datavorous/entrance-exam-dataset huggingface dataset:
A circular disc of radius $0.2 \mathrm{~m}$ is placed in a uniform magnetic field of induction $\frac{1}{\pi}\left(\frac{W b}{m^2}\right)$ in such a way that its axis makes an angle of $60^{\circ}$ with $\overrightarrow{\mathbf{B}}$. The magnetic flux linked with the disc is

Options:
A $0.02 \mathrm{~Wb}$
B $0.06 \mathrm{~Wb}$
C $0.08 \mathrm{~Wb}$
D $0.01 \mathrm{~Wb}$

The correct answer is:
$0.02 \mathrm{~Wb}$

Solution:
The magnetic flux $\phi$ passing surface of area $A$ placed in a u field $B$ is given by
$\phi=B A \cos \theta$
where $\theta$ is the angle between
$\begin{aligned}
& \text {Here, } \theta=60^{\circ}, B=\frac{1}{\pi} \mathrm{Wb} / \mathrm{m}^2, A=\pi(0.2)^2 \\
& \text {Therefore, } \phi=\frac{1}{\pi} \times \pi(0.2)^2 \times \cos 60^{\circ} \\
& =(0.2)^2 \times \frac{1}{2} \\
& =0.02 \mathrm{~Wb} \\
&
\end{aligned}$

Topic: Physics, Electromagnetic Induction, NEET, NEET 2008 (Screening)
A circular disc of radius R rolls without slipping along the horizontal surface with constant velocity v0. We consider a point A on the surface of the disc. Then, the acceleration of point A is

Options:
A Constant in magnitude
B Constant in direction
C Constant in magnitude as well as direction
D None of the above

The correct answer is:
Constant in magnitude

Solution:
With respect to centre ω, all points on circumference is moving in a circle with angular velocity ω . Its acceleration towards centre in ω 2 R and different point has direction.

Topic: Physics, Rotational Motion, NEET
A circular disc of the radius 0.2 m is placed in a uniform magnetic field of induction 1Ï Wbââmâ2 in such a way that its axis makes an angle of 60Â° with Bâ. The magnetic flux linked with the disc is

Options:
A 0.08  Wb
B 0.01  Wb
C 0.02ââWb
D 0.06  Wb

The correct answer is:
0.02  Wb

Solution:
B=1Ï Wb m-2
Area of the disc normal to B is ÏR2cos60â.
Flux = B x Area normal to the magnetic field
â´ Flux = 1 Ï Ã Ï Ã0.04Ã 1 2 = 0.02 Wb

Topic: Physics, Electromagnetic Induction, NEET
A circular disc rolls down an inclined plane. The ratio of rotational kinetic energy to total kinetic energy is

Options:
A 12
B 13
C 23
D 34

The correct answer is:
13

Solution:
The ratio of rotational kinetic energy to total kinetic energy is
KERKET=12 Iω212Iω2+12 Mv2
The moment of inertia of a ring
= 12MR2
∴ KERKET=12 12MR2vR212×12MR2vR2+12 Mv2
=1212Mv21212Mv2+12Mv2
KERKET=13

Topic: Physics, Rotational Motion, NEET
A circular disk of moment of inertia $I_t$ is rotating in a horizontal plane, about its symmetry axis, with a constant angular speed $\omega_i$. Another disk of moment of inertia $I_b$ is dropped coaxially onto the rotating disk. Initially the second disk has zero angular speed. Eventually both the disks rotate with a constant angular speed $\omega_{\mathrm{f}}$. The energy lost by the initially rotating disc due to friction is

Options:
A $\frac{1}{2} \frac{\mathrm{I}_{\mathrm{b}}^2}{\left(\mathrm{I}_{\mathrm{t}}+\mathrm{I}_{\mathrm{b}}\right)} \omega_{\mathrm{i}}^2$
B $\frac{1}{2} \frac{\mathrm{I}_{\mathrm{t}}^2}{\left(\mathrm{I}_{\mathrm{t}}+\mathrm{I}_{\mathrm{b}}\right)} \omega_{\mathrm{i}}^2$
C $\frac{1}{2} \frac{\mathrm{I}_{\mathrm{b}}-\mathrm{I}_{\mathrm{t}}}{\left(\mathrm{I}_{\mathrm{t}}+\mathrm{I}_{\mathrm{b}}\right)} \omega_{\mathrm{i}}^2$
D $\frac{1}{2} \frac{\mathrm{I}_{\mathrm{b}} \mathrm{I}_{\mathrm{t}}}{\left(\mathrm{L}_{\mathrm{t}}+\mathrm{I}_{\mathrm{b}}\right)} \omega_{\mathrm{i}}^2$

The correct answer is:
$\frac{1}{2} \frac{\mathrm{I}_{\mathrm{b}} \mathrm{I}_{\mathrm{t}}}{\left(\mathrm{L}_{\mathrm{t}}+\mathrm{I}_{\mathrm{b}}\right)} \omega_{\mathrm{i}}^2$

Solution:
Loss of energy, $\Delta \mathrm{E}=\frac{1}{2} \mathrm{I}_{\mathrm{t}} \omega_{\mathrm{i}}^2-\frac{1}{2} \frac{\mathrm{I}_{\mathrm{t}}^2 \omega_{\mathrm{i}}^2}{\left(\mathrm{I}_{\mathrm{t}}+\mathrm{I}_{\mathrm{b}}\right)}$
$$
=\frac{1}{2} \frac{\mathrm{I}_{\mathrm{b}} \mathrm{I}_{\mathrm{t}} \omega_{\mathrm{i}}^2}{\left(\mathrm{I}_{\mathrm{t}}+\mathrm{I}_{\mathrm{b}}\right)}
$$

Topic: Physics, Rotational Motion, NEET, NEET 2010 (Screening)
A circular loop with N turns has radius r. It lies in the x-y plane carrying current I in the anti-clockwise direction. If the magnetic field in the region is Bâ=B0i^, then find the torque (Ïâ) acting on the loop.

Options:
A B0 NI Ïr2j^
B B0IÂ Ïr2j^
C B0Â NIÂ Ïr2k^
D B0Â Ïr2Â k^

The correct answer is:
B0 NI Ïr2j^

Solution:
Magnetic moment M→=NIπr2k^
Torque τ→=M→×B→
=NIπr2k^×(B0i^)
=B0NI πr2 j^ N m

Topic: Physics, Magnetic Properties of Matter, NEET
A circular ring of radius R and uniform linear charge density +Î» C/m are kept in x-y plane with its centre at the origin. The electric field at a point 0,0,R2 is

Options:
A λ33ε0R
B 2λ33R2
C 233λε0R
D none of these

The correct answer is:
λ33ε0R

Solution:
Eaxis=14πε0qz(R2+z2)32 , at z=R2
⇒ Eaxis=23314πε0qR2
⇒Eaxis=233.14πε0.λ.2πRR2
⇒Eaxis=λ33ε0R

Topic: Physics, Electrostatics, NEET
A circular tube of radius R and cross-sectional radius r ( r<<R ) is filled completely with iron balls of the radius Ï. Iron balls are just fitting into the tubes. The tension in the tube when it is rotated about its axis perpendicular to its plane with angular velocity Ï

Options:
A 43πρω2r3R
B 43πρω2r2R2
C 23πρω2r3R
D 23πρω2r2R2

The correct answer is:
23πρω2r2R2

Solution:
∵ r<<R, mass/length can be given by
λ= m2r= 43πr3ρ2r=23πr2ρ
Now,
Tension in the tube is given by
T=λ .ω2RR=23πr2ρ ω2R.R
=23πρω2r2R2

Topic: Physics, Laws of Motion, NEET
A clear sheet of polaroid is placed on the top of similar sheet so that their axes make an angle sin-135 with each other. The ratio of intensity of the emergent light to that of unpolarised incident light is

Options:
A 16Â :25
B 9Â :25
C 4Â :5
D 8 :25

The correct answer is:
8 :25

Solution:
If l is the intensity of the incident unpolarized light, the intensity transmitted by the first is l2. This is the intensity of incident light on the second polaroid. Intensity transmitted by the second polaroid is l2cos2θ, where θ is the angle between the axes. Here sin⁡θ=35, cos⁡θ is therefore 45.
l2cos2θ=l2×452=825l
825 is the required ratio

Topic: Physics, Wave Optics, NEET
A clock that keeps correct time at 20 Â°C, is subjected to 40 Â°C. If the coefficient of linear expansion of the pendulum is 12Ã10-6 Â°C-1, the time gained or lost, per day, is

Options:
A 10.3 s day-1
B 20.6Â sÂ day-1
C 5Â sÂ day-1
D 20minâ¡day-1

The correct answer is:
10.3 s day-1

Solution:
Time period, T=2Ïlg
âTT=12 âll=12Î±âÎ¸
= 1 2 Ã12Ã 10 â6 Ã( 40â20 )=12Ã 10 â5
âT=TÃ12Ã10-5
= 24Ã60Ã60Ã12Ã10-5
=10.3  s   day −1

Topic: Physics, Thermal Properties of Matter, NEET
A clock which keeps correct time at 20°C has a pendulum rod made of brass. How many seconds will it gain or lose per day when temperature falls to 0° C? α=18×10-6/°C

Options:
A 155.5 s
B 15.55 s
C 25.55 s
D 18.55 s

The correct answer is:
15.55 s

Solution:
ΔT=12αTΔθ
ΔT=12×18×10-6×86400×20
ΔT=15.55 s

Topic: Physics, Thermal Properties of Matter, NEET
A clock which keeps correct time at 25Â°C, has a pendulum made of brass. The coefficient of linear expansion for brass is 0.000019 Â°C-1. How many seconds a day will it gain if the ambient temperature falls to 0 Â°C?

Options:
A 20.52 s
B 15.00 s
C 52.10 s
D 63.10 s

The correct answer is:
20.52 s

Solution:
Let L0 and L25 be the length of pendulum at 0 °C and 25 °C respectively.
We know that,
L25=L01+αT=L01+0.000019×25=1.000475L0
⇒  ΔL=L25−L0=0.000475L0Also,  T=2πLgFrom  method of errors⇒  ΔTT=12ΔLLhere,  ΔT=gain  in  time              T= Total  time  of  day                =24×60×60  sec⇒  ΔT= 12×0.000475L0L0×24×60×60  sec  ⇒  ΔT= 20.52  sec

Topic: Physics, Thermal Properties of Matter, NEET
A close relation between flower and pollinating agent is best exhibited by

Options:
A Cocos
B Vallisneria
C Yucca
D Avena

The correct answer is:
Yucca

Solution:
The plant and the insect co-evolved; that is, they developed an intimate connection with each other. This is called an obligate mutualism in which if one were to disappear, the other would follow the same path. Originally it was a relationship of exploitation, with the moth larvae feeding on the seeds of the yucca. But, eventually, the plant devised methods to take advantage of the situation and as a consequence, the moth also evolved to ensure food for its babies and preservation of its food plant.

Topic: Biology, Sexual Reproduction in Flowering Plants, NEET
A closed compartment containing gas is moving with some acceleration in horizontal direction. Neglect effect of gravity. Then, the pressure in the compartment is

Options:
A same everywhere
B lower in front side
C lower in rear side
D lower in upper side

The correct answer is:
lower in front side

Solution:
If a fluid (gas or liquid) is accelerated in positive x - direction, then pressure decreases in positive x - direction. Change in pressure has following differential equation.
dp dx=-pa
where p is the density of the fluid. Therefore, pressure is lower in front side.

Topic: Physics, Mechanical Properties of Fluids, NEET
A closed gas cylinder is divided into two parts by a piston held tight. The pressure and the volume of gas in two parts respectively are (p, 5V) and (10p, V). If now the piston is left free and the system undergoes the isothermal process, then the volume of the gas in two parts, respectively are

Options:
A 4V, 2V

B 5V, V

C 2V, 4V

D 3V, 3V

The correct answer is:
2V, 4V

Solution:
When the piston is allowed to move the gases are kept separated but the pressure has to be equal (p1=p2) and the final volumes are x and 6V-x.
∵p1=p2
P(5V)=n1RT
⇒n1=5PVRT
∴n1RTx=n2RT6V-x
⇒n1x=n26V-x
∵10pV=n2RT
⇒n2=10PVRT
⇒5PVx  RT=10 PVRT(6V-x)
⇒1x=26V-x
∴x=2V and 6V-x=4V

Topic: Physics, Thermodynamics, NEET
A closed loop PQRS carrying a current is placed in a uniform magnetic field. If the magnetic forces on segments $P S, S R$ and $R Q$ are $F_1, F_2$ and $F_3$ respectively and are in the plane of the paper and along the directions shown, the directions shown, the force on the segment $Q P$ is

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

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

Solution:
\(\begin{aligned}
& F_4 \sin \theta=F_2 \\
& F_4 \cos \theta=\left(F_3-F_1\right) \\
\therefore \quad & F_4=\sqrt{\left(F_3-F_1\right)^2+F_2^2}
\end{aligned}\)
For a closed loop there is no translation.

Topic: Physics, Magnetic Effects of Current, NEET, NEET 2008 (Screening)
A closed organ pipe of length L L4is in resonance with a tuning fork. If a hole is made in the pipe at a distance L/4 from closed-end, it will be in resonance again, when :

Options:
A Tuning fork is replaced by another of high frequency.
B Tuning fork is replaced by another of lower frequency.
C It will be resonance with same tuning fork.
D Now the pipe will never resonate with any tuning fork.

The correct answer is:
Now the pipe will never resonate with any tuning fork.

Solution:
Standing wave cannot be formed. It becomes combination of two pipes one closed and other open pipe. They will not satisfy the condition of standing wave simultaneously.
n 1 V 2( 3L 4 ) = n 2 V 4( L 4 )
n 1 2V 3L = n 2 V L = V 4L
n 1 = 3 8    n 2 = 1 4 Which are not integer so not possible

Topic: Physics, Waves and Sound, NEET
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 it is given that First overtone of closed pipe = First overtone of open pipe
⇒3( ν 4 l 1 )=2( ν 2 l 2 );where   l 1 and   l 2
are the lengths of closed and open organ pipes hence l 1 l 2 = 3 4

Topic: Physics, Waves and Sound, NEET
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 F1,F2 and F3 respectively and are in the plane of the paper and along the directions shown, the force on the segment QP is

Options:
A (F3–F1)2–F22
B F3-F1+F2
C F3-F1-F2
D (F3–F1)2+F22

The correct answer is:
(F3–F1)2+F22

Solution:
∵ The net force on the current-carrying closed loop is zero.
∴ FPQ=(F3–F1)2+F22

Topic: Physics, Magnetic Effects of Current, NEET
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 F1, F2 and F3 respectively and are in the plane of the paper and along the directions shown, the force on the segment QP is

Options:
A F3-F12-F22
B F3- F1+F2
C F3- F1- F2
D F3-F12+F22

The correct answer is:
F3-F12+F22

Solution:
∵ Net force on current carrying closed loop is zero
∴FPQ=F3-F12+F22

Topic: Physics, Magnetic Effects of Current, NEET
A closely packed coil having 1000 turns has an average radius of $62.8 \mathrm{~cm}$. If current carried by the wire of the coil is $1 \mathrm{~A}$, the value of magnetic field produced at the centre of the coil will be (permeability of free space $=4 \pi \times 10^7 \mathrm{H} / \mathrm{m}$ ) nearly:

Options:
A $10^{-1} \mathrm{~T}$
B $10^{-2} \mathrm{~T}$
C $10^2 \mathrm{~T}$
D $10^{-3} \mathrm{~T}$

The correct answer is:
$10^{-3} \mathrm{~T}$

Solution:
As, magnetic field at centre of coil,
$\begin{aligned}
B & =\frac{\mu_0 \mathrm{~N} i}{2 \mathrm{R}} \\
\mathrm{B} & =\frac{4 \pi \times 10^{-7} \times 1000 \times 1}{2 \times 62.8 \times 10^{-2}} \\
& =\frac{4 \times 3.14 \times 10^{-7} \times 10^3}{2 \times 62.8 \times 10^{-2}} \\
& =10^{-3} \mathrm{~T}
\end{aligned}$

Topic: Physics, Magnetic Effects of Current, NEET, NEET 2022 (Phase 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 2SR2ρg
B 2SR1ρg
C SR2-R1ρg
D 2SR2-R1ρg

The correct answer is:
2SR2-R1ρg

Solution:
Net upward force
= 2π R2S + 2πR1 S   contact angle=0°
∴ Capillary rise is given by
h=2πSR1+R2πR22-R12ρg
= 2SR2-R1ρg

Topic: Physics, Mechanical Properties of Fluids, NEET
A coil has 200 turns and area of 70 cm2. The magnetic field perpendicular to the plane of the coil is 0.3 Wb m-2 and takes 0.1 s to rotate through 180Â°. The value of the induced E.M.F. will be

Options:
A 8.4 V
B 84 V
C 42 V
D 4.2 V

The correct answer is:
8.4 V

Solution:
Induced EMF is,
E=-NΔϕΔt
E=-NBAcos⁡θ2-cos⁡θ1Δt
E=-NBAcos⁡180o-coso⁡0Δt
E=2NBAt
E=22000.370×10-40.1
E=8.4 V

Topic: Physics, Electromagnetic Induction, NEET
A coil has self inductance L=0.04 H and resistance R=12 Î©. When it is connected to 220 V, 50 Hz supply; what will be the current flowing through the coil?

Options:
A 12.7 A
B 14.7 A
C 11.7 A
D 10.7 A

The correct answer is:
12.7 A

Solution:
L=0.04 H
R=12 Î©
V=220 V, 50 Hz
Impedance z= R2+w2z2
w=2Ïf=2ÏÃ50=100Ï
z= 144+100Ï2Ã0.042
=17.38
i=vz=22017.38=12.7 A.

Topic: Physics, Alternating Current, NEET
A coil having 200 turns has a surface area of 0.15 m2. A magnetic field applied perpendicular to the coil changes from 0.2 T to 0.6 T in 0.4 s, then the induced emf [in V] is

Options:
A 45
B 30
C 15
D 60

The correct answer is:
30

Solution:
Ï=nÃAÃB
Where nâ¡ No. of turns
A â¡ Area of loop
Bâ¡ Magnetic field
â´ Îµ=ÎÏÎt=nAÎBÎt
=200Ã0.15Ã0.6-0.20.4=30V

Topic: Physics, Magnetic Effects of Current, NEET
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 t1t2 will be

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

The correct answer is:
1:2

Solution:
I=I01-e-tτ
910I0=I01-e-t1τ
⇒ e-t1τ=110
⇒ et1τ=10
⇒ t1=τln10 ............(1)
and 99100I0=I01-e-t2τ
⇒ e-t2τ=1100
⇒ t2=τln100=2τln10 ..............(2)

From equations (1) and (2), we have
⇒ t1t2=12

Topic: Physics, Alternating Current, NEET
A coil having N turns is wound tightly in the form of a spiral with inner and outer radii a and b respectively. When a current I passes through the coil, the magnetic field at the centre is

Options:
A μ0NIb
B 2μ0NIa
C μ0NI2b-alog⁡ba
D μ0IN2b-alog⁡ab

The correct answer is:
μ0NI2b-alog⁡ba

Solution:
An element is assumed at distance x from center whose width is dx. No. of turns in width b–a = N
∴ No. of turns is width dx=n=Nb-adx
dB=μ0ni2x=μ0iN2b-adxx
∴B=μ0in2b-a∫abdxx=μ0Ni2b-aloge⁡ba

Topic: Physics, Magnetic Effects of Current, NEET
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, NEET, NEET 2005
A coil, in the shape of an equilateral triangle of side l, is suspended between two pole pieces of a permanent magnet, such that the magnetic field B is in the plane of the coil. If due to current I in the triangle, a torque Ï acts on it, the side l of the triangle is

Options:
A 2Ï3BI12
B 23ÏBI
C 2ÏBI12
D 13ÏBI

The correct answer is:
2Ï3BI12

Solution:
Torque acting on an equilateral triangle in a magnetic field B is
τ=IAB sinθ
Area of ΔLMN:A=34l2
and θ=90o
Substituting the given values in the expression for torque, we have
τ=I×34l2 Bsin90°=34I l2B
∴l=2τ3BI12

Topic: Physics, Magnetic Effects of Current, NEET
A coil is suspended in a uniform magnetic field, with the plane of the coil parallel to the magnetic lines of force. When a current is passed through the coil, it starts oscillating. It is very difficult to stop, but if an aluminium plate is placed near to the coil, it stops. This is due to

Options:
A induction of electrical charge on the plate.
B shielding of magnetic lines of force as aluminium is a paramagnetic material.
C electromagnetic induction in the aluminium plate giving rise to electromagnetic damping.
D development of air current when the plate is placed.

The correct answer is:
electromagnetic induction in the aluminium plate giving rise to electromagnetic damping.

Solution:
When an aluminium plate is placed near to the coil, electromagnetic induction in the aluminium plate gives rise to electromagnetic damping that helps to stop the motion.

Topic: Physics, Electromagnetic Induction, NEET
A coil of 40 henry inductance is connected in series with a resistance of $8 \mathrm{ohm}$ and the combination is joined to the terminals of a 2 volt battery. The time constant of the circuit is:

Options:
A 5 seconds
B 1/5 seconds
C 40 seconds
D 20 seconds

The correct answer is:
5 seconds

Solution:
According to the question Time constant $L R$ circuit is $T=\frac{L}{P}$
$$
\therefore \quad T=5 \mathrm{sec}
$$

Topic: Physics, Electromagnetic Induction, NEET, NEET 2004
A coil of circular cross-section having 1000 turns and 4 cm2 face area is placed with its axis parallel to a magnetic field which decreases by 10-2 Wb m-2 in 0.01 s. The e.m.f. induced in the coil is :

Options:
A 0.4 mV
B 4 mV
C 200 mV
D 400 mV

The correct answer is:
400 mV

Solution:
As we know, Ï = B â Â· A â
Induced emf e= -dÏdt=-N.AdBdt
= 1 0 0 0 Ã 4 Ã 1 0 - 4 Ã 1
=400 mV

Topic: Physics, Electromagnetic Induction, NEET
A coil of inductance 0.2 H and 1.0 Î© resistance is connected to a 90 V source. At what rate does the current in the coil grow at the instant the coil is connected to the source?

Options:
A 450 A s-1
B 4.5 A s-1
C 45 A s-1
D 0.45 A s-1

The correct answer is:
450 A s-1

Solution:
From
dIdt=eL=900.2=450 A s-1

Topic: Physics, Electromagnetic Induction, NEET
A coil of inductance 5 H is joined to a cell of emf 6 V through a resistance of 10 Ω at t=0. The emf across the coil at time t=ln2 sec is

Options:
A 3 V
B 1.5 V
C 0.75 V
D 4.5 V

The correct answer is:
3 V

Solution:
i=εR1–e–tRL
∴ Emf across R is εR=ε1–e–tRL
⇒εR=6 V1–e–ln2 . 2
⇒εR=6V1–12=3 V
∴ Voltage across coil εc=ε–εR=3 volt

Topic: Physics, Electromagnetic Induction, NEET
A coil of inductive reactance 1/3 Î© and resistance 1 Î© is connected to a 200 V, 50 Hz AC supply. The time lag between maximum voltage and current is

Options:
A 1200 s
B 1300Â s
C 1500Â s
D 1600&nbsps

The correct answer is:
1600&nbsps

Solution:
Given,
Inductive reactance of coil XL=13 Î©
Resistance R=1 Î©
We know that,
tanâ¡Ï=ÏLR
tanâ¡Ï=XLR
tanâ¡Ï=1/31
Ï=tan-1â¡13
Ï=30o
â´ Ït=Ï6
t=Ï6Ï
t=Ï6Ã2Ïf
t=112f
Here, f=50 Hz
t=112Ã50=1600 s

Topic: Physics, Motion In One Dimension, NEET
A coil of inductive reactance $31 \Omega$ has a resistance of $8 \Omega$. It is placed in series with a condenser of capacitative reactance $25 \Omega$. The combination is connected to an a.c. source of $110 \mathrm{~V}$. The power factor of the circuit is:

Options:
A 0.33
B 0.56
C 0.64
D 0.80

The correct answer is:
0.80

Solution:
$X_L=31 \Omega, X_C=25 \Omega, R=8 \Omega$
$\therefore$ impedance of series LCR is
$$
\begin{aligned}
& \mathrm{Z}=\sqrt{\mathrm{R}^2+\left(\mathrm{X}_{\mathrm{L}}-\mathrm{X}_{\mathrm{C}}\right)^2} \\
& =\sqrt{8^2+(31-25)^2} \\
& =\sqrt{64+36}=\sqrt{100}=10 \Omega
\end{aligned}
$$
Power factor
$$
\cos \phi=\frac{R}{2}=\frac{8}{10}=0.8
$$

Topic: Physics, Alternating Current, NEET, NEET 2006
A coil of resistance $400 \Omega$ is placed in a magnetic field. If the magnetic flux $\phi(\mathrm{Wb})$ linked with the coil varies with time $t$ (sec) as $\phi=50 t^2+4$
The current in the coil at $t=2 \mathrm{~s}$ is

Options:
A $0.5 \mathrm{~A}$
B $0.1 \mathrm{~A}$
C $2 \mathrm{~A}$
D $1 \mathrm{~A}$

The correct answer is:
$0.5 \mathrm{~A}$

Solution:
Induced emf of coil $E=\left|-\frac{d \phi}{d t}\right|_t$ Given, $\phi=50 t^2+4$ and $R=400 \Omega$
$\begin{aligned}
E & =\left|-\frac{d \phi}{d t}\right|_{t=2} \\
& =|100 t|_{t=2}=200 \mathrm{~V}
\end{aligned}$
Current in the coil
$\begin{aligned}
i & =\frac{E}{R}=\frac{200}{400} \\
& =\frac{1}{2}=0.5 \mathrm{~A}
\end{aligned}$

Topic: Physics, Electromagnetic Induction, NEET, NEET 2012 (Screening)
A coil of self-inductance $L$ is connected in series with a bulb $B$ and an $A C$ source. Brightness of the bulb decreases when

Options:
A frequency of the $\mathrm{AC}$ source is decreased
B number of turns in the coil is reduced
C a capacitance of reactance $X_C=X_L$ is included in the same circuit
D an iron rod is inserted in the coil

The correct answer is:
an iron rod is inserted in the coil

Solution:
As $Z=\sqrt{R^2+X_L^2}=\sqrt{R^2+(2 \pi f V L)^2}$
As $/=\frac{V}{Z}, P=/^2 R$
i.e., $V \uparrow, L \uparrow \Rightarrow Z \uparrow, / \downarrow$ and $P \downarrow$

Topic: Physics, Alternating Current, NEET, NEET 2013 (All India)
A coil of wire of radius r has 600 turns and self-inductance of 108 mH. The self-inductance of a coil with the same radius and 500 turns is

Options:
A 80 mH

B 75 mH

C 108 mH

D 90 mH

The correct answer is:
75 mH

Solution:
Given L1=108 mH, N1=600 turns, N2=500 turns and L2=?
By self - inductance of a plane coil
L1=Î¼0ÏN12a12 ......(i)
L2=Î¼0ÏN22a22 ......(ii)
From the equations (i) and (ii), we get
L1L2=N1N22 âµ a1=a2
108L2=6005002
L2=108Ã2536
L2=3Ã25=75 mH

Topic: Physics, Electromagnetic Induction, NEET
A coil of wire of resistance 50 Ω is embedded in a block of ice and a potential difference of 210 V is applied across it. The amount of ice which melts in 1 second is [latent heat of fusion of ice = 80 cal g-1]

Options:
A 0.262Â g
B 2.62 g
C 26.2Â g
D 0.0262Â g

The correct answer is:
2.62 g

Solution:
Energy of melting is provided by electric energy therefore
mL=V2Rt
m(80)=(210)250(1)
mâ¡=2.62 g

Topic: Physics, Current Electricity, NEET
A coin is of mass 4.8 kg and radius 1 m is rolling on a horizontal surface without sliding, with an angular velocity of 600 rad min-1. The total kinetic energy of the coin is

Options:
A 360 J
B 1440Â Ï2 J
C 4000Â Ï2 J
D 600Â Ï2 J

The correct answer is:
360 J

Solution:
Angular velocity is given by, ω=600 rad min-1
ω=10 rad s-1
k=12Iω2+12mv2
=12×12mr2ω2+12×mωr2
=14×mω2r2+12mω2r2=14+12mω2r2
=360 J

Topic: Physics, Rotational Motion, NEET