datavorous/entrance-exam-dataset huggingface dataset:
A TV transmitting antenna is 98 m high and the receiving antenna is at the ground level. If the radius of the earth is 6400 km, the surface area covered by the transmitting antenna is approximately:

Options:
A 1240 km2
B 3942 km2
C 4868 km2
D 1549 km2

The correct answer is:
3942 km2

Solution:
The given data is

H=98 m

R=6400 km

The line of sight formula for an antenna is

d=2RH

⇒Area =πd2=π×2×R×h=2×6400×98×10-3×π=3940 km2

Topic: Physics, Communication System, JEE Main, JEE Main 2023 (08 Apr Shift 1)
A two metre wire is moving with a velocity of $1 \mathrm{~m} / \mathrm{sec}$ perpendicular to a magnetic field of 0.5 weber/m². The e.m.f. induced in it will be

Options:
A $0.5$ volt
B $0.1$ volt
C $1$ volt
D $2$ volt

The correct answer is:
$1$ volt

Solution:
$e=B v l=0.5 \times 2 \times 1=1 \mathrm{~V}$

Topic: Physics, Electromagnetic Induction, JEE Main
A two point charges 4q and –q are fixed on the x -axis at x=−d2 and x=d2, respectively. If the third point charge ‘q ’ is taken from the origin to x=d along the semicircle as shown in the figure, the energy of the charge will:

Options:
A Increase by 3q24π∈0d
B Increase by 2q23π∈0d
C decrease by q24π∈0d
D decrease by 4q23π∈0d

The correct answer is:
decrease by 4q23π∈0d

Solution:
Potential at origin O, the potential due to two charges is given by

V0=4q4πε0d2+−q4πε0d2=6q4πε0d

Potential at P at a distance, x=d is given by

VP=4q4πε03d2+−q4πε0d2=q6πε0d

Change in potential energy due the charge motion of charge q is given by

∆U=qΔV=qVf−Vi

⇒∆U=qVP−VO

∆U=qq6πε0d−3q2πε0d=−4q23πε0d

Topic: Physics, Electrostatics, JEE Main, JEE Main 2020 (04 Sep Shift 1)
A U238 nucleus originally at rest, decays by emitting an alpha particle having a speed u. Due to this emission, the residual nucleus recoils with a speed nu234. What is the value of n?

Options:

The correct answer is:
4

Solution:
Apply conservation of linear momentum
4uâ¡-234vâ¡=0
â vâ¡ = 4 uâ¡ 2 3 4
n=4

Topic: Physics, Nuclear Physics, JEE Main
A U-shaped conducting frame is fixed in space. A conducting rod CD lies at rest on the smooth frame as shown. The frame is in a uniform magnetic field B0, which is perpendicular to the plane of the frame. At the time t=0 , the magnitude of the magnetic field begins to change with time t as, B=B01+kt, where k is a positive constant. For no current to be ever induced in the frame, the speed with which rod should be pulled starting from time t=0 is (the rod CD should be moved such that its velocity must lie in the plane of frame and perpendicular to rod CD)

Options:
A ak
B bk
C a1+kt
D b1+kt

The correct answer is:
ak

Solution:
The magnetic flux must remain constant.
∴ ϕm=B bx=B01+ktbx
Where x is as shown
∴ x=a1+kt or v=dxdt=ak.

Topic: Physics, Magnetic Effects of Current, NEET
A U-tube has a liquid of density Ï 0 at 0 Â°C. In the two limbs they are kept to a length lâ¡ 1 and lâ¡ 2 at a constant temperature of t1 and t2 respectively. Then, the cubical expansion coefficient will be

Options:
A l⁡2-l⁡1l⁡1t1-l2t2
B l⁡1-l⁡2l⁡1t2-l2t1
C l⁡2-l⁡1l⁡1t2-l2t1
D l⁡2-l⁡1l⁡2t2-l1t1

The correct answer is:
l⁡2-l⁡1l⁡1t2-l2t1

Solution:
Pressure on each side has to be same or the level to be maintained.
â´ lâ¡ 1 Ï 1 g = lâ¡ 2 Ï 2 g
â´ lâ¡ 1 Ï 0 1 + Î³ t 1 g = lâ¡ 2 Ï 0 1 + Î³ t 2 g
â´ lâ¡ 1 + lâ¡ 1 t 2 Î³ = lâ¡ 2 + Î³ lâ¡ 2 t 1
Î³ lâ¡ 1 t 2 - lâ¡ 2 t 1 = lâ¡ 2 - lâ¡ 1
Î³ = lâ¡ 2 - lâ¡ 1 lâ¡ 1 t 2 - lâ¡ 2 t 1

Topic: Physics, Mechanical Properties of Fluids, JEE Main
A U tube of the uniform bore of cross-sectional area A is set up vertically with open ends. A liquid of mass M and density d is poured into it. The liquid column will oscillate with a period

Options:
A 2 π M g
B 2 π M A d g
C 2 π M A d g
D 2 Ï M 2 A d g

The correct answer is:
2 π M 2 A d g

Solution:
T = 2 Ï L g
Where L is the length of the liquid is one of the limbs. However, if L is taken to be the length of the liquid column, then the length of liquid in each limb is L 2.
â T = 2 Ï L 2 g
â M = AL d â L = M Ad
â T = 2 Ï M 2 Adg

Topic: Physics, Oscillations, JEE Main
A U tube with both ends open to the atmosphere is partially filled with water. Oil, which is immiscible with water, is poured into one side until it stands at a distance of 10 mm above the water level on the other side. Meanwhile, the water rises by 65 mm from its original level (see diagram). The density of the oil is____ρw=1000 kg m-3

Options:
A 650 kg m-3
B 425Â kgÂ m-3
C 800Â kgÂ m-3
D 928 kg m-3

The correct answer is:
928 kg m-3

Solution:
∵ Pressure in a stationary liquid at the same level is the same everywhere. As both ends of the tube are open, the pressure on both free surfaces should be equal.
hoil ρoil g=hwater ρwaterg
140×ρoil=130×ρwater
ρoil=928 kg m-3.

Topic: Physics, Mechanical Properties of Fluids, JEE Main
A uniform ball of radius r is placed on the top of a sphere of radius R=10r. It is given a slight push due to which it starts rolling down the sphere without slipping. The spin angular velocity of the ball when it breaks off from the sphere is Ï=pqgr, where g is the acceleration due to gravity and p and q are the smallest integers. What is the value of p+q?

Options:

The correct answer is:
127

Solution:
mv2R+r=mg cosθ

mgh=12mv2+12Iω2

mg R + r 1 - cos θ = 1 2 mv 2 + 1 5 mv 2 = 7 1 0 mv 2

1 0 7 mg 1 - cos θ = mg cos θ
mv 2 = 1 0 7 mg R + r 1 - cos θ 1 0 7 = 1 7 7 cos θ
or cos θ = 1 0 1 7
v = g R + r cos θ = 1 0 1 7 g R + r
and ω=vr=10gR+r17r2=110g17r

⇒p+q=127

Topic: Physics, Rotational Motion, JEE Main
A uniform ball of radius r rolls without slipping down from the top of a sphere of radius R. The spin angular velocity of the ball when it breaks away from the sphere is (Assume initial velocity negligible)

Options:
A 1 0 g R + r 1 7 r 2
B 1 0 g R - r 1 7 r 2
C 1 0 g R + r 1 7
D

1 0 R + r 17 r 2

The correct answer is:
1 0 g R + r 1 7 r 2

Solution:
mv2R+r=mg cosθ; mgh=12mv2+12Iω2
mg R + r 1 - cos θ = 1 2 mv 2 + 1 5 mv 2 = 7 1 0 mv 2

1 0 7 mg 1 - cos θ = mg cos θ
mv 2 = 1 0 7 mg R + r 1 - cos θ 1 0 7 = 1 7 7 cos θ
or cos θ = 1 0 1 7
v = g R + r cos θ = 1 0 1 7 g R + r
and ω = v r = 1 0 g R + r 1 7 r 2

Topic: Physics, Rotational Motion, JEE Main
A uniform but time-varying magnetic field is present in a circular region of radius R. The magnetic field is perpendicular and into the plane of the paper and the magnitude of the field is increasing at a constant rate α. There is a straight conducting rod of length 2R placed as shown in the diagram. The magnitude of induced E.M.F. across the rod is

Options:
A π R 2 α
B Ï R 2 Î± 4
C R 2 α 2
D π R 2 α 2

The correct answer is:
π R 2 α 4

Solution:
Considering a square loop as shown, E.M.F. across the loop is given by
e=∮E.dl=-dϕdt=-AdBdt (where A = area = π R 2 )
⇒ e = - π R 2 dB dt = - π R 2 α
∴ Magnitude of E.M.F. across one side = π R 2 α 4

Topic: Physics, Electromagnetic Induction, JEE Main
A uniform cable of mass M and length L is placed on a horizontal surface such that its 1nth part is hanging below the edge of the surface. To lift the hanging part of the cable upto the surface, the work done should be:

Options:
A MgL2n2
B MgLn2
C nMgL
D 2MgLn2

The correct answer is:
MgL2n2

Solution:
Mass of the part hanging is Mn. Centre of mass of this part lies L2n distance from the surface.
∴ Work done in lifting the hanging part to surface is
W=Mn·gL2n
=MgL2n2

Topic: Physics, Work Power Energy, JEE Main, JEE Main 2019 (09 Apr Shift 1)
A uniform chain of 6 m length is placed on a table such that a part of its length is hanging over the edge of the table. The system is at rest. The co-efficient of static friction between the chain and the surface of the table is 0.5, the maximum length of the chain hanging from the table is _____ m.

Options:

The correct answer is:
2

Solution:
Let, λ=mass per unit length of the chain and length of mass hanging through the table is x.

Normal force applied by the table on the chain will be, N=l-xλg.

Tension developed in the chain will balance the gravitational force acting in the downward direction, T=λxg.

Maximum limiting force possible for the part of the table, fmax=μN=μl-xλg

For equilibrium,

fmax=T,

⇒μl-xλg=λxg⇒12×6-x=x ⇒x=2 m

Topic: Physics, Laws of Motion, JEE Main, JEE Main 2022 (25 Jun Shift 1)
A uniform chain of length $2 \mathrm{~m}$ is kept on a table such that a length of $60 \mathrm{~cm}$ hangs freely from the edge of the table. The total mass of the chain is $4 \mathrm{~kg}$. What is the work done in pulling the entire chain on the table?

Options:
A
$7.2 \mathrm{~J}$

B
$3.6 \mathrm{~J}$

C
$120 \mathrm{~J}$

D
$1200 \mathrm{~J}$

The correct answer is:
$3.6 \mathrm{~J}$

Solution:
Work done $=\mathrm{mgh}=1.2 \times 0.3 \times 10=3.6 \mathrm{~J}$

Topic: Physics, Work Power Energy, JEE Main, JEE Main 2004
A uniform chain of length 3 m and mass 3 kg overhangs a smooth table with 2 m laying on the table. If K is the kinetic energy of the chain in J as it completely slips off the table, then the value of K is
Take g=10 m s-2

Options:

The correct answer is:
40

Solution:
Apply conservation of energy Ui+Ki=Uf+Kf

−1×10×12+0=−3×10×1.5+KE

K.E.=45−5

K.E.=40 J

Topic: Physics, Work Power Energy, JEE Main, JEE Main 2021 (26 Aug Shift 1)
A uniform chain of length L and mass M is lying on a smooth table and one-third of its length is hanging vertically down over the edge of the table. If g is the acceleration due to gravity, the work required to pull the hanging part on to the table is

Options:
A MgL
B MgL3
C MgL9
D MgL18

The correct answer is:
MgL18

Solution:
Mass of hanging portion is M3 h =L6 below the table top.
Therefore, the required work done is,
W=mgh =M3(g)L6=MgL18

Topic: Physics, Work Power Energy, JEE Main
A uniform chain of length L and mass M overhangs a horizontal table with its two-third part on the table. The friction coefficient between the table and the chain is Î¼. The work done by friction during the period, the chain slips of the table is

Options:
A -29μMgL
B -69μMgL
C -14μMgL
D -49μMgL

The correct answer is:
-29μMgL

Solution:
The linear mass density is ML. The small work done for the slippage on the small distance dl is given by
dW=-μMLgldl
Now, total work done
W=∫02L3-μMgLl dl
= -μMgLl2202L3=-μMg2L4L29-0
⇒ W=-29μMgL

Topic: Physics, Work Power Energy, JEE Main
A uniform chain of length $L$ is lying on the horizontal table. If the coefficient of friction between the chain and the table top is $\mu$, what is the maximum length of the chain that can hang over the edge of the table without disturbing the rest of the chain on the table?

Options:
A $\frac{L}{(1+\mu)}$
B $\frac{\mu L}{(1+\mu)}$
C $\frac{L}{(1-\mu)}$
D $\frac{\mu L}{(1-\mu)}$

The correct answer is:
$\frac{\mu L}{(1+\mu)}$

Solution:
Let $l^{\prime}$ part of the chain is hanging over the edge of table without sliding.
$\therefore \quad \mu=\frac{\text { Length hanging over the edge }}{\text { Length lying on the table }}$
(As the chain have uniform linear density)
$\begin{array}{ll}
\therefore & \mu=\frac{l^{\prime}}{\left(L-l^{\prime}\right)} \\
\Rightarrow & l^{\prime}=\frac{\mu L}{(1+\mu)}
\end{array}$

Topic: Physics, Laws of Motion, JEE Main
A uniform chain of length $L$ is lying partly on a table, the remaining part hanging down from the edge of the table. If the coefficient of friction between the chain and the table is $0.5$, what is the minimum length of the chain that should lie on the table, to prevent the chain from slipping down to the ground?

Options:
A $L / 3$
B $L / 2$
C $2 L / 3$
D $3 L / 4$

The correct answer is:
$2 L / 3$

Solution:
Let length of chain lying the table be $y$.
Mass per unit length of the chain $=\frac{M}{L}$
Mass of the chain lying on the table, $M^{\prime}=\frac{M}{L} \cdot y$
Mass of the length $(L-y)$ of hanging chain $=\frac{M}{L}(L-y)$
At equilibrium,
Friction force between table and chain $=$ Weight of hanging part of the chain
$\mu\left(\frac{M}{L} y\right) g=\frac{M}{L}(L-y) g$
$\Rightarrow \quad \mu y=L-y$
$\Rightarrow \quad 0.5 y=L-y \quad[\because \mu=0.5]$
$\Rightarrow \quad y=\frac{2 L}{3}$

Topic: Physics, Laws of Motion, COMEDK, COMEDK 2015
A uniform circular disc has radius \(r\). A square portion of diagonal \(r\) is cut from it. The centre of mass of the remaining disc from the centre of disc is

Options:
A \(\frac{r}{2-4 \pi}\)
B \(\frac{r}{3-3 \pi}\)
C \(\frac{r}{2-5 \pi}\)
D \(\frac{2 r}{1-2 \pi}\)

The correct answer is:
\(\frac{r}{2-4 \pi}\)

Solution:
Suppose, the mass of circular disc is \(M\). A square portion of diagonal \(r\) is cut from it. Assuming centre of mass of circle to be origin. Centre of mass of square portion will be at distance \(\frac{r}{2}\) from it.
\(\therefore\) Side of square \(=\frac{r}{\sqrt{2}}\)
Area of square \(=\left(\frac{r}{\sqrt{2}}\right)^2=\frac{r^2}{2}\)
Mass of square, \(m=M \times \frac{r^2 / 2}{\pi r^2}\)
\(m=\frac{M}{2 \pi}\)
Centre of a mass of remaining disc from the centre of disc,
\(\begin{aligned}
x & =M \times 0-\frac{m \times r / 2}{M-m}=\frac{-\frac{M}{2 \pi} \times \frac{r}{2}}{M-\frac{M}{2 \pi}}=\frac{-r}{4 \pi-2} \\
\Rightarrow \quad x & =\frac{r}{2-4 \pi}
\end{aligned}\)

Topic: Physics, Center of Mass Momentum and Collision, AP EAMCET, AP EAMCET 2020 (21 Sep Shift 2)
A uniform circular disc has radius R and mass m. A particle, also of mass m, is fixed at a point A on the edge of the disc as shown in the diagram. The disc can rotate freely about a fixed horizontal chord PQ that is at a distance R4 from the centre C of the disc. The line AC is perpendicular to PQ.
Initially, the disc is held vertical with point A at its highest position. It is then allowed to fall so that it starts rotating about PQ. Find the linear speed of the particle as it reaches its lowest position.

Options:
A - 2gR
B 3gR
C - 4gR
D 5gR

The correct answer is:
5gR

Solution:
As moment of inertia of a disc about a diameter is 1 2 1 2 mR 2 , the moment of inertia of the disc about the chord PQ by 'theorem of parallel axes' will be
I D PQ = 1 4 mR 2 + m 1 4 R 2 = 5 16 mR 2
and as particle of mass m is at a distance [R + (R/4) = (5/4)R] from PQ, the moment of inertia of the system about PQ
I = I D PQ + I P PQ = 5 16 mR 2 + m 5 4 R 2 = 15 8 mR 2
Now if ω is the angular speed of the system when A reaches the lowest point A' on rotation about the axis PQ, by 'conservation of mechanical energy',
1 2 I ω 2 = mg AA ′ + mgCC ′ = mg 2AD + 2CD
i.e., 1 2 × 15 8 mR 2 ω 2 = 2 mg R + 1 4 R + 1 4 R ,
i.e., ω = 4 g 5R
so v A ′ = r ω = R + 1 4 R × 4 g 5R = 5gR

Topic: Physics, Rotational Motion, JEE Main
A uniform circular disc of radius 50 cm at rest is free to rotate about an axis which is perpendicular to its plane and passes through its centre. It is subjected to a torque which produces a constant angular acceleration of 2.0 rad s-2. Its net acceleration in m s-2 at the end of 2.0 s is approximately:

Options:
A 8.0
B 7.0
C 6.0
D 3.0

The correct answer is:
8.0

Solution:
At the end of 2ââsec, w=w0+Î±t
w=0+22=4 rad/sec
Particle acceleration towards the center is =ac=rw2
ac=1242=8 m/s 2
Tangential acceleration a t =âÎ±âR
=ââ1ââ m/s 2
a net = a c 2 + a t 2
= 8 2 + 1 2
= 65 â8 (approximately)

Topic: Physics, Rotational Motion, JEE Main
A uniform circular disc of radius r placed on a rough horizontal plane has initial velocity v0 and angular velocity Ï0 as shown. The disc comes to rest after moving some distance in the direction of motion. Then

Options:
A The friction force acts in the forward direction
B The point of contact of disc with ground has zero velocity
C v0 must be equal to rÏ02 in magnitude
D v0 must be equal to 2rω0 in magnitude

The correct answer is:
v0 must be equal to rÏ02 in magnitude

Solution:
Conserving the angular momentum of the disc about its bottom-most point, we get

12mR2ω0-mv0R=0 (Since the disc finally comes to rest, it means its total angular momentum is zero)

v0=rω02

Topic: Physics, Rotational Motion, JEE Main
A uniform circular loop of radius a and resistance R is pulled at a constant velocity v out of a region of a uniform magnetic field whose magnitude is B. The plane of the loop and the velocity are both perpendicular to B. Then the electrical power in the circular loop at the instant when the arc (of the circular loop) outside the region of magnetic field subtends an angle Ï3 at the centre of the loop is

Options:
A B2a2v2R
B 2B2a2v2R
C B2a2v22R
D none of these

The correct answer is:
B2a2v2R

Solution:
The induced emf in the circular loop =BPQv=Bav
Therefore electrical power consumed by the circular loop
=emf2R=Bav2R

Topic: Physics, Electromagnetic Induction, JEE Main
A uniform conducting wire of length is 24a, and resistance R is wound up as a current carrying coil in the shape of an equilateral triangle of side a and then in the form of a square of side a. The coil is connected to a voltage source V0. The ratio of magnetic moment of the coils in case of equilateral triangle to that for square is 1:y where y is ____________.

Options:

The correct answer is:
3

Solution:
4an1=24a

thus n1=6 turns

3an2=24a⇒n2=8 turns

M1M2=n1iA1n2iA2=6×a28×a2×34=31

M2M1=13=1y

y=3

Topic: Physics, Magnetic Effects of Current, JEE Main, JEE Main 2021 (27 Aug Shift 1)
A uniform conductor of resistance $R$ is cut into 20 equal pieces. Half of them are joined in series and the remaining half of them are connected in parallel. If the two combinations are joined in series, the effective resistance of all the pieces is

Options:
A $R$
B $\frac{R}{2}$
C $\frac{101 R}{200}$
D $\frac{201 R}{200}$

The correct answer is:
$\frac{101 R}{200}$

Solution:
Resistance of each piece $=\frac{R}{20} \Omega$
Equivalent resistance of 10 such pieces connected in series
$$
R_1=10 \times \frac{R}{20}=\frac{R}{2}
$$
Similarly, equivalent resistance of 10 such pieces connected in parallel
$$
\begin{aligned}
\frac{1}{R_2} & =\frac{1}{R / 20}+\ldots 10 \text { times } \\
& =\frac{10}{R / 20}=\frac{200}{R} \\
R_2 & =\frac{R}{200}
\end{aligned}
$$
Now, $R_1$ and $R_2$ are connected in series so total resistance $=R_1+R_2$
$$
\begin{aligned}
& =\frac{R}{2}+\frac{R}{200} \\
& =\frac{100 R+R}{200} \\
& =\frac{101 R}{200}
\end{aligned}
$$

Topic: Physics, Current Electricity, AP EAMCET, AP EAMCET 2002
A uniform conductor of resistance $R$ is cut into 20 equal pieces. Half of them are joined in series and the remaining half of them are connected in parallel. If the two combinations are joined in series, the effective resistance of all the pieces is

Options:
A $R$
B $\frac{R}{2}$
C $\frac{101 R}{200}$
D $\frac{201 R}{200}$

The correct answer is:
$\frac{101 R}{200}$

Solution:
Resistance of each piece $=\frac{R}{20} \Omega$
Equivalent resistance of 10 such pieces connected in series
$$
R_1=10 \times \frac{R}{20}=\frac{R}{2}
$$
Similarly, equivalent resistance of 10 such pieces connected in parallel
$$
\begin{aligned}
\frac{1}{R_2} & =\frac{1}{R / 20}+\ldots 10 \text { times } \\
& =\frac{10}{R / 20}=\frac{200}{R} \\
R_2 & =\frac{R}{200}
\end{aligned}
$$
Now, $R_1$ and $R_2$ are connected in series so total resistance $=R_1+R_2$
$$
\begin{aligned}
& =\frac{R}{2}+\frac{R}{200} \\
& =\frac{100 R+R}{200} \\
& =\frac{101 R}{200}
\end{aligned}
$$

Topic: Physics, Current Electricity, TS EAMCET, TS EAMCET 2002
A uniform copper rod of 50 cm length is insulated on the sides and has its ends exposed to ice and steam respectively. If there is a layer of water 1 mm thick at each end, the temperature gradient in Â°C m-1 in the bar is (assume that the thermal conductivity of copper is 400 W m-1 K-1 and water is 0.4 W m-1 K-1 )

Options:

The correct answer is:
40

Solution:
Given, Kc=400 W m-1 K-1,
Kw=0.4 W m-1 K-1
θ1=100 °C, θ2=0 °C
In steady-state, the flow of heat will be the same throughout the whole system
KwA100-θ10-3=KcAθ-θ′50100
Or 100-θ=KcKw×10-3×θ-θ′0.5
=4000.4×10-3×θ-θ′0.5
100-θ=2θ-θ′
(100−θ) K w A 10 −3 = ( θ ′ −0) K w A 10 −3
100−θ= θ ′
Or 100-θ=4θ-200
⇒ 5θ=300⇒θ=60 °C
∴ θ′=100-θ=100-60=40 °C
Hence, the temperature gradient along the bar
=θ-θ′0.5=60-400.5= 40 °C m-1

Topic: Physics, Thermal Properties of Matter, JEE Main
A uniform cubical box of side a is placed on a rough floor and it is to be moved by applying minimum possible force F at a point which is at a distance b above its centre of mass. If the coefficient of friction is μ=0.4, then what is the maximum possible value of 100×ba for which the box will not topple before moving?

Options:

The correct answer is:
50.00

Solution:
For no toppling

Fa2+bâ¤mga2
âÎ¼a2+Î¼bâ¤a2
â0.2a+0.4bâ¤0.5a
â 0.4bâ¤0.3a
â bâ¤3a4

â bâ¤0.75a

(in limiting case )

But it is not possible as maximum value of b can be equal to 0.5a only

â´ 100bamax=50.00

Topic: Physics, Rotational Motion, JEE Main
A uniform current carrying ring of mass m and radius R is connected by a massless string as shown in diagram. A uniform magnetic field B0 exists in the region to keep the ring in horizontal position, then the current in the ring is ( lâ¡ length of string)

Options:
A mg π RB 0
B mg RB 0
C mg 3 π RB 0
D mg â π R 2 B 0

The correct answer is:
mg π RB 0

Solution:
Torque due to magnetic field τ mag = MB 0 = I π R 2 B 0 ...(i)
Torque due to weight about the point where string is connected τ weight = mgR ...(ii)
If ring remains horizontal, then τ mag = τ weight
I π R 2 B 0 = mgR ⇒ I = mg π RB 0

Topic: Physics, Magnetic Effects of Current, NEET
A uniform current is flowing along the length dinfinite, straight. thin. hollow cylinder dradius $R$ The magnetic field $B$ produced at perpendicular distance $d$ from the axis of the cylinder is plotied in a graph. Which of the following figures looks like the plot?

Options:
A
B
C
D

The correct answer is:

Solution:
$\because$ Magnetic field due to a hollow cylinder of radius $r$

$$
\begin{aligned}
B_{\text {inside }} &=0(\text { for } d < r) \\
B_{\text {outside }} &=\frac{\mu_{0}}{2 \pi} \cdot \frac{i}{d}(\text { for } d \geq r)
\end{aligned}
$$
So, graph between $B$ and $d$.

Topic: Physics, Magnetic Effects of Current, JEE Main
A uniform cylinder of length L and mass M having cross-sectional area A is suspended, with its length vertical, from a fixed point by a massless spring, such that it is half submerged in a liquid of density σ at equilibrium position. The extension x 0 of the spring when it is in equilibrium is :

Options:
A Mg k 1 - LA σ 2 M
B Mg k 1 + LA σ M
C Mg k
D Mg k 1 - LA σ M

The correct answer is:
Mg k 1 - LA σ 2 M

Solution:
kx 0 + F B = mg
kx 0 + σ L 2 Ag = Mg
x 0 = Mg - σ LAg 2 k
= Mg k 1 - σ LA 2 M

Topic: Physics, Laws of Motion, JEE Main, JEE Main 2013 (07 Apr)
A uniform cylinder of length $\mathrm{L}$ and mass $M$ having cross-sectional area $\mathrm{A}$ is suspended, with its length vertical, from a fixed point by a massless spring, such that it is half submerged in a liquid of density $\sigma$ at equilibrium position. When the cylinder is given a downward push and released, it starts oscillating vertically with a small amplitude. The time period $\mathrm{T}$ of the oscillations of the cylinder will be :

Options:
A
Smaller than $2 \pi\left[\frac{M}{(k+A \sigma g)}\right]^{1 / 2}$

B
$2 \pi \sqrt{\frac{M}{k}}$

C
Larger than $2 \pi\left[\frac{M}{(k+A \sigma g)}\right]^{1 / 2}$

D
$2 \pi\left[\frac{M}{(k+A \sigma g)}\right]^{1 / 2}$

The correct answer is:
Smaller than $2 \pi\left[\frac{M}{(k+A \sigma g)}\right]^{1 / 2}$

Solution:
No solution is given.

Topic: Physics, Oscillations, JEE Main, JEE Main 2013 (25 Apr Online)
A uniform cylinder of length L and mass M having cross-sectional area A is suspended, with its length vertical from a fixed point by a massless spring such that it is half-submerged in a liquid of density Ï at the equilibrium position. When the cylinder is given a small downward push and released it starts oscillating vertically with a small amplitude. If the force constant of the spring is k, the frequency of oscillation of the cylinder is

Options:
A 12Ï(k-AÏgM)1/2
B 12πk+AρgM1/2
C 12π(k+ρgL2M)1/2
D 12Ï(k+AÏgAÏg)1/2

The correct answer is:
12πk+AρgM1/2

Solution:
When cylinder is displaced by an amount x from its meanposition, spring force and upthrust both will increase. Hence,
Net restoring force = extra spring force + extra upthrust
Or F = - kx + Ax Ï g

Or a = - k + Ï Ag M x
Now, f = 1 2 Ï a x
= 1 2 Ï k + Ï Ag M

Topic: Physics, Oscillations, JEE Main