datavorous/entrance-exam-dataset huggingface dataset:
A spherically symmetric charge distribution is characterised by a charge density having the following variations:
$\rho(r)=\rho_o\left(1-\frac{r}{R}\right)$ for $r < R$
$\rho(\mathrm{r})=0$ for $r \geq \mathrm{R}$
Where $r$ is the distance from the centre of the charge distribution $\rho_{\mathrm{o}}$ is a constant. The electric field at an internal point $(r < R)$ is:

Options:
A
$\frac{\rho_{\mathrm{o}}}{4 \varepsilon_{\mathrm{o}}}\left(\frac{\mathrm{r}}{3}-\frac{\mathrm{r}^2}{4 \mathrm{R}}\right)$

B
$\frac{\rho_0}{\varepsilon_0}\left(\frac{r}{3}-\frac{r^2}{4 R}\right)$

C
$\frac{\rho_{\mathrm{o}}}{3 \varepsilon_{\mathrm{o}}}\left(\frac{\mathrm{r}}{3}-\frac{\mathrm{r}^2}{4 \mathrm{R}}\right)$

D
$\frac{\rho_{\mathrm{o}}}{12 \varepsilon_{\mathrm{o}}}\left(\frac{\mathrm{r}}{3}-\frac{\mathrm{r}^2}{4 \mathrm{R}}\right)$

The correct answer is:
$\frac{\rho_0}{\varepsilon_0}\left(\frac{r}{3}-\frac{r^2}{4 R}\right)$

Solution:
Let us consider a spherical shell of radius $x$ and thickness $\mathrm{dx}$.
Charge on this shell
$$
d q=\rho .4 \pi x^2 d x=\rho_0\left(1-\frac{x}{R}\right) \cdot 4 \pi x^2 d x
$$
$\therefore$ Total charge in the spherical region from centre to $r(r < R)$ is
$$
\begin{aligned}
&q=\int d q=4 \pi \rho_0 \int_0^r\left(1-\frac{x}{R}\right) x^2 d x \\
&=4 \pi \rho_0\left[\frac{x^3}{3}-\frac{x^4}{4 R}\right]_0^r \\
&=4 \pi \rho_0\left[\frac{r^3}{3}-\frac{r^4}{4 R}\right] \\
&=4 \pi \rho_0 r^3\left[\frac{1}{3}-\frac{r}{4 R}\right]
\end{aligned}
$$
$\therefore$ Electric field at $\mathrm{r}, \mathrm{E}=\frac{1}{4 \pi \varepsilon_0} \cdot \frac{\mathrm{q}}{\mathrm{r}^2}$
$$
\begin{aligned}
&=\frac{1}{4 \pi \varepsilon_0} \cdot \frac{4 \pi \rho_0 r^3}{r^2}\left[\frac{1}{3}-\frac{r}{4 R}\right] \\
&=\frac{\rho_0}{\varepsilon_0}\left[\frac{r}{3}-\frac{r^2}{4 R}\right]
\end{aligned}
$$

Topic: Physics, Electrostatics, JEE Main, JEE Main 2014 (12 Apr Online)
A sping balance is adjusted at zero. Elastic collisions are brought about by dropping particles of one gram each on the pan of the balance. They recoil upwards without change is their speeds. If the height of fall of particles is \( 2 \) meter and the rate of particle dropping is \( 100 \) per seconds, then the reading of the balance is -

Options:
A \( 128 \mathrm{gm} \mathrm{wt} \).
B \( 1252 \mathrm{gm} \mathrm{wt} \).
C \( 625 \mathrm{gm} \mathrm{wt} \).
D \( 1.25 \mathrm{gm} \mathrm{wt} \).

The correct answer is:
\( 128 \mathrm{gm} \mathrm{wt} \).

Solution:
Velocity of the particle after it has fallen 2m,
v= 2gh = 2×9.8×2  m/s
When a particles rebound from the pan, change in its momentum =2 mv
If n particles fall per second, then force exerted
=n.2mv
=100×2×1× 10 −13 × 4×9.8N
= 100×2×1× 10 −3 × 4×9.8  gm wt 9.8× 10 −3
=128 gm wt

Topic: Physics, Center of Mass Momentum and Collision, JEE Main
A spool of mass M=3 kg and radius R=20 cm has an axle of radius r=10 cm around which a string is wrapped. The moment of inertia about an axis perpendicular to the plane of the spool and passing through the centre is MR22. If the coefficient of friction between the surface and the spool is 0.4 then the maximum tension which can be applied to the string for which the spool doesn't slip, is g=10 m s-2

Options:
A 24 N
B 18 N
C 9 N
D 36 N

The correct answer is:
18 N

Solution:
Writing the torque about the instantaneous axis of rotation we get,

τ=T×R-r=MR22+MR2α

⇒T=3MR22R-raR

Also,

T-μMg=Ma

Substituting the values and solving for T we get,

T=18 N

Topic: Physics, Rotational Motion, JEE Main
A spool of mass M and radius 2R lies on an inclined plane as shown in figure. A light thread is wound around the connecting tube of the spool and its free end carries a weight of mass m. The value of m so that system is in equilibrium is

Options:
A 2Msinα
B Msinα
C 2Mtanα
D Mcosα

The correct answer is:
2Msinα

Solution:
Equilibrium of m give

T=mg (T = Tension in string)
Net torque about point of contact of the spool should be
zero. Hence,

(2R)(MgsinÎ±) = TR
or 2MgsinÎ± = mg
or m = 2MsinÎ±

Topic: Physics, Rotational Motion, JEE Main
A spring $40 \mathrm{~mm}$ long is stretched by the application of a force. If $10 \mathrm{~N}$ force required to stretch the spring through 1 $\mathrm{mm}$, then work done in stretching the spring through $40 \mathrm{~mm}$ is

Options:
A 84 J
B 68 J
C 23 J
D 8 J

The correct answer is:
8 J

Solution:
Here $k=\frac{F}{x}=\frac{10}{1 \times 10^{-3}}=10^4 \mathrm{~N} / \mathrm{m}$
$W=\frac{1}{2} k x^2=\frac{1}{2} \times 10^4 \times\left(40 \times 10^{-3}\right)^2=8 J$

Topic: Physics, Work Power Energy, JEE Main
A spring balance is attached to the ceiling of a lift. A man hangs his bag on the spring and the spring balance reads $49 \mathrm{~N}$, when the lift is stationary. If the lift moves downward with an acceleration of $5 \mathrm{~m} / \mathrm{s}^2$, the reading of the spring balance will be

Options:
A 74 N
B 15 N
C 24 N
D 49 N

The correct answer is:
24 N

Solution:
$\begin{aligned} & \mathrm{W}=49 \mathrm{~N} \\ & \therefore \mathrm{m}=\frac{\mathrm{W}}{\mathrm{g}}=\frac{49}{9.8}=5 \mathrm{~kg} \\ & \mathrm{~W}^{\prime}=\mathrm{W}-\mathrm{ma}=49-5 \times 5=49-25 \\ & =24 \mathrm{~N}\end{aligned}$

Topic: Physics, Laws of Motion, JEE Main
A spring balance is attached to the ceiling of a lift. A man hangs his bag on the spring and the spring reads 49 $\mathrm{N}$, when the lift is stationary. If the lift moves downward with an acceleration of $5 \mathrm{~m} / \mathrm{s}^2$, the reading of the spring balance will be

Options:
A
$24 \mathrm{~N}$

B
$74 \mathrm{~N}$

C
$15 \mathrm{~N}$

D
$49 \mathrm{~N}$

The correct answer is:
$24 \mathrm{~N}$

Solution:
Mass $=\frac{49}{9.8}=5 \mathrm{~kg}$. When lift is moving downward, apparent weight $=5(9.8-5)=5 \times 4.8=24 \mathrm{~N}$

Topic: Physics, Laws of Motion, JEE Main, JEE Main 2003
A spring balance is attached to the ceiling of a lift. A man hangs his bag on the spring and the spring reads $49 \mathrm{~N}$, when the lift is stationary. If the lift moves downward with an acceleration of $5 \mathrm{~m} / \mathrm{s}^2$, the reading of the spring balance will be

Options:
A $49 \mathrm{~N}$
B $24 \mathrm{~N}$
C $74 \mathrm{~N}$
D $15 \mathrm{~N}$

The correct answer is:
$24 \mathrm{~N}$

Solution:
When the lift is stationary $W=m g$
$\Rightarrow 49=m \times 9.8 \Rightarrow m=5 \mathrm{~kg} \text {. }$
When the lift is moving downward with an acceleration $R=m(9.8-a)=5[9.8-5]=24 \mathrm{~N}$

Topic: Physics, Laws of Motion, JEE Main
A spring is compressed between two blocks of masses $m_1$ and $m_2$ placed on a horizontal frictionless surface as shown in the figure. When the blocks are released, they have initial velocity of $v_1$ and $v_2$ as shown. The blocks travel distances $x_1$ and $x_2$ respectively before coming to rest. The ratio $\left(\frac{x_1}{x_2}\right)$ is

Options:
A
$\frac{m_2}{m_1}$

B
$\frac{m_1}{m_2}$

C
$\sqrt{\frac{m_2}{m_1}}$

D
$\sqrt{\frac{m_1}{m_2}}$

The correct answer is:
$\frac{m_2}{m_1}$

Solution:
No Solution Available

Topic: Physics, Center of Mass Momentum and Collision, JEE Main, JEE Main 2012 (12 May Online)
A spring mass system (mass m, spring constant k and natural length l ) rests in equilibrium on a horizontal disc. The free end of the spring is fixed at the centre of the disc. If the disc together with spring mass system rotates about it's axis with an angular velocity ω,k>>mω2 the relative change in the length of the spring is best given by the option:

Options:
A 23mω2k
B 2mω2k
C mω2k
D mω23k

The correct answer is:
mω2k

Solution:
mω2l0+x=kx
l0x+1=kmω2
x=l0mω2k-mω2
k≫mω2
So, xl0 is equal to mω2k

Topic: Physics, Motion In Two Dimensions, JEE Main, JEE Main 2020 (09 Jan Shift 2)
A spring of force constant 200 N m-1 has a block of mass 10 kg hanging at its one end and the other end of the spring is attached to the ceiling of an elevator. The elevator is rising upwards with an acceleration of g4 and the block is in equilibrium with respect to the elevator. When the acceleration of the elevator suddenly ceases, the block starts oscillating. What is the amplitude (in m) of these oscillations?

Options:

The correct answer is:
0.125

Solution:
When elevator is moving up, using NLM,
T−mg=mg4⇒T=5mg4
This tension elongates the spring by x,
T=kx⇒  x=5mg4k
The equilibrium position of the block is,
x0=mgk
So the amplitude is,
A=x−x0=5mg4k−mgk=mg4=0.125  m

Topic: Physics, Oscillations, JEE Main
A spring of force constant $800 \mathrm{~N} / \mathrm{m}$ has an extension of $5 \mathrm{~cm}$. The work done is extending it from $5 \mathrm{~cm}$ to $15 \mathrm{~cm}$ is

Options:
A
$16 \mathrm{~J}$
option 1 goes here

B
$8 \mathrm{~J}$

C
$32 \mathrm{~J}$

D
$24 \mathrm{~J}$

The correct answer is:
$8 \mathrm{~J}$

Solution:
$\mathrm{W}=\int_{x_1}^{x_2} F d x=\int_{x_1}^{x_2} K x d x=K\left[\frac{x^2}{2}\right]_{x_1}^{x_2}=\frac{K}{2}\left[x_2^2-x_1^2\right]=\frac{800}{2}\left[(0.15)^2-(0.05)^2\right]=8 \mathrm{~J}$

Topic: Physics, Work Power Energy, JEE Main
A spring of force constant $800 \mathrm{~N} / \mathrm{m}$ has an extension of $5 \mathrm{~cm}$. The work done is extending it from $5 \mathrm{~cm}$ to $15 \mathrm{~cm}$ is

Options:
A
$16 \mathrm{~J}$
option 1 goes here

B
$8 \mathrm{~J}$

C
$32 \mathrm{~J}$

D
$24 \mathrm{~J}$

The correct answer is:
$8 \mathrm{~J}$

Solution:
$\mathrm{W}=\int_{x_1}^{x_2} F d x=\int_{x_1}^{x_2} K x d x=K\left[\frac{x^2}{2}\right]_{x_1}^{x_2}=\frac{K}{2}\left[x_2^2-x_1^2\right]=\frac{800}{2}\left[(0.15)^2-(0.05)^2\right]=8 \mathrm{~J}$

Topic: Physics, Work Power Energy, JEE Main, JEE Main 2002
A spring of force constant $\mathrm{k}$ is cut into two equal halves. The force constant of each half is

Options:
A $\frac{\mathrm{k}}{\sqrt{2}}$
B $\mathrm{k}$
C $\frac{\mathrm{k}}{2}$
D $2 \mathrm{k}$

The correct answer is:
$2 \mathrm{k}$

Solution:
As $\quad \mathrm{K} \ell=$ constant
$K^{\prime}=2 K$

Topic: Physics, Laws of Motion, JEE Main
A spring of spring constant $5 \times 10^3 \mathrm{~N} / \mathrm{m}$ is stretched initially by $5 \mathrm{~cm}$ from the unstretched position. Then the work required to stretch it further by another $5 \mathrm{~cm}$ is

Options:
A
$12.50 \mathrm{~N}-\mathrm{m}$

B
$18.75 \mathrm{~N}-\mathrm{m}$

C
$25.00 \mathrm{~N}-\mathrm{m}$

D
$6.25 \mathrm{~N}-\mathrm{m}$

The correct answer is:
$18.75 \mathrm{~N}-\mathrm{m}$

Solution:
Required work done
$=\frac{1}{2} \mathrm{~K}\left(\mathrm{x}_2^2-\mathrm{x}_1^2\right)=\frac{1}{2} \times 5 \times 10^3\left[10^2-5^2\right] \times 10^{-4}$
$=\frac{1}{2} \times 5 \times 75 \times 10^3 \times 10^{-4}=18.75$

Topic: Physics, Work Power Energy, JEE Main, JEE Main 2003
A spring of unstretched length 1 has a mass $m$ with one end fixed to a rigid support. Assuming spring to be made of a uniform wire, the kinetic energy possessed by it if its free end is pulled with uniform velocity $v$ is:

Options:
A
$\frac{1}{2} m v^2$

B
$\mathrm{mv}^2$

C
$\frac{1}{3} \mathrm{mv}^2$

D
$\frac{1}{6} m v^2$

The correct answer is:
$\frac{1}{6} m v^2$

Solution:
No Solution Available

Topic: Physics, Work Power Energy, JEE Main, JEE Main 2014 (12 Apr Online)
A spring whose unstrentches length is l has a force constant k. The spring is cut into two pieces of unstretches lengths l1 and l2 where, l1=nl2 and n is an integer. The ratio k1/k2 of the corresponding force constants, k1 and k2 will be:

Options:
A 1n2
B n2
C n
D 1n

The correct answer is:
1n

Solution:
Spring constants of the spring will be inversely proportional to length.
k1k2=l2l1=1n

Topic: Physics, Laws of Motion, JEE Main, JEE Main 2019 (12 Apr Shift 2)
A spring with spring constant $k$ is extended from $x=0_{\mathrm{x}=0 \text { to } \mathrm{x}=\mathrm{x}_1} x=x_1$. The work done will be

Options:
A $k x_1^2$
B $\frac{1}{2} k x_1^2$
C $2 k x_1^2$
D $2 k x_1$

The correct answer is:
$\frac{1}{2} k x_1^2$

Solution:
we know that
work done = change in Energy
$\mathrm{W}=\Delta \mathrm{E}$
and energy of spring is defined as $\frac{1}{2} k x^2$ where $x$ is stretch in spring.
$\therefore \mathrm{W}=\frac{1}{2} \mathrm{kx}_1^2$

Topic: Physics, Work Power Energy, JEE Main
A square coil ACDE with its plane vertical is released from rest in a horizontal uniform magnetic field B→ which extends over a length 2L as shown in the figure. The acceleration of the coil is

Options:
A less than g for all the time till the loop crosses the magnetic field completely
B less than g when it enters the field and greater than g when it comes out of the field
C equal to g all the time
D less than g when it enters and comes out of the field but equal to g when it is within the field

The correct answer is:
less than g when it enters and comes out of the field but equal to g when it is within the field

Solution:
There will be three situations,
(i) When the coil is entering into the field, flux increases and from Lenz's law the induced emf or current opposes the change. Therefore a<g .
(ii) When the coil is completely within the field, flux remains constant and hence no emf is induced. Therefore a=g .
(iii) When the coil is coming out from the field, flux decreases and from Lenz's law the induced emf or current opposes the change. Therefore a<g .

Topic: Physics, Electromagnetic Induction, NEET
A square coil of edge I having n turns carries a current i. It is kept on a smooth horizontal plate. A uniform magnetic field B exists in a direction parallel to an edge. The total mass of the coil is m. The minimum value of B for which the coil will start tipping over

Options:
A mgnil2
B 2mgnil
C mg2 nil
D mgnil '

The correct answer is:
mg2 nil

Solution:
Deflecting torque =ni​ABsin90
=nil2B
Restoring torque =mgl2
For tipping nil2B≥mgl2
⇒B≥mg2nil
∴Bmin=mg2nil

Topic: Physics, Magnetic Effects of Current, JEE Main
A square coil of side 25 cm having 1000 turns is rotated with a uniform speed in a magnetic field about an axis perpendicular to the direction of the field. At an instant t, the EMF induced in the coil is E =200sin100Ït. The magnetic induction is

Options:
A 0.02 T
B 10-3Â T
C 0.1 T
D 0.01 T

The correct answer is:
0.01 T

Solution:
e=200sin⁡100 πt
We have,
e0=200
ω=100 π
∴BANω=e0
∴B=e0ANω
B=200(0.25×0.25)×1000×100π
Or B=0.01 T

Topic: Physics, Electromagnetic Induction, JEE Main
A square conducting loop is placed near an infinitely long current carrying wire with one edge parallel to the wire as shown in the figure. If the current in the straight wire is suddenly halved. which of the following statements will be true? "The loop will .......".

Options:
A stay stationary
B move towards the wire
C move away from the wire
D move parallel to the wire

The correct answer is:
move away from the wire

Solution:
As we know that, magnetic field due to a long wire at a distance $x$ from it is given by $B=\frac{\mu_{0} I}{2 \pi x},$ where, $I$ is the current flowing through the wire $\beta \propto I$
Magnetic flux associated with the square loop,
$$
\phi \propto \beta \propto I
$$
Now, if the current increase, then $\phi$ also increases. Direction of long wire will be $\otimes$.
This means, magnetic field due to induced current will be opposite to the existing magnetic field, i.e. according to Lenz's law, The induced current in the loop will be in the anti-clockwise direction. Now,

since, wires attract each other, if current flowing through them is in same direction and repel each other, if currents are in opposite direction.
$\therefore$ Part $C D$ of the loop will experience a force of repulsion, whereas part $A B$ will experience attraction. Parts $B C$ and $A D$ will not experience any force. Thus, the overall force will be a force of repulsion because $A B$ is closer to straight. The force between two current carrying conductors is inversely proportional to the distance between them
$\because$
$$
\begin{array}{l}
F \propto \frac{1}{r} \\
r_{1} < r_{2}
\end{array}
$$
So,
$$
F_{C D}>F_{A B}
$$
$$
F_{\text {net }}=F_{C D}-F_{A B}
$$
Hence, the loop will moves away from the wire.

Topic: Physics, Magnetic Effects of Current, JEE Main
A square current carrying loop is suspended in a uniform magnetic field acting in the plane of the loop. If the force on one arm of the loop is $\vec{F}$, the net force on the remaining three arms of the loop is

Options:
A $3 \overrightarrow{\mathrm{F}}$
B $-\overrightarrow{\mathrm{F}}$
C $-3 \vec{F}$
D $\vec{F}$

The correct answer is:
$-\overrightarrow{\mathrm{F}}$

Solution:
When a current carrying loop is placed in a magnetic field, the coil experiences a torque given by $\tau=\mathrm{NBiA} \sin \theta$. Torque is maximum when $\theta=90^{\circ}$, i. e., the plane of the coil is parallel to the field $\tau_{\max }=\mathrm{NBiA}$
Forces $\overrightarrow{\mathrm{F}}_1$ and $\overrightarrow{\mathrm{F}}_2$ acting on the coil are equal in magnitude and opposite in direction. As the forces $\vec{F}_1$ and $\vec{F}_2$ have the same line of action, their resultant effect on the coil is zero.
The two forces $\overrightarrow{\mathrm{F}}_3$ and $\overrightarrow{\mathrm{F}}_4$ are equal in magnitude and opposite in direction. As the two forces have different lines of action, they constitute a torque. Thus, if the force on one arc of the loop is $\overrightarrow{\mathrm{F}}$, the net force on the remaining three arms of the loop is $-F$.

Topic: Physics, Magnetic Effects of Current, NEET, NEET 2010 (Screening)
A square frame ABCD is made of insulated wires and there is a short dipole, with dipole moment p, fixed in the plane of the figure. The dipole is lying at the centre of the square, making an angle Î¸, as shown in the figure. If four point charges are placed at the four corners of the square, then the magnitude of force exerted by the dipole on the system of charges is

Options:
A 6Kpqr3
B 4Kpqr3
C 2Kpqr3
D 8Kpqr3

The correct answer is:
6Kpqr3

Solution:
Fy=3Kpqsin⁡θr3×2

Fx=3Kpqr3cos⁡θ×2

F=Fx2+Fy2=6Kpqr3

Topic: Physics, Electrostatics, JEE Main
A square frame of side 10 cm and a long straight wire carrying current 1 A are in the plane of the paper. Starting from close to the wire, the frame moves towards the right with a constant speed of 10 m s-1 (see figure). The e.m.f induced at the time the left arm of the frame is at x=10 cm from the wire is

Options:
A 0.75 μV
B 1 μV
C 2 μV
D 0.5 μV

The correct answer is:
1 μV

Solution:
dϕ=μ0i2πyady
ϕ=μ0ia2π∫xx+adyy=μ0ia2πlnx+a-lnx
E.m.f=-dϕdt=-μ0ia2π1x+adxdt-1xdxdt
=μ0ia2πa.vxx+a=μ02π.ia2vxx+a=2×10-7×1×0.12×100.1×0.2=1μV
Alternative method:
Consider two sides which are perpendicular to the velocity as case of motional e.m.f.

E=E1-E2=B1–B2V l
=μ0I2π1x-1x+aV l
=2×10-7×110.1-10.2×10×0.1
=10-6 V=1 μV

Topic: Physics, Electromagnetic Induction, JEE Main, JEE Main 2014 (19 Apr Online)
A square hole of side length $\ell$ is made at a depth of $\mathrm{h}$ and a circular hole of radius $\mathrm{r}$ is made at a depth of $4 \mathrm{~h}$ from the surface of water in a water tank kept on a horizontal surface. If $\ell< < h, r< < h$ and the rate of water flow from the holes is the same, then $r$ is equal to

Options:
A
$\frac{\ell}{\sqrt{2 \pi}}$

B
$\frac{\ell}{\sqrt{3 \pi}}$

C
$\frac{\ell}{3 \pi}$

D
$\frac{\ell}{2 \pi}$

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

Solution:
As $A_1 v_1=A_2 v_2$ (Principle of continuity) or, $\ell^2 \sqrt{2 g h}=\pi r^2 \sqrt{2 g \times 4 h}$
(Efflux velocity $=\sqrt{2 \mathrm{gh}})$
$\therefore \quad r^2=\frac{\ell^2}{2 \pi} \quad$ or $\quad r=\sqrt{\frac{\ell^2}{2 \pi}}=\frac{\ell}{\sqrt{2 \pi}}$

Topic: Physics, Mechanical Properties of Fluids, JEE Main, JEE Main 2012 (07 May Online)
A square is inscribed in the circle $x^{2}+y^{2}-6 x+8 y-103=0$ with its sides parallel to the coordinate axes. Then the distance of the vertex of this square which is nearest to the origin is:

Options:
A 6
B $\sqrt{137}$
C $\sqrt{41}$
D 13

The correct answer is:
$\sqrt{41}$

Solution:
The equation of circle is, $x^{2}+y^{2}-6 x+8 y-103=0$

$\Rightarrow(x-3)^{2}+(y+4)^{2}=(8 \sqrt{2})^{2}$

$C(3,-4), r=8 \sqrt{2}$

$\Rightarrow$ Length of side of square $=\sqrt{2} r=16$

$\Rightarrow P(-5,4), Q(-5,-12)$

$R(11,-12), S(11,4)$

$\Rightarrow$ Required distance $=O P$

$=\sqrt{(-5-0)^{2}+(-4-0)^{2}}=\sqrt{25+16}=\sqrt{41}$

Topic: Mathematics, Circle, JEE Main, JEE Main 2019 (11 Jan Shift 1)
A square lead slab of side \( 50 \mathrm{~cm} \) and thickness \( 10 \mathrm{~cm} \) is subjected to a shearing force (on its narrow face) of \( 9 \times 10^{4} \mathrm{~N} \). The lower edge is riveted to the floor. How much will the upper edge be displaced? (Shear modulus of lead = \( 5.6 \times 10^{9} \mathrm{~N} \mathrm{~m}^{-2} \) )

Options:
A \( 0.16 \mathrm{~mm} \)
B \( 1.6 \mathrm{~mm} \)
C \( 0.16 \mathrm{~cm} \)
D \( 1.6 \mathrm{~cm} \)

The correct answer is:
\( 0.16 \mathrm{~mm} \)

Solution:
The lead slab is fixed and force is applied parallel to the narrow face as shown in the figure. Area of the face parallel to which this force is applied is
A = 50 cm×10 cm=0.5 m×0.1 m=0.05 m2
If Δ L is the displacement of the upper edge of the slab due to tangential force F, then
η=FAΔLL or ΔL=FlηA
Substituting the given values, we get
ΔL=9×104 N(0.5 m)5.6×109 N m2×0.05 m2
=1.6×10–4 m=0.16 mm

Topic: Physics, Mechanical Properties of Solids, JEE Main
A square loop, carrying a steady current $I$, is placed in a horizontal plane near a long straight conductor carrying a steady current $I_1$ at a distance $d$ from the conductor as shown in figure. The loop will experience

Options:
A a net repulsive force away from the conductor
B a net torque acting upward perpendicular to the horizontal plane
C a net torque acting downward normal to the horizontal plane
D a net attractive force towards the conductor

The correct answer is:
a net attractive force towards the conductor

Solution:
$$
\begin{aligned}
& \mathbf{F}_2=-\mathbf{F}_4 \\
& \mathbf{F}_1=\frac{\mu_0 I_1 I l}{2 \pi d} \\
& \mathbf{F}_2=\frac{\mu_0 I_1 I l}{2 \pi(d+l)} \\
& \mathbf{F}_1>\mathbf{F}_3 \\
& F_{\text {net }}=F_1-F_3
\end{aligned}
$$
So, wire atrract loop.

Topic: Physics, Magnetic Effects of Current, NEET, NEET 2011 (Mains)
A square loop EFGH of side a, mass m and total resistance R is falling under gravity in a region of transverse non-uniform magnetic field given by B=B0ya, where B0 is a positive constant and y is the position of side EF of the loop. If at some instant the speed of the loop is v, then the total Lorentz force acting on the loop is

Options:
A F=B02a2v2R
B F=2B02a2vR
C F=B02a2vR
D zero

The correct answer is:
F=B02a2vR

Solution:
Motional emf in EH and FG = 0 as v → || I →

Motional emf in EF is e1=B0yaav⁡=B0yv⁡

Similarly, motional emf in GH will be

e2=B0y+aaav⁡=B0y+av⁡

Polarities of e1and e2 are shown in adjoining figures.

So the net emf is

e=e2-e1e = B0av
i = e R = B 0 a v⁡ R

F EF = B 0 a v⁡ R a B 0 y a (downwards)

F GH = B 0 a v⁡ R a B 0 y + a a

FGH=B02av⁡Ry+a (upwards)

Net Lorentz force on the loop
Fnet=FGH-FEF=B02a2v⁡R
F=B02a2vR

Topic: Physics, Electromagnetic Induction, JEE Main
A square loop is carrying a steady current I and the magnitude of its magnetic dipole moment is m . If this square loop is changed to a circular loop and it carries the same current, the magnitude of the magnetic dipole moment of circular loop will be:

Options:
A 4mπ
B 3mπ
C 2mπ
D mπ

The correct answer is:
4mπ

Solution:
Let the side of square loop be a.
The magnetic dipole is∴m=Ia2
a=mI
Radius of circular loop r related to a as circumference of circular loop is equal to perimeter of square loop.
2πr=4a=4mI
r=2πmI
∴ New dipole moment m'=π2πmI2I
=4mπ

Topic: Physics, Magnetic Effects of Current, JEE Main, JEE Main 2019 (10 Apr Shift 2)
A square loop of area 25 cm2 has a resistance of 10 Ω. The loop is placed in uniform magnetic field of magnitude 40.0 T. The plane of loop is perpendicular to the magnetic field. The work done in pulling the loop out of the magnetic field slowly and uniformly in 1.0 sec, will be

Options:
A 2.5×10-3 J
B 1.0×10-3 J
C 1.0×10-4 J
D 5×10-3 J

The correct answer is:
1.0×10-3 J

Solution:
As l2=25 cm2, then l=5 cm=0.05 m.

Given: t=1 s

Velocity of the square loop, v=0.051=0.05 m s-1.

Induced current,

i=VR=BlvR=40×0.05×0.0510=0.01 A

Now force acting on the side of the square loop, F=Bil=40×0.01×0.05

⇒F=0.02 N

Therefore, work done

W=Fl=0.02×l=0.02×0.05

⇒W=1×10-3 J

Topic: Physics, Magnetic Effects of Current, JEE Main, JEE Main 2023 (29 Jan Shift 2)