datavorous/entrance-exam-dataset huggingface dataset:
A closed and an open organ pipe have the same length. When they are vibrating simultaneously in their first overtone, they produce three beats. The length of the open pipe is now made one third the original length and one of its ends is closed. On the other hand, the length of the closed pipe is made three times the original length. The number of beats produced when they vibrate with fundamental frequencies will be

Options:
A 8
B 14
C 17
D 8

The correct answer is:
8

Solution:
V= speed of sound and L= lenght of pipes
For open pipe first overtone f1=VL
For closed pipe first overtone f'1=3V4L
â´âââf1âf'1=VLâ3V4L=3
â´ V4L=3
â´ VL=12
When we close one end of the open pipe and make its length one-third of the original length, then its fundamental frequency is
f=V4(L3)=3V4L
When the length of closed pipe is made 3 times, the fundamental frequency is
f'=V4(3L)=V12L
Beats produced =f-f'
=3V4L-V12L
=812.VL=812Ã12
=8

Topic: Physics, Waves and Sound, JEE Main
A closed circular tube of average radius 15 cm, whose inner walls are rough, is kept in vertical plane. A block of mass 1 kg just fit inside the tube. The speed of block is 22 m s-1, when it is introduced at the top of tube. After completing five oscillations, the block stops at the bottom region of tube. The work done by the tube on the block is ________ J. (Given g=10 m s-2).

Options:

The correct answer is:
245

Solution:
The data given is

r=15 cmm=1 kgv=22 m s-1

Total work done by the tube on the block is equal to loss in mechanical energy,

w=12mv2+mgh

⇒w=12×1×222+1×10×0.3

⇒w=245 J

Topic: Physics, Work Power Energy, JEE Main, JEE Main 2023 (10 Apr Shift 1)
A closed cylindrical can has to be made with 100 m2 of plastic. If its volume is maximum, then the ratio of its radius to the height is

Options:
A 1:1
B 1:2
C 2:1
D 2:1

The correct answer is:
1:2

Solution:
Let x be the radius and y be the height
âµ 2Ïx2+xy=100
â 50Ïx-x=y
Volume, V=Ïx2y=Ï50xÏ-x3
dVdx=0 â x=523Ï
and y=1023Ï
â´ xy=12

Topic: Mathematics, Application of Derivatives, JEE Main
A closed hollow insulated cylinder is filled with gas at $0^{\circ} \mathrm{C}$ and also contains an insulated piston of negligible weight and negligible thickness at the middle point. The gas in one side of the piston is heated to $100^{\circ} \mathrm{C}$. If the piston moves $5 \mathrm{~cm}$. The length of the hollow cylinder is

Options:
A $15.65 \mathrm{~cm}$
B $27.3 \mathrm{~cm}$
C $38.6 \mathrm{~cm}$
D $64.6 \mathrm{~cm}$

The correct answer is:
$64.6 \mathrm{~cm}$

Solution:
From Charle's law, $\frac{V_1}{V_2}=\frac{T_1}{T_2}$
Volume $(V)$ is proportional to the length of hollow cylinder.
$\begin{aligned} \frac{l_1}{l_2} & =\frac{T_1}{T_2} \\ \frac{l+5}{l-5} & =\frac{273+100}{273+0} \\ \frac{l+5}{l-5} & =\frac{373}{273} \\ \frac{2 l}{10} & =\frac{646}{100} \\ & =\frac{646 \times 10}{2 \times 100}=32.3 \mathrm{~cm}\end{aligned}$
$\begin{aligned} \therefore \text { Length of cylinder } & =2 l=2 \times 32.3 \\ & =64.6 \mathrm{~cm}\end{aligned}$

Topic: Physics, Kinetic Theory of Gases, JEE Main
A closed loop PQRS carrying a current is placed in a uniform magnetic field. If the magnetic forces on segments PS, SR and RQ are $F_1, F_2$ and $F_3$ respectively and are in the plane of the paper and along the directions shown, the force on the segment QP is

Options:
A $F_3-F_1+F_2$
B $F_3-F_1-F_2$
C $\sqrt{\left(F_3-F_1\right)^2+F_2^2}$
D $\sqrt{\left(F_3-F_1\right)^2-F_2^2}$

The correct answer is:
$\sqrt{\left(F_3-F_1\right)^2+F_2^2}$

Solution:
Since net force on current carrying loop in uniform magnetic field is zero therefore force on remaining segment will be equal and oppsoite to $\mathrm{F}$.

Topic: Physics, Magnetic Effects of Current, NEET, NEET 2008 (Mains)
A closed organ pipe 150 cm long gives 7 beats per second with an open organ pipe of length 350 cm, both vibrating in fundamental mode. The velocity of sound is ______ m s-1.

Options:

The correct answer is:
294

Solution:
The formula to calculate the fundamental frequency in a closed pipe is given by

fc=v4l1 ...1

And, that in an open pipe is given by

fo=v2l2 ...2

Given that

fc-fo=7 ...3

From equations (1), (2) and (3), it follows that

v4×150-v2×350=7⇒v600 cm-v700 cm=7⇒v6m-v7m=7⇒v142=7⇒v=42×7=294 m s-1

Topic: Physics, Waves and Sound, JEE Main, JEE Main 2024 (27 Jan Shift 2)
A closed organ pipe and an open pipe of the same length produce 4 beats when they are set into vibrations simultaneously. If the length of each of them were twice their initial lengths, then find the number of beats produced.

Options:

The correct answer is:
2

Solution:
νopen=V2L
νclosed=V4L
⇒νopen-νclosed=4
∴V2L-V4L=4
⇒V=16L

Now in 2nd case:-
ν′open=V4L
ν′closed=V8L
∴ν′open-ν′closed=V8L-V4L=V8L=16L8L
=2

Topic: Physics, Waves and Sound, JEE Main
A closed organ pipe has a fundamental frequency of 1.5 kHz. The number of overtones that can be distinctly heard by a person with this organ pipe will be (Assume that the highest frequency a person can hear is 20,000 Hz).

Options:
A 7
B 4
C 6
D 5

The correct answer is:
6

Solution:
For a closed organ pipe, the resonate frequency(harmonics) is an odd multiple of the fundamental frequency.

Therefore, 2n-1f0≤20000 Hz, where n=1,2,3...

f0=1.5 kHz⇒ fundamental frequency

Then, n=7

The total number of overtones that can be heard is 6. (2nd harmonic to 7th harmonic. )

Topic: Physics, Waves and Sound, JEE Main, JEE Main 2019 (10 Jan Shift 2)
A closed organ pipe is excited to vibrate in the third overtone. It is observed that there are

Options:
A three nodes and three antinodes.
B three nodes and four antinodes.
C four nodes and three antinodes.
D four nodes and four antinodes.

The correct answer is:
four nodes and four antinodes.

Solution:
The third overtone of the closed pipe is

7v4l

where v is the velocity of the sound and l is the length of the closed organ pipe.
The above figure is for the third overtone of the closed organ pipe. There are three complete loops and half-loop. Hence, there are four nodes and four antinodes.

Topic: Physics, Waves and Sound, JEE Main
A closed organ pipe of length L and an open organ pipe contain gases of densities ρ1 and ρ2 respectively. The compressibility of gases are equal in both the pipes. Both the pipes are vibrating in their first overtone with same frequency. The length of the open pipe is x3Lρ1ρ2, where x is _______.

(Round off to the Nearest Integer)

Options:

The correct answer is:
4

Solution:
fc=f0

3VC4L=2V02L'

3VC4L=V0L'

L'=4L3V0VC=4L3B.ρ1ρ2.B (B is bulk modulus)

=4L3ρ1ρ2

x=4

Topic: Physics, Waves and Sound, JEE Main
A closed organ pipe of length L and an open organ pipe contain gases of densities ρ1 and ρ2 respectively. The compressibility of gases are equal in both the pipes. Both the pipes are vibrating in their first overtone with same frequency. The length of the open pipe is x3Lρ1ρ2, where x is _______.

(Round off to the Nearest Integer)

Options:

The correct answer is:
4

Solution:
fc=f0

3VC4L=2V02L'

3VC4L=V0L'

L'=4L3V0VC=4L3B.ρ1ρ2.B (B is bulk modulus)

=4L3ρ1ρ2

x=4

Topic: Physics, Waves and Sound, JEE Main, JEE Main 2021 (16 Mar Shift 2)
A closed pipe and an open pipe have their first overtones identical in frequency. Their lengths are in the ratio

Options:
A 1:2
B 2:3
C 3:4
D 4:5

The correct answer is:
3:4

Solution:
If is given that
First over tone of closed pipe $=$ First over tone of open pipe $\quad 3\left(\frac{v}{4 l_1}\right)=2\left(\frac{v}{2 l_2}\right) ;$ where $l_1$ and $l_2$ are the lengths of closed and open organ pipes hence $\frac{I_1}{I_2}=\frac{3}{4}$

Topic: Physics, Waves and Sound, JEE Main
A closed pipe is suddenly opened and changed to an open pipe of same length. The fundamental frequency of the resulting open pipe is less than that of 3rd harmonic of the earlier closed pipe by $55 \mathrm{~Hz}$. Then, the value of fundamental frequency of the closed pipe is

Options:
A $165 \mathrm{~Hz}$
B $110 \mathrm{~Hz}$
C $55 \mathrm{~Hz}$
D $220 \mathrm{~Hz}$

The correct answer is:
$55 \mathrm{~Hz}$

Solution:
Frequency of first mode of vibration
$$
n_1=\frac{3 v}{4 /}
$$
(closed pipe) and frequency of first mode of vibration
$$
n_2=\frac{v}{2 l}
$$
(open pipe)
According to the question,
$$
n_1-n_2=\frac{3 v}{4 l}-\frac{v}{2 l}=55
$$
So $\quad \frac{3 v-2 v}{4 !}=55$
$$
\Rightarrow \quad \frac{v}{4 l}=55 \mathrm{~Hz}
$$

Topic: Physics, Waves and Sound, JEE Main
A closed pipe of length 22 cm, when excited by a 1875 Hz source forms standing waves. The number of pressure nodes formed in the pipe are [velocity of sound in air =330 m s-1]

Options:
A 1
B 2
C 4
D 3

The correct answer is:
3

Solution:
Velocity of sound =330 m s-1
Length of closed pipe =22 cm=22×10-2 m
The fundamental frequency of the stationary wave,
ν0=v4l=3304×22×10-2
=330×1024×22=30008
ν0=375 Hz
3ν0=375×3=1125 Hz
5ν0=5×375=1875 Hz
5th harmonic
Number of nodes =3
Note: In a closed pipe, odd harmonics are produced.

Topic: Physics, Waves and Sound, JEE Main
A closed rectangular tank is completely filled with water and is accelerated horizontally with an acceleration $a$ towards right. Pressure is (i) maximum at, and (ii) minimum at

Options:
A (i) $\mathrm{B}$ (ii) $\mathrm{D}$
B (i) $\mathrm{C}$ (ii) $\mathrm{D}$
C (i) $\mathrm{B}$ (ii) $\mathrm{C}$
D (i) $B$ (ii) $A$

The correct answer is:
(i) $\mathrm{B}$ (ii) $\mathrm{D}$

Solution:
Due to acceleration towards right, there will be a pseudo force in a left direction. So the pressure will be more on rear side (Points $A$ and $B$ ) in comparison with front side (Point $D$ and $C$ ).
Also due to height of liquid column pressure will be more at the bottom (points $B$ and C) in comparison with top (point $A$ and $D$ ).
So overall maximum pressure will be at point $B$ and minimum pressure will be at point D.

Topic: Physics, Mechanical Properties of Fluids, JEE Main
A closed tube filled with water is rotating uniformly in a horizontal plane about the axis OO' as shown in the figure. The manometers A and B which are fixed on the tube at distances r1 and r2, indicate pressures P1 and P2 respectively. The angular velocity (Ï) of the tube is

Options:
A ω=2P2-P1ρr22-r12
B ω=2P2+P1ρr22-r12
C ω=2P2+P1ρr22+r12
D ω=2P2-P1ρr22+r12

The correct answer is:
ω=2P2-P1ρr22-r12

Solution:
Consider the condition of equilibrium for the mass of water contained between cross section separated by x and x+dx from rotation axis.
dp=ÏÏ2xdx ;
â«p1p2dp=â«r1r2ÏÏ2xdx
p2-p1=ÏÏ2x22r1r2=ÏÏ2r22-r122
ω=2P2-P1ρr22-r12

Topic: Physics, Mechanical Properties of Fluids, JEE Main
A closed vessel contains 0.1 mole of a monoatomic ideal gas at 200 K. If 0.05 mole of the same gas at 400 K is added to it, the final equilibrium temperature (in K) of the gas in the vessel will be close to ______

Options:

The correct answer is:
267

Solution:
0.1 200+0.05 400=0.15T

T=20+200.15=8003=266.67

=266.67

Topic: Physics, Kinetic Theory of Gases, JEE Main, JEE Main 2020 (04 Sep Shift 1)
A closely wound solenoid of 2000 turns and area of cross-section $1.5 \times 10^{-4} \mathrm{~m}^2$ carries a current of $2.0 \mathrm{~A}$. It is suspended through its centre and perpendicular to its length, allowing it to turn in a horizontal plane in a uniform magnetic field $5 \times 10^{-2} \mathrm{~T}$ making an angle of $30^{\circ}$ with the axis of the solenoid. The torque on the solenoid will be

Options:
A $3 \times 10^{-3} \mathrm{~N}-\mathrm{m}$
B $1.5 \times 10^{-3} \mathrm{~N}-\mathrm{m}$
C $1.5 \times 10^{-2} \mathrm{~N}$-m
D $3 \times 10^{-2} \mathrm{~N}-\mathrm{m}$

The correct answer is:
$1.5 \times 10^{-2} \mathrm{~N}$-m

Solution:
Given, $\mathrm{N}=2000, \mathrm{~A}=1.5 \times 10^{-4} \mathrm{~m}^2$ $\mathrm{i}=2.0 \mathrm{AB}=5 \times 10^{-2} \mathrm{~T}$, and $\theta=30^{\circ}$
Torque, $\tau=$ NiBA $\sin \theta$
$$
\begin{aligned}
& =2000 \times 2 \times 5 \times 10^{-2} \times 1.5 \times 10^{-4} \times \sin 30^{\circ} \\
& =2000 \times 50 \times 10^{-6} \times \frac{1}{2} \\
& =1.5 \times 10^{-2} \mathrm{Nm}
\end{aligned}
$$

Topic: Physics, Magnetic Effects of Current, NEET, NEET 2010 (Mains)
A closely wounded circular coil of radius 5 cm produces a magnetic field of 37.68×10-4 T at its center. The current through the coil is _____ A. [Given, number of turns in the coil is 100 and π=3.14]

Options:

The correct answer is:
3

Solution:
Given here, N=100, R=5 cm and B=37.68×10-4 T

Magnetic field at the centre of circular coil is given by Bcentre =Nμ0l2R.

Putting the values,

37.68×10-4=100×4π×10-7×I2×5×10-2

Current through coil is I=3 A.

Topic: Physics, Magnetic Effects of Current, JEE Main, JEE Main 2022 (29 Jul Shift 1)
A coaxial cable consists of an inner wire of radius a surrounded by an outer shell of inner and outer radii b and c respectively. The inner wire carries an electric current i0 which is distributed uniformly across cross-sectional area. The outer shell carries an equal current in opposite direction and distributed uniformly. What will be the ratio of the magnetic field at a distance x from the axis when i x<a and ii a<x<b ?

Options:
A a2x2
B b2-a2x2
C x2b2-a2
D x2a2

The correct answer is:
x2a2

Solution:
B1×2πx=μ0I0(x)2a2

B2×2πx=μ0I0

B1B2=x2a2

Topic: Physics, Magnetic Effects of Current, JEE Main, JEE Main 2021 (27 Aug Shift 2)
A coaxial cylinder made of glass is immersed in a liquid of surface tension S. The radius of the inner and outer surface of the cylinder are R1 and R2 respectively. Height till which liquid will rise is (Density of liquid is ρ )

Options:
A 2SR2Ïg
B 2SR1Ïg
C SR2-R1Ïg
D 2SR2-R1ρg

The correct answer is:
2SR2-R1ρg

Solution:
Net upward force
=2πR2S+2πR1S
Contact angle=0o
∴ Capillary rise is given by
h=2πSR1+R2πR22-R12ρg
=2SR2-R1ρg

Topic: Physics, Mechanical Properties of Fluids, JEE Main
A coffee maker makes coffee by passing steam through a mixture of coffee powder, milk and water. If the steam is mixed at the rate of 50 g per minute in a mug containing 500 g of mixture, then it takes about t0 seconds to make coffee at 70°C when the initial temperature of the mixture is 25°C. The value of t0 is close to (ratio of latent heat of evaporation to specific heat of water is 540°C and specific heat of the mixture can be taken to be the same as that of water)

Options:
A 30
B 45
C 60
D 90

The correct answer is:
45

Solution:
Let m gram of steam is condensed in the process of heating mixture from 25°C to 70°C

Then,
Heat lost by steam = Heat gained by mixture

⇒ Heat of condensation of steam + Heat given by water formed = Heat gained by mixture

⇒ m·L+mswΔT=M·snΔT

=500×sw×(70-25)

⇒m=500×sw×45L+30sw

⇒m=500×45Lsw+30

⇒m=500×45(540+30)≈40 g

Now, in 1 min,50 g of steam is condensed.

∴40 g of steam will be condensed in time

t0

t0=40×6050 s=48 s

Nearest answer is 45 s.

Topic: Physics, Thermal Properties of Matter, JEE Main
A coil has 1,000 turns and $500 \mathrm{~cm}^2$ as its area. The plane of the coil is placed at right angles to a magnetic induction field of $2 \times 10^{-5} \mathrm{~Wb} / \mathrm{m}^2$. The coil is rotated through $180^{\circ}$ in 0.2 seconds. The average e.m.f. induced in the coil, in milli-volts, is

Options:
A $5$
B $10$
C $15$
D $20$

The correct answer is:
$10$

Solution:
By using $e=-\frac{N B A\left(\cos \theta_2-\cos \theta_1\right)}{\Delta t}$
$e=-\frac{1000 \times 2 \times 10^{-5} \times 500 \times 10^{-4}\left(\cos 180^{\circ}-\cos 0^{\circ}\right)}{0.2}$
$=10^{-2} \mathrm{vol} / \mathrm{t}=10 \mathrm{mV}$

Topic: Physics, Electromagnetic Induction, JEE Main
A coil has an inductance of 2 H and resistance of 4 Ω. A 10 V is applied across the coil. The energy stored in the magnetic field after the current has built up to its equilibrium value will be _____ ×10-2 J

Options:

The correct answer is:
625

Solution:
The data given is

L=2 HR=4 ΩV=10 V

Using Ohm's law, V=IR,

I=10 V4 Ω=52 A

The energy of an inductor is given by
E=12LI2=12×2×254=625×10-2 J

Topic: Physics, Electromagnetic Induction, JEE Main, JEE Main 2023 (11 Apr Shift 2)
A coil has resistance $30 \Omega$ and inductive reactance $20 \Omega$ at $50 \mathrm{~Hz}$ frequency. If an $\mathrm{AC}$ source of $200 \mathrm{~V}, 100 \mathrm{~Hz}$, is connected across the coil, the current in the coil will be

Options:
A $4.0 \mathrm{~A}$
B $8.0 \mathrm{~A}$
C $\frac{20}{\sqrt{13}} \mathrm{~A}$
D $2.0 \mathrm{~A}$

The correct answer is:
$4.0 \mathrm{~A}$

Solution:
If $\omega=50 \times 2 \pi$ then $\omega L=20 \Omega$
Similarly $\omega^{\prime}=100 \times 2 \pi$ then $\omega^{\prime} L=40 \Omega$
$$
\begin{aligned}
i & =\frac{200}{Z}=\frac{200}{\sqrt{R^2+\left(\omega^{\prime} L\right)^2}} \\
& =\frac{200}{\sqrt{(30)^2+(40)^2}}=4 \mathrm{~A}
\end{aligned}
$$

Topic: Physics, Alternating Current, NEET, NEET 2011 (Mains)
A coil has resistance 30 ohm and inductive reactance 20 ohm at $50 \mathrm{~Hz}$ frequency. If an ac source, of 200 volt, $100 \mathrm{~Hz},$ is connected across the coil, the current in the coil will be

Options:
A $4.0 \mathrm{~A}$
B $8.0 \mathrm{~A}$
C $7.2 \mathrm{~A}$
D $2.0 \mathrm{~A}$

The correct answer is:
$4.0 \mathrm{~A}$

Solution:
If $\omega=50 \times 2 \pi$ then $\omega \mathrm{L}=20 \Omega$ If $\omega^{\prime}=100 \times 2 \pi$ then $\omega^{\prime} \mathrm{L}=40 \Omega$
Current flowing in the coil is
$$
\mathrm{I}=\frac{200}{\mathrm{Z}}=\frac{200}{\sqrt{\mathrm{R}^{2}+\left(\omega^{\prime} \mathrm{L}\right)^{2}}}=4 \mathrm{~A}
$$

Topic: Physics, Alternating Current, JEE Main
A coil having 100 turns, area of $5 \times 10^{-3} \mathrm{~m}^2$, carrying current of $1 \mathrm{~mA}$ is placed in uniform magnetic field of $0.20 \mathrm{~T}$ such a way that plane of coil is perpendicular to the magnetic field. The work done in turning the coil through $90^{\circ}$ is _______ $\mu \mathrm{J}$.

Options:

The correct answer is:
100

Solution:
$\begin{aligned} & W=\Delta U=U_f-U_i \\ & W=(-\vec{\mu} . \vec{B})_f-(-\vec{\mu} . \vec{B})_i \\ & =0+(\vec{\mu} \cdot \vec{B})_i \\ & =\left(100 \times 5 \times 10^{-3} \times 1 \times 10^{-3}\right) \times 0.2 \mathrm{~J} \\ & =1 \times 10^{-4} \mathrm{~J}=100 \mu \mathrm{J}\end{aligned}$

Topic: Physics, Magnetic Effects of Current, JEE Main, JEE Main 2024 (06 Apr Shift 2)
A coil having inductance L and resistance R is connected to a battery of emf Îµ at t=0. If t1 and t2 are the time for 90% and 99% completion of the current growth in the circuit, then t1t2 will be-

Options:

The correct answer is:
0.5

Solution:
I=I01-e-tτ
910I0=I01-e-t1τ
⇒ e-t1τ=110
⇒ et1τ=10
⇒ t1=τln10
and 99100I0=I01-e-t2τ
⇒ e-t2τ=1100
⇒ t2=τln100=2τln10
⇒ t1t2=0.5

Topic: Physics, Alternating Current, JEE Main
A coil having inductance L and resistance R is connected to a battery of emf Îµ at t=0. If t1 and t2 are the times for 90% and 99% completion of the current growth in the circuit, then t 1 t 2 will be

Options:
A 1:2
B 2:1
C log e 10 2
D 2loge10

The correct answer is:
1:2

Solution:
I = I 0 1 - e - t τ
910I0=I01-e-t1τ
⇒ e - t 1 τ = 1 1 0
⇒ e t 1 τ = 1 0
⇒ t 1 = τ ln 1 0 ............(1)
and 99100I0=I01-e-t2τ
⇒ e - t 2 τ = 1 1 0 0
⇒ t 2 = τ ln 1 0 0 = 2 τ ln 1 0 ..............(2)

From equations (1) and (2), we have
⇒ t 1 t 2 = 1 2

Topic: Physics, Alternating Current, JEE Main
A coil having $n$ turns and resistance $4 \mathrm{R} \Omega$. This combination is moved in time $t$ seconds from a magnetic field $W_1$ weber to $W_2$ weber. The induced current in the circuit is

Options:
A
$-\frac{W_2-W_1}{5 R n t}$

B
$-\frac{\left(\mathrm{W}_2-\mathrm{W}_1\right)}{5 \mathrm{Rt}}$

C
$-\frac{W_2-W_1}{\text { Rnt }}$

D
$-\frac{\mathrm{n}\left(\mathrm{W}_2-\mathrm{W}_1\right)}{\mathrm{Rt}}$

The correct answer is:
$-\frac{\left(\mathrm{W}_2-\mathrm{W}_1\right)}{5 \mathrm{Rt}}$

Solution:
$\mathrm{I}=-\frac{\mathrm{I}}{\mathrm{R}^{\prime}} \frac{\mathrm{d} \phi}{\mathrm{dt}}$
or, I $=-\frac{1}{R^{\prime}} \mathrm{n}\left[\frac{\mathrm{W}_2-\mathrm{W}_1}{\mathrm{t}_2-\mathrm{t}_1}\right]$
( $W_1$ and $W_2$ are not the magnetic field, but the values of flux associated with one turn of coil) $I=\frac{-1}{(R+4 R)} \frac{n\left(W_2-W_1\right)}{t}$
or, $\mathrm{I}=-\frac{\mathrm{n}\left(\mathrm{W}_2-\mathrm{W}_1\right)}{5 \mathrm{Rt}}$

Topic: Physics, Electromagnetic Induction, JEE Main, JEE Main 2004
A coil having $N$ turns carry a current / as shown in the figure. The magnetic field intensity at point $P$ is

Options:
A $\frac{\mu_0 N / R^2}{2\left(R^2+x^2\right)^{3 / 2}}$
B $\frac{\mu_0 N I}{2 R}$
C $\frac{\mu_0 N I R^2}{(R+x)^2}$
D Zero

The correct answer is:
$\frac{\mu_0 N / R^2}{2\left(R^2+x^2\right)^{3 / 2}}$

Solution:
Magnetic field intensity at point $\mathrm{P}$, shown in figure while carrying current "i" and having "N" turns is given by Magnetic field $\mathrm{P}$ due to entire loop is
$\frac{\mu_0 \text { NIR }^2}{2\left(\mathrm{R}^2+\mathrm{x}^2\right)^{3 / 2}}$

Topic: Physics, Magnetic Effects of Current, JEE Main
A coil in the shape of an equilateral triangle of side 10 cm lies in a vertical plane between the pole pieces of permanent magnet producing a horizontal magnetic field 20mT. The torque acting on the coil when a current of 0.2 A is passed through it and its plane becomes parallel to the magnetic field will be x×10-5Nm. The value of x is

Options:

The correct answer is:
3

Solution:
τ=M→×B→

=IAB

=0.2×3a24×2×10-2=0.2×3×10-2×2×10-24

=3×10-5 N m

Topic: Physics, Magnetic Effects of Current, JEE Main, JEE Main 2021 (26 Aug Shift 2)
A coil in the shape of an equilateral triangle of side $l$ is suspended between the pole of a permanent magnet such that $B$ is in plane of the coil. If due to current $i$ in the triangle a torque $\tau$ acts on it, the side $l$ of the triangle is:

Options:
A $\frac{2}{\sqrt{3}}\left(\frac{\tau}{B i}\right)$
B $2\left(\frac{\tau}{\sqrt{3} B i}\right)^{1 / 2}$
C $\frac{2}{\sqrt{3}}\left(\frac{\tau}{B i}\right)^{1 / 2}$
D $\frac{1}{\sqrt{3}} \frac{\tau}{B i}$

The correct answer is:
$2\left(\frac{\tau}{\sqrt{3} B i}\right)^{1 / 2}$

Solution:
Since $I=M B=(I A) B=I \frac{\sqrt{3}}{4} l^2 B$
$$
\Rightarrow \quad l=2\left(\frac{I}{\sqrt{3} B I}\right)^{1 / 2}
$$

Topic: Physics, Magnetic Effects of Current, JEE Main
A coil is placed in a magnetic field B→ as shown below :
A current is induced in the coil because B→ is :

Options:
A parallel to the plane of coil and increasing with time
B outward and increasing with time
C outward and decreasing with time
D parallel to the plane of coil and decreasing with time

The correct answer is:
outward and decreasing with time

Solution:
Lenz's law state that the direction of the induced current is such that it opposes its cause of generation, so,

B→, must not be parallel to the plane of coil for non-zero flux and according to Lenz's law if B is outward it should be decreasing for anticlockwise induced current.

Topic: Physics, Electromagnetic Induction, JEE Main, JEE Main 2021 (31 Aug Shift 2)
A coil is placed in a time varying magnetic field. If the number of turns in the coil were to be halved and the radius of wire doubled, the electrical power dissipated due to the current induced in the coil would be (Assume the coil to be short circuited.)

Options:
A Doubled
B The same
C Quadrupled
D Halved

The correct answer is:
The same

Solution:
\(\mathrm{P}=\frac{\varepsilon^2}{R}\)
where, \(\varepsilon=\) inducedemf \(=-\frac{d \phi}{d t}\)
\(\phi=\mathrm{NBA}\)
\(\varepsilon=\mathrm{NA} \frac{\mathrm{dB}}{\mathrm{dt}}\)
also, \(R \propto \frac{1}{\mathrm{r}^2}\)
where, \(\mathrm{R}=\) resistance and \(\mathrm{r}=\) radius of wire
\(\begin{aligned} & \mathrm{P} \propto \frac{\mathrm{N}^2 \mathrm{r}^2}{\mathrm{H}^2} \\ & \frac{\mathrm{P}_1}{\mathrm{P}_2}=1 \\ & \mathrm{P}_1=\mathrm{P}_2 \end{aligned}\)
This means if the number of turns were quadrupled and the wire radius halved, the electrical power dissipated would be remains same.

Topic: Physics, Electromagnetic Induction, JEE Main, JEE Main 2022 (28 Jun Shift 2)