datavorous/entrance-exam-dataset huggingface dataset:
A computer producing factory has only two plants T1 and T2. Plant T1 produces 20% and plant T2 produces 80% of the total computers produced. 7% of computers produced in the factory turn out to be defective. It is known that,

P(computer turns out to be defective given that it is produced in plant T1)=10 P(computer turns out to be defective given that it is produced in plant T2).

Where, P(E) denotes the probability of an event E. A computer produced in the factory is randomly selected and it does not turn out to be defective. Then the probability that it is produced in plant T2 is

Options:
A 3673
B 4779
C 7893
D 7583

The correct answer is:
7893

Solution:
PT1=20100
PT2=80100
Let, PDT2=x (where, D represents defective units)
PDT1=10x
PD=7100 (given)
PT1 PDT1+PT2 PDT2=7100
20100×10x+80100×x=7100
x=140
PDT2=140
⇒PD-T2=3940= Probability of not defective, given that it is produced in plant T2
PDT1=1040 ⇒ PD-T1=3040
Now using Bayes' theorem
Probability of computer from T2 given that it is not defective :
PT2D-=80100×394020100×3040+80100×3940=7893

Topic: Mathematics, Probability, JEE Main
A concave mirror for face viewing has a focal length of 0.4 m. The distance at which you hold the mirror from your face in order to see your image upright with a magnification of 5 is

Options:
A 1.60Â m
B 0.16Â m
C 0.32 m
D 0.24Â m

The correct answer is:
0.32 m

Solution:
m=5=-vu ⇒v=-5u
f=-0.4 m
From mirror formula,
1v+1u=1f
1-5u+1u=1-0.4
-1+55u=1-0.4
u=45-0.4=-0.32 m
So, distance of object from mirror=0.32 m

Topic: Physics, Ray Optics, JEE Main, JEE Main 2019 (09 Apr Shift 1)
A concave mirror has radius of curvature of 40 cm. It is at the bottom of a glass that has water filled up to 5 cm (see figure). If a small particle is floating on the surface of water, its image as seen, from directly above the glass, is at a distance d from the surface of water. The value of d is close to: (Refractive index of water =1.33)

Options:
A 13.4Â cm
B 8.8 cm
C 6.7Â cm
D 11.7 cm

The correct answer is:
8.8 cm

Solution:
Given
u=-5 cm
f=-R2=-402=-20 cm
Now
1v+1u=1f
1v=1-20=1-5
∴ v=+203 cm
Then
Object distance d=5+203=353 cm (below water surface)
The image after refraction
d'=dμ2μ1
=35311.33
=8.77
d'≃8.8 cm

Topic: Physics, Ray Optics, JEE Main, JEE Main 2019 (12 Apr Shift 1)
A concave mirror is placed on a horizontal table, with its axis directed vertically upwards. Let O be the pole of the mirror and C it's centre of curvature. A point object is placed at C. It has a real image, also located at C (a condition called auto-collimation). If the mirror is now filled with water, the image will be:

Options:
A real, and will remain at C
B real, and located at a point between C and ∞
C virtual, and located at a point between C and O
D real, and located at a point between C and O

The correct answer is:
real, and located at a point between C and O

Solution:
Due to refraction at water surface object will appear beyond C. This image will act as an object for reflection by mirror. So image will be real, and located between C and O

Topic: Physics, Ray Optics, NEET
A concave mirror is used to focus the image of a flower on a nearby well $120 \mathrm{~cm}$ from the flower. If a lateral magnification of 16 is desired, the distance of the flower from the mirror should be

Options:
A $8 \mathrm{~cm}$
B $12 \mathrm{~cm}$
C $80 \mathrm{~cm}$
D $120 \mathrm{~cm}$

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

Solution:
Let distance $=u$. Now $\frac{v}{u}=16$ and $v=u+120$
$\therefore \frac{120+u}{u}=16 \Rightarrow 15 u=120 \Rightarrow u=8 \mathrm{~cm}$

Topic: Physics, Ray Optics, JEE Main
A concave mirror of focal length 20 cm and a convex lens of focal length 10 cm are kept with their optic axes parallel but separated by 0.5 cm as shown in the figure. The distance between the lens and mirror is 10 cm. An object of height 3 mm is placed on the optic axis of lens at a distance 15 cm from the lens. Find the length of the image formed by the mirror in mm.

Options:

The correct answer is:
3

Solution:
For image formed by the lens
1v1-1-15=1+10
⇒v1=+30 cm

i.e. 20 cm behind mirror
For mirror
1v2+120=1-20
⇒v2=-10 cm

Overall magnification = 30-15×1020=-1
Length of image = 1 × 3 = 3 mm

Topic: Physics, Ray Optics, JEE Main
A concave mirror of focal length $f_1$ is placed at a distance of $d$ from a convex lens of focal length $f_2$. A beam of light coming from infinity and falling on this convex lens concave mirror combination returns to infinity. The distance $d$ must be equal

Options:
A $f_1+f_2$
B $-f_1+f_2$
C $2 f_1+f_2$
D $-2 f_1+f_2$

The correct answer is:
$2 f_1+f_2$

Solution:
$d=2 f_1+f_2$

Topic: Physics, Ray Optics, JEE Main
A concrete sphere of radius R has a cavity of radius r which is packed with sawdust. The relative densities of concrete and sawdust are 2.4 and 0.3 respectively. For this sphere to float with its entire volume submerged under water, the ratio of the mass of concrete to the mass of sawdust will be

Options:
A 2
B 3
C 4
D 6

The correct answer is:
4

Solution:
43 πr3×0.3g+43π R3-r3×2.4 g=43πR3×1×g
or 0.3r3+2.4R3-2.4r3=R3 or 1.4R3=2.1r3
or R3r3=32 or R3-r3r3=12
Now, mass of concretemass of saw-dust=43 π R3-r3 2.443 πr3×0.3=12×8=4

Topic: Physics, Mechanical Properties of Fluids, JEE Main
A concrete sphere of radius $R$ has a cavity of radius $r$ which is packed with sawdust. The specific gravities of concrete and sawdust are respectively 2.4 and 0.3 for this sphere to float with its entire volume submerged under water. Ratio of mass of concrete to mass of sawdust will be

Options:
A 8
B 4
C 3
D Zero

The correct answer is:
4

Solution:
Let specific gravities of concrete and saw dust are $\rho_1$ and $\rho_2$ respectively.
According to principle of floatation weight of whole sphere = upthrust on the sphere
$\frac{4}{3} \pi\left(R^3-r^3\right) \rho_1 g+\frac{4}{3} \pi r^3 \rho_2 g=\frac{4}{3} \pi R^3 \times 1 \times g$
$\Rightarrow R^3 \rho_1-r^3 \rho_1+r^3 \rho_2=R^3$
$\Rightarrow R^3\left(\rho_1-1\right)=r^3\left(\rho_1-\rho_2\right) \Rightarrow \frac{R^3}{r^3}=\frac{\rho_1-\rho_2}{\rho_1-1}$
$\Rightarrow \quad \frac{R^3-r^3}{r^3}=\frac{\rho_1-\rho_2-\rho_1+1}{\rho_1-1}$
$\Rightarrow \quad \frac{\left(R^3-r^3\right) \rho_1}{r^3 \rho_2}=\left(\frac{1-\rho_2}{\rho_1-1}\right) \frac{\rho_1}{\rho_2}$
$\Rightarrow \frac{\text { Massof concrete }}{\text { Massof saw dust }}=\left(\frac{1-0.3}{2.4-1}\right) \times \frac{2.4}{0.3}=4$

Topic: Physics, Mechanical Properties of Fluids, JEE Main
A concrete sphere of the radius R has a cavity of radius r which is packed with sawdust. The specific gravities of concrete and sawdust are respectively 2.4 and 0.3 For this sphere to float with its entire volume submerged underwater, the ratio of the mass of concrete to the mass of sawdust will be

Options:

The correct answer is:
4

Solution:
Let specific gravities of concrete and sawdust are Ï1 and Ï2 respectively.
According to principle of floatation, weight of whole sphere = upthrust on the sphere
4 3 Ï R 3 - r 3 Ï 1 g + 4 3 Ï r 3 Ï 2 g = 4 3 Ï R 3 Ã 1 Ã g
â R 3 Ï 1 - r 3 Ï 1 + r 3 Ï 2 = R 3
â R 3 Ï 1 - 1 = r 3 Ï 1 - Ï 2 â R 3 r 3 = Ï 1 - Ï 2 Ï 1 - 1
â R 3 - r 3 r 3 = Ï 1 - Ï 2 - Ï 1 + 1 Ï 1 - 1 â R 3 - r 3 Ï 1 r 3 Ï 2 = ( 1 - Ï 2 Ï 1 - 1 ) Ï 1 Ï 2
â Mass of concrete Mass of saw dust = ( 1 - 0 . 3 2 . 4 - 1 ) Ã 2 . 4 0 . 3 = 4

Topic: Physics, Mechanical Properties of Fluids, JEE Main
A condenser of capacity $C$ is charged to a potential difference of $V_1$. The plates of the condenser are then connected to an ideal inductor of inductance $L$. The current through the inductor when the potential difference across the condenser reduces to $V_2$ is

Options:
A $\left(\frac{\mathrm{C}\left(\mathrm{V}_1-\mathrm{V}_2\right)^2}{\mathrm{~L}}\right)^{\frac{1}{2}}$
B $\frac{\mathrm{C}\left(\mathrm{V}_1^2-\mathrm{V}_2^2\right)}{\mathrm{L}}$
C $\frac{\mathrm{C}\left(\mathrm{V}_1^2+\mathrm{V}_2^2\right)}{\mathrm{L}}$
D $\left(\frac{\mathrm{C}\left(\mathrm{V}_1^2-\mathrm{V}_2^2\right)}{\mathrm{L}}\right)^{\frac{1}{2}}$

The correct answer is:
$\left(\frac{\mathrm{C}\left(\mathrm{V}_1^2-\mathrm{V}_2^2\right)}{\mathrm{L}}\right)^{\frac{1}{2}}$

Solution:
\(\begin{aligned} & \frac{\mathrm{CV}^2}{2}+\frac{1}{2} \mathrm{Li}^2=\text { constant } \\ & \text { Initially } \mathrm{i}=0, \mathrm{~V}=\mathrm{V}_1 \\ & \text {Total energy }=\frac{1}{2} \mathrm{CV}_1^2 \\ & \mathrm{~When } \mathrm{V} \rightarrow \mathrm{V}_2 \\ & \frac{1}{2} \mathrm{C}\left(\mathrm{V}_2\right)^2+\frac{1}{2} \mathrm{Li}^2=\frac{1}{2} \mathrm{CV}_1^2 \\ & \mathrm{Li}^2=\mathrm{C}\left(\mathrm{V}_1^2-\mathrm{V}_2^2\right)^2 \\ & \mathrm{i}^2=\frac{\mathrm{C}}{\mathrm{L}}\left(\mathrm{V}_1^2-\mathrm{V}_2^2\right) \\ & \mathrm{i}=\sqrt{\frac{\mathrm{C}}{\mathrm{L}}\left(\mathrm{V}_1^2-\mathrm{V}_2^2\right)}\end{aligned}\)

Topic: Physics, Capacitance, NEET, NEET 2010 (Mains)
A condition for a function $y=f(x)$ to have an inverse is that it should be

Options:
A Defined for all $x$
B Continuous everywhere
C Strictly monotonic and continuous in the domain
D An even function

The correct answer is:
Strictly monotonic and continuous in the domain

Solution:
It is a fundamental concept.

Topic: Mathematics, Functions, JEE Main
A conducting balloon of the radius a is charged to potential V0 and held at a large height above the earth ensures that charge distribution on the surface of the balloon remains unaffected by the presence of the earth. It is connected to the earth through a resistance R and a valve in the balloon is opened. The gas inside the balloon escapes from valve and the size of the balloon decreases. The rate of decrease in radius of the balloon is controlled in such a manner that potential of the balloon remains constant. Assume the electric permittivity of the surrounding air equals to that of free space Îµ0 and charge cannot leak to the surrounding air.

The rate at which radius r of the balloon changes with time is best represented by the equation

Options:
A drdt=14πε0R
B drdt= –14πε0R
C drdt=14πε0aR
D drdt=-r4πε0aR

The correct answer is:
drdt= –14πε0R

Solution:
For a sphere V0=q4πε0r, where r is the radius

∴q=V04πε0r
–dqdt=V0R
–V04πε0drdt=V0R
drdt= –14πε0Rââââ

Topic: Physics, Electrostatics, JEE Main
A conducting bar of length L is free to slide on two parallel conducting rails as shown in the figure

Two resistors R1 and R2 are connected across the ends of the rails. There is a uniform magnetic field B→ pointing into the page. An external agent pulls the bar to the left at a constant speed v.

The correct statement about the directions of induced currents I1 and I2 flowing through R1 and R2 respectively is :

Options:
A Both I1 and I2 are in anticlockwise direction
B Both I1 and I2 are in clockwise direction
C I1 is in clockwise direction and I2 is in anticlockwise direction
D I1 is in anticlockwise direction and I2 is in clockwise direction

The correct answer is:
I1 is in clockwise direction and I2 is in anticlockwise direction

Solution:
Given, the external agent pulls the bar towards left. According to Lenz's law and Fleming's right hand rule, there will be a motional EMF generated in the bar. The right-hand rule tells us that the magnetic force on positive charges will be in downward direction and the electric equivalent circuit can be drawn as shown below.

So, I1 is in the clockwise direction and I2 is in the anticlockwise direction.

Topic: Physics, Electromagnetic Induction, JEE Main, JEE Main 2021 (16 Mar Shift 1)
A conducting bubble of radius a and thickness t (t<<a) has potential V. Now, the bubble collapses into a droplet. The potential of the droplet is

Options:
A V'=Va3t13
B V'=Va3t12
C V'=Va3t
D None of these

The correct answer is:
V'=Va3t13

Solution:
Let q be the charge on the bubble, then
V = Kq a (Here K = 1 4 Ï Îµ 0 )
â´ q = Va K
Let after collapsing, the radius of droplet becomes R. Then equating the volume, we have
4 Ï a 2 t = 4 3 Ï R 3
â´ R = 3 a 2 t 1 /3
Now, potential of droplet will be
V â² = Kq R
Substituting the values, we have
V â² = K Va K 3 a 2 t 1 / 3
or V â² = V a 3 t 1 / 3

Topic: Physics, Electrostatics, JEE Main
A conducting circular loop is placed in a uniform magnetic field 0.04 T with its plane perpendicular to the magnetic field. The radius of the loop starts shrinking at 2 mm s-1. The induced emf in the loop, when the radius is 2 cm, is

Options:
A 4.8π μV
B 0.8π μV
C 1.6π μV
D 3.2π μV

The correct answer is:
3.2π μV

Solution:
Ï = BA cos Î¸ = B Ï r 2 cos 0 â = B Ï r 2
Îµ = d Ï dt = B Ï 2 r dr dt
= 0.04 Ã Ï Ã 2 Ã 2 Ã 1 0 - 2 Ã 2 Ã 1 0 - 3
= 3.2 Ï Ã 1 0 - 6 V = 3.2 Ï Î¼ V

Topic: Physics, Electromagnetic Induction, JEE Main
A conducting circular loop is placed in a uniform magnetic field $0.04 \mathrm{~T}$ with its plane perpendicular to the magnetic field. The radius of the loop starts shrinking at $2 \mathrm{~mm} / \mathrm{s}$. The induced emf in the loop when the redius is $2 \mathrm{~cm}$ is :

Options:
A $1.6 \pi \mu v$
B $3.2 \pi \mu v$
C $4.8 \pi \mu v$
D $0.8 \pi \mu v$

The correct answer is:
$3.2 \pi \mu v$

Solution:
$$
\begin{aligned}
\mathrm{e} & =\frac{\mathrm{d} \phi}{\mathrm{dt}}=\frac{\mathrm{d}}{\mathrm{dt}}\left(\mathrm{B} \pi \mathrm{r}^2\right) \\
& =2 \pi \mathrm{rB} \frac{\mathrm{dr}}{\mathrm{dt}} \\
& =2 \times \pi \times 2 \times 10^{-2} \times 4 \times 10^{-2} \times 2 \times 10^{-3} \\
& =3.2 \times 10^{-6} \pi \mathrm{Vol}=3.2 \pi \mu \mathrm{V}
\end{aligned}
$$

Topic: Physics, Electromagnetic Induction, NEET, NEET 2009 (Mains)
A conducting circular loop is placed in a uniform magnetic field of 0.4 T with its plane perpendicular to the field. Somehow, the radius of the loop starts expanding at a constant rate of 1 mm s-1. The magnitude of induced emf in the loop at an instant when the radius of the loop is 2 cm will be _______ μV.

Options:

The correct answer is:
50

Solution:
The data given is

B=0.4 TR=2×10-2 m

It is given that dRdt=10-3 m s-1

Area of the loop is A=πR2

So, dAdt=2πRdRdt

The flux is,

ϕ=2πRdRdt×B

So, the magnitude of emf is

ε=2×227×2×10-2×10-3×0.4

=5.028×10-5 V=50.28 μV≈50 μV

Topic: Physics, Electromagnetic Induction, JEE Main, JEE Main 2023 (12 Apr Shift 1)
A conducting circular loop is placed in X-Y plane in presence of magnetic field B→=3t3j^+3t2k^ in SI unit. If the radius of the loop is 1 m, the induced emf in the loop, at time, t=2 s is nπ V. The value of n is _____ .

Options:

The correct answer is:
12

Solution:
By the faraday's law in electromagnetic induction

εind=-dϕdt=-dB→·A→dt

Magnetic flux is ϕ=B→·A→=3t3j^+3t2k^·π12k^

So, ϕ=3t2π

Now, induced emf is εind=dϕdt=6tπ

At t=2, εind=62π=12π V.

Thus, the value of n=12.

Topic: Physics, Electromagnetic Induction, JEE Main, JEE Main 2022 (27 Jul Shift 2)
A conducting circular loop made of a thin wire has area 3.5×10-2 m2 and resistance 10 Ω It is placed perpendicular to a time-dependent magnetic field Bt=0.4 Tsin⁡50πt The field is uniform in space. Then the net charge flowing through the loop during t=0 s and t=10 ms is close to

Options:
A 1.4 mC
B 7Â mC
C 21Â mC
D 6Â mC

The correct answer is:
1.4 mC

Solution:
It is given that, the magnetic field,

B=0.4 sin50πt

and the area of the given circular conducting loop,

A=3.5×10-2 m2

Magnetic flux,

ϕ=BA ...(1)

Then, induced emf,
ε=dϕdt ...(2)

When resistance in the circuit is R, the differential charge accumulated, based on current i,
dQ=idt=εRdt=1Rdϕ⇒∆Q=1R ∫dφ=1R ∆ϕ

Substituting given values and using equation (1) and (2),
∆Q=1100.4sin50π×10×10-3-0×3.5×10-2⇒∆Q=0.4×3.5×10-210=1.4×10-3 C

⇒∆Q=1.4 mC

Topic: Physics, Electromagnetic Induction, JEE Main, JEE Main 2019 (09 Jan Shift 1)
A conducting circular loop of area 2.5×10-3 m2 and resistance 10 Ω is placed perpendicular to a uniform time varying magnetic field Bt=0.6sin50πt T. What is the net charge (in mC) flowing through the loop during t=0 and t=10 ms ?

Options:

The correct answer is:
1.5

Solution:
Net charge =ÎÏR=AB2-B1R

=2.5Ã10-3(0.6)sin50Ï(10)103-sin0

=1.5Ã10-3sinÏ2-sin0

=1.5Ã10-3C=1.5mC

Topic: Physics, Electromagnetic Induction, JEE Main
A conducting circular loop of radius \( \mathrm{r} \) carries a constant current i. It is placed in a uniform magnetic field \( \overrightarrow{\mathrm{B}_{0}} \) such that \( \overrightarrow{\mathrm{B}_{0}} \) is perpendicular to the plane of the loop. The magnetic force acting on the loop is

Options:
A ir \( \overrightarrow{\mathrm{B}_{0}} \)
B \( 2 \pi \) ir \( \overrightarrow{\mathrm{B}_{0}} \)
C Zero
D \( \pi \) ir \( \overrightarrow{\mathrm{B}_{0}} \)

The correct answer is:
Zero

Solution:
The magnetic field is perpendicular to the plane of the paper. Let us consider two diametrically opposite elements on the loop. Applying Fleming’s left-hand rule on element AB, the direction of force will be leftwards and the magnitude will be
dF=Idl Bsin90o=IdlB
On element CD, the direction of force will be rightwards on the plane of the paper and the magnitude will be dF=IdlB.

Hence, the net force on the loop due to current elements will be zero, which implies that the total force on the loop will be zero.

Topic: Physics, Magnetic Properties of Matter, JEE Main
A conducting loop (as shown) has total resistance R. A uniform magnetic field B=γt is applied perpendicular to plane of the loop where γ is a constant and t is time. The induced current flowing through loop is

Options:
A b2+a2γtR
B b2-a2γR
C b2-a2γtR
D b2+a2γR

The correct answer is:
b2-a2γR

Solution:
Given,
Uniform magnetic field B=Î³t
Total flux, B1â. A1â+B2â.A2â
=Bb2cosâ¡0o+Ba2cosâ¡180o
=Bb2-Ba2
Ï=B(b2-a2)
Ï=Î³t(b2-a2)
We know that,
Induced current (i) =eR=dÏdtR
From Equation (i),
i=ddt[Î³tb2-a2]R
i=b2-a2Î³R

Topic: Physics, Electromagnetic Induction, JEE Main
A conducting loop of radius 10π cm is placed perpendicular to a uniform magnetic field of 0.5 T. The magnetic field is decreased to zero in 0.5 s at a steady rate. The induced emf in the circular loop at 0.25 s is:

Options:
A emf=1mV
B emf=10mV
C emf=100mV
D emf=5mV

The correct answer is:
emf=10mV

Solution:
Given: B=0.5 T at t=0 & B=0, at t=0.5 s.

Magnetic field will decrease linearly as given below. We can say, value of magnetic field at t=0.25 will be B=0.25.

Therefore,

εind =ΔϕΔt=ΔBAΔt=AΔBΔt

=π×10π2×10-4×0.250.25

=10-2×1

=0.01 V=10 mV

Topic: Physics, Electromagnetic Induction, JEE Main, JEE Main 2023 (24 Jan Shift 1)
A conducting loop which is in the shape of a right-angled isosceles triangle of height 10 cm is kept such that the 90Â° vertex is very close to an infinitely long conducting wire (see the figure). The wire is electrically insulated from the loop. The hypotenuse of the triangle is parallel to the wire. The current in the triangular loop is in a counterclockwise direction and increased at a constant rate of 10 A s-1. The emf induced in the wire is

Options:
A μ0π
B 2μ0π
C 2μ0π
D μ0π

The correct answer is:
μ0π

Solution:
by direction of induced electric field, current in wire is in same direction of current along the hypotenuse.
Flux through triangle if wire have current i = â«00.1Î¼0i2Ïx 2xdx=Î¼0i10Ï
â Mutual inductance =Î¼010Ï
Induced emf in wire =Î¼010Ïdidt=Î¼010ÏÃ10=Î¼0Ï

Topic: Physics, Electromagnetic Induction, JEE Main
A conducting metal circular-wire-loop of radius r is placed perpendicular to a magnetic field which varies with time as B=B0e-tτ, where B0 and τ are constants at time t=0. If the resistance of the loop is R, then the heat generated in the loop after a long time t→∞ is

Options:
A π2r4B042τR
B π2r4B022τR
C Ï2r4B02RÏ
D Ï2r4B02ÏR

The correct answer is:
π2r4B022τR

Solution:
Here, B=B0e-tτ
Area of the circular loop, A=πr2
Flux linked with the loop at any time t ,
ϕ=BA=πr2B0e-tτ
Emf induced in the loop, ε=-dϕdt=πr2B01τe-tτ
Net heat generated in the loop
∫0∞ε2Rdt=π2r4B02τ2R∫0∞e-2tτdt
=π2r4B02τ2R×1-2τ×e-2tτ0∞
=-π2r4B022τ2R×τ0-1=π2r4B022τR

Topic: Physics, Magnetic Effects of Current, JEE Main, JEE Main 2016 (10 Apr Online)
A conducting ring of circular cross-section with inner and outer radii a and b is made out of a material of resistivity Ï. The thickness of the ring is h. It is placed coaxially in a vertical cylindrical region of a magnetic field B=krt, where k is a positive constant, r is the distance from the axis and t is the time. If the current through the ring is I=khÎ±p b3-a3, then what is the value of Î±?

Options:

The correct answer is:
9

Solution:
The flux through the area of radius r is
ϕ=∫0rkrt2πrdr=2πktr33
E.M.F. induced,
ϵ=dϕdt=2πkr33
Net induced current is
I=∫ab2πr3khdr3ρ2πr=kh9ρ b3-a3
∴ α=9

Topic: Physics, Electromagnetic Induction, JEE Main
A conducting ring of radius r and resistance R is placed in a region of uniform time varying magnetic field B which is perpendicular to the plane of the ring. If the magnetic field is changing at a rate Î±, then the current induced in the ring is

Options:
A πr2α2R
B 2πrαR
C πrαR
D πr2αR

The correct answer is:
πr2αR

Solution:
The magnitude of the induced emf in the ring is
ε=dϕdt=ddtBA=AdBdt=πr2α
The induced current in the ring, I=εR=πr2αR

Topic: Physics, Electromagnetic Induction, JEE Main
A conducting rod of length $l$ is falling with a constant velocity $v$ perpendicular to a uniform horizontal magnetic field $B$. The potential difference between its two ends will be

Options:
A $2 B l v$
B $B l v$
C $\frac{1}{2} B l v$
D $B^2 l^2 v^2$

The correct answer is:
$B l v$

Solution:
If \(e\) is the induced emf, then according to Faraday's law \(e=-\frac{d \phi}{d t}=-\mathrm{Blv}\)
So, the potential difference between two ends of conducting rods will be \(B v l\).

Topic: Physics, Electromagnetic Induction, JEE Main
A conducting rod of length $L$ lies in $X Y$-plane and makes an angle $30^{\circ}$ with $X$-axis. One end of the rod lies at origin initially. A magnetic field also exists in the region pointing along positive $Z$-direction. The magnitude of the magnetic field varies with $y$ as $B_0\left(\frac{y}{L}\right)^3$, where, $B_0$ is a constant. At some instant the rod starts moving with a velocity $v_0$ along $X$-axis. The emf induced in the rod is

Options:
A $\frac{B_0 V_0 L}{64}$
B $\frac{B_0 V_0 L}{16}$
C $B_0 V_0 L$
D $64 B_0 v_0^L$

The correct answer is:
$\frac{B_0 V_0 L}{64}$

Solution:
Length along $Y$-axis would be $l \sin 30^{\circ}=\frac{l}{2}$
length $=L$
Velocity of rod $=v_0$, along $X$-axis
variable magnet is field, $B=B_0\left(\frac{y}{L}\right)^3$, along $Z$-axis.
Now, induced emf (formula)
$$
E=B l v
$$
where, $B, l_{\perp}$ and $v$ are perpendicular to each other, $l=L / 2$
Now, consider a small element of length $d y$ at a distance $y$ from origin along $Y$-axis, then emf of element
$$
d E=B(d y) v_0=B_0 \frac{y^3}{L^3} v_0 \cdot d y
$$
Emf of the rod,
$$
\begin{aligned}
& E=\frac{B_0 v_0}{L^3} \int_0^{L / 2} y^3 \cdot d y \\
& E=\frac{B_0 v_0}{L^3}\left\lceil\frac{y^4}{4}\right\rceil^{L / 2} \\
& E=\frac{B_0 v_0}{L^3}\left\lfloor\frac{(L / 2)^4}{4}\right\rfloor \Rightarrow E=\frac{B_0 v_0 L}{64}
\end{aligned}
$$

Topic: Physics, Electromagnetic Induction, JEE Main
A conducting rod of resistance r moves uniformly with a constant speed v in a uniform magnetic field. If the rod keeps moving uniformly, then the amount of force required is

Options:
A vB2l2R
B 2vB2l2R+r
C vB2l2R+r
D zero

The correct answer is:
vB2l2R+r

Solution:
F → =i∫ dl → × B →
⇒F=ilB
i=ein(R+r)
When conducting rod moves in crossed magnetic field constant velocity v , emf is induced =Bvl

⇒ i=BvlR+r
∴ F =vBlR+rlB
=vB2l2R+r

Topic: Physics, Electromagnetic Induction, JEE Main
A conducting soap bubble having a radius a, thickness t<<a is charged to a potential V. If now, the bubble collapses and forms a droplet, then the potential of the droplet is

Options:
A Vdroplet=Va3t13
B Vdroplet=Vat13
C Vdroplet=V3at13
D Vdroplet=V3a4t13

The correct answer is:
Vdroplet=Va3t13

Solution:
Let q be the charge on the bubble, then
V = Kq a (Here K = 1 4 Ï Îµ 0 )
â´ q = Va K
Let after collapsing, the radius of droplet becomes R, then equating the volume, we have
4 Ï a 2 t = 4 3 Ï R 3
â´ R = 3 a 2 t 1 /3
Now, potential of droplet will be
V â² = Kq R
Substituting the values, we have
V â² = K Va K 3 a 2 t 1 / 3
or V â² = V a 3 t 1 / 3

Topic: Physics, Electrostatics, JEE Main