datavorous/entrance-exam-dataset huggingface dataset:
A cubical block is floating in a liquid with half of its volume immersed in the liquid. When the whole system accelerates upwards with acceleration of $g / 3$, the fraction of volume immersed in the liquid will be

Options:
A $\frac{1}{2}$
B $\frac{3}{8}$
C $\frac{2}{3}$
D $\frac{3}{4}$

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

Solution:
Fraction of volume immersed in the liquid $V_{i n}=\left(\frac{\rho}{\sigma}\right) V$ i.e. it depends upon the densities of the block and liquid.
So there will be no change in it if system moves upward or downward with constant velocity or some acceleration.

Topic: Physics, Mechanical Properties of Fluids, JEE Main
A cubical block of density ρ is floating on the surface of water. Out of its height L, what fraction x is submerged in water? The vessel is in an elevator accelerating upward with acceleration a. Which of the following statements is true regarding the fraction of the block submerged?

Options:
A The fraction of the block submerged increases with the acceleration of the elevator.
B The fraction of the block submerged decreases with the acceleration of the elevator.
C The fraction of the block submerged remains constant regardless of the acceleration of the elevator.
D The fraction of the block submerged is zero in an accelerating elevator.

The correct answer is:
The fraction of the block submerged remains constant regardless of the acceleration of the elevator.

Solution:
Correct Option is : (C)
The fraction of the block submerged remains constant regardless of the acceleration of the elevator.

Topic: Chemistry, Mechanical Properties of Fluids, NEET
A cubical block of side 0.5 m floats on water with 30% of its volume under water. What is the maximum weight that can be put on the block without fully submerging it under water?
[Take, density of water =103 kg/m3 ]

Options:
A 87.5 kg
B 65.4 kg
C 30.1 kg
D 46.3 kg

The correct answer is:
87.5 kg

Solution:
In first case, when 30% of the cube inside water,
ρwVin g=ρcVg
VinV=ρcρw=0.3
Additional weight we can put w is,
Vρwg=ρcVg+w
w=Vρw-ρcg
=0.53×0.7×103×g
=87.5g N
Addition of mass that we can put =87.5 kg

Topic: Physics, Mechanical Properties of Fluids, JEE Main, JEE Main 2019 (10 Apr Shift 2)
A cubical block of side 30 cm is moving with velocity 2 m s-1 on a smooth horizontal surface. The surface has a bump at a point O as shown in the figure. The angular velocity (in rad/s) of the block immediately after it hits the bump, is :

Options:
A 13.3
B 5.0
C 9.4
D 6.7

The correct answer is:
5.0

Solution:
Conservation of angular momentum about point O,
Lbefore=Lafter
⇒mvR=Iω
⇒ω=m×2×a2m a26+a22
⇒ω=128a=32×0.3=102=5 rad s-1

Topic: Physics, Rotational Motion, JEE Main, JEE Main 2016 (09 Apr Online)
A cubical block of side moving with a velocity v on a horizontal smooth plane as shown. It hits a ridge at point O and sticks to it (collision is perfectly inelastic). The angular speed of the block after it hits O is

Options:
A 3v4a

B 3v2a
C 3v2a
D Zero

The correct answer is:
3v4a

Solution:
Net torque about O is zero.
Therefore, angular momentum L about point O will be conserved
or, Li=Lf
Mva2=I0ω=Icom+Mr2ω
=Ma26+Ma22ω=23Ma2ω
∴ ω=3v/4a

Topic: Physics, Rotational Motion, JEE Main
A cubical block of wood, of length $10 \mathrm{~cm}$, floats at the interface between oil of density $800 \mathrm{~kg} / \mathrm{m}^3$ and water. The lower surface of the block is $1.5 \mathrm{~cm}$ below the interface. If the depth of water is $10 \mathrm{~cm}$ below the interface and oil is upto $10 \mathrm{~cm}$ above the interface then the difference in pressure at the lower and the upper face of the wooden block is
(Assume density of water, $\rho=1000 \mathrm{~kg} / \mathrm{m}^3$ and acceleration of gravity, $g=10 \mathrm{~m} / \mathrm{s}^2$ )

Options:
A $850 \mathrm{~Pa}$
B $780 \mathrm{~Pa}$
C $800 \mathrm{~Pa}$
D $830 \mathrm{~Pa}$

The correct answer is:
$830 \mathrm{~Pa}$

Solution:
Given, length of cubical block, $l=10 \mathrm{~cm}=0.1 \mathrm{~m}$
density of oil, $\rho_0=800 \mathrm{~kg} / \mathrm{m}^3$
and density of water, $\rho_w=10^3 \mathrm{~kg} / \mathrm{m}^3$
If $V_1$ and $V_2$ be the volume of wooden block in water and oil respectively, then
$$
\begin{aligned}
V_1: V_2 & =1.5: 8.5 \\
\frac{V_1}{V_2} & =\frac{1.5}{8.5}=\frac{3}{17} \\
\therefore \quad V_1 & =\frac{3}{17} V_2
\end{aligned}
$$
$\therefore V_1+V_2=V$, where, $V$ is the volume of the block.
From Eq. (i), we get
$$
\begin{array}{rlrl}
& \frac{3}{17} V_2+V_2 & =V \Rightarrow V_2=\frac{17}{20} \mathrm{~V} \\
\therefore \quad V_1 & =\frac{3}{20} \mathrm{~V}
\end{array}
$$
If $\rho$ be the density of the block.
$\therefore$ weight of block $=$ weight of oil displaced + weight of water displaced
$$
\begin{gathered}
V \rho G=V_1 \rho_\omega g+V_2 \rho_{\text {oil }} g \\
V \rho g=\frac{3}{20} V \rho_w g+\frac{17}{20} V \rho_{\text {oil }} g \\
\rho=\frac{3}{20} \rho_w+\frac{17}{20} \rho_{\text {oil }}=\frac{3}{20} \times 1000+\frac{17}{20} \times 800 \\
=830 \mathrm{~kg} / \mathrm{m}^3
\end{gathered}
$$
$\therefore$ Pressure difference
$$
p=\rho g h=830 \times 10 \times 0.1=830 \mathrm{~Pa}
$$

Topic: Physics, Mechanical Properties of Fluids, JEE Main
A cubical box with porous walls containing an equal number of O2 and H2 molecules is placed in a larger evacuated chamber. The entire system is maintained at constant temperature T. The ratio of root mean square speed of O2 molecules to that of the H2 molecules is

Options:
A 122
B 14
C 12
D 2

The correct answer is:
14

Solution:
∵Vrms=γRTM
⇒ VrmsO2VrmsH2=M0H2M0O2
=232=14

Topic: Physics, Kinetic Theory of Gases, NEET
A cubical vessel of height \(1 \mathrm{~m}\) is full of water. What is the amount of work done in pumping water out of the vessel? (Take \(\mathrm{g}=\) \(10 \mathrm{~m} \mathrm{~s}^{-2}\) )

Options:
A \(1250 \mathrm{~J}\)
B \(5000 \mathrm{~J}\)
C \(1000 \mathrm{~J}\)
D \(2500 \mathrm{~J}\)

The correct answer is:
\(5000 \mathrm{~J}\)

Solution:
Hints:
\(\begin{aligned}
& \mathrm{V}=\ell^3=1 \mathrm{~m}^3 \\
& \mathrm{~m}=1 \times 1000=1000 \mathrm{~kg} ; \mathrm{W}=\mathrm{mgh}=1000 \times 10 \times \frac{1}{2}=5000 \mathrm{~J}
\end{aligned}\)

Topic: Physics, Work Power Energy, JEE Main
A cup of coffee cools from 90°C to 80°C in t min, when the room temperature is 20°C. The time taken by a similar cup of coffee to cool from 80°C to 60°C at a room temperature same at 20°C is :

Options:
A 1310t
B 1013t
C 135t
D 513t

The correct answer is:
135t

Solution:
From Newton's Law of cooling,

ΔTΔt=kT-Ts

where,

⇒T2-T1t-kT1+T22-TsTs

Where, T1= Initial temperature of body

T2= Final temperature of body

Ts= surrounding temperature

t= time taken

For case1,

90-80t=k90+802-20

⇒10t=65k

⇒k=1065t …i

For case 2 ,

80-60t'=k80+602-20

⇒20t'=50k

Put the value k from eq.i,

⇒20t'=50×1065t

⇒t'=20×65500t

⇒t'=135t

Topic: Physics, Thermal Properties of Matter, JEE Main
A cup of tea cools from 80 Â°C to 60 Â°C in 40 seconds. The ambient temperature is 30 Â°C. In cooling from 60 Â°C to 50 Â°C, It will take time :

Options:
A 35 s

B 30 s

C 32 s
D 48 s

The correct answer is:
32 s

Solution:
From Î¸ 2 - Î¸ 1 t = k Î¸ 1 + Î¸ 2 2 - Î¸ 0
we have 8 0 â - 6 0 â 4 0 â = k 8 0 â + 6 0 â 2 - 3 0 â
â k = 1 8 0
Now 6 0 â - 5 0 â t = k 6 0 â + 5 0 â 2 - 3 0 â
â 1 0 t = 1 8 0 2 5 â t = 3 2 s

Topic: Physics, Thermal Properties of Matter, JEE Main
A cup of tea cools from 65.5 oC to 62.5 oC in 1âmin in a room at 22.5 oC. How long will it take to cool from 46.5 oC to 40.5 oC in the same room?

Options:
A 4 min

B 2 min
C 1 min
D 3 min

The correct answer is:
4 min

Solution:
According to Newton's law of cooling,
Î¸1-Î¸2t=Î±Î¸1+Î¸22-Î¸0
âµÎ±=universal constant
65.5-62.51=Î±65.5+62.52-22.5 .....(i)
46.5-40.5t=Î±46.5+40.52-22.5 .....(ii)
Solving Equations. (i) and (ii),
we get t=4 min

Topic: Physics, Thermal Properties of Matter, JEE Main
A current carrying conductor obeys Ohm's law $(V=R I)$. If the current passing through the conductor is $I=(5 \pm 0.2)$ A and voltage developed is $V=(60 \pm 6) \mathrm{V}$, then find the percentage of error is resistance, $R$

Options:
A 18
B 6
C 14
D 2

The correct answer is:
14

Solution:
Given, current passing through the conductor,
$I=(5 \pm 0.2) \mathrm{A}, \Delta I=0.2 \mathrm{~A}$
Voltage developed,
$\begin{array}{rlrl}
& & V & =(60 \pm 6) \mathrm{V} \\
\Rightarrow \quad & \Delta V & =6 \mathrm{~V}
\end{array}$
By Ohm's law,
$\begin{aligned}
V & =R I \\
R & =\frac{V}{I} \\
\frac{\Delta R}{R} & =\frac{\Delta V}{V}+\frac{\Delta I}{I} \\
& =\frac{6}{60}+\frac{0.2}{5} \\
& =0.1+0.04 \\
\frac{\Delta R}{R} & =0.14 \\
\therefore \quad \frac{\Delta R}{R} \times 100 & =0.14 \times 100=14 \%
\end{aligned}$

Topic: Physics, Mathematics in Physics, JEE Main
A current carrying loop is placed in a uniform magnetic field 'B' in different orientations I, II, III and IV as shown in the figure. The correct order of decreasing potential energy is
( $\hat{\mathrm{n}}$-unit vector normal to the plane of the loop)

Options:
A I, III, II, IV
B I, II, III, IV
C I, IV, II, III
D III, IV, I, II

The correct answer is:
I, IV, II, III

Solution:
Potential energy in a uniform magnetic field, $\mathrm{u}=-\mathrm{mB} \cos \theta$
(i) For orientation I $\theta=180^{\circ} \therefore \mathrm{u}=-\mathrm{mB} \cos 180^{\circ}=$ $\mathrm{mB}$
(ii) For orientation II $\theta=90^{\circ} \therefore \mathrm{u}=-\mathrm{mB} \cos 90^{\circ}=0$
(iii) For orientation III $\theta=0^{\circ} < \theta < 90^{\circ} \therefore-\mathrm{mB} < \mathrm{u} < 0$
(iv) For orientation IV $\theta=90^{\circ} < \theta < 180^{\circ} \therefore 0 < \mathrm{u} < $ $\mathrm{mB}$

Topic: Physics, Magnetic Effects of Current, JEE Main
A current carrying loop is placed in a uniform magnetic field in four different orientations I, II, III and IV, arrange them in the decreasing order of potential energy.
(I)
(II)
(III)
(IV)

Options:
A \( \mathrm{I}>\mathrm{III}>\mathrm{II}>\mathrm{IV} \)
B \( \mathrm{I}>\mathrm{II}>\mathrm{III}>\mathrm{IV} \)
C \( \mathrm{I}>\mathrm{IV}>\mathrm{II}>\mathrm{III} \)
D \( \mathrm{III}>\mathrm{IV}>\mathrm{I}>\mathrm{II} \)

The correct answer is:
\( \mathrm{I}>\mathrm{IV}>\mathrm{II}>\mathrm{III} \)

Solution:
We know potential energy of a magnetic dipole in external magnetic field B→ is

U=-M→.B→=-MB cosθ

where M→ is magnetic moment of the loop, and θ is angle between M→ and B→.

We can see that U is maximum when θ=180° and minimum when θ=0°.

So, as θ decreases from 180° to 0°, potential energy also decreases.

Topic: Physics, Magnetic Effects of Current, JEE Main
A current carrying rectangular loop PQRS is made of uniform wire. The length PR=QS=5 cm and PQ=RS=100 cm. If ammeter current reading changes from I to 2I, the ratio of magnetic forces per unit length on the wire PQ due to wire RS in the two cases respectively fPQI:fPQ2I is:

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

The correct answer is:
1:4

Solution:
Force acting per unit length between two parallel wires is given by, f=μ0I1I22πd.

Clearly, f∝I1I2

Therefore, fPQI:fPQ2I=I×I2I×2I=1:4

Topic: Physics, Magnetic Effects of Current, JEE Main, JEE Main 2023 (30 Jan Shift 2)
A current I=10sin100πt A. is passed in first coil, which induces a maximum e.m.f of 5 π volt in second coil. The mutual inductance between the coils is –

Options:
A 10 mH
B 15 mH
C 25 mH
D 5 mH

The correct answer is:
5 mH

Solution:
Let I = I 0 sin ω t
where I 0 = 1 0 , ω = 1 0 0
then ε = M dI dt
= M d dt I 0 sin ω t
= M I 0 ω cos ω t
∴ ε max = M I 0 ω
5 π = M × 1 0 × 1 0 0 π
M = 5 mH

Topic: Physics, Electromagnetic Induction, JEE Main
A current I ampere flows along an infinitely long straight thin-walled tube, then the magnetic induction at any point inside the tube is

Options:
A
infinite

B
zero

C
$\frac{\mu_0}{4 \pi} \frac{2 \mathrm{i}}{r}$ tesla

D
$2 n^2 \mathrm{~B}$

The correct answer is:
zero

Solution:
Considering Ampere's loop (shown by dotted line), no current is enclosed by this loop. Therefore, the magnetic field will be zero inside the tube.

Topic: Physics, Magnetic Effects of Current, JEE Main, JEE Main 2004
A current $\mathrm{I}$ flows along the length of an infinitely long, straight, thin walled pipe. Then

Options:
A
the magnetic field is zero only on the axis of the pipe

B
the magnetic field is different at different points inside the pipe

C
the magnetic field at any point inside the pipe is zero

D
the magnetic field at all points inside the pipe is the same, but not zero

The correct answer is:
the magnetic field at any point inside the pipe is zero

Solution:
Use Ampere's law

Topic: Physics, Magnetic Effects of Current, JEE Main, JEE Main 2007
A current I flows in an infinitely long wire with cross section in the form of a semicircular ring of radius $\mathrm{R}$. The magnitude of the magnetic induction along its axis is

Options:
A
$\frac{\mu_0 I}{2 \pi^2 R}$

B
$\frac{\mu_0 I}{2 \pi R}$

C
$\frac{\mu_0 I}{4 \pi^2 R}$

D
$\frac{\mu_0 l}{\pi^2 R}$

The correct answer is:
$\frac{\mu_0 l}{\pi^2 R}$

Solution:
$\mathrm{d} \overrightarrow{\mathrm{B}}=\frac{\mu_0 \mathrm{di}}{2 \pi \mathrm{R}}[-\cos \theta \hat{\mathbf{i}}-\sin \theta \hat{\mathbf{j}}]$
$\mathrm{di}=\frac{\mathrm{T}}{\pi \mathrm{R}} \mathrm{Rd} \theta$
$=\frac{\mathrm{I}}{\pi} \mathrm{d} \theta$
$\mathrm{d} \overrightarrow{\mathrm{B}}=\frac{\mu_0 \mathrm{I}}{2 \pi^2 \mathrm{R}}(-\cos \theta \hat{\mathrm{i}}-\sin \theta \hat{\mathrm{j}})$
$\vec{B}=-\frac{\mu_0 I}{\pi^2 R} \hat{j}$

Topic: Physics, Magnetic Effects of Current, JEE Main, JEE Main 2011
A current ' $I^{\prime}$ is flowing along an infinite, straight wire, in the positive $Z$ -direction and the same current is flowing along a similar parallel wire $5 \mathrm{m}$ apart, in the negative Z-direction. A point $P$ is at a perpendicular distance $3 \mathrm{m}$ from the first wire and $4 \mathrm{m}$ from the second. What will be magnitude of the magnetic field B of $P ?$

Options:
A $\frac{5}{12}\left(u_{0} I\right)$
B $\frac{7}{24}\left(\mu_{0} I\right)$
C $\frac{5}{24}\left(\mu_{0} I\right)$
D $\frac{25}{288}\left(\mu_{0} I\right)$

The correct answer is:
$\frac{5}{24}\left(\mu_{0} I\right)$

Solution:
According to the question.

Magnetic field due to first wire $(A)$ is $B_{1}=\frac{\mu_{0} \times I}{2 \pi \times 3}$
Magnetic field due to second wire $(B)$ is $B_{2}=\frac{\mu_{0} \times 1}{2 \pi \times 4}$
Net magnitude of magnetic field $B=\sqrt{B_{1}^{2}+B_{2}^{2}}$
$$
B=\frac{\mu_{0} I}{2 \pi} \sqrt{\frac{1}{9}+\frac{1}{16}}=\frac{\mu_{0} I \times 5}{2 \pi \times 3 \times 4}
$$
Magnetic field $B=\frac{5}{24} \times \frac{\mu_{0} I}{\pi}$

Topic: Physics, Magnetic Effects of Current, JEE Main
A current \( I \) is flowing along an infinite, straight wire, in the positive \( Z- \) direction and the same current is flowing along a similar parallel wire \( 5 \mathrm{~m} \) apart, in the negative \( Z- \) direction. A point \( P \) is at a perpendicular distance \( 3 \mathrm{~m} \) from the first wire and \( 4 \mathrm{~m} \) from the second. What will be the magnitude of the magnetic field \( B \) of \( P \) ?

Options:
A \( \frac{5}{12}\left(\frac{\mu_{0} I}{\pi}\right) \)
B \( \frac{7}{24}\left(\frac{\mu_{0} I}{\pi}\right) \)
C \( \frac{5}{24}\left(\frac{\mu_{0} I}{\pi}\right) \)
D \( \frac{25}{288}\left(\frac{\mu_{0} I}{\pi}\right) \)

The correct answer is:
\( \frac{5}{24}\left(\frac{\mu_{0} I}{\pi}\right) \)

Solution:
According to the question,

Magnetic field due to first wireA is B1=μ0×I2π×3

Magnetic field due to second wire (B) is B2=μ0×I2π×4

Net magnitude of magnetic field B=B12+B22

B=μ0I2π19+116=μ0I×52π×3×4

Magnetic field B=524×μ0Iπ

Topic: Physics, Magnetic Effects of Current, JEE Main
A current $i$ is flowing in a straight conductor of length $L$. The magnetic induction at a point on its axis at a distance $\frac{L}{4}$ from its centre will be :

Options:
A
Zero

B
$\frac{\mu_0 i}{2 \pi L}$

C
$\frac{\mu_0 i}{\sqrt{2} L}$

D
$\frac{4 \mu_0 i}{\sqrt{5} \pi L}$

The correct answer is:
Zero

Solution:
Magnetic field at any point lies on axial position of current carrying conductor $\mathrm{B}=0$

Topic: Physics, Magnetic Effects of Current, JEE Main, JEE Main 2013 (22 Apr Online)
A current ' $i$ ' is flowing through a wire of length ' $L$ '. If it is made into a circular loop of one turn, then its magnetic moment is

Options:
A $\frac{\mathrm{L}^2 \mathrm{i}}{4 \pi}$
B $\frac{\mathrm{L}^2}{4 \pi}$
C $\frac{4 \pi}{\mathrm{L}^2 \mathrm{i}}$
D $4 \pi \mathrm{L}^2 \mathrm{i}$

The correct answer is:
$\frac{\mathrm{L}^2 \mathrm{i}}{4 \pi}$

Solution:
Length, $L=2 \pi r \Rightarrow r=\frac{L}{2 \pi}$
Magnetic moment, $\mathrm{M}=\mathrm{iA}=\mathrm{i} \pi\left(\frac{\mathrm{L}}{2 \pi}\right)^2$
$=\mathrm{i} \times \pi \times \frac{\mathrm{L}^2}{4 \pi^2} \quad \therefore \mathrm{M}=\frac{\mathrm{IL}^2}{4 \pi}$

Topic: Physics, Magnetic Effects of Current, JEE Main
A current i is flowing through the loop. The direction of the current and the shape of the loop are as shown in the figure. The magnetic field at the centre of the loop is
(MA=R, MB=2R, â DMA=90Â°)

Options:
A 516Î¼oiR, but out of the plane of the paper
B 5 16 μ 0 i R , but into the plane of the paper
C 716Î¼oiR, but out of the plane of the paper
D 716Î¼oiR, but into the plane of the paper

The correct answer is:
716μoiR, but into the plane of the paper

Solution:
(i) The magnetic field at the centre due to the curved portion DA=μ0i4πR 3π2
According to the right-hand screw rule, the magnetic field will be into the plane of the paper.
(ii) The magnetic field at M due to AB is zero.
(iii) The magnetic field at the centre due to the curved portion BC is μ0i4π2R π2.

According to the right-hand screw rule, the magnetic field will be into the plane of the paper.
(iv) The magnetic field at M due to DC is zero.
Hence, the resultant magnetic field at M
BM=Bsmall loop+Bbig loop
Both are perpendicular downward
BM=μOi4πR32π+μOi4π2Rπ2
BM=716μOiR ⨂

Topic: Physics, Magnetic Effects of Current, JEE Main
A current $/$ is passing through a wire having two sections $P$ and $Q$ of uniform diameters $d$ and $d / 2$ respectively. If the mean drift velocity of electrons in sections $P$ and $Q$ is denoted by $v_P$ and $v_Q$ respectively, then

Options:
A $v_{\mathrm{p}}=v_{\mathrm{Q}}$
B $v_p=\frac{1}{2} v_Q$
C $v_p={ }^{\frac{1}{4}} v_Q$
D $v_{\mathrm{p}}=2 v_Q$

The correct answer is:
$v_p={ }^{\frac{1}{4}} v_Q$

Solution:
Drift velocity $v_d=\frac{i}{n e A} \Rightarrow v_d \propto \frac{1}{A}$ or $v_d \propto \frac{1}{d^2}$
$\Rightarrow \frac{v_P}{v_Q}=\left(\frac{d_Q}{d_p}\right)^2=\left(\frac{d / 2}{d}\right)^2=\frac{1}{4} \Rightarrow v_P=\frac{1}{4} v_Q$.

Topic: Physics, Current Electricity, JEE Main
A current loop consists of two identical semicircular parts each of radius $R$, one lying in the $x-y$ plane and the other in $x-z$ plane. If the current in the loop is $i$. The resultant magnetic field due to the two semicircular parts at their common centre is

Options:
A $\frac{\mu_0 i}{2 \sqrt{2} R}$
B $\frac{\mu_0 \mathrm{i}}{2 R}$
C $\frac{\mu_0 \mathrm{i}}{4 \mathrm{R}}$
D $\frac{\mu_0 \mathrm{i}}{\sqrt{2} \mathrm{R}}$

The correct answer is:
$\frac{\mu_0 i}{2 \sqrt{2} R}$

Solution:
The magnetic field at centre $O$ for each semicircular parts each of radius $R$,
$$
\mathrm{B}_1=\mathrm{B}_2=\frac{\mu_0 \mathrm{i}}{4 \mathrm{R}}
$$
The magnetic field at their common centre
$$
\begin{aligned}
\overrightarrow{\mathrm{B}} & =\overrightarrow{\mathrm{B}}_1+\overrightarrow{\mathrm{B}}_2 \\
\mathrm{~B} & =\sqrt{\mathrm{B}_1^2+\mathrm{B}_2^2} \\
& =\sqrt{\left(\frac{\mu_0 \mathrm{i}}{4 \mathrm{R}}\right)^2+\left(\frac{\mu_0 \mathrm{i}}{4 \mathrm{R}}\right)^2} \\
\mathrm{~B} & =\frac{\mu_0 \mathrm{i}}{2 \sqrt{2} \mathrm{R}}
\end{aligned}
$$

Topic: Physics, Magnetic Effects of Current, NEET, NEET 2010 (Mains)
A current of 1.5 A is flowing through a triangle, of side 9 cm each. The magnetic field at the centroid of the triangle is :
(Assume that the current is flowing in the clockwise direction.)

Options:
A 23×10-5 T, inside the plane of triangle
B 23×10-7 T, outside the plane of triangle
C 3×10-7 T, outside the plane of triangle
D 3×10-5 T, inside the plane of triangle

The correct answer is:
3×10-5 T, inside the plane of triangle

Solution:
B1=μ0I4πR2sin60=4π×10-7×1.54π.332×10-2×2×32=1×10-5
Bnet=3B1=3×10-5T⊗

Topic: Physics, Magnetic Effects of Current, JEE Main, JEE Main 2021 (31 Aug Shift 2)
A current of 1.93 ampere is passed through 200 mL of 0.5 M Zinc sulphate (aq.) solution for 50 min with a current efficiency of 80%. If volume of solution remain constant, then [Zn2+] after deposition of Zn2+ is:

Options:
A 0.38 M
B 0.26 M
C 0.35 M
D 0.076 M

The correct answer is:
0.38 M

Solution:
No. of Faraday =1.93Ã50Ã6096500=0.06
â´ Moles of Zn2+ deposited =80100Ã0.062=0.024
∴  Zn 2+  =  0.5 × 0.2 − 0.024 0.2 = 0.38 M

Topic: Chemistry, Electrochemistry, JEE Main
A current of $0.1 \mathrm{~A}$ circulates around a coil of 100 turns and having a radius equal to $5 \mathrm{~cm}$. The magnetic field set up at the celutre of the coil is
$\left(\mu_0=4 \pi \times 10^{-7}\right.$ weber $/$ ampere - metre $)$

Options:
A $4 \pi \times 10^{-5}$ tesla
B $8 \pi \times 10^{-5}$ tesla
C $4 \times 10^{-5}$ tesla
D $2 \times 10^{-5}$ tesla

The correct answer is:
$4 \pi \times 10^{-5}$ tesla

Solution:
$B=\frac{\mu_0 N i}{2 r}=\frac{4 \pi \times 10^{-7} \times 100 \times 0.1}{2 \times 5 \times 10^{-2}}=4 \pi \times 10^{-5} \text { Tes/a }$

Topic: Physics, Magnetic Effects of Current, JEE Main
A current of $0.1$ A flows through a $25 ~\Omega$ resistor represented by the circuit diagram. The current in the $80 ~\Omega$ resistor is

Options:
A $0.1 \mathrm{~A}$
B $0.2 \mathrm{~A}$
C $0.3 \mathrm{~A}$
D $0.4 \mathrm{~A}$

The correct answer is:
$0.3 \mathrm{~A}$

Solution:
$0.1 \times\left(25+\frac{20 \times 60}{20+60}\right)=i_{2} \times 20$
$I_{2}=0.2 \mathrm{~A}$
Hence, $\mathrm{i}$ through $80 \Omega$
$0.1+0.2=0.3 \mathrm{~A}$

Topic: Physics, Current Electricity, JEE Main
A current of 1 A is flowing along positive $x$ -axis through a straight wire of length $0.5 \mathrm{m}$ placed in a region of a magnetic field given by $\mathbf{B}=(2 \mathbf{i}+4 \mathbf{j}) \mathrm{T} .$ The magnitude and the direction of the force experienced by the wire respectively are

Options:
A $\sqrt{18} \mathrm{N},$ along positive $z$ -axis
B $\sqrt{20} \mathrm{N}$, along positive $x$ -axis
C $2 \mathrm{N}$, along positive $z$ -axis
D $4 \mathrm{N},$ along positive $y$ -axis

The correct answer is:
$2 \mathrm{N}$, along positive $z$ -axis

Solution:
The force experienced by the wire placed in magnetic field, $F=B il$
$$
\begin{array}{l}
=(2 \hat{i}+4 \hat{j})\left(1 \hat{i} \times \frac{1}{2}\right) \\
=\hat{i} \times \hat{i}+4 \times \frac{1}{2}(\hat{i} \times \hat{j})=2 \hat{k} N
\end{array}
$$
Direction of this force can be find out by Fleming's left hand rule which is along pasitive z-axis

Topic: Physics, Magnetic Effects of Current, JEE Main
A current of $1 \mathrm{~A}$ is flowing on the sides of an equilateral triangle of side $4.5 \times 10^{-2} \mathrm{~m}$. The magnetic field at the centre of the triangle will be:

Options:
A
$4 \times 10^{-5} \mathrm{~Wb} / \mathrm{m}^2$

B
Zero

C
$2 \times 10^{-5} \mathrm{~Wb} / \mathrm{m}^2$

D
$8 \times 10^{-5} \mathrm{~Wb} / \mathrm{m}^2$

The correct answer is:
$4 \times 10^{-5} \mathrm{~Wb} / \mathrm{m}^2$

Solution:
Here, side of the triangle, $l=4.5 \times 10^{-2} \mathrm{~m}$, current, $\mathrm{I}=1 \mathrm{~A}$
magnetic field at the centre of the triangle ' $O$ ' $B=$ ?
From figure, $\tan 60^{\circ}=\sqrt{3}=\frac{1}{2 d}$
$$
\Rightarrow d=\frac{l}{2 \sqrt{3}}=\left(\frac{4.5 \times 10^{-2}}{2 \sqrt{3}}\right) \mathrm{m}
$$
Magnetic field, $B=\frac{\mu_0 i}{4 \pi d}\left(\cos \theta_1+\cos \theta_2\right)$
Putting value of $\mu=4 \pi \times 10^{-7}$ and $\theta_1$ and $\theta_2$ we will get net magenetic field $=3 \times B=4 \times 10^{-5} \mathrm{~Wb} / \mathrm{m}^2$

Topic: Physics, Magnetic Effects of Current, JEE Main, JEE Main 2018 (15 Apr Shift 2 Online)