datavorous/entrance-exam-dataset huggingface dataset:
A rocket is intended to leave the Earth's gravitational field. The fuel in its main engine is a little less than the amount that is necessary and an auxiliary engine, (only capable of operating for a short time) has to be used as well. When is it best to switch on the auxiliary engine?

Options:
A at take-off
B when the rocket has nearly stopped with respect to the Earth.
C It doesn’t matter.
D Can’t say

The correct answer is:
at take-off

Solution:
The rocket has to reach the highest possible total energy. If the zero level of gravitational potential energy is 'infinitely' far away, then the energy of the rocket standing on the surface of the earth is negative. The energy released during the operation of the engines increases the total energy of the rocket, and the rocket can leave, the earth's gravitational field if the sum of its potential and kinetic energies becomes positive.
The energy released in the course of the operation of the principal and auxiliary engines increases the total energy of the rocket and its ejected combustion products by a fixed value; this increase is independent of the moment when the engines are scratched on. However, the speed at which the combustion products fall back to the earth does depend on the timing of the rocket's operations. Indeed, if the auxiliary engine starts working when the rocket is at a greater height, the combustion products fall further and their speed and total energy are higher when they hit the ground. This means that the sooner the auxiliary engine is switched on, the higher the energy ultimately acquired by the rocket. The same argument is valid for the principal engine, and if the only energy consideration apply, it is best to operate the engines for the shortest time and at the highest thrust.

Topic: Physics, Gravitation, JIPMER, JIPMER 2017
A rocket is launched straight up from the surface of the earth. When its altitude is \(\frac{1}{3}\) of the radius of the earth, its fuel runs out and therefore it coasts. If the rocket has to escape from the gravitational pull of the earth, the minimum velocity with which it should coast is (Escape velocity on the surface of the earth is \(11.2 \mathrm{kms}^{-1}\).)

Options:
A \(11.2 \mathrm{kms}^{-1}\)
B \(10.7 \mathrm{kms}^{-1}\)
C \(9.7 \mathrm{kms}^{-1}\)
D \(8.7 \mathrm{kms}^{-1}\)

The correct answer is:
\(9.7 \mathrm{kms}^{-1}\)

Solution:
Given, escape velocity on the surface of the earth.
\(v_e=\sqrt{2 g R_e}=11.2 \mathrm{~km} / \mathrm{s}\)...(i)
where, \(R\) is the radius of earth.
Height of the rocket from the surface of earth, \(h=\frac{R_e}{3}\). If the rocket has to escape from the gravitational pull of the earth from \(h=\frac{R_\epsilon}{3}\). Then, the potential energy at the altitude,
\(\begin{aligned}
\frac{R_e}{3} & =\text { kinetic energy } \\
\frac{G M_e m}{R_e+h} & =\frac{1}{2} m v_{e_1}^2 \Rightarrow \frac{G M_e}{R_e+\frac{R_e}{3}}=\frac{1}{2} v_{e_1}^2
\end{aligned}\)
\(\left[v_{e_1}=\right.\) escape velocity of rocket from \(\left.\frac{R_e}{3}\right]\) \(\left(\begin{array}{rl}
M_e & =\text { mass of the earth and } \\
m & =\text { mass of the rocket }
\end{array}\right)\)
\(\begin{array}{rlrl}
\frac{g R_e^2}{4 R_e} & =\frac{1}{2} v_{e_1}^2 & \\
\Rightarrow \quad \frac{3}{4} g R_e & =\frac{v_{e_1}^2}{2} \quad \left[\because g=\frac{G M_e}{R_e^2}\right] \\
\frac{3}{4} \cdot 2 g R_e & =v_{e_1}^2 \\
\Rightarrow \quad \frac{\sqrt{3}}{2} \cdot \sqrt{2 g R_e} & =v_{e_1} \\
\frac{\sqrt{3}}{2} 11.2 & =v_{e_1} \quad \text {[From Eq. (i)] } \\
\Rightarrow \quad 9.7 & =v_{e_1} \\
\Rightarrow \quad v_{e_1} & =9.7 \mathrm{~km} / \mathrm{s}
\end{array}\)

Topic: Physics, Gravitation, JEE Main
A rocket is launched with velocity $10 \mathrm{~km} / \mathrm{s}$. If radius of earth is $R$, then maximum height attained by it will be

Options:
A $2 R$
B $3 R$
C $4 R$
D $5 R$

The correct answer is:
$4 R$

Solution:
Given,
$\mathrm{v}=10 \mathrm{~km} / \mathrm{s}$
$\mathrm{R}=$ radius of the earth
When a rocket is launched, kinetic energy is converted into potential energy.
At maximum height, potential energy is equal to the kinetic energy.
$\begin{aligned}& \text { P.E }=K \cdot E \\& m g(R+H)=\frac{m v^2}{2} \\& R+H=\frac{v^2}{2 g} \\& R+H=\frac{10 \times 1000 \times 10 \times 1000}{2 \times 10} \\& R+H=5 R\left(\text { as } R=6.5 \times 10^6 \mathrm{~m}\right) \\
& H=4 R\end{aligned}$

Topic: Physics, Gravitation, JEE Main
A rocket takes off from the earth and continues to move in a circular orbit with the thrusters on. What can be said about the angular velocity of the rocket?

Options:
A It increases
B It decreases
C It remains constant
D It changes abruptly

The correct answer is:
It increases

Solution:
Concept: When the rocket is in orbit with the thrusters activated, it feels the tangential force.
This force will cause the tangential velocity to increase.
The angular velocity will increase since it is exactly related to the tangential velocity.

Topic: Physics, Laws of Motion, JEE Main
A rocket with a lift-off mass $3.5 \times 10^4 \mathrm{~kg}$ is blasted upwards with an initial acceleration of $10 \mathrm{~m} / \mathrm{s}^2$. Then the initial thrust of the blast is

Options:
A
$3.5 \times 10^5 \mathrm{~N}$

B
$7.0 \times 10^5 \mathrm{~N}$

C
$14.0 \times 10^5 \mathrm{~N}$

D
$1.75 \times 10^5 \mathrm{~N}$

The correct answer is:
$3.5 \times 10^5 \mathrm{~N}$

Solution:
Thrust $=$ Mass $\times$ Acceleration $=3.5 \times 10^4 \times 10=3.5 \times 10^5 \mathrm{~N}$

Topic: Physics, Laws of Motion, JEE Main, JEE Main 2003
A rod AB of length L and mass M is free to move on a frictionless horizontal surface. It is moving with a velocity v, as shown in figure. End B of rod AB strikes the end of the wall. Assuming elastic impact, the angular velocity of the rod AB, just after impact, is

Options:
A v2L
B 3vL
C 3v2L
D vL

The correct answer is:
3vL

Solution:
From the conservation of angular momentum,

âMvcmL2+ML212Ï=MvL2âvcm+ÏL6=v
From the conservation of linear momentum,

â ÏL2-vcm=vâÏL2+L6=2v
â´ Ï=3vL

Topic: Physics, Rotational Motion, JEE Main
A rod AB of mass M and length L is lying on a horizontal frictionless surface. A particle of mass m travelling along the surface hits the end A of the rod with a velocity v0 in a direction perpendicular to AB. The collision is perfectly elastic and after the collision, the particle comes to rest. The ratio mM is

Options:
A 14
B 13
C 12
D 1

The correct answer is:
14

Solution:
Using the principle of conservation of linear momentum

mv0=Mvcm

Using the principle of conservation of angular momentum

mv0L2=ML212ω

Since the collision is completely elastic, the velocity of approach is equal to the velocity of separation

v0=vcm+ωL2

⇒v0=mMv0+3mMv0

⇒mM=14

Topic: Physics, Work Power Energy, JEE Main
A rod length AB is moving with ends remaining in contact with frictionless wall and floor. If at the instant shown, the velocity of the end B is 3 m s-1 towards the negative x-direction, then the magnitude of the velocity of the end A will be

Options:
A 3 m s-1
B 3 m s-1
C 1.5 m s-1
D 2 m s-1

The correct answer is:
3 m s-1

Solution:
From geometry, it is clear that x=lcosâ¡Î¸ and y=lsinâ¡Î¸
â´ vx=dxdv=-lsinâ¡Î¸dÎ¸dt and vy=dydt=lcosâ¡Î¸dÎ¸dt
â´ vyvx=-cotâ¡Î¸ or vy=-vxcotâ¡Î¸
Since, vx=-3 m s-1, hence
vy=--3cotâ¡60Â°=3Ã13=3 m s-1

Topic: Physics, Laws of Motion, NEET
A rod \( \mathrm{AB} \) of mass \( \mathrm{M} \) and length \( \mathrm{L} \) is lying on a horizontal frictionless surface. A particle of mass \( \mathrm{m} \) travelling along the surface hits the end \( \mathrm{A} \) of the rod with a velocity \( \mathrm{v}_{0} \) in a direction perpendicular to \( \mathrm{AB} \). The collision is perfectly elastic and after the collision, the particle comes to rest. The ratio \( \frac{m}{M} \) is

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

The correct answer is:
\( \frac{1}{4} \)

Solution:
Using the principle of conservation of linear momentum

mv0=Mvcm

Using the principle of conservation of angular momentum

mv0L2=ML212ω

Since the collision is completely elastic, the velocity of approach is equal to the velocity of separation

v0=vcm+ωL2

⇒v0=mMv0+3mMv0

⇒mM=14

Topic: Physics, Rotational Motion, JEE Main
A rod of length $10 \mathrm{~cm}$ lies along the principal axis of a concave mirror of focal length $10 \mathrm{~cm}$ in such a way that its end closer to the pole is $20 \mathrm{~cm}$ away from the mirror. The length of the image is

Options:
A $10 \mathrm{~cm}$
B $15 \mathrm{~cm}$
C $2.5 \mathrm{~cm}$
D $5 \mathrm{~cm}$

The correct answer is:
$5 \mathrm{~cm}$

Solution:
By mirror formula, image distance of $A$
$$
\frac{1}{v}+\frac{1}{u}=\frac{1}{f}
$$
$$
\begin{aligned}
\frac{1}{v_A}+\frac{1}{u} & =\frac{1}{f} \\
\frac{1}{v_A}+\frac{1}{(-30)} & =\frac{1}{-10} \\
v_A & =-15 \mathrm{~cm}
\end{aligned}
$$
Also image distance of $C$
$$
v_C=-20 \mathrm{~cm}
$$
The length of image $=\left|v_A-v_C\right|$
$$
=|-15-(-20)|
$$
$$
=5 \mathrm{~cm}
$$

Topic: Physics, Ray Optics, JEE Main
A rod of length $10 \mathrm{~cm}$ lies along the principal axis of a concave mirror of focal length $10 \mathrm{~cm}$ in such a way that its end closer to the pole is $20 \mathrm{~cm}$ away from the mirror. The length of the image is

Options:
A $10 \mathrm{~cm}$
B $15 \mathrm{~cm}$
C $2.5 \mathrm{~cm}$
D $5 \mathrm{~cm}$

The correct answer is:
$5 \mathrm{~cm}$

Solution:
By mirror formula, image distance of $A$
$$
\frac{1}{v}+\frac{1}{u}=\frac{1}{f}
$$
$$
\begin{aligned}
\frac{1}{v_A}+\frac{1}{u} & =\frac{1}{f} \\
\frac{1}{v_A}+\frac{1}{(-30)} & =\frac{1}{-10} \\
v_A & =-15 \mathrm{~cm}
\end{aligned}
$$
Also image distance of $C$
$$
v_C=-20 \mathrm{~cm}
$$
The length of image $=\left|v_A-v_C\right|$
$$
=|-15-(-20)|
$$
$$
=5 \mathrm{~cm}
$$

Topic: Physics, Ray Optics, NEET, NEET 2012 (Mains)
A rod of length $20 \mathrm{~cm}$ is rotating with angular speed of $100 \mathrm{rps}$ in a magnetic field of strength $0.5 T$ about it's one end. What is the potential difference between two ends of the rod

Options:
A $2.28 \mathrm{~V}$
B $4.28 \mathrm{~V}$
C $6.28 \mathrm{~V}$
D $2.5 \mathrm{~V}$

The correct answer is:
$6.28 \mathrm{~V}$

Solution:
$e=\left.\frac{1}{2} B\right|^2 \omega=B I^2 \pi v$
$\Rightarrow e=0.5\left(20 \times 10^{-2}\right)^2 \times 3.14 \times 100=6.28 \mathrm{~V}$

Topic: Physics, Electromagnetic Induction, JEE Main
A rod of length 50 cm is pivoted at one end. It is raised such that it makes an angle of 30o from the horizontal as shown and released from rest. Its angular speed when it passes through the horizontal (in rad s-1 ) will be g=10 ms-2

Options:
A 203
B 30
C 302
D 302

The correct answer is:
30

Solution:
From conservation of mechanical energy
Loss in gravitational PE= gain in KE
mgl2sin⁡30o=12Iω2
mgl2sin⁡30o=12ml23ω2
ω=3g2l=3×102×50×10-2=30 rad/s

Topic: Physics, Rotational Motion, JEE Main, JEE Main 2019 (09 Jan Shift 2)
A rod of length lâ¡ = 2 m is maintained to rotate with a constant angular velocity Ï=1 rad s-1 about a vertical axis passing through one end (diagram). There is a spring constant k=1 N m-1 which just encloses rod inside it in natural length. One end of the spring is attached to the axis of rotation. S is a sleeve of mass m=1 kg which can just fit on rod. All surfaces are smooth. With what minimum kinetic energy (in j) sleeve should be projected so that it enters on the rod without jerk and completely compresses the spring.

Options:

The correct answer is:
6

Solution:
For entering without jerk v2=lâ¡Ï0=2 m s-1
Using work-energy theorem on the sleeve after entering in the frame of rod
W spring + W centrifugal = Î K
- 1 2 k lâ¡ 2 - 1 2 m Ï 2 lâ¡ 2 = 0 - 1 2 mv 1 2
â v 1 2 = 8
Now, K = 1 2 m v 1 2 + v 2 2 = 6

Topic: Physics, Rotational Motion, JEE Main
A rod of length l and cross-section area A has a variable thermal conductivity given by K=Î±T, where Î± is a positive constant and T is the temperature in Kelvin. Two ends of the rod are maintained at temperatures T1 and T2 (T1>T2) . Heat current flowing through the rod will be -

Options:
A AαT12-T223l
B AαT12+T22l
C AαT12+T223l
D AαT12-T222l

The correct answer is:
AαT12-T222l

Solution:
Heat current,
i=-KAdTdX
i dx=-KA dT
i ∫0ldX=-Aα∫T1T2T dT
⇒il=-AαT22-T122
⇒i=AαT12-T222l

Topic: Physics, Thermal Properties of Matter, JEE Main
A rod of length $l$ and negligible mass is suspended at its two ends by two wires of steel (wire $A$ ) and aluminium (wire $B$ ) of equal lengths (figure).
The cross-sectional areas of wires $A$ and $B$ are $1.0 \mathrm{~mm}^2$ and $2.0 \mathrm{~mm}^2$, respectively.
$\left(Y_{\mathrm{Al}}=70 \times 10^9 \mathrm{Nm}^{-2}\right.$ and $\left.Y_{\text {steel }}=200 \times 10^9 \mathrm{Nm}^{-2}\right)$

Options:
A
Mass $m$ should be suspended close to wire $A$ to have equal stresses in both the wires

B
Mass $m$ should be suspended close to $B$ to have equal stresses in both the wires

C
Mass $m$ should be suspended at the middle of the wires to have equal stresses in both the wires

D
Mass $m$ should be suspended close to wire $A$ to equal strain in both wires

The correct answers are:
Mass $m$ should be suspended close to $B$ to have equal stresses in both the wires
,
Mass $m$ should be suspended close to wire $A$ to equal strain in both wires

Solution:
Consider the given diagram, and a rod is balanced by weight $(m g)$, i.e. (rod remain horizontal)
Let the mass is placed at $x$ from the end $B$
Let $T_A$ and $T_B$ be the tensions in wire $A$ and wire $B$ respectively.
For the rotational equilibrium of the system, $\Sigma \tau=0 \quad($ Total torque $=0$ )
$T_B(x)-T_A(l-x)=0$ (taking moment about point A) $\frac{T_B}{T_A}=\frac{l-x}{x}$
Stress in wire $A=S_A=\frac{T_A}{a_A}$
Stress in wire $B=S_B=\frac{T_B}{a_B}$
where $a_A$ and $a_B$ are cross-sectional areas of wire $A$ and $B$ respectively.
$$
a_B=2 a_A \quad \text { (given) }
$$
As the stress on steel $\left(S_A\right)=$ Stress on $A 1\left(S_B\right)$
So, $S_A=S_B$
$$
\begin{aligned}
&\frac{T_A}{a_A}=\frac{T_B}{a_B} \Rightarrow \frac{T_B}{T_A}=\frac{a_B}{a_A}=2 \\
&\frac{l-x}{x}=\frac{2}{1} \Rightarrow \frac{l}{x}-1=2 \Rightarrow x=\frac{l}{3} \quad(\text { from } B)
\end{aligned}
$$
distance of $m$ from $A$
$$
\Rightarrow \quad l-x=l-l / 3=\frac{2 l}{3}
$$
Hence, mass $m$ should be placed closer to $B$.
(b) Strain : Strain in both wire is equal because rod remain horizontal in balanced condition.
For equal strain,
$(\text { strain })_A=(\text { strain })_B$
$$
\frac{Y_A}{S_A}=\frac{Y_B}{S_B}
$$
(where $Y_A$ and $Y_B$ are Young modulii)
$$
\begin{aligned}
&\frac{Y_{\text {stecl }}}{T_A / a_A}=\frac{Y_{\mathrm{Al}}}{T_B / a_B} \\
&\frac{Y_{\text {steel }}}{Y_{\mathrm{Al}}}=\frac{T_A}{T_B} \times \frac{a_B}{a_A}=\left(\frac{x}{1-x}\right)\left(\frac{2 a_A}{a_A}\right) \\
&\frac{200 \times 10^9}{70 \times 10^9}=\frac{2 x}{1-x} \Rightarrow \frac{20}{7}=\frac{2 x}{1-x} \\
&\Rightarrow 14 x=20 l-20 x \Rightarrow 34 x=20 l \\
&\text { or } 17 x=10 l \Rightarrow x=\frac{10 l}{17} \quad \text { (from } B \text { ) } \\
&(l-x) \text { from } A=\left[l-\frac{10 l}{17}\right]=\frac{7 l}{17} \quad \text { (from } A \text { ) }
\end{aligned}
$$
Hence for equal strain, mass $m$ should be closer to wire $A$.

Topic: Physics, Mechanical Properties of Solids
A rod of length 'L' and negligible mass is suspended by two identical strings $A B$ and $C D$ as shown in the figure. $A$ mass ' $M$ ' is suspended from point ' $O$ ' which is at a distance ' $x$ ' from $\mathrm{B}$. If the frequency of the first harmonic of $\mathrm{AB}$ is equal to the frequency of the second harmonic of $C D$, then the value of ' $x$ ' is

Options:
A $\frac{L}{5}$
B $\frac{2 L}{7}$
C $\frac{3 L}{10}$
D $\frac{L}{9}$

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

Solution:
Frequency of Harmenic motion is given by:
$$
\mathrm{f}=\frac{\mathrm{n}}{2 \ell} \sqrt{\frac{\mathrm{T}}{\mathrm{m}}}
$$
Frequency of first Harmonic of $\mathrm{AB}$ is given by:
$$
\mathrm{f}_{\mathrm{A}}=\frac{1}{2 \ell} \sqrt{\frac{\mathrm{T}_{\mathrm{A}}}{\mathrm{m}}}
$$
Frequency of second Harmonic of $\mathrm{CD}$ is given by:
$$
\mathrm{f}_{\mathrm{C}}=\frac{2}{2 \ell} \sqrt{\frac{\mathrm{T}_{\mathrm{C}}}{\mathrm{m}}}=\frac{1}{\ell} \sqrt{\frac{\mathrm{T}_{\mathrm{C}}}{\mathrm{m}}}
$$
Since frequencies are equal.
$$
\begin{aligned}
& \mathrm{f}_{\mathrm{A}}=\mathrm{f}_{\mathrm{C}} \\
& \frac{1}{2 \ell} \sqrt{\frac{\mathrm{T}_{\mathrm{A}}}{\mathrm{m}}}=\frac{1}{\ell} \sqrt{\frac{\mathrm{T}_{\mathrm{C}}}{\mathrm{m}}} \\
& \frac{1}{4}\left(\mathrm{~T}_{\mathrm{A}}\right)=\mathrm{T}_{\mathrm{C}} \\
& \mathrm{T}_{\mathrm{A}}=4 \mathrm{~T}_{\mathrm{C}}
\end{aligned}
$$
Since the system is in equilibrium the torque about point $\mathrm{O}$ is same.
$$
\begin{aligned}
& \mathrm{T}_{\mathrm{A}}(\mathrm{x})=\mathrm{T}_{\mathrm{C}}(\mathrm{L}-\mathrm{x}) \\
& \Rightarrow 4 \mathrm{~T}_{\mathrm{C}}(\mathrm{x})=\mathrm{T}_{\mathrm{C}}(\mathrm{L}-\mathrm{x}) \\
& \Rightarrow 4 \mathrm{x}=\mathrm{L}-\mathrm{x} \\
& \Rightarrow 5 \mathrm{x}=\mathrm{L} \\
& \mathrm{x}=\frac{\mathrm{L}}{5}
\end{aligned}
$$

Topic: Physics, Oscillations, JEE Main
A rod of length $l$ and radius $r$ is joined to a rod of length $l / 2$ and radius $r / 2$ of same material. The free end of small rod is fixed to a rigid base and the free end of larger rod is given a twist of $\theta^{\circ}$, the twist angle at the joint will be

Options:
A $\theta / 4$
B $\theta / 2$
C $5 \theta / 6$
D $8 \theta / 9$

The correct answer is:
$8 \theta / 9$

Solution:
$\tau=C . \theta=\frac{\pi w^4 \theta}{2 L}=$ Constant
$\Rightarrow \frac{\pi \eta r^4\left(\theta-\theta_0\right)}{2 l}=\frac{\pi \eta(r / 2)^4\left(\theta_0-\theta^{\prime}\right)}{2(l / 2)}$
$\Rightarrow \frac{\left(\theta-\theta_0\right)}{2}=\frac{\theta_0}{16} \Rightarrow \theta_0=\frac{8}{9} \theta$

Topic: Physics, Mechanical Properties of Solids, JEE Main
A rod, of length L at room temperature and uniform area of cross section A, is made of a metal having coefficient of linear expansion α /°C It is observed that an external compressive force F is applied on each of its ends, prevents any change in the length of the rod when its temperature rises by ∆T K Young's modulus, Y for this metal is:

Options:
A FAα∆T-273
B 2FAÎ±âT
C FAÎ±âT
D F2AÂ Î±âT

The correct answer is:
FAα∆T

Solution:
Since the length of the rod does not change, the increment due to increase in temperature will be balanced by the compression due to applied force.
Y=FAΔll ⇒ Δll=FAY
Also Δll=αΔT
∴ FAY=α ΔT
Y=FAαΔT

Topic: Physics, Mechanical Properties of Solids, JEE Main, JEE Main 2019 (09 Jan Shift 1)
A rod of length l has non-uniform linear mass density given by ρx=a+bxl2, where a and b are constants and 0≤x≤l The value of x for the centre of mass of the rod is at:

Options:
A 32a+b2a+bL
B 342a+b3a+bL
C 43a+b2a+3bL
D 322a+b3a+bL

The correct answer is:
342a+b3a+bL

Solution:
Xcm=1M∫0ιx.dM
dM=ρ.dx=a+bxı2.dx
xcm=∫xdM∫dm=∫xρdx∫ρdx=∫0ιxa+bx2ι2dx∫0ιa+bx2ι2dx
=ax220ι+bι2x440ιax0ι+bι2x330ι
=aι22+bι24aι+bι3=2a+b3a+bι4×3
=3ι42a+b3a+b

Topic: Physics, Center of Mass Momentum and Collision, JEE Main, JEE Main 2020 (09 Jan Shift 2)
A rod of length, \( l \) is given two velocities, \( v_{1} \) and \( v_{2} \) in opposite directions at its two ends at right angles to the length. The distance of the instantaneous axis of rotation from \( v_{1} \) is

Options:
A \( 0 \)
B \( \frac{v_{1}}{v_{1}+v_{2}} l \)
C \( \frac{v_{2}}{v_{1}+v_{2}} l \)
D \( \frac{1}{2} \)

The correct answer is:
\( \frac{v_{1}}{v_{1}+v_{2}} l \)

Solution:
Let, the distance of the instantaneous axis from v1=x.

About the instantaneous axis of rotation, angular velocity of each point will be the same.

So, ωv1=ωv2

v1x=v2l-x

x=v1v1+v2l.

Topic: Physics, Rotational Motion, JEE Main
A rod of length 'L' is hung from its one end and a mass 'm' is attached to its free
end. What tangential velocity must be imparted to 'm', so that it reaches the top of
the vertical circle? (g = acceleration due to gravity)

Options:
A $4 \sqrt{\mathrm{gL}}$
B $2 \sqrt{\mathrm{gL}}$
C $5 \sqrt{\mathrm{gL}}$
D $3 \sqrt{\mathrm{gL}}$

The correct answer is:
$2 \sqrt{\mathrm{gL}}$

Solution:
$2 \sqrt{g L}$
$h=2 L$
$\frac{1}{2} \mathrm{mv}^{2}=\mathrm{mg} 2 \mathrm{L}$
$v_{2}=4 g L$
$v=2 \sqrt{g L}$

Topic: Physics, Work Power Energy, JEE Main
A rod of length l is placed along x -axis. One of its ends is at the origin. The rod has a non-uniform charge density Î»=ax , a being a positive constant. Electric potential at P as shown in the figure is

Options:
A a4πε0lnb+lb
B a4πε0lnlb
C a4πε0lnbl
D zero

The correct answer is:
a4πε0lnb+lb

Solution:
V =∫bb+lkdqx=∫bb+lkλdxx= ka ∫bb+ldxx
V=Ka lnb+lb
V =a4πε0lnb+lb

Topic: Physics, Electrostatics, JEE Main
A rod of length l is placed along x-axis. One of its ends is at the origin. The rod has a non-uniform charge density Î»=ax, a being a positive constant. The electric potential at the point P (origin) as shown in the figure is

Options:
A V=a4πε0lbb+l
B V=a4πε0blb+l
C V=a4πε0bl
D V=a4πε0lb

The correct answer is:
V=a4πε0lbb+l

Solution:
V =∫bb+lkdqx=∫bb+lkλdxx= ka ∫bb+ldxx2
V=ka 1b-1b+l
V=a4πε0lbb+l

Topic: Physics, Electrostatics, JEE Main
A rod of length l (laterally thermally insulated) of the uniform cross-sectional area A consists of a material whose thermal conductivity varies with temperature as K = K 0 a + bT, where K0, a and b are constants. T1 and T2 < T1 are the temperatures of two ends of the rod. Then, the rate of flow of heat across the rod is

Options:
A AK0bla+bT1a+bT2
B AK0bla+bT2a+bT1
C K0Abl⁡lna+bT1a+bT2
D AK0bllna+bT2a+bT1

The correct answer is:
K0Abl⁡lna+bT1a+bT2

Solution:
As, we know that,
â dQ dt = - KA dT dx
â dQ dt = - K 0 A a + bT dT dx
On integrating both sides within the proper limits.
dQdtâ«0lâ¡dx=-K0Aâ«T1T2dTa+bT
This gives,
dQdt=K0Ablâ¡lna+bT1a+bT2

Topic: Physics, Thermal Properties of Matter, JEE Main
A rod of length l with thermally insulated lateral surface consists of material whose heat conductivity coefficient varies with temperature as Kx=αT (here α is a positive constant). The ends of the rod are kept at temperature T1 and T2 (T1>T2).

Find the heat current per unit cross-sectional area in the rod

Options:
A αllnT1+T2T1
B αllnT1T2
C αlln2T1T1+T2
D αllnT1+T2T2

The correct answer is:
αllnT1T2

Solution:
dQdt=-KATx+dx-TxdxâI0=-Î±TAdTdx
â â«0lI0AÎ± dx=-â«T1T2dTT
â´ I0lAÎ±=lnT1T2
â I0A=Î±llnT1T2

Topic: Physics, Thermal Properties of Matter, JEE Main
A rod of mass m and length 2l is connected to two massless strings as shown in figure. A wooden disk of same mass m is fixed at the mid-point of the rod in such a way that the plane of disk is perpendicular to the rod. A massless wire is wrapped around disk and having current I. There is vertically upward magnetic field B at the location of disk. (Sense of current in the loop when seen from string A side is clockwise)

The tension in the string P is

Options:
A mg+IπR2B2l
B mg-IπR2B2l
C mg
D 2mg

The correct answer is:
mg+IπR2B2l

Solution:
net torque about disc is zero
TA×l=TB×l+IπR2B
And TA+TB=2mg⇒ T A =mg+IπR2B2l

Topic: Physics, Magnetic Effects of Current, JEE Main
A rod of mass M and length 2L is performing SHM as torsional pendulum in the horizontal plane. Two blocks, each of mass m, are put at distance L2 from the centre. The frequency after putting blocks of mass m is 20% of initial frequency. Then, the ratio of mM will be

Options:
A 12
B 14
C 16
D 18

The correct answer is:
16

Solution:
T=2π IC
f=12π CI
finitial=12π 3CML2
ffinal=12πCML23+2mL24
0.2 12π 3CML2=12πCML23+2ML24
=0.043M=124M+6M
4M=0.16M+0.24 M
mM=3.840.24=16

Topic: Physics, Oscillations, JEE Main
A rod of mass M and length 2L is suspended at its middle by a wire. It exhibits torsional oscillations. If two masses, each of mass m, are attached at a distance L/2 from its centre on both sides, it reduces the oscillation frequency by 20%. The value of ratio m/M is close to

Options:
A 0.17
B 0.77
C 0.57
D 0.37

The correct answer is:
0.37

Solution:
Let C be the torsional constant of the wire.
f=12ÏCM2L212=12Ï3CM. L2
After masses are attached,
fâ²=12Ï CM.2L212+mL24Ã2
â0.8f=12Ï CM3+m2L2
â0.64Ã3CM=CM3+m2
0.64M+0.64Ã32m=M
âmM=0.37

Topic: Physics, Oscillations, JEE Main