datavorous/entrance-exam-dataset huggingface dataset:
A rectangle is inscribed in a circle with a diameter lying along the line 3y=x+7. If the two adjacent vertices of the rectangle are -8, 5 and 6, 5, then the area of the rectangle (in sq. units) is:

Options:
A 72
B 98
C 56
D 84

The correct answer is:
84

Solution:
Given, the two adjacent vertices of the rectangle are -8, 5 and 6, 5, the y-co-ordinates of both of these points is same this means they lie on the line y=5, which is parallel to the x-axis.

And, we know that the sides of a rectangle are perpendicular to each other, hence the other two sides passing through these vertices are parallel to y-axis.

Also, we know that the equation of a line parallel y-axis is x=k.

Therefore, the equation of the sides of the rectangle passing through -8, 5 and 6, 5 are respectively x=-8 and x=6.

And, hence the other two vertices of the rectangle can be taken as 6, a and -8, b.

And since the other side is parallel to x-axis, hence a=b.

Here, diagonal subtends right angle at the circumference of circle, so its mid-point is the same as centre of circle, which lies on the given diameter.

The mid-point of a line segment joining the points x1, y1 and x2, y2 is x1+x22, y1+y22, hence, the mid-point of the diagonal of the rectangle is -8+62, a+52=-1, a+52

This point lie on the diameter 3y=x+7

So, 3a+52=-1+7
⇒a+5=263

⇒a=-1

Hence, the vertices are -8, 5, 6, 5, 6, -1 and -8, -1.

The distance between the points x1, y1 and x2, y2 is x1-x22+y1-y22

Thus, the length of sides are 6 and 14 units.

Hence, area =l×b=14×6=84 sq. units

Topic: Mathematics, Circle, JEE Main, JEE Main 2019 (09 Apr Shift 2)
A rectangle with its sides parallel to the X-axis and Y-axis is inscribed in the region bounded by the curves y=x2-4 and 2y=4-x2. The maximum possible area of such a rectangle is closest to the integer.

Options:
A 10
B 9
C 8
D 7

The correct answer is:
9

Solution:
Equation of parabola

y=x2-4 and 2y=4-x2

Let point B=h,4-h22

A=-h,4-h22

C=h1h2-4

D=-h,h2-4

Area of rectangle

ABCD=AB×BC=2h×4-h22-h2+4

⇒A=12h-3h3⇒dAdh=12-9h2

For maxima or minima

dAdh=0⇒12-9h2=0

h=±23 maximum at h=23

∴A=1223-3233=243-83

=163=1633

A-16×1733=9.22

A=9.22=9

Topic: Mathematics, Application of Derivatives, JEE Main
A rectangle with sides 2m-1 and 2n-1 is divided into squares of unit length. The number of rectangles that can be formed with both the sides of odd length is

Options:
A m2n2
B mnm+1(n+1)
C 4m+n-1
D None of these

The correct answer is:
m2n2

Solution:
Number of rectangles possible is
= (Number of ways to choose a horizontal side of odd length).(Number of ways to choose a vertical side of odd length)
1 + 3 + 5 + ..... + 2 m - 1 1 + 3 + 5 + ..... + 2 n - 1 = m 2 n 2

Topic: Mathematics, Permutation Combination, JEE Main
A rectangular, a square, a circular and an elliptical loop, all in the $(x-y)$ plane, are moving out of a uniform magnetic field with a constant velocity, $\vec{V}=v \cdot \hat{i}$. The magnetic field is directed along the negative z-axis direction. The induced emf, during the passage of these loops, come out of the field region, will not remain constant for :

Options:
A any of the four loops
B The rectangular, circular and elliptical loops
C The circular and the elliptical loops
D Only the elliptical loop

The correct answer is:
The circular and the elliptical loops

Solution:
Out of the four structures, when the circular and elliptical loops come out from the field, equal are is not traced in equal interval of time. So any induced in both is not constant.

Topic: Physics, Electromagnetic Induction, NEET, NEET 2009 (Mains)
A rectangular, a square, a circular and an elliptical loop, all in the $(x-y)$ plane, are moving out of a uniform magnetic field with a constant velocity, $\vec{v}=$ vi. The magnetic field is directed along the negative $z$-axis direction. The induced emf, during the passage of these loops, out of the field region, will not remain constant for

Options:
A the rectangular, circular and elliptical loops
B the circular and the elliptical loops
C only the elliptical loop
D any of the four loops

The correct answer is:
the circular and the elliptical loops

Solution:
Area coming out per second from the magnetic field is not constant for elliptical and circular loops, so induced emf, during the passage of these loops, out of the field resion will not remain constant for the circular and the elliptical loops.

Topic: Physics, Electromagnetic Induction, JEE Main
A rectangular, a square, a circular and an elliptical loop, all in the $(x-y)$ plane, are moving out of a uniform magnetic field with a constant velocity, $\vec{v}=$ vi. The magnetic field is directed along the negative $z$-axis direction. The induced emf, during the passage of these loops, out of the field region, will not remain constant for

Options:
A the rectangular, circular and elliptical loops
B the circular and the elliptical loops
C only the elliptical loop
D any of the four loops

The correct answer is:
the circular and the elliptical loops

Solution:
Area coming out per second from the magnetic field is not constant for elliptical and circular loops, so induced emf, during the passage of these loops, out of the field resion will not remain constant for the circular and the elliptical loops.

Topic: Physics, Electromagnetic Induction, NEET, NEET 2009 (Screening)
A rectangular block is $5 \mathrm{~cm} \times 5 \mathrm{~cm} \times 10 \mathrm{~cm}$ in size. The block is floating in water with $5 \mathrm{~cm}$ side vertical. If it floats with $10 \mathrm{~cm}$ side vertical, what change will occur in the level of water?

Options:
A No change
B It will rise
C It will fall
D It may rise or fall depending on the density of block

The correct answer is:
No change

Solution:
Since no change in volume of displaced water takes place, hence level of water remains same.

Topic: Physics, Mechanical Properties of Fluids, JEE Main
A rectangular block of mass m and area of cross-section A floats in a liquid of density ρ. If it is given a vertical displacement from equilibrium, it undergoes oscillation with a time period T. Then

Options:
A T∝ρ
B T∝1A
C T∝1ρ
D T∝1m

The correct answer is:
T∝1A

Solution:
Let block is displaced through x m, then weight of displaced water or upthrust (upwards)

=-Axρg

Where, A is area of a cross-section of the block and $\rho$ is density. This must be equal to force =ma applied where m is mass of the block and a is acceleration.

∴ ma=-Axρg

or a=-Aρgmx=-ω2x

This is equation of SHM.

The time period of oscillation,

T=2πω=2πmAρg

⇒ T∝1A

Topic: Physics, Oscillations, JEE Main
A rectangular block of mass ' $\mathrm{m}$ ' and crosssectional area A, floats on a liquid of density ' $\rho$ '. It is given a small vertical displacement from equilibrium, it starts oscillating with frequency ' $n$ ' equal to ( $g=$ acceleration 'due to gravity)

Options:
A $\frac{1}{2 \pi} \sqrt{\frac{\mathrm{A} \rho \mathrm{g}}{\mathrm{m}}}$
B $2 \pi \sqrt{\frac{\mathrm{A} \rho \mathrm{g}}{\mathrm{m}}}$
C $\frac{1}{2 \pi} \sqrt{\frac{\mathrm{m}}{\mathrm{A} \rho \mathrm{g}}}$
D $2 \pi \sqrt{\frac{\mathrm{m}}{\mathrm{A} \rho \mathrm{g}}}$

The correct answer is:
$\frac{1}{2 \pi} \sqrt{\frac{\mathrm{A} \rho \mathrm{g}}{\mathrm{m}}}$

Solution:
The formula for the time period is given as
$\mathrm{T}=2 \pi \sqrt{\frac{l}{\mathrm{~g}}}$
The mass of displaced fluid is
$\text { Mass }=\text { density } \times \text { volume }$
$\mathrm{m}=\rho \times \mathrm{Al}$
At equilibrium,
Weight of the block = Weight of the displaced liquid
$\begin{aligned}
& \therefore \quad \mathrm{mg}=\mathrm{A} / \mathrm{pg} \\
& \therefore \quad l=\frac{\mathrm{m}}{\mathrm{Ap}}
\end{aligned}$
Substituting the values in the equation
$\begin{aligned}
& \mathrm{T}=2 \pi \sqrt{\frac{l}{\mathrm{~g}}} \\
& \mathrm{~T}=2 \pi \sqrt{\frac{\mathrm{m}}{\mathrm{A} \rho \mathrm{g}}}
\end{aligned}$
The frequency $f=\frac{1}{T}$
$\therefore \quad \mathrm{f}=\frac{1}{2 \pi} \sqrt{\frac{\mathrm{Apg}}{\mathrm{m}}}$

Topic: Physics, Oscillations, JEE Main
A rectangular box lies on a rough inclined surface. The coefficient of friction between the surface and the box is $\mu$. Let the mass of the box be $m$.
(a) At what angle of inclination $\theta$ of the plane to the horizontal will the box just start to slide down the plane?
(b) What is the force acting on the box down the plane, if the angle of inclination of the plane is increased to $\alpha>\theta$ ?
(c) What is the force needed to be applied towards along the plane to make the box either remain stationary or just move up with uniform speed?
(d) What is the force needed to be applied upwards along the plane to make the box move up the plane with acceleration $a$ ?

Options:

None

Solution:
(a) Force of friction on the box will act up the plane.
For the box to just starts sliding down $(m g)$ the plane then
$$
\begin{aligned}
&\sin \theta=f=\mu N=\mu m g \cos \theta \\
&\text { or } \tan \theta=\mu \Rightarrow \theta=\tan ^{-1}(\mu) \\
&\text { or } \tan \theta=\mu \Rightarrow \theta=\tan ^{-1}(\mu)
\end{aligned}
$$
(b) If angle of inclination is increased to $\alpha>\theta$. The angle of inclination of plane with horizontal it will slide down, then net force acting on the box, down the plane is
$$
\begin{aligned}
F_1 &=m g \sin \alpha-f=m g \sin \alpha-\mu N \\
&=m g \sin \alpha-\mu m g \cos \alpha \\
&=m g(\sin \alpha-\mu \cos \alpha) .
\end{aligned}
$$
(c) To keep the box either stationary or just move it up with uniform velocity $(a=0)$ upward force needced.
$$
\begin{aligned}
F_2 &=m g \sin \alpha+f \\
&=m g \sin \alpha+\mu N=m g \sin \alpha+\mu m g \cos \alpha \\
&=m g(\sin \alpha+\mu \cos \alpha)
\end{aligned}
$$
In this case, friction would act down the plane.
(d) The force applied $F_3$ to move the box with an upward acceleration $a$, then upward force needed.
$$
\begin{aligned}
&F_3-m g \sin \alpha-\mu m g \cos \alpha=m a \\
&F_3=m g(\sin \alpha+\mu \cos \alpha)+m a .
\end{aligned}
$$

Topic: Physics, Laws of Motion
A rectangular coil $A B C D$ is rotated anticlockwise with a uniform angular velocity about the axis shown in diagram below. The axis of rotation of the coil as well as the magnetic field $B$ are horizontal. The induced e.m.f. in the coil would be maximum when

Options:
A The plane of the coil is horizontal
B The plane of the coil makes an angle of $45^{\circ}$ with the magnetic field
C The plane of the coil is at right angles to the magnetic field
D The plane of the coil makes an angle of $30^{\circ}$ with the magnetic field

The correct answer is:
The plane of the coil is horizontal

Solution:
Emf $=e=e_0 \sin \theta ;$ e will be maximum when $\theta$ is $90^{\circ}$ i.e. plane of the coil will be horizontal

Topic: Physics, Electromagnetic Induction, JEE Main
A rectangular coil (Dimension 5 cm × 2.5 cm ) with 100 turns, carrying a current of 3 A in the clock-wise direction, is kept centered at the origin and in the X-Z plane. A magnetic field of 1 T is applied along X-axis . If the coil is tilted through 45° about Z-axis , then the torque on the coil is:

Options:
A 0.42 N m
B 0.55 N m
C 0.38 N m
D 0.27 N m

The correct answer is:
0.27 N m

Solution:
Torque= M→×B→
=MBsinθ
=niABsinθ
=100×3×5×10-2×2.5×10-2×1×sin45o
=0.27 N m

Topic: Physics, Magnetic Effects of Current, JEE Main, JEE Main 2019 (09 Apr Shift 1)
A rectangular conducting loop carrying a current i1 is situated near a long straight wire carrying a steady current i2. The wire is parallel to one of the sides of the loop and is placed in the same plane as that of the loop as shown in the figure. Then, the current loop will

Options:
A move away from the wire
B move towards the wire
C remain stationary
D rotate about an axis parallel to the wire

The correct answer is:
move towards the wire

Solution:
F=μ02πi1i2dl
where d is distance between conductors

∴ F∝1d

Hence, lesser the distance, more is the force In the given circuit force of attraction is more between EF and AD that of repulsion between EF and BC, hence current loop will move towards the wire.

F=μ02πi1i2dl

∴ F∝1d

Topic: Physics, Magnetic Effects of Current, NEET
A rectangular film of liquid is extended from (4 cmÃ2 cm) to (5 cmÃ4 cm). If the work done is 3Ã10-4 J, the value of the surface tension of the liquid is

Options:
A 0.250 N m-1
B 0.125 N m-1
C 0.2 N m-1
D 8.0 N m-1

The correct answer is:
0.125 N m-1

Solution:
W=2Af-AiT
âT=WAf-AiÃ2
=3Ã10-4 J25Ã4Ã10-4-4Ã2Ã10-4
=0.125 N m-1

Topic: Physics, Mechanical Properties of Fluids, JEE Main
A rectangular frame ABCD made of a uniform metal wire has a straight connection between E & F made of the same wire as shown in the figure. AEFD is a square of side 1 m & EB=FC=0.5 m. The entire circuit is placed in a steadily increasing uniform magnetic field directed into the plane of the paper. The rate of change of the magnetic field is 1 T s-1, the resistance per unit length of the wire is 1 Ω m-1. If the currents in the segments AE and EF are IAE and IEF respectively, then what is the value of IAEIEF?

Options:

The correct answer is:
7

Solution:
e1=AdBdt=1×1×1=1 V
e2 = 0.5 V
From loop AEFDA

â1I1-I2-1+3I1=0⇒4I1-I2=1 ...(1)
From loop FCBEF
0.5-2I2+I1-I2=0⇒-I1+3I2=0.5 ...(2)
From (1) and (2)
I1=722 A,I2=311 A,I1-I2=122 A
IAE=722 A, IBE=311 A , IEF=122 A

IAEIEF=7

Topic: Physics, Electromagnetic Induction, JEE Main
A rectangular glass slab ABCD, of refractive index, n 1 , is immersed in water of refractive index n 2 ( n 1 > n 2 ). A ray of light is incident at the surface AB of the slab, as shown. The maximum value of the angle of incidence Î± max , such that the ray comes out only from the other surface CD, is given by

Options:
A

sinâ1[n1n2cosâ(sinâ1ân2n1)]

B

sinâ1[n1cosââ(sinâ11n2)]

C

sinâ1(n1n2)

D

sinâ1(n2n1)

The correct answer is:

sinâ1[n1n2cosâ(sinâ1ân2n1)]

Solution:
Consider the figure. The ray will come out from CD if it suffers a total internal reflection at surface AD, i.e., it strikes the surface AD at critical angle C (the limiting case).

Applying Snellâs law at P, we get

n 1 ââsinC= n 2 ââorââsinââC= n 2 n 1

Applying snellâs law at Q, we get

n 2 ââsinÎ±= n 1 cosC;â

ââââââsinÎ±= n 1 n 2 cos{ sin â1 ( n 2 n 1 ) }

or Î±ââ=ââ sin â1 â[ n 1 n 2 âcosâ{ sin â1 â( n 2 n 1 ) } ]

Topic: Physics, Ray Optics, JEE Main
A rectangular loop carrying a current $i_1$, is situated near a long straight wire carrying a steady current $i_2$. The wire is parallel to one of the sides of the loop and is in the plane of the loop as shown in the
figure. Then the current loop will

Options:
A move away from the wire
B move towards the wire
C remain stationary
D rotate about an axis parallel to the wire.

The correct answer is:
move towards the wire

Solution:
As two like currents attract and unlike currents repel each other, the wire will attract the side $A$ of the loop near the wire and parallel to it and the far side of the loop again parallel to the wire will be repelled. But the force of attraction will be more as that side of the loop is closer to the wire and so effectively the current loop will more towards the wire.

Topic: Physics, Magnetic Effects of Current, AIIMS, AIIMS 2003
A rectangular loop carrying a current i is situated near a long straight wire such that the wire is parallel to one of the sides of the loop and is in the plane of the loop. If the steady current i is established in the wire as shown in the figure, the loop will

Options:
A Rotate about an axis parallel to the wire
B Move away from the wire
C Move towards the wire
D Remain stationary

The correct answer is:
Move towards the wire

Solution:
Straight wire will produce a non-uniform field to the right of it. F â bc and F â ad will be calculated but integration of these two forces will cancel each other. Further force on wire ab will be towards the long wire and on wire cd will be away from the long wire. But since the wire ab is nearer to the long wire, force of attraction towards the long wire will be more. Hence, the loop will move towards the wire.

Topic: Physics, Magnetic Effects of Current, NEET
A rectangular loop has a sliding connector PQ of length $\ell$ and resistance $\mathrm{R} \Omega$ and it is moving with a speed $v$ as shown. The set-up is placed in a uniform magnetic field going into the plane of the paper. The three currents $I_1, I_2$ and $I$ are

Options:
A
$\mathrm{I}_1=-\mathrm{I}_2=\frac{\mathrm{B} \ell \mathrm{v}}{\mathrm{R}}, \mathrm{I}=\frac{2 \mathrm{~B} \ell \mathrm{v}}{\mathrm{R}}$

B
$\mathrm{I}_1=\mathrm{I}_2=\frac{\mathrm{B} \ell \mathrm{v}}{3 \mathrm{R}}, \mathrm{I}=\frac{2 \mathrm{~B} \ell \mathrm{v}}{3 \mathrm{R}}$

C
$l_1=I_2=I=\frac{B \ell v}{R}$

D
$I_1=I_2=\frac{B \ell v}{6 R}, I=\frac{B \ell v}{3 R}$

The correct answer is:
$\mathrm{I}_1=\mathrm{I}_2=\frac{\mathrm{B} \ell \mathrm{v}}{3 \mathrm{R}}, \mathrm{I}=\frac{2 \mathrm{~B} \ell \mathrm{v}}{3 \mathrm{R}}$

Solution:
A moving conductor is equivalent to a battery of emf $=\mathrm{v} \mathrm{B} \ell \quad$ (motion emf)
Equivalent circuit
$$
\mathrm{I}=\mathrm{l}_1+\mathrm{l}_2
$$
applying Kirchoff's law
$$
\begin{aligned}
& \mathrm{I}_1 R+I R-v B \ell=0 \\
& \mathrm{I}_2 R+I R-v B \ell=0
\end{aligned}
$$
adding (1) \& (2)
$$
\begin{aligned}
& 2 \mathrm{IR}+\mathrm{IR}=2 \mathrm{vB} \ell \\
& \mathrm{I}=\frac{2 \mathrm{vB} \ell}{3 \mathrm{R}} \\
& \mathrm{I}_1=\mathrm{I}_2=\frac{\mathrm{vB} \ell}{3 \mathrm{R}}
\end{aligned}
$$

Topic: Physics, Electromagnetic Induction, JEE Main
A rectangular loop has a sliding connector PQ of length $\ell$ and resistance $\mathrm{R} \Omega$ and it is moving with a speed $v$ as shown. The set-up is placed in a uniform magnetic field going into the plane of the paper. The three currents $I_1, I_2$ and $I$ are

Options:
A
$\mathrm{I}_1=-\mathrm{I}_2=\frac{\mathrm{B} \ell \mathrm{v}}{\mathrm{R}}, \mathrm{I}=\frac{2 \mathrm{~B} \ell \mathrm{v}}{\mathrm{R}}$

B
$\mathrm{I}_1=\mathrm{I}_2=\frac{\mathrm{B} \ell \mathrm{v}}{3 \mathrm{R}}, \mathrm{I}=\frac{2 \mathrm{~B} \ell \mathrm{v}}{3 \mathrm{R}}$

C
$l_1=I_2=I=\frac{B \ell v}{R}$

D
$I_1=I_2=\frac{B \ell v}{6 R}, I=\frac{B \ell v}{3 R}$

The correct answer is:
$\mathrm{I}_1=\mathrm{I}_2=\frac{\mathrm{B} \ell \mathrm{v}}{3 \mathrm{R}}, \mathrm{I}=\frac{2 \mathrm{~B} \ell \mathrm{v}}{3 \mathrm{R}}$

Solution:
A moving conductor is equivalent to a battery of emf $=\mathrm{v} \mathrm{B} \ell \quad$ (motion emf)
Equivalent circuit
$$
\mathrm{I}=\mathrm{l}_1+\mathrm{l}_2
$$
applying Kirchoff's law
$$
\begin{aligned}
& \mathrm{I}_1 R+I R-v B \ell=0 \\
& \mathrm{I}_2 R+I R-v B \ell=0
\end{aligned}
$$
adding (1) \& (2)
$$
\begin{aligned}
& 2 \mathrm{IR}+\mathrm{IR}=2 \mathrm{vB} \ell \\
& \mathrm{I}=\frac{2 \mathrm{vB} \ell}{3 \mathrm{R}} \\
& \mathrm{I}_1=\mathrm{I}_2=\frac{\mathrm{vB} \ell}{3 \mathrm{R}}
\end{aligned}
$$

Topic: Physics, Electromagnetic Induction, JEE Main, JEE Main 2010
A rectangular loop of length 2.5 m and width 2 m is placed at 60° to a magnetic field of 4 T. The loop is removed from the field in 10 sec. The average emf induced in the loop during this time is

Options:
A -2 V
B +2 V
C +1 V
D -1 V

The correct answer is:
+1 V

Solution:
The formula to calculate the change in the magnetic flux associated with the given loop can be written as

∆ϕ=-BAcosθ=-Blbcosθ ...1

The formula to calculate the induced emf is given by

ε=-∆ϕ∆t=Blbcosθ∆t ...2

From equation (2), it follows that

ε=4×2.5×2×cos60°10 V=1 V

Topic: Physics, Electromagnetic Induction, JEE Main, JEE Main 2024 (27 Jan Shift 1)
A rectangular loop of metallic wire is of length a and breadth b and carries a current i. The magnetic field at the centre of the loop is -

Options:
A μ0 i4π 8a2+b2ab
B μ0 i4π4a2+b2ab
C μ0 i4π2a2+b2ab
D μ0 i4πa2+b2ab

The correct answer is:
μ0 i4π 8a2+b2ab

Solution:
BAB=BCD
=μ0i4π b/2 sin⁡ϕ1+sin⁡ϕ1
=μ0 4π . 4ib aa2+b2
BBC=BDA=μ04π .4 ia .ba2+b2
∴ B=BAB+BBC+BCD+BDA
=μ04π .4 ia2+b2 ab+ba+ab+ba
=μ04π . 8 ia2+b2ab

Topic: Physics, Magnetic Effects of Current, JEE Main
A rectangular loop of sides 10 cm and 5 cm, carrying a current I of 12 A, is placed in different orientations as shown in the figure below.
(a) (b)
(c) (d)

If there is a uniform magnetic field of 0.3 T in the positive z direction, in which orientations the loop would be in (i) stable equilibrium and (ii) unstable equilibrium?

Options:
A (b) and (c), respectively.
B (a) and (b), respectively.
C (a) and (c), respectively.
D (b) and (d), respectively.

The correct answer is:
(b) and (d), respectively.

Solution:
For a magnetic dipole placed in a uniform magnetic field, the torque is given by τ→=M→×B→, and potential energy U is given as,
U=-M→·B→=-MBcosθ
When M→ is in the same direction as B→ then τ→=0 and U is minimum = -MB as θ=0o.⇒ Stable equilibrium is (b).

When M→ the opposite direction of B→, then τ→=0 and U is max = +MB
As, θ=180o ⇒Unstable equilibrium in (d)

Topic: Physics, Magnetic Effects of Current, JEE Main, JEE Main 2015 (04 Apr)
A rectangular loop of wire is placed in the $X Y$-plane with its side of length $3 \mathrm{~cm}$ parallel to the $X$-axis and the side of length $4 \mathrm{~cm}$ parallel to the $Y$-axis. It is moving in the positive $X$-direction with the speed $10 \mathrm{~cm} / \mathrm{s}$. A magnetic field exists in the space with its direction parallel to the $Z$-axis. The field decreases by $2 \times 10^{-3} \mathrm{~T} / \mathrm{cm}$ along the positive $X$-axis and increases in time by $2 \times 10^{-2} \mathrm{~T} / \mathrm{s}$. The induced emf in the wire is

Options:
A $-4.8 \times 10^{-5} \mathrm{~V}$
B $4.8 \times 10^{-5} \mathrm{~V}$
C 0
D $3.6 \times 10^{-5} \mathrm{~V}$

The correct answer is:
0

Solution:
Induced emf in wire = Rate of change of flux
$$
\begin{aligned}
& =\frac{d}{d t}(B A)=A\left\{\left(\frac{d}{d t} B\right)+v \frac{d}{d x} B\right\} \\
& =12 \times 10^{-4}\left(2 \times 10^{-2}+10 \times 10^{-2} \times 2 \times 10^{-3}\right) \\
& =12 \times 10^{-4}\left(2 \times 10^{-2}+2 \times 10^{-4}\right) \\
& =24 \times 10^{-4}\left(10^{-2}+10^{-4}\right)=24 \times 10^{-5} \mathrm{~V} \\
& =0.000024 \mathrm{~V} \approx 0 \mathrm{~V}
\end{aligned}
$$

Topic: Physics, Electromagnetic Induction, JEE Main
A rectangular loop of wire, supporting a mass $m$, hangs with one end in a uniform magnetic field $\vec{B}$ pointing out of the plane of the paper. A clockwise current is set up such that $i>m g / B a$, where $a$ is the width of the loop. Then :

Options:
A
The weight rises due to a vertical force caused by the magnetic field and work is done on the system.

B
The weight do not rise due to vertical for caused by the magnetic field and work is done on the system.

C
The weight rises due to a vertical force caused by the magnetic field but no work is done on the system.

D
The weight rises due to a vertical force caused by the magnetic field and work is extracted from the magnetic field.

The correct answer is:
The weight rises due to a vertical force caused by the magnetic field but no work is done on the system.

Solution:
No Solution Available

Topic: Physics, Magnetic Effects of Current, JEE Main, JEE Main 2013 (23 Apr Online)
A rectangular loop $\mathrm{PQMN}$ with movable arm $\mathrm{PQ}$ of length $12 \mathrm{~cm}$ and resistance $2 \Omega$ is placed in a uniform magnetic field of $0.1 \mathrm{~T}$ acting perpendicular to the plane of the loop as shown in figure. The resistance of the arms MN, NP and MQ are negligible. The current induced in the loop when arm $\mathrm{PQ}$ is moved with velocity $20 \mathrm{~ms}^{-1}$ is

Options:
A $0.12 \mathrm{~A}$
B $0.06 \mathrm{~A}$
C $0.24 \mathrm{~A}$
D $0.18 \mathrm{~A}$

The correct answer is:
$0.12 \mathrm{~A}$

Solution:
$\mathrm{B}=0.1 \mathrm{~T}, \ell=0.12 \mathrm{~m}, \mathrm{v}=20 \mathrm{~m} / \mathrm{s}, \mathrm{R}=2 \Omega$
emf induced $\mathrm{e}=\mathrm{B} \ell \mathrm{v}=0.1 \times 0.12 \times 20=0.24 \mathrm{~V}$
$\mathrm{I}=\frac{\mathrm{V}}{\mathrm{R}}=\frac{0.24}{2}=0.12 \mathrm{~A}$

Topic: Physics, Electromagnetic Induction, JEE Main