datavorous/entrance-exam-dataset huggingface dataset:
A charge ' $q$ ' is placed at the centre of the line joining two equal charges ' $Q$ '. The system of three charges will be in equilibrium if ' $q$ ' is equal to:

Options:
A $-Q / 4$
B $Q / 4$
C $-Q / 2$
D $Q / 2$

The correct answer is:
$-Q / 4$

Solution:
$$
\text { According to the question }
$$
for equilibrium, net force on charge $\quad Q=0$
$$
\therefore \frac{1}{4 \pi \epsilon_0} \frac{\mathrm{Qq}}{r^2}+\frac{1}{4 \pi \epsilon_0} \cdot \frac{\mathrm{Q}^2}{\sigma\left(\frac{r}{2}\right)^2}=0
$$
$$
\begin{aligned}
\Rightarrow & & \frac{1}{4 \pi \epsilon_0} \frac{\mathrm{Q}_2}{r^2} & =-\frac{1}{4 \pi \epsilon_0} \frac{4 \mathrm{Q} q}{r^2} \\
\Rightarrow & & Q & =-4 q \\
\Rightarrow & & q & =-\frac{Q}{4}
\end{aligned}
$$

Topic: Physics, Electrostatics, NEET, NEET 2013 (Karnataka)
A charge q is placed at the centre of the line joining two equal point charges each equal to Q. The system of three charges will be in equilibrium if q is equal to

Options:
A + Q⁡ 4
B - Q⁡ 2
C + Q⁡ 2
D - Q⁡ 4

The correct answer is:
- Q⁡ 4

Solution:
For equilibrium of system of charges,
force produced between q and Q + force produced between Q and Q = 0

1 4 π ε 0 qâ¡ Qâ¡ râ¡ / 2 2 + 1 4 π ε 0 Qâ¡ Qâ¡ râ¡ 2 = 0
⇒ 4 qâ¡ Qâ¡ râ¡ 2 + Qâ¡ Qâ¡ râ¡ / 2 2 = 0 ⇒ qâ¡ = - Qâ¡ 4

Topic: Physics, Electrostatics, NEET
A charge Q is uniformly distributed over the surface of two conducting concentric spheres of radii R and r(R>r). Then potential at the common centre of these spheres

Options:
A kQR+rRr
B kQR+rR2+r2
C kQR2+r2
D kQ1R-1r

The correct answer is:
kQR+rR2+r2

Solution:
Let the surface charge density is Ï
â´ Q = q A + q B
or Q = 4 Ï r 2 Ï + 4 Ï R 2 Ï
â´ Ï = Q 4 Ï r 2 + R 2
â´ V C = V A + V B
â´ V C = q A 4 Ï Îµ 0 r + q B 4 Ï Îµ 0 R
1 4 Ï Îµ 0 q A r + q B R = k 4 Ï r 2 Ï r + 4 Ï R 2 Ï R
k = 4 Ï r Ï + 4 Ï R Ï
= 4 Ï k Qr 4 Ï R 2 + r 2 + QR 4 Ï r 2 + R 2
= kQ R + r R 2 + r 2 Here , k = 1 4 Ï Îµ 0

Topic: Physics, Electrostatics, NEET
A charge Q is uniformly distributed over the surface of two conducting concentric spheres of radii R and r R>r. Then potential at the common centre of these spheres

Options:
A kQÂ (R+r)Rr
B kQ (R+r)(R2+r2)
C kQR2+r2
D kQ1R-1r

The correct answer is:
kQ (R+r)(R2+r2)

Solution:
Let the surface charge density is Ï.
And, q A and q B are charges on shell A and B
â´ Q=qA+qB
or Q=4Ïr2Ï+4ÏR2Ï
â´ Ï=Q4Ï r2+R2
â´ VC=VA+VB
â´ VC=qA4ÏÏµ0r+qB4ÏÏµ0R
= 1 4π ϵ 0 [ q A r + q B R ]=k ( 4π r 2 σ r + 4π R 2 σ R )
=k 4ÏrÏ+4ÏRÏ
=4Ïk Qr4Ï R2+r2+QR4Ï r2+R2
=kQ (R+r)(R2+r2) Here, k=14ÏÏµ0

Topic: Physics, Electrostatics, NEET
A charge q moves in a region where electric field and magnetic field both exist, then force on it is

Options:
A qv→×B→
B qEâ+qÂ BâÂ Ãvâ
C qBâ+qÂ EâÂ Ãvâ
D qEâ+qÂ vâÂ ÃBâ

The correct answer is:
qE→+q v→ ×B→

Solution:
F→=FE→+FB→
=q E→+v→×B→

Topic: Physics, Magnetic Effects of Current, NEET
A charge ' $q$ ' moves in a region where the electric field and the magnetic field both exist, then the force on it is:

Options:
A $q(\vec{v} \times \vec{B})$
B $q \vec{E}+q(\vec{v} \times \vec{B})$
C $q \vec{E}+q(\vec{B} \times \vec{v})$
D $q \vec{E}+q(\vec{B} \times \vec{v})$

The correct answer is:
$q \vec{E}+q(\vec{v} \times \vec{B})$

Solution:
Lorentz force $\vec{F}_v=\vec{F}_e+\vec{F}_m$
$=q \vec{E}+q(\vec{V} \times \vec{B})$

Topic: Physics, Magnetic Effects of Current, NEET, NEET 2002
A charge q, moving with velocity v , enters a uniform magnetic field. The charge keeps on revolving along a closed circular path in the magnetic field. The frequency of revolution does not depend upon

Options:
A mass of the charge particle
B velocity of the charge particle
C magnitude of charge particle
D intensity of the magnetic field

The correct answer is:
velocity of the charge particle

Solution:
mv2r=qvB
â´vr=qBm
Frequency f=v2Ïr=qB2Ï.m
f does not depend upon velocity of the charge particle.

Topic: Physics, Magnetic Effects of Current, NEET
A charge Q placed at the centre of a metallic spherical shell with inner and outer radii R1 and R2 respectively. The normal component of the electric field at any point on the Gaussian surface with radius between R1 and R2 will be

Options:
A Zero
B Q4πR12
C Q4πR22
D Q4πR1-R22

The correct answer is:
Zero

Solution:
The induced charges will appear at the inner and outer surface of the metallic spherical shell. Also there is no charge in between the metal of the shell. So, the normal component of the electric field at any point on the Gaussian surface with radius between R1 and R2 will be zero.

Topic: Physics, Electrostatics, NEET
A charged capacitor C=30 Î¼F is connected to an inductor L=27 mH, then the angular frequency of the system is

Options:
A 9.1×103 rad s-1
B 3×103 rad s-1
C 1.1×103 rad s-1
D 0.3×103 rad s-1

The correct answer is:
1.1×103 rad s-1

Solution:
Given, C=30 Î¼F, =30Ã10-6F, L=27 mH=27Ã10-3H
Since, we know, angular frequency,
Ï=1LC=127Ã10-3Ã30Ã10-6
=19Ã10-4=1.1Ã103 rad s-1

Topic: Physics, Alternating Current, NEET
A charged oil drop falls with terminal velocity v0 in the absence of electric field. An electric field E keeps it stationary. The drop acquires charge 3q, it starts moving upwards with velocity v0. The initial charge on the drop is

Options:
A q2
B q
C 32q
D 2q

The correct answer is:
32q

Solution:
When drop is stationary, then
q1E=6Ï Î·r v0 or q1=6Ï Î·r v0/E
When drop moves upwards, then
3q=6Ï Î·rv0+v0E=2Ã6Ï Î·rv0E=2q1
â´ q1=32q

Topic: Physics, Electrostatics, NEET
A charged particle (charge $q$) is moving in a circle of radius $R$ with uniform speed $v$. The associated magnetic moment $\mu$ is given by:

Options:
A $q v R^2$
B $q v R^2 / 2$
C $q v R$
D $q v R / 2$.

The correct answer is:
$q v R / 2$.

Solution:
Magnetic moment $\mu=I A$
Since $T=\frac{2 \pi R}{v}$
also $I=\frac{q}{T}=\frac{q v}{2 \pi R}$
$\begin{aligned}
\therefore \quad \mu & =\left(\frac{q v}{2 \pi R}\right)\left(\pi R^2\right) \\
& =\frac{q v R}{2}
\end{aligned}$

Topic: Physics, Magnetic Effects of Current, NEET, NEET 2007
A charged particle (charge q) is moving in a circle of the radius R with uniform speed v. The associated magnetic moment μ is given by

Options:
A qvR2
B qvR2
C qvR22
D qvR

The correct answer is:
qvR2

Solution:
As revolving charge is equivalent to a current, so
I=q f=q×ω2π
But ω=vR
Where R is radius of circle and v is uniform speed of charged particle.
Therefore, I=qv2πR
Now, magnetic moment associated with charged particle is given by
μ=IA=I×πR2
or μ=qv2πR ×πR2=12 qvR

Topic: Physics, Magnetic Effects of Current, NEET
A charged particle enters a uniform magnetic field with a velocity vector at an angle of 45° with the magnetic field. The pitch of the helical path followed by the particle is p. The radius of the helix will be

Options:
A p2π
B 2p
C p2π
D 2pπ

The correct answer is:
p2π

Solution:
Pitch of helical path is
p=vcos⁡θ×2πmqB
⇒p=2πmvcos⁡45°qB
⇒p=2πmv2qB
Also radius of helix is,
r=mvsin⁡θqB
⇒r=mv2qB
∴ rp=mv2qB×2qB2πmv
rp=12π
or r=p2π

Topic: Physics, Magnetic Effects of Current, NEET
A charged particle enters a uniform magnetic field with a velocity vector at an angle of 45° with the magnetic field. The pitch of the helical path followed by the particle is p. The radius of the helix will be

Options:
A p2π
B 2p
C p2π
D 2pπ

The correct answer is:
p2π

Solution:
Pitch of helical path is,
p=vcos⁡θ×2πmqB=2πmvcos⁡45oqB
p=2πmv2qB
Also radius of helix is,
r=mvsin⁡θqB=mv2qB
∴rp=mv2qB×2qB2πmv=12π
or r=p2π

Topic: Physics, Magnetic Effects of Current, NEET
A charged particle having a mass of 1.6Ã10â26 kg comes out of accelerator tube with kinetic energy 2Ã103 eV. Calculate the smallest magnitude of the magnetic field that should be applied in vertically downwards direction to just prevent the charged particle from colliding the plate (Assume charge on particle = charge of the proton)

Options:
A 2 T
B 4 T

C 0.02 T
D 0.04 T

The correct answer is:
2 T

Solution:
r=d ⇒ 2mkqB=d ⇒B=2mkq2d2
⇒B=2×1.6×10–26×2×103×ee2×10–4
⇒B=2×1.6×10–26×2×1031.6×10–19×10–4=2 T

Topic: Physics, Magnetic Effects of Current, NEET
A charged particle having drift velocity of 7.5×10−4 m s−1 in an electric field of 3×10−10 V m−1, has a mobility in m2 V–1 s–1 of:

Options:
A 2.5×106
B 2.5×10–6
C 2.25×10–15
D 2.25×1015

The correct answer is:
2.5×106

Solution:
Drift velocity is given by, vd=E×μ

⇒μ=vdE=7.5×10−43×10−10

μ=2.5×106 m2 V-1 s-1

Topic: Physics, Current Electricity, NEET, NEET 2020 (Phase 1)
A charged particle is moving in a uniform magnetic field in a circular path. The energy of the particle is doubled. The initial radius of the new circular path will be

Options:
A R2
B 2R
C 2R
D R2

The correct answer is:
2R

Solution:
r=mvqB,r2=m2v2q4B2=2m(KE)q2B2r∝KE⇒r2r1=KE2KE1

⇒ r2=2KE1KE1R⇒ r2=2R

Topic: Physics, Magnetic Effects of Current, NEET
A charged particle moving in a magnetic field experiences a resultant force

Options:
A In the direction opposite to the field
B In the direction of field
C In the direction perpendicular to both the field and its velocity
D None of the above

The correct answer is:
In the direction perpendicular to both the field and its velocity

Solution:
As F → =q( v → × B → )
Thus, force perpendicular to both the vectors velocity and magnetic field.

Topic: Physics, Magnetic Effects of Current, NEET
A charged particle of mass m and charge q is projected into a uniform magnetic field of induction B with speed v which is perpendicular to B . The width of the magnetic field is d . The impulse imparted to the particle by the field is

d << mvqB

Options:
A qBv
B mvqB
C qBd
D 2 mv2qB

The correct answer is:
qBd

Solution:
As d <<< mvqB
d=r sinâ¡Î¸
Impulse is change in momentum
sinâ¡Î¸=dr
where r=mvqB
â´ d <<< r
â´sinâ¡Î¸=Î¸
Î¸=dr
Change in momentum
=mv Î¸
=mv Ã dr=mvÃdÃqBmv
Impulse=qBd

Topic: Physics, Magnetic Effects of Current, NEET
A charged particle q1 is at position 2,-1,3 .The electrostatic force on another charge particle q2 at 0,0,0 is

Options:
A q1q256πε02i^–j^+3k^
B q1q25614ε02i^–j^–3k^
C q1q256πε0i^–2j^–3k^
D q1q25614πε0-2i^+j^–3k^

The correct answer is:
q1q25614πε0-2i^+j^–3k^

Solution:
Aq1 →2,–1,3
Bq2→0,0,0
F→BA=kqAqB|rB→–rA→|3rB→–rA→
rA→=2i^–j^+3k^; rB→=0
Putting all the values in the formula,
We get F→BA=q1q25614 πε0-2i^+j^–3k^

Topic: Physics, Electrostatics, NEET
A charged particle q enters a region of uniform magnetic field B (out of page) and is deflected distance d after travelling a horizontal distance a. The magnitude of the momentum of the particle is

Options:
A qB2a2d+d
B qB2
C Zero
D 2qB

The correct answer is:
qB2a2d+d

Solution:
From the figure given below we have, AB=a2+d2=h
AB=a2+d2=h
In ΔOAD, cosθ=h2R
In ΔACB,sinπ2-θ=dh
⇒h2R=dh
⇒h2=2Rd
⇒a2+d2=2Rd

⇒R=a22d+d2
∴ Momentum, P=mv
=qBR=qB2a2d+d∵R=mvqB

Topic: Physics, Magnetic Effects of Current, NEET
A charged particle q is shot towards another charged particle Q (which is fixed) with a speed v. It approaches Q up to the closest distance r and then returns. If q were given a speed 2v, the closest distance of approach would be

Options:
A r
B 2r
C r2
D r4

The correct answer is:
r4

Solution:
From conservation of mechanical energy
12mv2=kqQr
In second case
12m(2vâ¡)2=kqQr′
r′=r4

Topic: Physics, Electrostatics, NEET
A charged soap bubble having surface charge density σ and radius r. If the pressure inside and outside the soap bubble is the same, then the surface tension of the soap solution is

Options:
A T=σ2R8ε0
B T=σ2R4ε0
C T=σ2R2ε0
D T=σ2Rε0

The correct answer is:
T=σ2R8ε0

Solution:
Excess of Pressure inside liquid bubble P1=4TR
electrostatic pressure due to charge P2=σ22ε0
Comparing them 4TR=σ22ε0
Therefore T=σ2R8ε0

Topic: Physics, Mechanical Properties of Fluids, NEET
A child has blood group 'O'. If the father has the blood group 'A' and the mother has the blood group 'B'. What will be the genotypes of the parents?

Options:
A IAIA and IBi
B IAi and IBi
C IAi and ii
D ii and IBIB

The correct answer is:
IAi and IBi

Solution:
When crossing two plants homozygous for contrasting pair of characteristics results in a hybrid showing complete expression of both phenotypes is called as codominance. It is another deviation from Mendel's law of dominance. For example, when blood type A (homozygous) and blood type B (homozygous) are crossed, the resultant offspring has blood type AB where both allele for A and allele for B are expressed completely without any interference from each other.
Blood group O is expressed in homozygous recessive condition - ii and the child has blood group O.
Each parent provides one gene each for expression of the blood group, thus, each parent has the allele 'i'in their genotype. Father is having blood group A and the genotype will be IAi and similarly for mother the genotype with blood group B will be IBi.

Topic: Biology, Principles of Inheritance and Variation, NEET
A child is standing at one end of a long trolley moving with a speed v on a smooth horizontal track. If the child starts running towards the other end of the trolley with a speed u , the centre of mass of the system (trolley + child) will move with a speed

Options:
A zero
B v+u
C v-u
D v

The correct answer is:
v

Solution:
Internal forces can not change the velocity of the center of mass of the system. Therefore it continues to move with the same speed.

Topic: Physics, Center of Mass Momentum and Collision, NEET
A childless couple can be assisted to have a child through a technique called GIFT. The full form of this techniques is:

Options:
A Gamete intra fallopian transfer
B Gamete internal fertilization and transfer
C Germ cell internal fallopian transfer
D Gamete inseminated fallopian transfer

The correct answer is:
Gamete intra fallopian transfer

Solution:
GAMETE INTRA-FALLOPIAN TRANSFER (GIFT):
It is also one of the assisted methods of conception carried out to defects like blockage of the opening in the uterus and fallopian tube. In this method, both the sperms and ova are collected from male and female partners and directly transferred into one of the fallopian tubes separately by artificial method. In fallopian tube fertilization and early development of fertilized egg takes place naturally (In-vivo fertilization) and followed by transportation of blastocyst into the uterine cavity for implantation to cause pregnancy.

Topic: Biology, Reproductive Health, NEET, NEET 2015 (Phase 2)
A chimeric DNA is

Options:
A DNA which contain uracil
B DNA synthesized from RNA
C recombinant DNA
D DNA which contain single strand

The correct answer is:
recombinant DNA

Solution:
Chimeric DNA is the recombinant DNA. It is also known as hybrid DNA as it carries DNA segments from other sources. The steps in making chimeric DNA involve isolating DNA, digestion of DNA with restriction enzymes, joining of DNA fragments, inserting DNA into host cells and then screening and selection of the transformed cells.

Topic: Biology, Biotechnology - Principles and Processes, NEET
A chloride dissolves appreciably in cold water. When placed on a platinum wire in Bunsen flame, no distinctive colour is noticed. Then the cation is

Options:
A Mg2+
B Ba2+
C Ag+
D Ca2+

The correct answer is:
Mg2+

Solution:
Chlorides of Ag are insoluble, while chlorides Ba and Ca imparts colour to Bunsen flame. Thus option Mg is correct.

Topic: Chemistry, s Block Elements, NEET
A chromosome in which the centromere is situated close to its end so that one arm is veryshort and the other very long is:

Options:
A Acrocentric
B Metacentric
C Sub-metacentric
D Telocentric

The correct answer is:
Acrocentric

Solution:
A chromosome in which centromere is present close to its end is called as acrocentric. These chromosome shows maximum arm ratio. Chromosome with centromere at the end is called as telocentric.

Topic: Biology, Principles of Inheritance and Variation, NEET
A chronic disorder in which alveolar walls are damaged is

Options:
A Asthma
B Emphysema
C Occupational respiratory disorders
D Pneumonia

The correct answer is:
Emphysema

Solution:
Emphysema is a chronic disorder in which alveolar walls are damaged due to which respiratory surface is decreased. One of the major causes of this is cigarette smoking.

Topic: Biology, Breathing and Exchange of Gases, NEET
A circuit ABCD is held perpendicular to the uniform magnetic field of B = 5 Ã 1 0 -2 T extending over the region PQRS and directed into the plane of the paper. The circuit is moving out of the field at a uniform speed of 0.2 ms-1 for 1.5 s. During this time, the current in the 5 Î© resistor is

Options:
A 0.6 mA from B to C
B 0.9 mA from B to C
C 0.9 mA from C to B
D 0.6 mA from C to B

The correct answer is:
0.6 mA from B to C

Solution:
I = B â v R
I = 5 Ã 1 0 - 2 Ã 0.3 Ã 0.2 5 A = 0.6 mA.
Area and flux are decreasing. So, current flows to increase the flux. Clearly, current should be clockwise. So, it flows from B to C through 5 Î©.

Topic: Physics, Electromagnetic Induction, NEET
A circuit contains an ammeter, a battery of 30 v and a resistance 40.8 ohm all connected in series. If the ammeter has coil of resistance 480 ohm and a shunt of 20 ohm, the reading in the ammeter will be:

Options:
A 0.25 A
B 2 A
C 1 A
D 0.5 A

The correct answer is:
0.5 A

Solution:
Resistance of ammeter =480×20480+20=19.2Ω
i=3040.8+19.2=0.5 A

Topic: Physics, Current Electricity, NEET, NEET 2015 (Phase 2)
A circuit has a self-inductance of 1 H and carries a current of 2 A. To prevent sparking when the circuit is switched off, a capacitor which can withstand 400 V is used. The least capacitance of capacitor connected across the switch must be equal to

Options:
A 50 μF
B 25 μF
C 100 μF
D 12.5 μF

The correct answer is:
25 μF

Solution:
Energy stored in capacitor is equal to energy stored in inductance
12CV2=12LI2
C=LI2V2
=1Ã224002
=25 Î¼F

Topic: Physics, Electrostatics, NEET
A circuit is connected as shown in the figure with switch S open. When the switch is closed, the total amount of charge that flows from Y to X is

Options:
A zero
B 54 μC
C 81 μC
D 27 μC

The correct answer is:
27 μC

Solution:
From Y to X charge flows to plates a and b.
qa+qbi=0, qa+qbf=27 μC
∴ 27 μC charge flows from Y to X.

Topic: Physics, Current Electricity, NEET
A circuit when connected to an AC source of $12 \mathrm{~V}$ gives a current of $0.2 \mathrm{~A}$. The same circuit when connected to a DC source of $12 \mathrm{~V}$, gives a current of $0.4 \mathrm{~A}$. The circuit is

Options:
A series LR
B series $\mathrm{RC}$
C series LC
D series LCR

The correct answer is:
series LR

Solution:
When the circuit is connected to AC source, Voltage, $\mathrm{V}=12 \mathrm{~V}$
Current, $\mathrm{I}=0.2 \mathrm{~A}$
$\Rightarrow$ Impedance $\mathrm{Z}=\frac{\mathrm{V}}{\mathrm{I}}=\frac{12}{0.2}=60 \Omega$
When it is connected to DC source,
Voltage, $\mathrm{V}=12 \mathrm{~V}$
Current, $\mathrm{I}=0.4 \mathrm{~A}$
$\Rightarrow$ Resistance, $\mathrm{R}=\frac{\mathrm{V}}{\mathrm{I}}=\frac{12}{0.4}=30 \Omega$
As in case of DC supply, the capacitor act as an open circuit and no current flows through the circuit. So the given circuit will not have capacitor in series combination. Therefore the circuit should be a series LR circuit.

Topic: Physics, Alternating Current, NEET, NEET 2019 (Odisha)
A circular beam of light of diameter d=2 cm falls on a plane surface of glass. The angle of incidence is 60o and refractive index of glass is Î¼=32. The diameter of the refracted beam is

Options:
A 4.0Â cm
B 3.0Â cm
C 3.26 cm
D 2.52Â cm

The correct answer is:
3.26 cm

Solution:
From the diagram, diameter of incident beam is d1=PQ cos60o and diameter of refracted beam is d2=PQcosr.
∴ d1d2=12cosr
⇒d2=2d1cosr
From Snell's law:
1×sini=μsinr
sinr=sin60°μ=13
⇒cosr=23
∴ d2=2×2×23=3.26 cm

Topic: Physics, Ray Optics, NEET
A circular beam of light, of diameter d=2 cm, falls on a plane surface of a glass. The angle of incidence is 60o and refractive index of glass is Âµ=32. Find the approximate diameter of the refracted beam (in cm).

Options:
A 3
B 6
C 5
D 8

The correct answer is:
3

Solution:
d1=PQcosâ¡60o
d2=PQcosâ¡r
d1d2=12cosâ¡r
d2=2d1cosâ¡r
1Ãsinâ¡i=Âµsinâ¡r
sinâ¡r=sinâ¡60oÎ¼=13âcosâ¡r=23
d2=2Ã2Ã23=3.26 cm

Topic: Physics, Ray Optics, NEET
A circular coil carrying current I is placed in a region of uniform magnetic field acting perpendicular to the coil as shown in the figure. Mark the correct option.

Options:
A coil expands
B coil contracts
C coil moves left
D coil moves right

The correct answer is:
coil expands

Solution:
According to Fleming left hand rule at a small element, it is found that force is acting outward. Hence coil expands as at all points, force is acting outwards.

Topic: Physics, Magnetic Effects of Current, NEET
A circular coil of 100 turns and effective diameter 20 cm carries a current of 0.5 A. It is to be turned in a magnetic field of B=2 T from a position in which the normal to the plane of the coil making an angle Î¸ with magnetic field changes from Î¸=0Â° to Î¸=180o . The work required in this process is

Options:
A Ï J
B 2Ï J
C 4Ï J
D 8Ï J

The correct answer is:
2Ï J

Solution:
We know that work done is given by :
W=NiAB( cos 0 o −cos 180 o )    = 100×0.5×π× ( 0.1 ) 2 ×2( 1−cos180° )    = 2π J

Topic: Physics, Magnetic Effects of Current, NEET
A circular coil of radius 4 cm having 50 turns carries a current of 2 A. It is placed in a uniform magnetic field of the intensity of 0.1 Weber m-2. The work done to rotate the coil from the equilibrium position by 180Â° is

Options:
A 0.1 J
B 0.2Â J
C 0.4Â J
D 0.8Â J

The correct answer is:
0.1 J

Solution:
R=4 cm
N=50
I=2 A
B=0.1
W=NIABcos⁡θ-cos⁡180°
W=50×2×π4×10-22×0.1×1--1
W=100×π×16×10-4×12×2
W=32π×10-3
W=0.1 J

Topic: Physics, Magnetic Effects of Current, NEET
A circular coil of radius 10 cm and 100 turns carries a current 1 A. What is the magnetic moment of the coil?

Options:
A 3.142Ã104 A m2
B 104 A m2
C 3.142 A m2
D 3 A m2

The correct answer is:
3.142 A m2

Solution:
Given, Radius of coil r=10Ã10-2 m
Number of turns
N = 100 turn, Current in coil i = 1 A
Magnetic moment M = ?
By magnetic dipole moment of current loop
M=NiA, where A=Ïr2
M=Ni.Ïr2
=100Ã1Ã3.142Ã10Ã10-22
=100Ã3.142Ã100Ã10-4
=3.142 A m2

Topic: Physics, Magnetic Properties of Matter, NEET
A circular coil of radius 20 cm and 20 turns, is mounted vertically with its plane in the magnetic meridian. A small magnetic needle (free to rotate about vertical axis) is placed at the centre of the coil. It is deflected through 45 o when a current passes through the coil and is in equilibrium (Horizontal component of earth's field is BH=0.34×10−4 T). The current in the coil is

Options:
A 1 7 1 0 Ï A
B 6âA
C 6Ã 10 â3 âA
D 3 5 0 A

The correct answer is:
1 7 1 0 Ï A

Solution:
WWe know magnetic field due to coil B=Î¼0Ni2r
The needle is in equlibrium , BH tanÎ¸=Î¼0Ni2r
the current in the coil , i=0.34Ã10-4Ã2Ã0.24ÏÃ10-7Ã20=1710Ï A

Topic: Physics, Magnetic Properties of Matter, NEET
A circular current carrying coil has a radius R. The distance from the centre of the coil on the axis, where the magnetic induction will be 18th to its value at the centre of the coil is

Options:
A R3
B R3
C 23 R
D 23R

The correct answer is:
R3

Solution:
BcentreBaxis=1+x2R32
Baxis=18 Bcentre
81=1+x2R232
4=1+x2R2
3=x2R2
x2=3R2
x= 3R

Topic: Physics, Magnetic Effects of Current, NEET
A circular disc A of radius r is made from an iron plate of thickness t and another circular disc B of radius 4r is made from an iron plate of thickness t4. The relation between the moments of inertia IA and IB is

Options:
A IA>IB
B IA=IB
C IA<IB
D depends on the actual values of t and r.

The correct answer is:
IA<IB

Solution:
Disc	A	B

Radius	r	4r

Thickness	t	t4

IAIB=12mArA212mBrB2=mArA2mBrB2
IAIB=ÏrA2tArA2ÏrB2tBrB2
IAIB=rA4tArB4tB=14441
IAIB=164
â´ IA<IB

Topic: Physics, Rotational Motion, NEET
A circular disc is to be made by using iron and aluminium so that it acquire moment of inertia about the geometrical axis. It is possible with:

Options:
A aluminium at interior and iron surrounding it
B iron at interior and aluminium surrounding it
C using iron and aluminium lives in an alternate order
D sheet of iron is used at the external surface and aluminium sheet as internal layers

The correct answer is:
aluminium at interior and iron surrounding it

Solution:
Density of Iron is more than Aluminum.

Topic: Physics, Rotational Motion, NEET, NEET 2002