datavorous/entrance-exam-dataset huggingface dataset:
Two sources of light emit with powers of 100 W. One emits X-rays of wavelength 1 nm, and the other emits visible light at 500 nm. What is the ratio of the number of photons of the X-rays to the number of photons of visible light at the given wavelength?

Options:
A 1/500
B 500
C 1
D 1/250

The correct answer is:
1/500

Solution:
Correct Option is : (A)
1/500

Topic: Chemistry, Dual Nature of Matter, NEET
Two sources of sound placed close to each other, are emitting progressive waves given by
$y_1=4 \sin 600 \pi t \text { and } y_2=5 \sin 608 \pi t$
An observer located near these two sources of sound will hear

Options:
A 4 beats per second with intensity ratio $25: 16$ between waxing and waning
B 8 beats per second with intensity ratio $25: 16$ between waxing and waning
C 8 beats per second with intensity ratio $81: 1$ between waxing and waning
D 4 beats per second with intensity ratio $81: 1$ between waxing and waning

The correct answer is:
4 beats per second with intensity ratio $81: 1$ between waxing and waning

Solution:
Given, $y_1=4 \sin 600 \pi t$ and $y_2=5 \sin 608 \pi t$
Comparing with general equation $y=a \sin 2 \pi f t$
we get, $f_2=300 \mathrm{~Hz}$ and $f_2=304 \mathrm{~Hz}$
Number of beats $=f_2-f_1=4 \mathrm{~s}^{-1}$
$\begin{aligned}
\frac{I_{\max }}{I_{\min }} & =\left(\frac{a_1+a_2}{a_1-a_2}\right)^2 \\
& =\left(\frac{4+5}{4-5}\right)^2=\frac{81}{1}
\end{aligned}$

Topic: Physics, Waves and Sound, NEET, NEET 2012 (Screening)
Two special character of a family are given, select the correct option for the character and the name of the family.

Options:
A Papilionaceous corolla, diadelphous stamen - Solanaceae
B Papilionaceous corolla, diadelphous stamen - Fabaceae
C Cruciform corolla, diadelphous stamen - Fabaceae
D Papilionaceous corolla , monodelphous stamen - Solanaceae

The correct answer is:
Papilionaceous corolla, diadelphous stamen - Fabaceae

Solution:
Papilionaceous corolla is character of Fabaceae family and it has vexillary aestivation. Cohesion of stamens in which filaments are united and anther lobes are free is called adelphous condition and when the fusion results in formation of two bundles it is called diadelphous condition. In Fabaceae the stamens are united forming (9) + 1 condition.

Topic: Biology, Morphology of Flowering Plants, NEET
Two spheres A and B have diameters in the ratio 1:2, densities in the ratio 2:1 and specific heat capacities in the ratio 1:3. The ratio of their thermal capacities is

Options:
A 1:6
B 1:12
C 1:3
D 1:4

The correct answer is:
1:12

Solution:
H1H2=m1C1m2C2=43πr13.ρ1.C143πr23.ρ2.C2
=r13r23×ρ1ρ2×C1C2 = ( 1 2 ) 3 ×( 2 1 )×( 1 3 )= 1 12

Topic: Physics, Thermal Properties of Matter, NEET
Two spheres A and B have diameters in the ratio 1:2, densities in the ratio 2:1 and specific heats in the ratio 1:3. Find the ratio of their thermal capacities.

Options:
A 1:6
B 1:12
C 1:3
D 1:4

The correct answer is:
1:12

Solution:
H1H2=m1c1m2c2
H1H2=43πr13⋅ρ1⋅C143πr23⋅ρ2⋅C2
H1H2=r13r23×ρ1ρ2×C1C2
H1H2=1323×21×13=112

Topic: Physics, Thermal Properties of Matter, NEET
Two spheres $A$ and $B$ of masses $m_1$ and $m_2$ respectively collide. $A$ is at rest initially and $B$ is moving with velocity $y$ along $x$-axis. After collision $B$ has a velocity $\frac{v}{2}$ in a direction perpendicular to the original direction. The mass $A$ moves after collision in the direction

Options:
A same as that of $B$
B opposite to that of $B$
C $\theta=\tan ^{-1}\left(\frac{1}{2}\right)$ to the $x$-axis
D $\theta=\tan ^{-1}\left(\frac{-1}{2}\right)$ to the $x$-axis

The correct answer is:
$\theta=\tan ^{-1}\left(\frac{1}{2}\right)$ to the $x$-axis

Solution:
$\text {Here, } \mathbf{p}_i=m_2 v \mathbf{i}+m_2 \times 0$

$\mathbf{p}_f=m_2 \frac{v}{2} \mathbf{j}+m_1 \times \mathbf{v}_2$
Law of conservation of momentum
$\begin{gathered}
\mathbf{p}_i=\mathbf{p}_f \\
m_2 v \mathbf{i}=m_2 \frac{v}{2} \mathbf{j}+m_1 \times \mathbf{v}_1 \\
\mathbf{v}_1=\frac{m_2}{m_1} v \mathbf{i}+\frac{m_2}{m_1} \frac{v}{2} \mathbf{j}
\end{gathered}$
From this equation we can find
$\begin{aligned}
\tan \theta & =\frac{1}{2} \\
\theta & =\tan ^{-1}\left(\frac{1}{2}\right) \text { to the } x \text {-axis. }
\end{aligned}$

Topic: Physics, Center of Mass Momentum and Collision, NEET, NEET 2012 (Screening)
Two spheres are made of same metal and have same mass. One is solid and the other is hollow. When heated to the same temperature, which of the following statements is correct about the percentage increase in their diameters

Options:
A it will be more for hollow sphere
B it will more for solid sphere
C it will be same for both spheres
D it may be more or less depending on the ratio of the diameters of the two spheres

The correct answer is:
it will be same for both spheres

Solution:
Percentage increase in diameter is given by,
ΔDD×100=αΔθ×100
As two spheres are made of same metal hence have same coefficient of linear thermal expansion. Therefore for same rise in temperature the percentage increase in their diameters will be same for both.

Topic: Physics, Thermal Properties of Matter, NEET
Two spheres of equal masses, one of which is a thin spherical shell and the other a solid, have the same moment of inertia about their respective diameters. The ratio of their radii will be

Options:
A 5:7
B 3:5
C 3: 5
D 3: 7

The correct answer is:
3: 5

Solution:
Let the radii of the thin spherical and the solid sphere are R1 and R2 respectively.
Then the moment of inertia of the spherical shell about their diameter
I=23 MR12 ...(i)
and the moment of inertia of the solid sphere is given by
I=25MR22 ...(ii)
Given that the masses and moment of inertia for both the bodies are equal, then from Eqs. (i) and (ii)
23MR12=25MR22⇒R12R22=35
⇒ R1R2=35 ⇒R1:R2=3:5

Topic: Physics, Rotational Motion, NEET
Two spheres of masses $m$ and $M$ are situated in air and the gravitational force between them is $F$. The space around the masses is now filled with a liquid of specific gravity 3 . The gravitational force will now be:

Options:
A $3 F$
B $F$
C $\frac{F}{3}$
D $\frac{F}{9}$

The correct answer is:
$F$

Solution:
The gravitational force does not depend upon the medium in which objects are placed.

Topic: Physics, Gravitation, NEET, NEET 2003
Two spheres of masses m and M are situated in air and the gravitational force between them is F. The space around the masses is now filled with a liquid of specific gravity 3. The gravitational force will now be -

Options:
A 3F
B F
C F/3
D F/9

The correct answer is:
F

Solution:
Gravitational force between two masses does not depends on medium between them.

Topic: Physics, Gravitation, NEET
Two spheres of radii 2 cm and 3 cm are charged to same potential. If Ï1 and Ï2 be respectively the values of surface charge density on the conductors, then the ratio Ï1Ï2 will be

Options:
A 49
B 23
C 32
D 94

The correct answer is:
32

Solution:
V1=σ1rε0=σ1×2ε0
V2=σ2Rε0=σ2×3ε0
As the spheres are charged to same potential,
∴ σ1×2ε0=σ2×3ε0
∴ σ1σ2=32

Topic: Physics, Electrostatics, NEET
Two spheres of radii 2 cm and 3 cm are charged to the same potential. If Ï1 and Ï2 be respectively the values of surface charge density on the conductors, then the ratio Ï1Ï2 will be

Options:
A 49
B 23
C 32
D 94

The correct answer is:
32

Solution:
V=σ1rε0=σ1×2ε0    V=σ2Rε0=σ2×3ε0
∴ σ1×2ε0=σ2×3ε0
∴ σ1σ2=32

Topic: Physics, Electrostatics, NEET
Two spheres of the same material have radii 1 m and 4 m and temperature 4000 K and 2000 K respectively. The ratio of the energy radiated per second by the first sphere to that by the second is

Options:
A 1:1
B 16:1
C 4:1
D 1:9

The correct answer is:
1:1

Solution:
Energy radiated
E = Ï T 4 Ã 4 Ï R 2 Ã time Ã e
E 1 E 2 = 4000 4 Ã 1 2 Ã 1 Ã 4 Ï Ï e 2000 4 Ã 4 2 Ã 1 Ã 4 Ï Ï e = 1 1 .

Topic: Physics, Thermal Properties of Matter, NEET
Two spheres of the same radius are made from the same material. One is hollow and the other is solid. If they are heated together from 20 Â°C to 100 Â°C .

Options:
A both will expand equally
B hollow sphere will expand more
C solid sphere will expand more
D the relative expansion of solid and hollow sphere depends on the material of sphere

The correct answer is:
both will expand equally

Solution:
both will expand equally as thermal expansion depends on size, material and temperature, which are same in all the two.

Topic: Physics, Thermal Properties of Matter, NEET
Two spherical bodies of mass M and 5 M and radii R and 2 R respectively are released in free space with initial separation between their centres equal to 12 R . If they attract each other due to gravitational force only, then distance covered by the smaller body just before collision is

Options:
A 2.5 R
B 4.5 R
C 1.5 R
D 7.5 R

The correct answer is:
7.5 R

Solution:
Let the spheres collide after time t , when the smaller sphere covered distance x 1 and bigger sphere covered distance x 2 .
The gravitational force acting between two spheres depends on the
distance which is variable quantity.
The gravitational force, F x = GM Ã 5 M 12 R - x 2
Accelration of smaller body, a 1 x = G Ã 5 M 12 R - x 2
Acceleration of bigger body, a 2 x = G M 12 R - x 2
From equation of motion,
x 1 = 1 2 a 1 x t 2 and x 2 = 1 2 a 2 x t 2
â x 1 x 2 = a 1 x a 2 x = 5 â x 1 = 5 x 2
We know that x 1 + x 2 = 9 R

x 1 + x 1 5 = 9 R â´ x 1 + 45 R 5 = 7 . 5 R
Therefore, the two spheres collide when the smaller sphere covers a distance of 7.5 R.

Topic: Physics, Gravitation, NEET
Two spherical bodies of masses m and 6m and radii R and 2R respectively are released in free space with initial separation between their centres equal to 10R . If they attract each other due to gravitational force only, then the distance covered by smaller sphere just before collisions will be

Options:
A 6R
B 7.5R
C 2.5R
D 9R

The correct answer is:
6R

Solution:
Their relative displacement till collision will be 7R . Therefore distance covered by smaller sphere is
x1=m2drelm1+m2
x1=6m7Tm+6m
x1=6R

Topic: Physics, Center of Mass Momentum and Collision, NEET
Two springs A and B are identical except that A is stiffer than B, i.e., kA>kB. If the two springs are stretched by the same force, then

Options:
A more work is done on B, i.e. WB>WA
B more work is done on A, i.e. WA>WB
C work done on A and B are equal
D work done depends upon the way in which they are stretched

The correct answer is:
more work is done on B, i.e. WB>WA

Solution:
F=kAxA=kBxB
xA⁡xB⁡=kB⁡kA⁡…i
W⁡A⁡=12kA⁡xA⁡2;W⁡B⁡=12kB⁡xB⁡2
W⁡A⁡W⁡B⁡=kA⁡kB⁡xA⁡2xB⁡2=kA⁡kB⁡kB⁡2kA⁡2=kB⁡kA⁡…Usingi
Hence WB>WA. So more work is done on B.

Topic: Physics, Work Power Energy, NEET
Two springs are joined and attached to a mass of 16 kg. The system is then suspended vertically from a rigid support. The spring constant of the two springs are k1 and k2 respectively. The period of vertical oscillations of the system will be

Options:
A 8πk1+k2k1k2
B π2k1k2
C k1+k28π
D πk1-k22

The correct answer is:
8πk1+k2k1k2

Solution:
Here two springs are in series
â´ Effective spring constant is given by 1k=1k2+1k2
â k=k1k2k1+k2
â´ Time period of oscilliation.
T=2Ïmk=2Ï 16 k 1 k 2 .( k 1 + k 2 )
=8Ïk1+k2k1k2

Topic: Physics, Oscillations, NEET
Two springs are joined and attached to a mass of 16 kg. The system is then suspended vertically from a rigid support. The spring constant of the two springs are k1 and k2 respectively. The period of vertical oscilliations of the system will be

Options:
A 8πk1+k2k1k2
B π2k1k2
C k1+k28π
D πk1-k22

The correct answer is:
8πk1+k2k1k2

Solution:
Here two springs are in series
â´ Effective spring constant is given by 1k=1k2+1k2
â k=k1k2k1+k2
â´ Time period of oscilliation.
T=2Ïmk=2Ï 16 k 1 k 2 .( k 1 + k 2 )
=8Ïk1+k2k1k2

Topic: Physics, Oscillations, NEET
Two springs of force constants K1 and K2 , respectively, are connected to a mass m, as shown. The frequency of oscillation of the mass is f . If both K1 and K2 are made four times their original values, the frequency of oscillation becomes

Options:
A f / 2
B f / 4
C 4 f
D 2 f

The correct answer is:
2 f

Solution:
In the given figure, two springs are connected in parallel. Therefore, the effective spring constant is given by
Keff=K1+ K2
Frequency of oscillation,
f=12ÏKeffm=12ÏK1+K2m ...........(i)
As K1 and K2 are increased by four times
New frequency,
f'=12Ï4K1+K2m=2f ( using (i) )

Topic: Physics, Oscillations, NEET
Two springs of spring constant $k_1$ and $k_2$ are joined in series. The effective spring constant of the combination is given by:

Options:
A $\sqrt{k_1 k_2}$
B $\left(k_1+k_2\right) / 2$
C $k_1+k_2$
D $k_1 k_2 /\left(k_1+k_2\right)$

The correct answer is:
$k_1 k_2 /\left(k_1+k_2\right)$

Solution:
When the spring joined in series the total extension of spring is:
$$
\begin{aligned}
& \Rightarrow y=y_1+y_2=\frac{-F}{k_1}-\frac{F}{k_2} \\
& \Rightarrow \quad y=-\mathrm{F}\left[\frac{1}{k_1}+\frac{1}{k_2}\right] \\
& \therefore \text { Spring constant } k=\frac{k_1 k_2}{k_1+k_2}
\end{aligned}
$$

Topic: Physics, Laws of Motion, NEET, NEET 2004
Two springs P and Q having stiffness constants K1 and K2(<K1) respectively, are stretched equally. Then:

Options:
A more work is done to stretch Q
B more work is done to stretch P
C their force constants will become equal
D equal work is done on both the springs

The correct answer is:
more work is done to stretch P

Solution:
Since K1>K2, therefore, P will develop greater restoring force than Q. So the applied force on P will be more than the applied force on Q. Extension is same in both the cases. So, more work is done on P.

Topic: Physics, Work Power Energy, NEET
Two springs with negligible masses and force constant of k1=200âN mâ1 and k2=160âN mâ1 are attached to the block of mass m=10âkg as shown in the figure. Initially, the block is at rest, at the equilibrium position in which both springs are neither stretched nor compressed. At time t=0, the sharp impulse of 50 N s is given to the block with a hammer along with the spring.

Options:
A Period of oscillations for the mass m is Ï 6 s
B Maximum velocity of the mass m during its oscillation is 10 m s-1
C Data are insufficient to determine maximum velocity
D Amplitude of oscillation is 0.83 m

The correct answer is:
Amplitude of oscillation is 0.83 m

Solution:
T= period
= 2 Ï m K 1 + K 2 = 2 Ï 1 0 3 6 0 = Ï 3 s
The maximum velocity is always at equilibrium position since at any other point there will be a restoring force attempting to slow the mass.
â´ V max = impulse mass = 5 0 1 0 = 5 m s - 1
A = amplittude = V max Ï = 5 2 Ï Ï 3 = 5 6 = 0.83 m

Topic: Physics, Oscillations, NEET
Two stars radiate maximum energy at wavelengths 3.6 Ã 10-5 cm and 4.8 Ã 10-5 cm respectively. The ratio of their temperatures is

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

The correct answer is:
4 3

Solution:
Let T and T' be the temperatures of the two stars and λmax and λmax' be their respective wavelengths of maximum emission.
Here, λmax = 3.6 × 10-5 cm and λmax' = 4.8 × 10-5 cm
Now, T T ′ = λ max ′ λ max = 4.8 × 10 - 5 3.6 × 10 - 5 = 4 3

Topic: Physics, Thermal Properties of Matter, NEET
Two stationary sources of sound, S1 and S2 having an equal frequency are fixed some distance apart. The position A is left of S1, position B in the middle of the two sources and position C is to the right of S2. An observer starts moving with velocity V0â¡ from position A towards S1, then

Options:
A beats for three position A,B and C will be heard
B beats will be heard from A and C but not in case of B
C beats will be not heard for A and C but will be heard for B
D beats will be not heard for three position of A,B and C

The correct answer is:
beats will be not heard for A and C but will be heard for B

Solution:
At point, A and C two sounds are approaching from both the two sources and the listener is either moving towards sources or going away from sources.
But at point B cases are not same as mentioned in the previous condition at point A or C, hence beat is produced as the listener is moving away from S1 and approaching towards S2

Topic: Physics, Waves and Sound, NEET
Two stations A and B are 110 km apart on a straight line. One train starts from A at 7 a.m. and travels towards B at 20 kmph. Another train starts from B at 8 a.m. and travels towards A at a speed of 25 kmph. At what time will they meet?

Options:
A 9 a.m.

B 10 a.m.

C 10.30 a.m.

D 11 a.m.

The correct answer is:
10 a.m.

Solution:
Suppose they meet x hours after 7 a.m.
Distance covered by A in x hours = 20x km.
Distance covered by B in (x – 1) hours = 25 (x – 1) km .
∴ 20x + 25 (x- 1) = 110 ⇔ 45x = 135 ⇒ x = 3 .
So, they meet at 10 a.m.

Topic: Physics, Motion In One Dimension, NEET
Two stones of masses m and 2m are whirled in horizontal circles, the heavier one in radius r2 and the lighter one in radius r. The tangential speed of lighter stone is n times that of the tangential speed of the heavier stone. They are reported to experience the same centripetal force. The value of n is,

Options:
A 3
B 4
C 1
D 2

The correct answer is:
2

Solution:
Fcplight=Fcpheavy
So, equating the centripetal forces,
mv12r=2mv22r2
So, v1=2v2

Topic: Physics, Motion In Two Dimensions, NEET, NEET 2015 (Phase 2)
Two stretched wires are in unison. If the tension in one of the wire is increased by 1%, 3 beats are produced in 2 s. The initial frequency of each wire is

Options:
A 150 Hz
B 200Â Hz
C 300 Hz
D 450Â Hz

The correct answer is:
300 Hz

Solution:
n1=kT, n2=k1.01T
âµ n2-n1=32
k1.01T-T=1.5
kT1+0.0112-1=1.5
kT1+12Ã0.01-1=1.5
kT=1.5Ã20.01
=1.5Ã20.01
=300 Hz

Topic: Physics, Waves and Sound, NEET
Two strings of copper are stretched to the same tension. If their cross-section area are in the ratio 1 : 4 , then the respective wave velocities will be

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

The correct answer is:
2 : 1

Solution:
v=Tμ=TρA
where, ρ= density
A= Area of cross-section
μ=ρA= mass of unit length
v1v2=A2A1
v1v2=41
v1v2=21

Topic: Physics, Waves and Sound, NEET
Two tangent galvanometers having coils of the same radius are connected in series. A current flowing in them produces deflections of 60° and 45° respectively. The ratio of the number of turns in the coil is -

Options:
A (4/3)
B (3+1)
C 3+13â1
D 3

The correct answer is:
3

Solution:
In series, current is same in tangent galvanometer.
B = BH tanθ⇒μ0ni2R= BH tanθ
∵ i is same
∴ n∝ tanθ⇒n1n2=tan60°tan45°=31

Topic: Physics, Experimental Physics, NEET
Two thermometers X and Y have ice points marked at 15o and 25o and steam points marked as 75o and 125o, respectively. When thermometer X measures the temperature of a bath as 60o on it, what would thermometer Y read when it is used to measure the temperature of the same bath?

Options:
A 60o
B 75o
C 100o
D 90o

The correct answer is:
100o

Solution:
60-1575-15=Y-25125-25
4560=Y-25100
Y=10060×45+25=100o

Topic: Physics, Thermal Properties of Matter, NEET
Two thin dielectric slabs of dielectric constants K1 and K2K1<K2 are inserted between plates of a parallel plate capacitor, as shown in the figure. The variation of electric field ‘ E ’ between the plates with distance ‘ d ’ as measured from plate P is correctly shown by:

Options:
A
B
C
D

The correct answer is:

Solution:
Here we have to calculate the graph between E and d in the given situation

We know that in dielectric E becomes E/k

where

E = Electric fjeld

k = Dielectric constants

Since K1<K2, The third option will be correct.

Topic: Physics, Capacitance, NEET, NEET 2014
Two thin equiconvex lenses each of focal length 0.2 m are placed coaxially with their optic centres 0.5 m apart. Then the focal length of the combination is

Options:
A -0.4 m

B 0.4 m

C -0.1 m

D 0.1 m

The correct answer is:
-0.4 m

Solution:
Equivalent focal length (F) of two lenses separated by distance d is given by
1F=1f1+1f2-df1f2
=10.2+10.2-0.50.20.2
=5+5-0.5×5×5
=10-12.5
=-2.5
F= -12.5=-0.4 m

Topic: Physics, Ray Optics, NEET
Two thin lenses are of same focal lengths f, but one is convex and the other one is concave. When they are placed in contact with each other, the equivalent focal length of the combination will be:

Options:
A Infinite
B Zero
C f4
D f2

The correct answer is:
Infinite

Solution:
The formula to calculate the effective focal length of the combination can be written as

1feff=1fconvex+1fconcave ...1

Substitute f for fconvex and -f for fconcave into equation (1) and solve to obtain the effective focal length of the combination.

1feff=1f-1f=0⇒feff=∞

Topic: Physics, Ray Optics, NEET, NEET 2023 (All India)
Two thin lenses have a combined power of +9 D. When they are separated by a distance of 20 cm, their equivalent power becomes +275 D. Their individual powers (in dioptre) are

Options:
A 4, 5
B 3, 6
C 2, 7
D 1, 8

The correct answer is:
3, 6

Solution:
P1+P2=9
P= P1+P2-dP1P2
279=9-20100ÃP1P2
The above equation is correct only for P1=3 and P2=6

Topic: Physics, Ray Optics, NEET
Two thin lenses of focal length 20 cm and 25 cm are in contact. The effective power of the combination is

Options:
A 4.5 D

B 18 D

C 45 D

D 9 D

The correct answer is:
9 D

Solution:
According to the question
P1=10020=5 D, P2=10025=4D
Effective power P=P1+P2
= 5 + 4 = 9 D

Topic: Physics, Ray Optics, NEET
Two thin lenses of focal lengths $f_1$ and $f_2$ are in contact and coaxial. The power of the combination is

Options:
A $\sqrt{\frac{f_1}{f_2}}$
B $\sqrt{\frac{f_2}{f_1}}$
C $\frac{f_1+f_2}{f_1 f_2}$
D $\frac{f_1+f_2}{f_1 f_2}$

The correct answer is:
$\frac{f_1+f_2}{f_1 f_2}$

Solution:
If two thin lenses of focal lengths $f_1, f_2$ are placed in contact coaxially, then equivalent focal length of combination is
$$
\frac{1}{F}=\frac{1}{f_1}+\frac{1}{f_2}-\frac{0}{f_1 f_2}=\frac{1}{f_1}+\frac{1}{f_2}
$$
Power for the combination is
$$
P=\frac{1}{F}=\frac{1}{f_1}+\frac{1}{f_2}=\frac{f_1+f_2}{f_1 f_2}
$$

Topic: Physics, Ray Optics, NEET, NEET 2008 (Screening)
Two thin uniform rings made of the same material and of radii R and 4R are joined as shown. The mass of the smaller ring is m . Find the moment of inertia of the system about an axis passing through the centre of mass of bigger ring and perpendicular to the plane.

Options:
A 85 mR2
B 90 mR2
C 70 mR2
D 35 mR2

The correct answer is:
90 mR2

Solution:
I=Ilarger+Ismaller
I=4m(4R)2+mR2+m(5R)2=90mR2

Topic: Physics, Rotational Motion, NEET
Two toroids 1 and 2 have total number of turns 200 and 100 respectively with average radii $40 \mathrm{~cm}$ and $20 \mathrm{~cm}$ respectively. If they carry same current $i$, the ratio of the magnetic fields along the two loops is

Options:
A $1: 1$
B $4: 1$
C $2: 1$
D $1: 2$

The correct answer is:
$1: 1$

Solution:
The magnetic field within the turns of toroid is
$$
\mathrm{B}=\frac{\mu_0 \mathrm{NI}}{2 \pi \mathrm{r}}
$$
where, $\mathrm{N}=$ number of turns, $\mathrm{I}=$ current in loops and
$\mathrm{r}=$ radius of each turn
Here, $\mathrm{N}_1=200, \mathrm{~N}_2=100, \mathrm{r}_1=40 \mathrm{~cm}, \mathrm{r}_2=20 \mathrm{~cm}$
and current $\mathrm{I}$ is same, then
$$
\frac{\mathrm{B}_1}{\mathrm{~B}_2}=\frac{\mu_0 \mathrm{~N}_1 \mathrm{I}}{2 \pi \mathrm{r}_1} \times \frac{2 \pi \mathrm{r}_2}{\mu_0 \mathrm{~N}_2 \mathrm{I}}
$$
Substituting the given values in the above relation, we get
$$
\begin{aligned}
\frac{\mathrm{B}_1}{\mathrm{~B}_2} & =\left(\frac{\mathrm{N}_1}{\mathrm{~N}_2}\right)\left(\frac{\mathrm{r}_2}{\mathrm{r}_1}\right) \\
& =\left(\frac{200}{100}\right)\left(\frac{20}{40}\right)=2 \times \frac{1}{2}=1 \\
\therefore \quad \mathrm{B}_1: \mathrm{B}_2 & =1: 1
\end{aligned}
$$

Topic: Physics, Magnetic Effects of Current, NEET, NEET 2019 (Odisha)
Two trains A and B of length 400 m each are moving on two parallel tracks with a uniform speed of 72 km h-1 in the same direction, with ahead of B. The driver of B decides to overtake A and accelerates by 1 m s-2 . If after 50 s, the guard of B just brushes past the driver of A, what was the original distance between them?

Options:
A 1000 m

B 1150 m

C 1300 m

D 1250 m

The correct answer is:
1250 m

Solution:
Given that,
uA=uB=72 km h-1=72×518=20 m s-1
Using the relations, S=ut+12 at2 , we get
SB=uBt+12 at2=20×50+12×1×502
SB=1000+1250=2250 m
Also, let SA be the distance covered by the train A,
then SA=uA×t
=20×50=1000 m
Original distance between the two trains
= SB-SA
= 2250-1000=1250 m

Topic: Physics, Motion In One Dimension, NEET