datavorous/entrance-exam-dataset huggingface dataset:
A scientist wants to insert a modified DNA directly into the nucleus of the animal cell. Which of the techniques will be best suited?

Options:
A Gene gun
B Disarmed pathogen
C Cloning vector
D Microinjection

The correct answer is:
Microinjection

Solution:
Microinjection is a technique of developing genetically modified organisms where the recombinant DNA is directly injected into the cell nucleus of the cell. A glass pipette is used to deliver foreign DNA into the cell and making it transgenic.

Topic: Biology, Biotechnology - Principles and Processes, NEET
A screen is placed 90 cm from an object. The image of an object on the screen is formed by a convex lens at two different locations separated by 20 cm . The focal length of the lens is

Options:
A 18 cm
B 21.4 cm
C 60 cm
D 85.6 cm

The correct answer is:
21.4 cm

Solution:
From displacement method
D = 30 cm and d = 20 cm
f=D2-d24D⇒f=902-2024×90
f=110×704×90⇒f=21.4 cm

Topic: Physics, Ray Optics, NEET
A screen is placed at a distance 40 cm away from an illuminated object. A converging lens is placed between the source and the screen and an attempt is made to form the image of the source on the screen. If no position of the lens could be found, the focal length of the lens

Options:
A must be less than 10Â cm
B must be greater than 10 cm
C must not be greater than 20Â cm
D must not be less than 10Â cm

The correct answer is:
must be greater than 10 cm

Solution:
Distance between object and image or screen is d=40 cm
If a converging lens makes real image of a real object then
dâ¥4f
fâ¤d4
fâ¤404
fâ¤10 cm
This is a condition for obtaining an image. As no image is found therefore focal length must be greater than 10 cm.

Topic: Physics, Ray Optics, NEET
A screw gauge has least count of 0.01 mm and there are 50 divisions in its circular scale. The pitch of the screw gauge is:

Options:
A 0.25 mm
B 0.5 mm
C 1.0 mm
D 0.01 mm

The correct answer is:
0.5 mm

Solution:
The least count on screw gauge is given by,L.C.=Pitch Number of division on circular scale

⇒0.01 mm=Pitch50

⇒Pitch=0.5 mm

Topic: Physics, Experimental Physics, NEET, NEET 2020 (Phase 1)
A scrubber in the exhaust of a chemical industrial plant removes

Options:
A particulate matter of the size 2.5 micrometer or less
B gases like sulphur dioxide
C particulate matter of the size 5 micrometer or above
D gases like ozone and methane

The correct answer is:
gases like sulphur dioxide

Solution:
Electrostatic precipitator is widely used in industries for removing dust and gases from polluted air. This device has electrode wires kept at several thousand volts that releases electrons by making a corona. Dust particles get a net negative charge by attaching to these electrons. The collecting plates are grounded and attract the charged dust particles.
Scrubber is an electrostatic precipitator used for cleaning dirty air by capturing the gases like SO2 and other oxides.

Topic: Biology, Environmental Issues, NEET
A scrubber in the exhaust of a chemical industrial plant removes

Options:
A Gases like SO2.
B Particulate matter of the size 5 μm or above.
C Gases like ozone and methane.
D Particulate matter of the size 2.5 μm or less.

The correct answer is:
Gases like SO2.

Solution:
Scrubbers are a type of system that is used to remove harmful gases such as SO2 from the industrial exhaust. Particulate matter is separated by electrostatic precipitators.

Topic: Biology, Environmental Issues, NEET
A second's pendulum clock having steel wire is calibrated at 20 Â°C. When the temperature is increased to 30 Â°C, how much times does the clock lose or gain in the week? Î±steel=1.2Ã10-5 Â°C-1

Options:
A 0.3628Â s
B 3.626Â s
C 362.8Â s
D 36.28 s

The correct answer is:
36.28 s

Solution:
T=2Ïlg
ÎTT=12Îll
ÎTT=12Î±ÎÎ¸
ÎT=12Î±ÎÎ¸T
ÎT=121.2Ã10-530-203600Ã24Ã7
ÎT=36.28 s

Topic: Physics, Oscillations, NEET
A section of fixed smooth circular track of radius R in vertical plane is shown in the figure. A block is released from position A and leaves the track at B . The radius of curvature of its trajectory when it just leaves the track at B is-

Options:
A R
B R/4
C R/2
D none

The correct answer is:
R/2

Solution:
In figure
OC=R cosâ¡53Â°=3R5; OE=R cosâ¡37Â°=4R5
So CE=R5
From A to B,
12mv2=mgR5 [Energy conservation]
v=2gR5
Let radius of curvature at B is r
g cosâ¡37Â°=v2r
r=v2g cosâ¡37Â°=2gR5Ã10Ã4/5=R2

Topic: Physics, Laws of Motion, NEET
A sedentary sea anemone gets attached to the shell lining of hermit crab. The association is

Options:
A ectoparasitism
B symbiosis
C commensalism
D amensalism

The correct answer is:
symbiosis

Solution:
This type of mutualisn to called protocooperation. In this type, the sea anemone grows on the back of the hermit crab. It protects the crab with the help of its nematocysts. Ectoparasites live on the outside of host, e.g., human body buse. In this interaction, the parasite gets the benefit at the expense of the host.
Commensalism is an assotiation between organisms in which one or both the species are benefitted and neither species is harmed. In amensalism one species is harmed, whereas the other is unaffected. Predation, parasitism and commensalism share a common characteristic, the interacting species live chosely together.

Topic: Biology, Organisms and Populations, NEET, NEET 2013 (All India)
A selectable marker is used to

Options:
A help in eliminating the non-transformants, so that the transformants can be regenerated
B identify the gene for a desired trait in an alien organism
C select a suitable vector for transformation in a specific crop
D mark a gene on a chromosome for isolation using restriction enzyme

The correct answer is:
help in eliminating the non-transformants, so that the transformants can be regenerated

Solution:
To facilitate cloning into a vector, the vector requires a selectable marker, which helps in identifying and eliminating non-transformants and selectively permitting the growth of the transformants.

Topic: Biology, Biotechnology - Principles and Processes, NEET, NEET 2019 (Odisha)
A self-fertilizing trihybrid plant forms:

Options:
A 8 different gametes and 64 different zygotes
B 4 different gametes and 16 different zygotes
C 8 different gametes and 16 different zygotes
D 8 different gametes and 32 different zygotes.

The correct answer is:
8 different gametes and 64 different zygotes

Solution:
The genotype of trihybrid would be $\mathrm{AaBbCc}$.
Number of gametes $=$ (no. of genes studied) ${ }^n$
Where $n=$ no. of heterozygous alleles
Thus, Number of gametes $=(2)^3=8$
No. of Zygotes $=(2)^{2 \mathrm{n}}=(2)^{2 \times 3}=(2)^6$ $=64$

Topic: Biology, Principles of Inheritance and Variation, NEET, NEET 2004
A self-fertilizing trihybrid plant forms

Options:
A 8 different gametes and 16 zygotes
B 4 different gametes and 16 zygotes
C 4 different gametes and 64 zygotes
D 8 different gametes and 64 zygotes

The correct answer is:
8 different gametes and 64 zygotes

Solution:
Number of different types of gametes formed by an organism is given by formula (2)n, where n= no. of genes/hybrid forms involved.
Number of gametes for a trihybrid will be = (2)3 = 8
Number of zygotes formed by an organism is given by formula (4)n, where n= no. of genes/hybrid forms involved.
Number of zygotes for a trihybrid will be = (4)3â = 64

Topic: Biology, Principles of Inheritance and Variation, NEET
A sequence of how many nucleotides in messenger RNA makes a codon for an amino acid?

Options:
A three
B four
C one
D two

The correct answer is:
three

Solution:
A codon is a sequence of three DNA or RNA nucleotides that corresponds with a specific amino acid or stop signal during protein synthesis.
Related Theory
There are 3 stop codons in the genetic code $U A G, U A A$, and $U G A$.

Topic: Chemistry, Biomolecules, NEET, NEET 2004
A sequential expression of a set of human genes occurs when a steroid molecule binds to the:

Options:
A messenger RNA
B DNA sequence
C ribosome
D transfer RNA.

The correct answer is:
DNA sequence

Solution:
A sequential expression of a set of human genes is the DNA sequence because geneis the functional part of DNA sequence. The steroid hormone receptor-protein complex activates the transcription of the target gene by binding to a specific DNA sequence.
Related Theory
Steroid hormones interact with intracellular receptors and regulate gene expression and protein synthesis. It results in physiological and development effects. The hormonereceptor complex interacts with the genome driving subsequent $m R N A$ and protein synthesis, which brings about tissue growth and differentiation. (Ref. to Fig. 25.3)

Topic: Biology, Chemical Coordination and Integration, NEET, NEET 2007
A series combination of $n_1$ capacitors, each of value $\mathrm{C}_1$, is charged by a source of potential difference $4 \mathrm{~V}$. When another parallel combination of $n_2$ capacitors, each of value $C_2$, is charged by a source of potential difference $V$, it has the same (total) energy stored in it, as the first combination has. The value of $\mathrm{C}_2$, in terms of $\mathrm{C}_1$, is then

Options:
A $\frac{2 \mathrm{C}_1}{\mathrm{n}_1 \mathrm{n}_2}$
B $16 \frac{\mathrm{n}_2}{\mathrm{n}_1} \mathrm{C}_1$
C $2 \frac{\mathrm{n}_2}{\mathrm{n}_1} \mathrm{C}_1$
D $\frac{16 C_1}{n_1 n_2}$

The correct answer is:
$\frac{16 C_1}{n_1 n_2}$

Solution:
P.D. across each capacitor is \(V_1=\frac{4 \mathrm{~V}}{\mathrm{n}_1}\)
Energy stored across the combination is \(\mathrm{U}_1=\mathrm{n}_1 \times \frac{1}{2} \mathrm{C}_1 \frac{16 \mathrm{~V}^2}{\mathrm{n}_1^2}=\frac{16 \mathrm{C}_1 \mathrm{~V}^2}{2 \mathrm{n}_1}\)

\(\begin{aligned}
& \text {Energy stored across parallel combination is } \mathrm{U}_2=\mathrm{n}_2 \times \frac{1}{2} \mathrm{C}_2 \mathrm{~V}^2\\
& \frac{16 C_1 v^2}{2 n_1}=\frac{n_2 C_1 v^2}{2} \\
& \Rightarrow C_2=\frac{16 C_1}{n_1 n_2}
\end{aligned}\)

Topic: Physics, Capacitance, NEET, NEET 2010 (Screening)
A series combination of N 1 is charged to a potential difference 3âV. Another parallel combination of N 2 (each of capacity C 2 is charged to a potential difference V. The total energy stored in both the combination is the same. The value of C 1 in terms of C 2 is

Options:
A C2N1N29
B C2N12N229
C C2N19N2
D C2N29N1

The correct answer is:
C2N1N29

Solution:
C eq = C 1 N 1
E= 1 2 C V 2
= 1 2 C 1 N 1 9 V 2
= 9 2 C 1 N 1 V 2
C eq = N 2 C 2
E= 1 2 C V 2 = 1 2 C 2 N 2 V 2
9 2 C 1 N 1 ââ V 2 = C 2 N 2 V 2 2
C 1 = C 2 N 2 N 1 9

Topic: Physics, Electrostatics, NEET
A series L-C-R circuit containing 5.0 H inductor, 80 μF capacitor and 40 Ω resistor is connected to 230 V variable frequency ac source. The angular frequencies of the source at which power transferred to the circuit is half the power at the resonant angular frequency are likely to be:

Options:
A 25 rad s-1 and 75 rad s-1
B 50 rad s-1 and 25 rad s-1
C 46 rad s-1 and 54 rad s-1
D 42 rad s-1 and 58 rad s-1

The correct answer is:
46 rad s-1 and 54 rad s-1

Solution:
L=5H,C=80μF,R=40ω; V=230 V

Resonance angular frequency =ω0=1LC=15×80×10-6=50 rad s-1

Half power angular frequency =ω0±R2L=50±402×5=50±4=54 or 46

We get,

ω1=46 rad s-1

And ω2=54 rad s-1

Topic: Physics, Alternating Current, NEET, NEET 2021
A seriesLCR circuit is connected to an ac voltage source.
When L is removed from the circuit, the phase difference between current and voltage is π3.
If instead C is removed from the circuit, the phase difference is again π3 between current and voltage. The power factor of the circuit is:

Options:
A 0.5
B 1.0
C –1.0
D zero

The correct answer is:
1.0

Solution:
As phase angle contribution by capacitor and inductor are equal in magnitude, so
\(\begin{gathered}
\mathrm{X}_{\mathrm{L}}=\mathrm{X}_{\mathrm{c}} \\
\Rightarrow \omega \mathrm{L}=\frac{1}{\omega \mathrm{C}}
\end{gathered}\)
It means that the RLC circuit is in resonance condition.
So, impedance at resonance, \(Z=\sqrt{R^2+\left(X_L-X_C\right)^2}=\sqrt{R^2+\left(X_L-X_L\right)^2}=\sqrt{R^2}=R\)
Thus, power factor, \(\cos \phi=\frac{R}{Z}=\frac{R}{R}=1\)

Topic: Physics, Alternating Current, NEET, NEET 2020 (Phase 1)
A series LCR circuit with inductance 10 H, capacitance 10 μF, resistance 50 Ω is connected to an AC source of voltage, V=200sin100t volt. If the resonant frequency of the LCR circuit is ν0 and the frequency of the AC source is ν, then

Options:
A ν0=ν=50πHz
B ν0=50πHz, ν=50 Hz
C ν=100 Hz; ν0=100πHz
D ν0=ν=50 Hz

The correct answer is:
ν0=ν=50πHz

Solution:
From the given equation V=200sin100t, the angular frequency of the source is

ω=100 rad s-1

⇒ν=ω2π=1002π=50πHz

For resonance

ν0=12πLC=12π10×10×10-6

⇒ν0=1002π=50πHz

Topic: Physics, Alternating Current, NEET, NEET 2022 (Phase 1)
A series R-C circuit is connected to an alternating voltage source. Consider two situations:
a. When capacitor is air filled.
b. When capacitor is mica filled.
Current through resistor is i and voltage across capacitor is V then:

Options:
A Va>Vb
B ia>ib
C Va=Vb
D Va<Vb

The correct answer is:
Va>Vb

Solution:
âââââââ
i=vR2+1Cω2
VC=vR2+1Cω2×1Cω
VC=VRCω2+1
If we fill a di-electric material
C↑⇒VC↓

Topic: Physics, Alternating Current, NEET, NEET 2015 (Phase 2)
A set of 6 test tubes having same temperature but different pH were setup to conduct an experiment. To each of the tubes, starch and amylase were added in a mix of sterile water. The concentration was kept the same. The following table notes the reaction completion time for each of the six tubes.

pH	1	3	5	7	9	11

Time taken to complete in seconds(s).	94	64	32	5	35	66

Using the above data, determine the pH where amylase has shown highest rate of activity.

Options:
A pH 1
B pH 5
C pH 7
D pH 11

The correct answer is:
pH 7

Solution:
Enzymes are biocatalyst that help in speeding up the reaction rate exponentially. Temperature, pH , substrate concentration are few of the factors that affect the activity of the enzyme. These molecules shows highest activity work in a particular pH called as optimum pH . Below or above that pH range the enzyme activity will reduce or become inactive. Since the enzyme amylase took least time to break starch in the range of pH 7 compared to other pH values, it is showing highest activity at pH 7 . Thus, pH 7 is the optimum pH .

Topic: Biology, Digestion and Absorption, NEET
A set of 56 tuning forks are arranged in series of increasing frequencies. If each fork gives 4 beats with preceding one and the frequency of the last is twice than that of first, then frequency of the first fork is

Options:
A 110 Hz
B 220 Hz
C 224 Hz
D 448 Hz

The correct answer is:
220 Hz

Solution:
n56=n1+56-14
Also, n56=2n1
â´2n1=n1+55Ã4
â´n1=220 Hz

Topic: Physics, Waves and Sound, NEET
A set of n equal resistances, of value R each, are connected in series to a battery of emf E and internal resistances R . The current drawn is I . Now, the n resistance are connected in parallel to the same battery. Then the current drawn from battery becomes 10I . The value of n is

Options:
A 20
B 11
C 10
D 9

The correct answer is:
10

Solution:
In the first case, it is given that 'n' resistances are arranged in series, each having a resistance R.

In that case, the nett reistance of the external circuit is 'nR'. The internal resistance itself is R. Hence, the resultant resistance of the circuit is nR+R=(n+1)R.

∵I= εnR+R
⇒ I=E(n+1)R .... (i)

In the case of parallel combination, these external n resistances are arranged in parallel,thereby their nett external resistance is Rn. This is in series with the internal resistance R.

Hence,
10I =ERn+R=nER+nR .... (ii)
From (i) and (ii);
nER+nR=10EnR+R
n=10

Topic: Physics, Current Electricity, NEET, NEET 2018
A sewage treatment process in which anaerobic decomposer bacteria present in the waste is used to further carry out the breakdown of organic matter is called:

Options:
A Cyclic treatment
B Activated sludge treatment
C Primary treatment
D Tertiary treatment

The correct answer is:
Activated sludge treatment

Solution:
Sewage treatment process consists of two steps, first is primary treatment which is a physical process and second is biological treatment in which aerobic heterotrophic bacteria carry out breakdown of organic matter present in water to form flocs, which forms activated sludge. Anaerobic decomposer bacteria carries out further breakdown of activated sludge to form biogas.

Topic: Biology, Microbes in Human Welfare, NEET
A sheet is placed on a horizontal surface in front of a strong magnetic pole. A force is needed to:
A. hold the sheet there if it is magnetic.
B. hold the sheet there if it is non-magnetic.
C. move the sheet away from the pole with uniform velocity if it is conducting.
D. move the sheet away from the pole with uniform velocity if it is both, non-conducting and non-polar.
Choose the correct statement(s) from the options given below:

Options:
A A and C only
B A, C and D only
C C only
D B and D only

The correct answer is:
A and C only

Solution:
A. A magnetic pole will repel or attract magnetic sheet so force is need.
B. If sheet is non-magnetic, no force needed.
C. If it is conducting, then there will be addy current in sheet, which opposes the motion. So forces is needed move sheet with uniform speed.
D. The non-conducting and non-polar sheet do not interact with magnetic field of magnet.

Topic: Physics, Magnetic Properties of Matter, NEET, NEET 2024
A shell of mass 0.020 kg is fired by a gun of mass 100 kg. If the muzzle speed of the shell is 80 m s-1. What is the recoil speed of the gun?

Options:
A 1.6 cm s-1
B 0.5 cm s-1
C 2 cm s-1
D 3 cm s-1

The correct answer is:
1.6 cm s-1

Solution:
Here, the mass of gun, M=100 kg
Mass of shell, m=0.02 kg
Speed of shell, v= 80 m s-1 and let recoil speed of the gun =V
∴ Speed of shell =v+V
As initially both the gun and the shell are the rest before firing, so the initial momentum Pi of the system = 0
Final momentum Pf of the system after firing
=MV+mv=100V+0.02× ( 80+V )
According to the law of conservation of linear momentum,
Pi=Pf
0=100 V+1.6+0.02V
⇒ V≈-1.6100=0.016 m s-1
∴ Recoil speed of gun, V=0.016 m s-1=1.6 cm s-1

Topic: Physics, Center of Mass Momentum and Collision, NEET
A shell of mass 10 kg is moving with a velocity of 10 ms-1 when it blasts and forms two parts of mass 9 kg and 1 kg respectively. If the 1st mass is stationary, the velocity of the 2nd is

Options:
A 1 ms-1
B 10 ms-1
C 100 ms-1
D 1000 ms-1

The correct answer is:
100 ms-1

Solution:
Given that,
v1=10 ms-1; m1=10 kg
v2=0; m2=9 kg
v3=v; m3=1 kg
According to conservation of momentum
m1v1=m2v2+m3v3
10×10=9×0+1×v⟹v=100 ms-1

Topic: Physics, Laws of Motion, NEET
A shell of mass 20 kg at rest explodes into two fragments whose masses are in the ratio 2 : 3. The smaller fragment moves with a velocity of 6 m s-1. The kinetic of the larger fragment is

Options:
A 96 J
B 216 J
C 144 J
D 360 J

The correct answer is:
96 J

Solution:
From conservation of momentum,
∴m1v1=m2v2 ( m1=3kg & m2=12kg )
∴8×6=12×v2
⇒ v2=4 m/s
∴K2=121242=96 J

Topic: Physics, Laws of Motion, NEET
A shell of mass $200 \mathrm{~g}$ is ejected from a gun of mass $4 \mathrm{~kg}$ by an explosion that generates $1.05 \mathrm{~kJ}$ of energy. The initial velocity of the shell is

Options:
A $100 \mathrm{~ms}^{-1}$
B $80 \mathrm{~ms}^{-1}$
C $40 \mathrm{~ms}^{-1}$
D $120 \mathrm{~ms}^{-1}$

The correct answer is:
$100 \mathrm{~ms}^{-1}$

Solution:
Key Idea: In the given problem conservation of linear momentum and energy hold good.
Conservation of momentum yields.
$\begin{aligned}
& m_1 v_1+m_2 y_2=0 \\
& \text{or } 4 y_1+0.2 y_2=0
\end{aligned}$
Conservation of energy yields
$\begin{array}{l}
& \frac{1}{2} m_1 y_1^2+\frac{1}{2} m_2 y_2^2=1050 \\
& \text {or } \frac{1}{2} \times 4 y_1^2+\frac{1}{2} \times 0.2 \times y_2^2=1050 \\
& \text {or } 2 v_1^2+0.1 v_2^2=1050
\end{array}$
Solving Eqs. (i) and (ii), we have
$v_1=100 \mathrm{~m} / \mathrm{s}$

Topic: Physics, Center of Mass Momentum and Collision, NEET, NEET 2008 (Screening)
A shell of mass m is at rest initially. It explodes into three fragments having mass in the ratio 2:2:1. If the fragments having equal mass fly off along mutually perpendicular directions with speed v, the speed of the third (lighter) fragment is:

Options:
A 2v
B 22v
C 32v
D v

The correct answer is:
22v

Solution:
The speed of the third particle will be along angle bisector of the other two velocities. Applying the conservation of momentum,

mv'=2mvcos45°+2mvcos45°⇒v'=22v

Topic: Physics, Center of Mass Momentum and Collision, NEET, NEET 2022 (Phase 1)
A ship A is moving Westwards with a speed of 10 kmh-1 and a ship B 100 km south of A , is moving Northwards with a speed of 10 km h-1 . The time after which the distance between them becomes shortest, is:

Options:
A 0 h
B 5 h
C 52 h
D 102 h

The correct answer is:
5 h

Solution:
Relative velocity of ship B is,
V→BA=102 +102=102 kmph
Let shortest distance between ship A & B is AO,
Distance OB=100cos⁡45o=502 km
Hence time taken to reach the shortest distance,
t=502102=5hrs

Topic: Physics, Motion In One Dimension, NEET, NEET 2015 (Phase 1)
A short bar magnet has a length 2l and a magnetic moment 10 A m2. Find the magnetic field at a distance of z=0.1 m from its centre on the axial line. Here, l is negligible as compared to z.

Options:
A 4×10-3 T
B 1Ã10-3 T
C 3Ã10-3 T
D 2Ã10-3 T

The correct answer is:
2Ã10-3 T

Solution:
If the point lies on the axial line of bar magnet

B=μ. 2Md4πd2-l22

If l< <d (as given in the question)

B=μ04π×2Md3

=μ04π×20×103

μ04π=10-7

=2×10-3 T

Topic: Physics, Magnetic Properties of Matter, NEET
A short bar magnet has a magnetic moment of 0.48 J T-1. Find the magnetic field produced by the magnet at a distance of 10 cm from the centre of the magnet on the axis.

Options:
A 0.96×10-4 T

B 2.16×10-4 T

C 0.96×10-3 T

D 2.16×10-3 T

The correct answer is:
0.96×10-4 T

Solution:
Given, the magnetic moment of the bar magnet M=0.48 J T-1
Distance from the centre of the magnet d=10 cm=0.1 m
When the point lies on the axial line. The magnetic field at point P

B = μ 0 4 π · 2 M d 3
= 1 0 - 7 × 2 × 0 . 4 8 0 . 1 3 = 0 . 9 6 × 1 0 - 4 T
The direction of the magnetic field is along the direction of the magnetic moment. We know that the direction of magnetic moment is from S to N pole. Thus, the direction of the magnetic field is from S to N pole of the magnet.

Topic: Physics, Magnetic Properties of Matter, NEET
A short bar magnet is placed in a uniform magnetic field of 0.25 T, with its axis at an angle of 30° with the field. If it experiences a torque of magnitude 4.5×10-2 N m, then its magnetic moment is

Options:
A 0.36 N m T-1
B 0.25 N m T-1
C 0.5 N m T-1
D 1.25 N m T-1

The correct answer is:
0.36 N m T-1

Solution:
The magnitude of torque acting on a magnetic dipole in a uniform magnetic field is

τ=MBsinθ=4.5×10-2 N mM×0.25×sin30°=4.5×10-2 N m⇒M=0.36 N m T-1

Topic: Physics, Magnetic Effects of Current, NEET
A short bar magnet of magnetic moment 0.4 J T-1 is placed in a uniform magnetic field of 0.16 T. The magnet is in stable equilibrium when the potential energy is

Options:
A -0.064 J
B zero
C -0.082Â J
D 0.064Â J

The correct answer is:
-0.064 J

Solution:
For stable equilibrium
U= -MB
= -0.40.16
= -0.064 J

Topic: Physics, Magnetic Properties of Matter, NEET
A short bar magnet of magnetic moment m=0.32 J T-1 is placed in a uniform magnetic field of 0.15 T. If the bar is free to rotate in the plane of the field, What is the potential energy of the magnet in unstable position?

Options:
A 4.8×10-2 J

B 9.6Ã10-2 J
C 2.4Ã10-2 J
D 1.2Ã10-2 J

The correct answer is:
4.8×10-2 J

Solution:
For the unstable equilibrium, the angle between the magnetic moment and magnetic field is 180o. ( ∵ In this position it will be in a direction to magnetic field thus no torque will act on it.)
θ = 1 8 0 o
The potential energy of the magnet
U=-mB cos 180°
=- 0.32×0.15 (- 1) = 4.8×10-2 J
Thus, for the unstable equilibrium, the potential energy is 4.8×10-2 J.

Topic: Physics, Magnetic Properties of Matter, NEET