{0} \mathrm{R} \)

The correct answer is:
\( 5 \mathrm{v}_{0}=2 \omega_{0} \mathrm{R} \)

Solution:
friction between the sphere and the horizontal surface,f=μN,⇒f=μmg,acceleration,a=μgwhere,μ-coefficient of kinetic friction, N-normal reaction=mg m-mass of the sphere
we know that,torqueΤ=Iα,⇒α=ΤIwhere, α-angular acceleration, I-moment of inertia of the sphere α = μ mgR 2 5 mR 2 = 5 2 g μ R

Now, using equation of motion, v=u+at,⇒0=v0-at⇒t=v0a t = v 0 a = ω 0 α ,where v0=Rω, and a=Rα, R-radius of the sphere
or a α = v 0 ω 0
∴ 2 R 5 = v 0 ω 0
∴ 5 v 0 = 2 ω 0 R

Topic: Physics, Rotational Motion, JEE Main
A uniform sphere of weight $W$ and radius $5 \mathrm{~cm}$ is being held by a string as shown in the figure. The tension in the string will be :

Options:
A
$12 \frac{W}{5}$

B
$5 \frac{W}{12}$

C
$13 \frac{W}{5}$

D
$13 \frac{\mathrm{W}}{12}$

The correct answer is:
$13 \frac{\mathrm{W}}{12}$

Solution:
$\begin{aligned}
& \mathrm{PQ}=\sqrt{\mathrm{OP}^2+\mathrm{OQ}^2} \\
& =\sqrt{13^2+5^2}=12
\end{aligned}$
Tension in the string $\mathrm{T}$, so, $\mathrm{T} \cos \theta=\mathrm{w}$
$\Rightarrow \mathrm{T}=\frac{\mathrm{w}}{\cos \theta}=\frac{13}{12} \mathrm{~W}$

Topic: Physics, Rotational Motion, JEE Main, JEE Main 2013 (09 Apr Online)
A uniform square plate of mass m is supported with its plane vertical as shown in the figure. Assume the centre of mass is on a horizontal line passing through A. If the cable at B breaks then just after that the angular acceleration of the plate is

Options:
A 3g22b
B 3g2b
C g2b
D 2g3b

The correct answer is:
3g22b

Solution:
The plate will rotate about the hinged point ( point A ) after breaking of the cable

∴ using τ=Iα for the plate

Mgb2=23Mb2α ( Moment of inertia about point A is Mb26+Mb22=23Mb2 )
∴ α=3g22b

Topic: Physics, Rotational Motion, JEE Main
A uniform string of length 20 m is suspended from a rigid support. A short wave pulse is introduced at its lowest end. It starts moving up the string. The time taken to reach the support is

(Take, g=10 m s-2)

Options:
A 22 s
B 2 s
C 2π2 s
D 2 s

The correct answer is:
22 s

Solution:
Velocity in a string, at a distance x, is given by
v=µgxµ=gx
⇒ dx dt = gx
⇒∫020dxx=g∫0tdt
On integrating, we get,

t=22  s

Topic: Physics, Waves and Sound, JEE Main, JEE Main 2016 (03 Apr)
A uniform string of length $L$ and mass $M$ is fixed at both ends while it is subject to a
tension $T$. It can vibrate at frequencies $(v)$ given by the formula (where $n=1,2,3, \ldots)$

Options:
A $v=\frac{n}{2} \sqrt{\frac{T}{M L}}$
B $v=\frac{n}{2 L} \sqrt{\frac{T}{M}}$
C $v=\frac{1}{2 n} \sqrt{\frac{T}{M L}}$
D $v=\frac{n}{2} \sqrt{\frac{T L}{M}}$

The correct answer is:
$v=\frac{n}{2} \sqrt{\frac{T}{M L}}$

Solution:
The formula of frequency of 3 vibrations produced in tensed wire is
$$
v=\frac{n}{2} \sqrt{\frac{T}{m}}
$$
$m=$ Mass per unit length of the wire $=\frac{M}{L}$
$M=$ Mass of wire
$L=$ Length of wire $n=$ Number of loops produced in wire $v=\frac{n}{2 L} \sqrt{\frac{T}{M / L}}=\frac{n}{2} \sqrt{\frac{T}{M L}}$

Topic: Physics, Waves and Sound, JEE Main
A uniform thin bar of mass 6 kg and length 2.4 meter is bent to make an equilateral hexagon. The moment of inertia about an axis passing through the centre of mass and perpendicular to the plane of hexagon is ___________ ×10-1 kg m2.

Options:

The correct answer is:
8

Solution:
61=2.4 ℓ=0.4 m

sin 60°=rℓ

r=1 sin 60°=ℓ32

MOI, I=mℓ212+mr26

=mℓ212+mℓ3226

=5 ml2

=5×1×0.16

=0.8

I=8×10-1 kg m2

Topic: Physics, Rotational Motion, JEE Main, JEE Main 2021 (24 Feb Shift 2)
A uniform, thin cylindrical shell and solid cylinder roll horizontally without slipping. The speed of the cylindrical shell is v. The solid cylinder and the hollow cylinder encounter an incline that they climb without slipping. If the maximum height they reach is the same, find the initial speed v' of the solid cylinder.

Options:
A 43v
B 34v
C 35v
D 53v

The correct answer is:
43v

Solution:
KE=12mv2+12Iω2
34mv12=mv22
v1v2=43

Topic: Physics, Rotational Motion, JEE Main
A uniform thin hemispherical shall is kept at rest and in equilibrium on an inclined plane of angle of inclination θ=30o as shown in figure. If the surface of the inclined plane is sufficiently rough to prevent sliding then the angle α made by the plane of hemisphere with inclined plane is:

Options:
A Value of μ is needed
B 30o
C 45o
D 60o

The correct answer is:
60o

Solution:
O is the centre of mass of the hollow hemisphere and is R2 from C.
f=mgsin⁡θ ...(i)
N-mgcos⁡θ ...(ii)
N×R2sin⁡α=RR2cos⁡αf ...(iii)
∴tan⁡θ=sin⁡α2-cos⁡α⇒α=60o

Topic: Physics, Rotational Motion, JEE Main
A uniform thin hemispherical shell is kept at rest and in equilibrium on an inclined plane of angle of inclination Î¸=30Â° as shown in the figure. If the surface of the inclined plane is sufficiently rough to prevent sliding then the angle Î± (in degree) made by the plane of hemisphere with inclined plane is

Options:

The correct answer is:
60

Solution:
O is the centre of mass of the hollow hemisphere and is at R2 from C.
f=mgsinâ¡Î¸
N=mgcosâ¡Î¸
NÃR2sinâ¡Î±=R-R2cosâ¡Î±f
â´tanâ¡Î¸=sinâ¡Î±2-cosâ¡Î±âÎ±=60o

Topic: Physics, Rotational Motion, JEE Main
A uniform thin rod AB of length L and mass m is undergoing rotation about a fixed axis passing through end A and perpendicular to the rod, such that end A remains stationary as shown. The kinetic energy of section AP of the rod is equal to the kinetic energy of section BP of the rod at an instant. Then what is the value of ABAp3?(AB and AP are lengths of respective parts of the rod)

Options:

The correct answer is:
2

Solution:
The KE of given section AP and PB will be equal if the moment of inertia of each section AP and section PB about A is same.
Moment of inertia of segment AP w.r.t given axis I1=Î»xx23
Moment of inertia of segment PB w.r.t given axis I2=I0-I1=Î»LL23-Î»xx23
(Moment of inertia or rod AB w.r.t. given axis is I0 )
â´I1=I2âÎ»x33=Î»L33-Î»x33
Or x3=L32
ABAP3=L3L32=2

Topic: Physics, Rotational Motion, JEE Main
A uniform thin rod AB of length L has linear mass density μx=a+bxL, where x is measured from A. If the CM of the rod lies at a distance of 712L from A, then a and b are related as:

Options:
A 2a=b
B a=2b
C a=b
D 3a=2b

The correct answer is:
2a=b

Solution:
Consider an infinitesimal small element of length dx and mars density λ
xcm=∫dmx∫dm=∫λ dxx∫λ dx
=∫0La+bLxxdx∫0La+bLxdx
=∫0Laxdx+∫0LbLx2dx∫0Ladx+∫0LbLxdx
=ax220L+bL x330Lax0L+bLx220L
=a2L2-02+b3L L3-03aL-0+b2LL2-02
=a2L2+bL23aL+bL2=a2L+bL3a+b2
∵xcm=7L12=a2+b3La+b2⇒712= 3a+2b62a+b2
⇒712=3a+2b3×12a+b⇒72a+b=43a+2b
⇒14a+7b=12a+8b⇒2a=b

Topic: Physics, Center of Mass Momentum and Collision, JEE Main, JEE Main 2015 (11 Apr Online)
A uniform thin rod AB of length L has linear mass density μx=a+bxL, where x is measured from A. If the CM of the rod lies at a distance of 712L from A, then a and b are related as:

Options:
A 2a=b
B a=2b
C a=b
D 3a=2b

The correct answer is:
2a=b

Solution:
Consider an infinitesimal small element of length dx and mars density λ
xcm=∫dmx∫dm=∫λ dxx∫λ dx
=∫0La+bLxxdx∫0La+bLxdx
=∫0Laxdx+∫0LbLx2dx∫0Ladx+∫0LbLxdx
=ax220L+bL x330Lax0L+bLx220L
=a2L2-02+b3L L3-03aL-0+b2LL2-02
=a2L2+bL23aL+bL2=a2L+bL3a+b2
∵xcm=7L12=a2+b3La+b2⇒712= 3a+2b62a+b2
⇒712=3a+2b3×12a+b⇒72a+b=43a+2b
⇒14a+7b=12a+8b⇒2a=b

Topic: Physics, Center of Mass Momentum and Collision, JEE Main
A uniform thin rod AB of length L has linear mass density μx=a+bxL, where x is measured from A. If the CM of the rod lies at a distance of 712L from A, then a and b are related as:

Options:
A 2a=b
B a=2b
C a=b
D 3a=2b

The correct answer is:
2a=b

Solution:
Consider an infinitisemal small element of length dx and mars density Î»
xcm=â«dmxâ«dm=â«Î» dxxâ«Î» dx
=â«0La+bLxxdxâ«0La+bLxdx
=â«0Laxdx+â«0LbLx2dxâ«0Ladx+â«0LbLxdx
=ax220L+bL x330Lax0L+bLx220L
=a2L2-02+b3L L3-03aL-0+b2LL2-02
=a2L2+bL23aL+bL2=a2L+bL3a+b2
âµxcm=7L12=a2+b3La+b2â712= 3a+2b62a+b2
â712=3a+2b3Ã12a+bâ72a+b=43a+2b
â14a+7b=12a+8bâ2a=b

Topic: Physics, Center of Mass Momentum and Collision, JEE Main
A uniform thin rod of length $2 \mathrm{~L}$ and mass $\mathrm{m}$ lies on a horizontal table. A horizontal inpulse $\mathrm{J}$ is given to the rod at one end. There is no friction. The total K.E. of the rod just after the impulse will be :

Options:
A $\mathrm{J}^{2} / 2 \mathrm{~m}$
B $\mathrm{J}^{2} / \mathrm{m}$
C $2 \mathrm{~J}^{2} / \mathrm{m}$
D $6 \mathrm{~J}^{2} / \mathrm{m}$

The correct answer is:
$2 \mathrm{~J}^{2} / \mathrm{m}$

Solution:
$\mathrm{J}=\mathrm{mv} \quad \mathrm{J} . \mathrm{L}=\frac{\mathrm{M}(2 \mathrm{~L})^{2}}{12} \omega$
$v^{v}=\mathrm{J} / \mathrm{M}$
$\mathrm{Cm}$
$J . L=\frac{M L^{2}}{3} \omega$
$\omega=\frac{3 J}{M L}$
$\mathrm{ke}=\frac{\mathrm{J}^{2}}{2 \mathrm{~m}}+\frac{3 \mathrm{~J}^{2}}{2 \mathrm{~m}}=2 \mathrm{~J}^{2} / \mathrm{m}$

Topic: Physics, Work Power Energy, JEE Main
A uniform thin rope of length 12 m and mass 6 kg hangs vertically from a rigid support and a block of mass 2 kg is attached to its free end. A transverse short wave train of wavelength 6 cm is produced at the lower and the rope. What is the wavelength of the wave train (in cm ) when it reaches the top of the rope?

Options:
A 3
B 6
C 12
D 9

The correct answer is:
12

Solution:
V=fλ

V1λ1=V2λ2

λ2=V2V1λ1=T2T1λ1T2=8gTop

8g2gλ1

T1=2gBottom=2λ1=12cm

Topic: Physics, Waves and Sound, JEE Main, JEE Main 2020 (03 Sep Shift 1)
A uniform tube of length $60.5 \mathrm{~cm}$ is held vertically with its lower end dipped in water. A sound source of frequency $500 \mathrm{~Hz}$ sends sound waves into the tube. When the length of tube above water is $16 \mathrm{~cm}$ and again when it is $50 \mathrm{~cm}$, the tube resonates with the source of sound. Two lowest frequencies (in $\mathrm{Hz}$ ), to which tube will resonate when it is taken out of water, are (approximately).

Options:
A
281,562

B
281,843

C
276,552

D
272,544

The correct answer is:
272,544

Solution:
Two lowest frequencies to which tube will resonates are $272 \mathrm{~Hz}$ and $544 \mathrm{~Hz}$.

Topic: Physics, Waves and Sound, JEE Main
A uniform tube of length $60.5 \mathrm{~cm}$ is held vertically with its lower end dipped in water. A sound source of frequency $500 \mathrm{~Hz}$ sends sound waves into the tube. When the length of tube above water is $16 \mathrm{~cm}$ and again when it is $50 \mathrm{~cm}$, the tube resonates with the source of sound. Two lowest frequencies (in $\mathrm{Hz}$ ), to which tube will resonate when it is taken out of water, are (approximately).

Options:
A
281,562

B
281,843

C
276,552

D
272,544

The correct answer is:
272,544

Solution:
Two lowest frequencies to which tube will resonates are $272 \mathrm{~Hz}$ and $544 \mathrm{~Hz}$.

Topic: Physics, Waves and Sound, JEE Main, JEE Main 2012 (19 May Online)
A uniform wire of length L and diameter D and density ρ is stretched under a tension T. The correct relation between its fundamental frequency f, the length L and the diameter D is

Options:
A f∝1LD2
B f∝1D2
C f∝1LD
D f∝1LD

The correct answer is:
f∝1LD

Solution:
Since, we know,
Frequency, f=n2L Tm
Where, T= tension in the string,
L= Length of the string
And m= linear mass density of the string.
â´m=ML=ÏD22âLâÏL=ÏD2Ï4
Thus, f=12LâTÏD2Ï4
For fundamental frequency, n=1
f=12LD2 TÏâÏ=1LDâTÏâÏ
fâ1LD

Topic: Physics, Waves and Sound, JEE Main
A uniform wire of length l and mass M is stretched between two fixed points, keeping a tension F. A sound of frequency μ is impressed on it. Then the maximum vibrational energy is existing in the wire when μ =

Options:
A 1 2 ML F
B FL M
C 2 × FM L
D 1 2 F ML

The correct answer is:
1 2 F ML

Solution:
We should check it dimensionally
Let μ = K F a M b L c
μ = F a M b L c
= MLT 2 a M b L c
M 0 L 0 T - 1 = M a + b L a + c T - 2 a
a + b = 0 ⇒ b = - a
a + c = 0 ⇒ c = - a
- 2 a = - 1
⇒ a = 1 2
b = - 1 2
c = - 1 2
⇒ μ = K F 1 2 M - 1 2 L - 1 2
μ = K F ML
So only option (iv) satisfies it dimensionally.

Topic: Physics, Units and Dimensions, JEE Main
A uniform wire of length l and radius r has a resistance of 100 Ω . It is recast into a wire of radius r2. The resistance of new wire will be-

Options:
A 1600 Î©
B 100 Î©
C 200 Î©
D 400 Î©

The correct answer is:
1600 Î©

Solution:
R=ρlA
Al=V
R=ρVA2
ρ→ constant
V→ constant
R∝1A2 ∝1r22
R∝1r4
R2=16R1=1600 Ω

Topic: Physics, Current Electricity, JEE Main, JEE Main 2017 (09 Apr Online)
A uniform wire of resistance $R$ is uniformly compressed along its length, until its radius becomes $n$ times the original radius. Now resistance of the wire becomes

Options:
A $\frac{R}{n^4}$
B $\frac{R}{n^2}$
C $\frac{R}{n}$
D $n R$

The correct answer is:
$\frac{R}{n^4}$

Solution:
$\frac{R_1}{R_2}=\left(\frac{r_2}{r_1}\right)^4 \Rightarrow \frac{R}{R_2}=\left(\frac{n r}{r}\right)^4 \Rightarrow R_2=\frac{R}{n^4}$.

Topic: Physics, Current Electricity, JEE Main
A uniform wire (Young's modulus $2 \times 10^{11} \mathrm{Nm}^{-2}$ ) is subjected to longitudinal tensile stress of $5 \times 10^7 \mathrm{Nm}^{-2}$. If the overall volume change in the wire is $0.02 \%$, the fractional decrease in the radius of the wire is close to:

Options:
A
$1.0 \times 10^{-4}$

B
$1.5 \times 10^{-4}$

C
$0.25 \times 10^{-4}$

D
$5 \times 10^{-4}$

The correct answer is:
$0.25 \times 10^{-4}$

Solution:
Given, $y=2 \times 10^{11} \mathrm{Nm}^{-2}$
$$
\begin{aligned}
& \text { Stress }\left(\frac{\mathrm{F}}{\mathrm{A}}\right)=5 \times 10^7 \mathrm{Nm}^{-2} \\
& \Delta \mathrm{V}=0.02 \%=2 \times 10^{-4} \mathrm{~m}^3 \\
& \frac{\Delta \mathrm{r}}{\mathrm{r}}=?
\end{aligned}
$$
$$
\begin{aligned}
& \gamma=\frac{\text { stress }}{\text { strain }} \Rightarrow \operatorname{strain}\left(\frac{\Delta \ell}{\ell_0}\right)=\frac{\gamma}{\text { stress }} \\
& \Delta \mathrm{V}=2 \pi \mathrm{r} \ell_0 \Delta \mathrm{r}-\pi \mathrm{r}^2 \Delta \ell
\end{aligned}
$$
From eqns (i) and (ii) putting the value of $\Delta \ell, \ell_0$ and $\Delta \mathrm{V}$ and solving we get
$$
\frac{\Delta \mathrm{r}}{\mathrm{r}}=0.25 \times 10^{-4}
$$

Topic: Physics, Mechanical Properties of Solids, JEE Main, JEE Main 2013 (22 Apr Online)
A uniformly charged disc of radius R having surface charge density σ is placed in the xy plane with its center at the origin. Find the electric field intensity along the z-axis at a distance Z from origin:

Options:
A E=2ε0σ1Z2+R21/2+Z
B E=σ2ε01+ZZ2+R21/2
C E=σ2ε01-ZZ2+R21/2
D E=σ2ε01Z2+R2+1Z2

The correct answer is:
E=σ2ε01-ZZ2+R21/2

Solution:
E=∫0R14πε0×σ×2πxdx×xx2+Z232=σ2ε01-ZZ2+R21/2

Topic: Physics, Electrostatics, JEE Main, JEE Main 2021 (27 Aug Shift 1)
A uniformly charged ring of radius 3a and total charge q is placed in x‐y plane centred at origin. A point charge q is moving towards the ring along the z- axis and has speed v at z=4a . The minimum value of v such that it crosses the origin is:

Options:
A 2m115q24πϵ0a1/2
B 2m415q24πϵ0a1/2
C 2m15q24πϵ0a1/2
D 2m215q24πϵ0a1/2

The correct answer is:
2m215q24πϵ0a1/2

Solution:
From the conservation of energy,
14πε0q23a2+4a2+12mv2=14πε0q23a
v=q24πε0a×215×2m
=2m215q24πε0a1/2

Topic: Physics, Electrostatics, JEE Main, JEE Main 2019 (10 Apr Shift 1)