datavorous/entrance-exam-dataset huggingface dataset:
A current carrying closed loop in the form of a right angle isosceles triangle $A B C$ is placed in a uniform magnetic field acting along $A B$. If the magnetic force on the $\operatorname{arm} B C$ is $\mathbf{F}$, the force on the arm $A C$ is

Options:
A $-\mathbf{F}$
B $\mathbf{F}$
C $\sqrt{2} \mathbf{F}$
D $-\sqrt{2} \mathbf{F}$

The correct answer is:
$-\mathbf{F}$

Solution:
$$
\begin{gathered}
\overrightarrow{\mathbf{F}}_{A B}=0 \\
\overrightarrow{\mathbf{F}}_{A B}+\overrightarrow{\mathbf{F}}_{B C}+\overrightarrow{\mathbf{F}}_{C A}=0 \\
\overrightarrow{\mathbf{F}}_{B C}+\overrightarrow{\mathbf{F}}_{C A}=0 \\
\overrightarrow{\mathbf{F}}_{C A}=-\overrightarrow{\mathbf{F}}_{B C}=-\overrightarrow{\mathbf{F}}
\end{gathered}
$$

Topic: Physics, Magnetic Effects of Current, NEET, NEET 2011 (Screening)
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, NEET
A current I=10sin100Ït A is passed in first coil, which induces a maximum EMF of 5Ï V in second coil. The mutual inductance between the coils is

Options:
A 5 mH
B 10 mH
C 15Â mH
D 20Â mH

The correct answer is:
5 mH

Solution:
Let current is I=I0 sin ωt,
Where I0 =10, ω=100 π
Then induced e.m.f. is given by
ω=-MdIdt=-MddtI0
ω=-MI0ωcos⁡ωt
∴ Maximum e.m.f. is
ω_max⁡=MI0ω
5π = M × 10 × 100π
M = 5 mH

Topic: Physics, Electromagnetic Induction, NEET
A current I=20sin100Ït A is passed in the first coil, which induces a maximum emf of 10Ï V in the second coil. The mutual inductance for the pair of coils is

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

The correct answer is:
5 mH

Solution:
Let I=I0sinωt,

Where I0=20 A, ω=100π rad s-1

Then, ε=MdIdt=MdI0sinωtdt=MI0ωcosωt
⇒εmax=MI0ω=10π

M=10π20×100π

M=0.005 H=5 mH

Topic: Physics, Electromagnetic Induction, NEET
A current I flows along the length of an infinitely long, straight, thin-walled pipe. Then

Options:
A the magnetic field at all point inside the pipe is the same, but not zero
B the magnetic field at any point inside the pipe is zero
C the magnetic field is zero only on the axis of the pipe
D the magnetic field is different at different points inside the pipe

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

Solution:
Using Ampere’s circuital law over a circular loop of any radius less than the radius of the pipe, we can see that net current inside the loop is zero. Hence, magnetic field at every point inside the loop will be zero.

Topic: Physics, Magnetic Effects of Current, NEET
A current 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:
Using Ampere's circuital law the magnetic field at any point inside the pipe is zero.

Topic: Physics, Magnetic Effects of Current, NEET
A current I flows in a circular arc of wire which subtends an angle 3Ï2 radians at its centre, whose radius is R. The magnetic induction B at the centre is

Options:
A μ0iR
B μ0i2R
C 2μ0iR
D 3μ0i8R

The correct answer is:
3μ0i8R

Solution:
For a loop, magnetic induction at centre
B=μ04π×2πir
When loop subtends angle θ at centre, then
B=μ04π ×θir (Here angle is θ instead of 2π )
In the given problem, θ=3π2 and r=R
∴ B=μ04π ×3π2 ×iR=3μ0i8R

Topic: Physics, Magnetic Effects of Current, NEET
A current I is flowing through the loop, as shown in the figure. The magnetic field at centre O is

Options:
A 7Î¼0I16 Râ
B 7Î¼0I16 Râ¨
C 5Î¼0I16 Râ
D 5Î¼0I16 Râ¨

The correct answer is:
7Î¼0I16 Râ

Solution:
B=B1+B2
=Î¼0I2r1 3Ï/22Ï+Î¼0I2r2 Ï/22Ï
=38 Î¼0IR+Î¼0I22 R Ã 14
=716 Î¼0IRâ

Topic: Physics, Magnetic Properties of Matter, NEET
A current i is uniformly distributed over the cross-section of a long hollow cylindrical wire of inner radius R1 and outer radius R2. Magnetic field B varies with distance from the axis of the cylinder as

Options:
A
B
C
D

The correct answer is:

Solution:
Using Ampere's circuital law
B = 0 for râ¤R1
B=Î¼0i2ÏR22-R12r2-R12r
For R1â¤râ¤R2
and B=Î¼02Ïir r â¥ R 2
The corresponding B-r graph will be as shown in

Topic: Physics, Magnetic Effects of Current, NEET
A current i is uniformly distributed over the cross-section of a long hollow cylindrical wire of inner radius R1 and outer radius R2. Magnetic field B varies with distance r from the axis of the cylinder as

Options:
A
B
C
D

The correct answer is:

Solution:
r < R1, B=0
r > R2 , b=Î¼0i2Ïr
R1 < r < R2, B=Î¼0ir2-R122ÏrR22-R12

Topic: Physics, Electromagnetic Induction, NEET
A current loop ABCD is held fixed on the plane of the paper as shown in the figure. The arcs BC (radius = b) and DA (radius = a) of the loop are joined by two straight wires AB and CD. A steady current I is flowing in the loop. Angle made by AB and CD at the origin O is 30Â°. Another straight thin wire with steady current I1 flowing out of the plane of the paper is kept at the origin.

The magnitude of the magnetic field (B) due to loop ABCD at the origin (O) is

Options:
A zero
B Î¼0Ib-a24ab
C μ 0 I 4 π b - a ab
D μ 0 I 4 π 2 b - a + π 3 a + b

The correct answer is:
μ0Ib-a24ab

Solution:
O is along the line CD and AB. They do not contribute to the magnetic induction at O. The field due to DA is positive or out of the paper and that due to BC is into the paper or negative.
The total magnetic field due to loop ABCD at O is B = B AB + B BC + B CD + B DA

â B=0-Î¼0I4ÏbÃÏ6+0+Î¼0I4ÏaÃÏ6
â B=Î¼0I24abb-a, out of the paper or positive.

Topic: Physics, Magnetic Effects of Current, NEET
A current loop in a magnetic field

Options:
A experiences a torque whether the field is uniform or non-uniform in all orientations
B can be in equilibrium in one orientations
C can be equilibrium in wo orientations, both the equilibrium states are unstable
D can be in equilibrium in two orientations, one stable while the other is unstable

The correct answer is:
can be in equilibrium in two orientations, one stable while the other is unstable

Solution:
For parallel $\mathbf{M}$ is stable and for antiparallel is unstable.

Topic: Physics, Magnetic Effects of Current, NEET, NEET 2013 (All India)
A current of 1 A is flowing through a straight conductor of length 16 cm. The magnetic induction (in tesla) at a point 10 cm from the either end of the wire is:

Options:
A 83×10-6
B 162×10-5
C 163×10-6
D 36×10-6

The correct answer is:
83×10-6

Solution:
B=μ0I4πrsin⁡ϕ1+sin⁡ϕ2

Topic: Physics, Magnetic Effects of Current, NEET
A current of 2.34 A flows in resistance of 11.111111 Î©. The potential difference across the given resistance with due regard for the significant figure is

Options:
A 26.000 V

B 26.00 V

C 26.0 V

D 26 V

The correct answer is:
26.0 V

Solution:
V=2.34×11.11111 volt =26.0 volt.
[Because the final result should contain three significant figure.
Why three because out of two multiplying data least number of significant number is three which is in 2.34]

Topic: Physics, Units and Dimensions, NEET
A current of 2 A flows through a 2 Î© resistor when connected across a battery. The same battery supplies a current of 0.5 A when connected across a 9 Î© resistor. The internal resistance of the battery is

Options:
A 13 Î©
B 14 Ω
C 5Â Î©
D 0.5Â Î©

The correct answer is:
13 Î©

Solution:
Current I=ER+r
⇒2=E2+r .... (i)
⇒0.5=E9+r .... (ii)
On dividing equation (i) by equation (ii), we get
20.5=9+r2+r
⇒4=9+r2+r
⇒3r=1
⇒r=13 Ω

Topic: Physics, Current Electricity, NEET
A current of 2 A flows through a $2 \Omega$ resistor when connected across a battery. The same battery supplies a current of 0.5 A when connected across a $9 \Omega$ resistor. The internal resistance of the battery is

Options:
A $1 / 3 \Omega$
B $1 / 4 \Omega$
C $1 \Omega$
D $0.5 \Omega$

The correct answer is:
$1 / 3 \Omega$

Solution:
$R=\frac{E}{z+r}$
$$
\begin{aligned}
& 2=\frac{E}{2+r} \\
& 0.5=\frac{E}{9+r}
\end{aligned}
$$
From Eqs. (i) and (ii), We have
$$
\begin{aligned}
\frac{2}{0.5} & =\frac{9+r}{2+r} \\
4 & =\frac{9+r}{2+r} \\
3 r & =1 \\
r & =\frac{1}{3} \Omega
\end{aligned}
$$

Topic: Physics, Current Electricity, NEET, NEET 2011 (Screening)
A current of 2A is increasing at a rate of 4 A/s through a coil of inductance 2 H. The energy stored in the inductor per unit time in given instant is

Options:
A 2 J/s
B 1 J/s
C 16 J/s
D 4 J/s

The correct answer is:
16 J/s

Solution:
Potential difference across coil is V = L di dt
or V = (2) (4) = 8V
Now energy stored per unit time
= power = Vi
= (8) (2) = 16 J/s

Topic: Physics, Electromagnetic Induction, NEET
A current of 3 A flows through the $2 \Omega$ resistor shown in the circuit. The power dissipated in the $5 \Omega$ resistor is

Options:
A $4 \mathrm{~W}$
B $2 \mathrm{~W}$
C $1 \mathrm{~W}$
D $5 \mathrm{~W}$

The correct answer is:
$5 \mathrm{~W}$

Solution:
Key Idea : Voltage across $2 \Omega$ is same as voltage across arm containing $1 \Omega$ and $5 \Omega$ resistances. Voltage across $2 \Omega$ resistance,
$V=2 \times 3=6 \mathrm{~V}$
So, voltage across lowest arm,
$V_1=6 \mathrm{~V}$
Current across $5 \Omega, I=\frac{6}{1+5}=1 \mathrm{~A}$
Thus, power across $5 \Omega$,
$P=I^2 R=(1)^2 \times 5=5 \mathrm{~W}$

Topic: Physics, Current Electricity, NEET, NEET 2008 (Screening)
A current of $2.5 \mathrm{~A}$ flows through a coil of inductance $5 \mathrm{H}$. The magnetic flux linked with the coil is:

Options:
A $0.5 \mathrm{~Wb}$
B $12.5 \mathrm{~Wb}$
C zero
D $2 \mathrm{~Wb}$

The correct answer is:
$12.5 \mathrm{~Wb}$

Solution:
According to the question
Current $I=2.5 \mathrm{~A}$ and Induction $L=5 \mathrm{H}$
$\therefore$ Magnetic flux linked with coil is
$$
\begin{aligned}
\phi & =L I=(5 \mathrm{H}) \\
(2.5 \mathrm{~A}) & =12.5 \mathrm{~Wb}
\end{aligned}
$$

Topic: Physics, Electromagnetic Induction, NEET, NEET 2013 (Karnataka)
A current-carrying circular coil is bent so as to convert it into a double loop. Both the loops are concentric and are carrying current in the same direction. If initially, B is the magnetic field at the centre of the coil, then finally magnetic field at the centre will be

Options:
A zero
B B
C 2B
D 4B

The correct answer is:
4B

Solution:
Let's assume that the current in the coil is Initially the field at the centre is i and the radius of the coil is r, then the initial magnetic field at the centre is

Binitial=μ0i2r

When we make two loops out of one, we are basically making a coil of half the radius and double the number of turns, so
Bfinal=2μ0i2r2=2μ0ir⇒Bfinal=4Binitial=4B

Topic: Physics, Magnetic Effects of Current, NEET
A current-carrying coil is subjected to a uniform magnetic field. The coil will orient so that its plane becomes

Options:
A Inclined at 45° to the magnetic field
B Inclined at any arbitrary angle to the magnetic field
C Parallel to the magnetic field
D Perpendicular to the magnetic field

The correct answer is:
Perpendicular to the magnetic field

Solution:
Torque on the coil will be zero when the plane of the coil is perpendicular to the magnetic field.

Topic: Physics, Magnetic Effects of Current, NEET
A current-carrying square loop is placed near an infinitely long current-carrying wire as shown in the figure. The torque acting on the loop is

Options:
A Î¼02Ï i1i2r
B Î¼02ÏÂ i1i2logeâ¡2
C Î¼02Ïi1i2r2
D Zero

The correct answer is:
Zero

Solution:
The force acting on the loop PQRSP due to current i1 through the long wire are acting in the plane of loop. Due to which, the torque on loop is zero.

Topic: Physics, Magnetic Effects of Current, NEET
A current-carrying straight wire is kept along the axis of a circular loop carrying a current. The straight wire

Options:
A Will exert an inward force on the circular loop
B Will exert an outward force on the circular loop
C Will exert a force on the circular loop parallel to itself
D Will not exert any force on the circular loop

The correct answer is:
Will not exert any force on the circular loop

Solution:
According to Fleming's left hand rule, direction of magnetic field at the centre O.
The force on the current carrying conductor is F=idlâÃ Bâ
As angle between dl and B is zero
F=0
The straight wire will not exert any force on the circular loop.

Topic: Physics, Magnetic Effects of Current, NEET
A current-carrying straight wire is kept along the axis of a circular loop carrying a current. The straight wire

Options:
A Will exert an inward force on the circular loop
B Will exert an outward force on the circular loop
C Will exert a force on the circular loop parallel to itself
D Will not exert any force on the circular loop

The correct answer is:
Will not exert any force on the circular loop

Solution:
According to Fleming's left hand rule, direction of magnetic field at the centre O.
The force on the current carrying conductor is F=idlÃB
As angle between dl and B is zero
â´ F=0
The straight with will not exert any force on the circular loop.

Topic: Physics, Magnetic Effects of Current, NEET
A curved road of diameter 1.8 km is banked so that no friction is required at a speed of 30  m s −1 . What is the banking angle?

Options:
A tan-10.1
B tan-10.3
C tan-10.9
D tan-11.5

The correct answer is:
tan-10.1

Solution:
As, tanâ¡Î¸=v2rg
Where, Î¸= the angle of banking.
â tanâ¡Î¸=30Ã3010Ã0.9Ã103=0.1
â Î¸=tan-1â¡0.1

Topic: Physics, Laws of Motion, NEET
A cycle followed by an engine (made of one mole of an ideal gas in a cylinder with a piston) is shown in Fig. 12.11. Find heat exchanged by the engine, with the surroundings for section BC of the cycle. (Cv = (3/2) R)
AB : constant volume BC : constant pressure CD : adiabatic DA : constant pressure

Options:
A (5/2)PB(VC+VA)
B (5/2)PB(VC-VA)
C UBC+WBC
D UBC-WBC

The correct answer is:
UBC+WBC

Solution:
Correct Option is : (C)
UBC+WBC

Topic: Chemistry, Thermodynamics, NEET
A cycle wheel of radius $0.5 \mathrm{~m}$ is rotated with constant angular velocity of $10 \mathrm{rad} / \mathrm{s}$ in a region of magnetic field of $0.1 \mathrm{~T}$ which is perpendicular to the plane of the wheel. The EMF generated between its centre and the rim is

Options:
A $0.25 \mathrm{~V}$
B $0.125 \mathrm{~V}$
C $0.5 \mathrm{~V}$
D zero

The correct answer is:
$0.125 \mathrm{~V}$

Solution:
When a conducting disc or wheel of radius $r$ rotates with constant angular velocity of $\omega$ about its axis in a uniform magnetic field perpendicular to its plane and parallel to its axis of rotation, then,
Induced emf is given by, $\mathrm{e}=\frac{1}{2}{\mathrm{~B} \omega \mathrm{r}^2}^2$
Here, $B=0.1 \mathrm{~T}, \omega=10 \mathrm{rad} / \mathrm{s}, \mathrm{r}=0.5 \mathrm{~m}$
Substituting these values in Eq. (i), we get
$$
\therefore \quad \mathrm{e}=\frac{1}{2} \times 0.1 \times 10 \times(0.5)^2=\frac{1}{8} \mathrm{~V}=0.125 \mathrm{~V}
$$

Topic: Physics, Electromagnetic Induction, NEET, NEET 2019 (Odisha)
A cyclic process ABCA shown in the V-T diagram is performed with a constant mass of an ideal gas. The same process in the P-V diagram is

Options:
A
B
C
D

The correct answer is:

Solution:
From the V-T diagram, process A to B is an isobaric expansion which is only in option A.
Process B to C is isochoric with a decrease in temperature and hence the pressure.
Process C to A is an isothermal compression.

Topic: Physics, Thermodynamics, NEET
A cyclic process is shown in the figure. Work done during the cyclic process ABCDA is

Options:
A 160 J
B 150 J
C 600 J
D 900 J

The correct answer is:
150 J

Solution:
Work done during cyclic process ABCDA
= area enclosed by ABCDA
= AB + CD 2 × AD
= 2 + 1 2 × 1 0 2
= 1 . 5 × 1 0 2 J
= 150 J

Topic: Physics, Thermodynamics, NEET
A cyclic process is shown in the p-T diagram. Which of the following curves shows the same process on the p-V diagram?

Options:
A
B
C
D

The correct answer is:

Solution:
∵  pVT= constant (for a given mass of an ideal gass)
AB: Isochoric process, V= constant
T↑  and  p↑
BC: Isobaric process
P = constant
T↓ hence V↓
CA: Isothermal process
p is decreasing hence, V is increasing, it means that p-V diagram will be rectangular hyperbola.

Topic: Physics, Thermodynamics, NEET
A cyclic process is shown on the V-T diagram. The same process on a P-T diagram is shown by

Options:
A
B
C
D

The correct answer is:

Solution:
AB Isobaric expansion
BC Isothermal expansion
CD Isobaric compression
DA Isothermal compression
So, when the graph is plotted between P & T

Topic: Physics, Thermodynamics, NEET
A cyclist taking a turn, bends inwards while a car passenger taking the same turn is thrown outwards. The reason is

Options:
A car is heavier than cycle
B car has four wheels while cycle has only two
C difference in the speed of the two
D cyclist has to counteract the centrifugal force while in the case of car only the passenger is thrown by this force

The correct answer is:
cyclist has to counteract the centrifugal force while in the case of car only the passenger is thrown by this force

Solution:
Cyclist has to counteract the centrifugal force while in the case of car only the passenger is thrown by this force

Topic: Physics, Laws of Motion, NEET
A cyclist turns around a curve at 15 miles hr-1. If he turns at double the speed, the tendency to overturn is

Options:
A Doubled
B Quadrupled
C Halved
D Unchanged

The correct answer is:
Quadrupled

Solution:
F=mv2r ⇒F∝v2. If v becomes double then F (tendency to overturn) will become four times.

Topic: Physics, Laws of Motion, NEET
A cyclotron is accelerating proton, where the applied magnetic field is 2 T and the potential gap is 100 keV. To acquire kinetic energy of 20 MeV, the number of turns, the proton has to move between the dees is

Options:
A 200
B 300
C 150
D 100

The correct answer is:
100

Solution:
Energy gained in one movement across the gap is 100 keV
Therefore the energy gained in one turn is 200 keV
N=20 MeV200 keV=20×106200×103=100

Topic: Physics, Magnetic Effects of Current, NEET
A Cylinder containing an ideal gas (0.1 mol of 1.0dm3 ) is in thermal equilibrium with a large volume of 0.5 molal aqueous solution of ethylene glycol at its freezing point. If the stoppers S1 and S2 (as shown in the figure) are suddenly withdrawn, the volume of the gas in litres after equilibrium is achieved will be
(Given, Kf (water) = 2.0 K kg mol-1 , R = 0.08 dm3 atm K-1mol-1 )

Options:
A 1.14
B 0.57
C 2.18
D 4.36

The correct answer is:
2.18

Solution:
ΔTf=Kf×i×m
ΔTf=2.0×1×0.5
ΔTf=1
273-Ti=1
Ti=272K
P=nRTV
P=0.1×0.08×2721
P=2.176 atm
Apply Boyle's law
P1V1=P2V2
2.176×1=1×V2
V2=2.17

Topic: Chemistry, Solutions, NEET
A cylinder contains hydrogen gas at pressure of 249 kPa and temperature 27oC. Its density is: R=8.3 JK Mol−1 K−1

Options:
A 0.2 kg m-3
B 0.1 kg m-3
C 0.02 kg m-3
D 0.5 kg m-3

The correct answer is:
0.2 kg m-3

Solution:
For an ideal gas sample
\(\begin{aligned}
& \frac{\mathrm{P}}{\rho}=\frac{\mathrm{RT}}{\mathrm{MW}} \\
& \rho=\frac{\mathrm{PMW}}{\mathrm{RT}}=\frac{249 \times 10^3 \times 2 \times 10^{-3}}{8.314 \times 300}=0.199 \\
& \rho=0.2 \mathrm{~kg} / \mathrm{m}^3
\end{aligned}\)

Topic: Physics, Kinetic Theory of Gases, NEET, NEET 2020 (Phase 1)
A cylinder of mass 10 kg and radius 15 cm is rolling perfectly on a plane of inclination 30Â° and length 1 m. If the coefficient of static friction is Î¼=0.25, then the work done against friction during rolling is

Options:
A 0
B 2532 J
C 5032 J
D 7532 J

The correct answer is:
0

Solution:
The work done by friction force in case of pure rolling is always zero because the point of application of friction force doesn't slip

Topic: Physics, Rotational Motion, NEET
A cylinder rolls down an inclined plane of inclination 30°, the acceleration of cylinder is

Options:
A g3
B g
C g2
D 2g3

The correct answer is:
g3

Solution:
Let a plane be inclined at an angle Î¸ and a cylinder rolls down then the acceleration of the cylinder of mass m, radius R, and I as moment of inertia is given by
a=gsinâ¡Î¸1+ImR2
Moment of inertia (I) of a cylinder =mR22
â´ a=gsinâ¡Î¸mR21+2mR2=23gsinâ¡30Â°
â a=g3

Topic: Physics, Rotational Motion, NEET
A cylinder rolls up an inclined plane, reaches some height, and then rolls down (without slipping throughout these motions). The directions of the frictional force acting on the cylinder are

Options:
A up the incline while ascending and down the incline while descending
B up the incline while ascending as well as descending
C down the incline while ascending and up the incline while descending
D down the incline while ascending as well as descending

The correct answer is:
up the incline while ascending as well as descending

Solution:
When cylinder rolls up the incline angular velocity decreases for that torque of friction should be in opposite direction of angular velocity. Therefore friction must be acting up the incline.
When cylinder rollsdown the incline angular velocity increases for that torque of friction should be in same direction of angular velocity. Therefore friction must be acting up the incline.

Topic: Physics, Rotational Motion, NEET
A cylindrical conductor of diameter 0.1âmm carries a current of 90 mA. The current density (in A m-2 ) is Ïâ3

Options:
A 1.2×107
B 3×106
C 6×106
D 2.4×107

The correct answer is:
1.2×107

Solution:
Given,
The diameter of the cylindrical conductor D=0.01 mm
Radius r=D2=0.12 mm
Current I=90 mA=90Ã10-3A
We know that,
Current density J=IA
=IÏr2
=90Ã10-3227Ã0.1Ã10-3 22
=90Ã10-3 3.14Ã0.1Ã10-3 22
=12000Ã103
=1.2Ã107 A m-2

Topic: Physics, Current Electricity, NEET
A cylindrical conductor of radius R is carrying a constant current. The plot of the magnitude of the magnetic field B with the distance d from the centre of the conductor is correctly represented by the figure.

Options:
A
B
C
D

The correct answer is:

Solution:
Inside (d<R)

Magnetic field inside conductor

B=Kd ..(i)

This is straight line passing through origin

At surface (d=R)

B=μ02π1d

Maximum at surface

Outside (d>R)

B=μ02π1d

or B∝1d ∴ Hyperbolic

Hence the correct option is (B)

Topic: Physics, Magnetic Effects of Current, NEET, NEET 2019 (All India)
A cylindrical metallic rod in thermal contact with two reservoirs of heat at its two ends conducts an amount of heat $Q$ in time t. The metallic rod is melted and the material is formed into a rod of half the radius of the original rod. What is the amount of heat conducted by the new rod when placed in thermal contact with the two reservoirs in time $\mathrm{t}$ ?

Options:
A $Q / 4$
B $Q / 16$
C $2 \mathrm{Q}$
D $Q / 2$

The correct answer is:
$Q / 16$

Solution:
In steady state the amount of heat flowing from one face to the other face in time $t$ is given by $Q=\frac{K A\left(\theta_1-\theta_2\right) t}{l}$ where $K$ is coefficient of thermal conductivity of material of rod $\Rightarrow \quad \frac{\mathrm{Q}}{\mathrm{t}} \propto \frac{\mathrm{A}}{\mathrm{l}} \propto \frac{\mathrm{r}^2}{\mathrm{l}}$
As the metallic rod is melted and the material is formed into a rod of half the radius
$$
\begin{aligned}
\mathrm{V}_1 & =\mathrm{V}_2 \\
\Rightarrow \quad \pi \mathrm{r}_1^2 \mathrm{l}_1 & =\pi \mathrm{r}_2^2 \mathrm{l}_2 \\
\quad \quad \quad \mathrm{l}_1 & =\frac{\mathrm{l}_2}{4}
\end{aligned}
$$
Now, from Eqs. (i) and (ii)
$$
\begin{aligned}
\Rightarrow \quad & \frac{Q_1}{Q_2}=\frac{r_1^2}{l_1} \times \frac{l_2}{r_2^2}=\frac{r_1^2}{l_1} \times \frac{4 l_1}{\left(r_1 / 2\right)^2} \\
\Rightarrow \quad Q_1 & =16 Q_2
\end{aligned}
$$

Topic: Physics, Thermal Properties of Matter, NEET, NEET 2010 (Screening)