datavorous/entrance-exam-dataset huggingface dataset:
Two planets A and B have the same average density. Their radii RA and RB are such that RA :RB=3:1. If gA and gB are the acceleration due to gravity at the surfaces of the planets, the gA :gB equals

Options:
A 3 :1
B 1 :3
C 1 :9
D 3 :1

The correct answer is:
3 :1

Solution:
Given,
RARB=31 .....(i)
and ρA=ρB ....(ii)
∵ Average density, ρ=3g4πRG
∴ From Equation (ii),
⇒ 3gA4πRAG=3gB4πRBG ⇒ gAgB=RARB
From Equation (i),
gAgB=31

Topic: Physics, Gravitation, NEET
Two planets A and B have the same average density. Their radii RA and RB are such that RA :RB=3:1 . If gA and gB are the acceleration due to gravity at the surfaces of the planets, the gA :gB equals

Options:
A 3 :1
B 1 :3
C 1 :9
D 3 :1

The correct answer is:
3 :1

Solution:
Given,
RARB=31 .....(i)
and ρA=ρB ....(ii)
∵ Average density, ρ=3g4πRG
∴ From Equation (ii),
⇒ 3gA4πRAG=3gB4πRBG ⇒ gAgB=RARB
From Equation (i),
gAgB=31

Topic: Physics, Gravitation, NEET
Two planets are at distances R1 and R2 from the Sun. Their time periods are T1 and T2 then T1T22 is equal to

Options:
A R1R2
B R1R22
C R1R23
D R2R13

The correct answer is:
R1R23

Solution:
Using Kepler's law of the period, T2∝R3
∴T1T22=R1R23

Topic: Physics, Gravitation, NEET
Two planets of radii in the ratio 2:3 are made from the material of density in the ratio 3:2. Then, the ratio of acceleration due to gravity g1g2 at the surface of the two planets will be

Options:
A 1
B 2.25
C 0.50
D 0.12

The correct answer is:
1

Solution:
The acceleration due to gravity (g) is given by
g=GMR2
where M is mass, G the gravitational constant and R the radius.
Since, planets have a spherical shape
V=43Ïr3
Also, mass M=volumeVÃdensityÏ
g=G43ÏR3ÏR2
â¹ g=4GÏÏR3
Given, R1:R2=2 :3
Ï1:Ï2=32
â´ g1g2=Ï1R1Ï2R2=32Ã23=1

Topic: Physics, Gravitation, NEET
Two plano-convex lenses of focal lengths 20 cm and 30 cm are placed together to form a double convex lens. The final focal length will be

Options:
A 12 cm
B 60 cm
C 20 cm
D 30 cm

The correct answer is:
12 cm

Solution:
Equivalent focal length
1F=1f1+1f2
=120+130
â´F=20Ã3020+30
=60050=12 cm

Topic: Physics, Ray Optics, NEET
Two platinum electrodes were immersed in a solution of cupric sulphate and electric current passed through the solution. After sometime it was found that the colour of copper sulphate disappeared with evolution of gas at the electrode. The colourless solution contains

Options:
A platinum sulphate
B copper hydroxide
C copper sulphate
D sulphuric acid

The correct answer is:
sulphuric acid

Solution:
Anode:- 2 H 2 O→ O 2 +4 H + +4 e −
Cathode:- Cu +2 aq +2 e − →Cu s
Solution will left with SO 4 −2 &  H + ions
So, we can say it is aq solutions of H 2 SO 4

Topic: Chemistry, Electrochemistry, NEET
Two point charges A=+3 nC and B=+1 nC are placed 5 cm apart in the air. The work done to move charge B towards A by 1 cm is

Options:
A 2.0Ã10-7 J
B 2.7Ã10-7J
C 12.1Ã10-7J
D 1.35Ã10-7 J

The correct answer is:
1.35Ã10-7 J

Solution:
Given,
A=+ 3 nC=3Ã10-9 C
B=+ 1 nC=1Ã10-9 C
Distance, r1=5 cm=0.05 m=5Ã10-2 m
Work done W=UB-UA
=kq1q2r2-kq1q2r1
Here, r2=r1-1
r2=5-1=4 cm=0.04 m=4Ã10-2 m
=kq1q21r2-1r1
=9Ã109Ã3Ã10-9Ã1Ã10-9
14Ã10-2-15Ã10-2
=27Ã10-9Ã15Ã4Ã10-2
=2720Ã10-7
=1.35Ã10-7J

Topic: Physics, Electrostatics, NEET
Two point charges A and B, having charges +Q and -Q respectively, are placed at certain distance apart and force acting between them is F. If 25% charge of A is transferred to B, then the force between the charges becomes

Options:
A F.
B 9F16.
C 16F9.
D 4F3.

The correct answer is:
9F16.

Solution:
F=KQ2r2
After transferring 25% of charge, q1=3Q4 and q2=-3Q4
F1=916KQ2r2=916F

Topic: Physics, Electrostatics, NEET, NEET 2019 (All India)
Two point charges placed at a distance r in air exert a force F on each other. The distance at which they experience force 4F when placed in a medium of dielectric constant K=16 is -

Options:
A r
B r8
C r4
D r2

The correct answer is:
r8

Solution:
Force in air at a distance r,
F= q1q24πϵ0r2 ......(i)
Let r’ be the distance in the medium, therefore
4F= q1q24πϵ016(r’)2 ......(ii)
From (i) ÷ (ii)
14 = 16(r’)2r2
∴r’=r8

Topic: Physics, Electrostatics, NEET
Two point charges q and -q are at positions (0, 0, d) and (0, 0, -d) respectively. What is the electric field at (a, 0, 0)?

Options:
A 2qd4πε0(d2+a2)3/2k^
B qd4πε0(d2+a2)3/2k^
C -2qd4πε0(d2+a2)3/2k^
D -qd4πε0(d2+a2)3/2k^

The correct answer is:
-2qd4πε0(d2+a2)3/2k^

Solution:
At P there are two fields
(1) Due to A=EA=kqr2
(2) Due to B=EB=kqr2
r =(a2+ d2)1/2 âââââat angle 180-2âÎ¸
Enet= 2EA cosâ12 (180-2Î¸)â= 2EAcos(90 -âÎ¸)
=2EAâsinâÎ¸ =2kqr2d(r)
=2kqd(a2+d2)3/2 in -z direction.

Topic: Physics, Electrostatics, NEET
Two point charges -q and +q are placed at a distance of L, as shown in the figure.

The magnitude of electric field intensity at a distance RR≫L varies as

Options:
A 1R3
B 1R4
C 1R6
D 1R2

The correct answer is:
1R3

Solution:
The given arrangement of charges forms a dipole.
Electric field due to a dipole.

E∝1R3

Topic: Physics, Electrostatics, NEET, NEET 2022 (Phase 1)
Two point charges q1 and q2 , of magnitude +${{10}^{-8}}$C and –${{10}^{-8}}$C, respectively, are placed 0.1 m apart. Calculate the electric field at point A as shown in Fig.
.

Options:
A 7.2 × ${{10}^{4}}$N ${{C}^{-1}}$
B 6.5 × ${{10}^{4}}$ N ${{C}^{-1}}$
C 5.2 × ${{10}^{4}}$ N ${{C}^{-1}}$
D 9.1 × ${{10}^{4}}$ N ${{C}^{-1}}$

The correct answer is:
7.2 × ${{10}^{4}}$N ${{C}^{-1}}$

Solution:
Correct Option is : (A)
7.2 × ${{10}^{4}}$N ${{C}^{-1}}$

Topic: Chemistry, Electrostatics, NEET
Two point masses connected by an ideal string are placed on a smooth horizontal surface as shown in the diagram. A sharp impulse of 10 kg-m s-1 is given to the 5 kg mass. The velocity of the 10 kg mass will be

Options:
A 23 m s-1
B 13 m s-1
C 2 m s-1
D zero

The correct answer is:
13 m s-1

Solution:
Apply impulse equation, along string
Impulse = Îp
10 Ã 12=10 + 5v
v=13m s-1

Topic: Physics, Center of Mass Momentum and Collision, NEET
Two point masses of 3.0 kg and 1.0 kg are attached to opposite ends of a horizontal spring whose spring constant is 300 N m-1 as shown in the figure. The natural vibration frequency of the system is of the order of :

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

The correct answer is:
3 Hz

Solution:
T=2ÏÎ¼k
Î¼=m1m2m1+m2
Î¼=1Ã31+3=34
Î½=1T=12ÏkÎ¼
Î½=12Ï300Ã43=10Ïâ3

Topic: Physics, Oscillations, NEET
Two point white dots are 1 mm apart on a black paper. They are viewed by the eye of pupil diameter 3 mm. Approximately, what is the maximum distance at which these dots can be resolved by the eye? [ Take wavelength of light = 500 mm ]

Options:
A 6 m

B 3 m
C 1 m
D 5 m

The correct answer is:
5 m

Solution:
Resolution limit = 1.22 Î» d
Again resolution limit = sin Î¸ = Î¸ = y D
â´ y D = 1 Â· 2 2Î» d
or D = y d 1.22 Î»
or D = 10 - 3 Ã 3 Ã 10 - 3 1.22 Ã 5 Ã 10 - 7 = 30 6.1 â 5 m .

Topic: Physics, Wave Optics, NEET
Two points are located at a distance of $10 \mathrm{~m}$ and $15 \mathrm{~m}$ from the source of oscillation. The period of oscillation is $0.05 \mathrm{~s}$ and the velocity of the wave is $300 \mathrm{~m} / \mathrm{s}$. What is the phase difference between the oscillations of two points?

Options:
A $\frac{\pi}{3}$
B $\frac{2 \pi}{3}$
C $\pi$
D $\frac{\pi}{6}$

The correct answer is:
$\frac{2 \pi}{3}$

Solution:
Key Idea : Phase difference
$=\frac{2 \pi}{\lambda} \times \text { path difference }$
Path difference between two points,
$\Delta x=15-10=5 \mathrm{~m}$
Time period, $T=0.05 \mathrm{~s}$
$\Rightarrow$ frequency $v=\frac{1}{T}=\frac{1}{0.05}=20 \mathrm{~Hz}$
Velocity, $y=300 \mathrm{~m} / \mathrm{s}$
$\therefore$ Wavelength, $\lambda=\frac{v}{v}=\frac{300}{20}=15 \mathrm{~m}$
Hence, phase difference
$\begin{aligned}
\Delta \phi & =\frac{2 \pi}{\lambda} \times \Delta x \\
& =\frac{2 \pi}{15} \times 5=\frac{2 \pi}{3}
\end{aligned}$

Topic: Physics, Waves and Sound, NEET, NEET 2008 (Screening)
Two points p and Q lie on either side of an axis XY as shown. It is desired to produce an image of p at Q using a spherical mirror, with XY as the optic axis. The mirror must be

Options:
A Converging and positioned to the left of p
B Diverging and positioned to the left of p
C Converging and positioned to the right of Q
D Diverging and positioned to the right of Q

The correct answer is:
Converging and positioned to the left of p

Solution:
The image is enlarged by mirror so it should be concave or converging mirror.
â´ 2 and 4 are wrong.
Now image is enlarged and real for a mirror. so the object should be between focus and centre of curvature.
â´ Mirror should be left like
â´ Mirror should be left of object(P).

Topic: Physics, Ray Optics, NEET
Two polaroids are kept crossed to each other. If one of them is rotated an angle 60°, the percentage of incident light now transmitted through the sytem is

Options:
A 10%
B 20%
C 25%
D 12.5%

The correct answer is:
12.5%

Solution:
Let I0 be the intensity of unpolarised light, so the intensity of first polaroid is I02. On rotating through 60o , the intensity of light from second polaroid is
I=I02 cosâ¡602
=I02 14=I08=0.125I0
So, percentage of incident light transmitted through the system is 12.5%

Topic: Physics, Wave Optics, NEET
Two Polaroids are placed in the path of an unpolarized beam of intensity I0 such that no light is emitted from the second polaroid. If a third Polaroid whose polarization axis makes an angle θ with the polarization axis of first Polaroid, is placed between these Polaroids then the intensity of light emerging from the last Polaroid will be

Options:
A I08sin22θ
B I04sin22θ
C I02cos4θ
D I0cos4â¡Î¸

The correct answer is:
I08sin22θ

Solution:
If no light comes from 2nd polaroid, i.e. angle between them =90Â°
If III polaroid has polatization axis at angle Î¸
I2=I1cos2â¡Î¸
I2=I02cos2â¡Î¸
As Polarising axis of I & II are mutually perpendicular
Hence angle between polarizing axis of II & III is 90Â°-Î¸
â´ I3=I2cos2â¡90Â°-Î¸
â I3=I02cos2â¡Î¸sin2â¡Î¸
â I3=I02Ã44cos2â¡Î¸sin2â¡Î¸
â I3=I082sinâ¡Î¸cosâ¡Î¸2
â I3=I08sin2â¡2Î¸

Topic: Physics, Wave Optics, NEET
Two Polaroids P1 and P2 are placed with their axis perpendicular to each other. Unpolarised light with intensity I0 is incident on P1. A third polaroid P3 is kept in between P1 and P2 such that its axis makes an angle 45o with that of P1. The intensity of transmitted light through P2 is

Options:
A I02
B I04
C I08
D I016

The correct answer is:
I08

Solution:
By Malus law, the intensity of transmitted light via P1 is I1=I02
The intensity of light transmitted through P3 as obtained from the above figure is I2=I02cos2⁡ 45o
=I02×12
=I04
Intensity of light transmitted via P2 is I3=I04cos2⁡45o
I3=I08

Topic: Physics, Wave Optics, NEET, NEET 2017
Two polypeptides of functional insulin in human are linked together by

Options:
A disulphide bridges between two methionine amino acids.
B covalent bonds.
C disulphide bridges between two cysteine amino acids.
D hydrogen bonds.

The correct answer is:
disulphide bridges between two cysteine amino acids.

Solution:
In humans, insulin is produced by ß-cells of the pancreas. It is synthesised as a prohormone in which two polypeptides are synthesised with an extra stretch of 'C' polypeptide. During maturation extra stretch of 'C' polypeptide is separated and two polypeptide chains (A and B) are linked together by disulphide linkages (bridges)s between two cysteine amino acids.

Topic: Biology, Biotechnology and Its Applications, NEET
Two positive ions, each carrying a charge $q$, are separated by a distance $d$. If $\mathrm{F}$ is the force of repulsion between the ions, the number of electrons missing from each ion will be ( $e$ being the charge on an electron)

Options:
A $\frac{4 \pi \varepsilon_0 \mathrm{Fd}^2}{\mathrm{e}^2}$
B $\sqrt{\frac{4 \pi \varepsilon_0 \mathrm{Fe}^2}{\mathrm{~d}^2}}$
C $\sqrt{\frac{4 \pi \varepsilon_0 \mathrm{Fd}^2}{\mathrm{e}^2}}$
D $\frac{4 \pi \varepsilon_0 \mathrm{Fd}^2}{\mathrm{q}^2}$

The correct answer is:
$\sqrt{\frac{4 \pi \varepsilon_0 \mathrm{Fd}^2}{\mathrm{e}^2}}$

Solution:
Two positive ions each carrying a charge $q$ are kept at a distance $\mathrm{d}$, then it is found that force of repulsion between them is
$$
\begin{aligned}
\mathrm{F} & =\frac{\mathrm{kqq}}{\mathrm{d}^2} \\
& =\frac{1}{4 \pi \varepsilon_0} \frac{q q}{\mathrm{~d}^2}
\end{aligned}
$$
where
$$
\begin{array}{ll}
\text { where } & \mathrm{q}=\mathrm{ne} \\
\therefore & \mathrm{F}=\frac{1}{4 \pi \varepsilon_0} \frac{\mathrm{n}^2 \mathrm{e}^2}{\mathrm{~d}^2} \\
\Rightarrow & \mathrm{n}=\sqrt{\frac{4 \pi \varepsilon_0 \mathrm{Fd}_2}{\mathrm{e}^2}}
\end{array}
$$

Topic: Physics, Electrostatics, NEET, NEET 2010 (Screening)
Two positive ions, each carrying a charge q are separated by a distance d . If F is the force of repulsion between the ions, the number of electrons missing from each ion will be (e being the charge on an electron)

Options:
A 4πε0Fd2e2
B 4πε0Fe2d2
C 4πε0Fd2e2
D 4πε0Fd2q2

The correct answer is:
4πε0Fd2e2

Solution:
From Coulomb's law force of repulsion is
F=kq2d2⇒F=14πε0q2d2
Where, q=ne
∴ F=14πε0n2e2d2⇒n=4πε0Fd2e2

Topic: Physics, Electrostatics, NEET
Two possible stereo-structures of CH3CHOHCOOH, which are optically active, are called:

Options:
A Diastereomers
B Atropisomers
C Enantiomers
D Mesomers

The correct answer is:
Enantiomers

Solution:
Two possible stereo-structures of CH3CHOHCOOH, which are optically active, are called enantiomers.

They are non-superimposable mirror images of each other and are shown in the image.

Both are enantiomers.

Topic: Chemistry, General Organic Chemistry, NEET, NEET 2015 (Phase 2)
Two processes are mentioned- (a) Throwing a stone from the ground to roof. (b) (½) H2(g) + (½) Cl2(g) ⇌ HCl(g) ; ∆rH⊖= –92.32 kJ mol–1In which of the processes potential energy/enthalpy change is contributing factor to the spontaneity?

Options:
A Amongst those two processes, in the reaction [A], enthalpy change (i.e. ΔH) is a contributing factor to the spontaneity, as negative value of enthalpy change favours the negative value of gibbs free energy change, which led to the spontaneous reaction.
B In both the processes,potential energy/enthalpy change is contributing factor to the spontaneity
C Amongst those two processes, in the reaction [B], enthalpy change (i.e. ΔH) is a contributing factor to the spontaneity, as negative value of enthalpy change favours the negative value of gibbs free energy change, which led to the spontaneous reaction.
D In none of the processes,potential energy/enthalpy change is contributing factor to the spontaneity.

The correct answer is:
Amongst those two processes, in the reaction [B], enthalpy change (i.e. ΔH) is a contributing factor to the spontaneity, as negative value of enthalpy change favours the negative value of gibbs free energy change, which led to the spontaneous reaction.

Solution:
Correct Option is : (C)
Amongst those two processes, in the reaction [B], enthalpy change (i.e. ΔH) is a contributing factor to the spontaneity, as negative value of enthalpy change favours the negative value of gibbs free energy change, which led to the spontaneous reaction.

Topic: Chemistry, Thermodynamics (C), NEET
Two projectiles thrown from the same point at angles 60° and 30° with the horizontal attain the same height. The ratio of their initial velocities is

Options:
A 1
B 2
C 3
D 13

The correct answer is:
13

Solution:
hmax.=v2sin2θ2g
as hmax. is same in both the cases
v12sin2θ12g=v22sin2θ22g
But θ1=60ο, θ2=30ο
v 1 2 3 4 g = v 2 2 1 4 g

v 1 v 2 = 1 3

Topic: Physics, Motion In Two Dimensions, NEET
Two radioactive materials X1 and X2 have decay constants 10Î» and Î» respectively. If initially, they have the same number of nuclei, then the ratio of the number of nuclei of X1 to that of X2 will be 1e after a time

Options:
A 110λ
B 111λ
C 1110λ
D 19λ

The correct answer is:
19λ

Solution:
Here, Nx1(t)Nx2(t)=1e
or N0e-10λtN0e-λt=1e
(Because initially, both have the same number of nuclei, N0).
or e=e-λte-10λt=e9λt
9λt=1
t=19λ

Topic: Physics, Nuclear Physics, NEET
Two radioactive materials $X_1$ and $X_2$ have decay constants $5 \lambda$ and $\lambda$ respectively. If initially they have the same number of nuclei, then the ratio of the number of nuclei of $X_1$ to that of $X_2$ will be $\frac{1}{e}$ after a time

Options:
A $\lambda$
B $\frac{1}{2} \lambda$
C $\frac{1}{4 \lambda}$
D $\frac{e}{\lambda}$

The correct answer is:
$\frac{1}{4 \lambda}$

Solution:
If $N$ is the number of radioactive nuclei present at some instant, then
$N=N_0 e^{-\lambda t}$
The constant $N_0$ represents the number of radioactive nuclei at $t=0$
Now,
$\begin{aligned}
& \frac{N_1}{N_2}=\frac{e^{-\lambda_1 t}}{e^{-\lambda_2 t}} \\
& \frac{N_1}{N_2}=\frac{e^{-5 \lambda t}}{e^{-\lambda t}}=e^{-4 \lambda t} \\
& \frac{N_1}{N_2}=\frac{1}{e}
\end{aligned}$
but $\frac{N_1}{N_2}=\frac{1}{e}$ (as provided)
Therefore, $\frac{1}{e}=\frac{1}{e^{4 \lambda t}}$
or $4 \lambda t=1$
or $t=\frac{1}{4 \lambda}$

Topic: Physics, Nuclear Physics, NEET, NEET 2008 (Screening)
Two radioactive substances $\mathrm{A}$ and $\mathrm{B}$ have decay constants $5 \lambda$ and $\lambda$ respectively. At $t=0$ they have the same number of nuclei. The ratio of number of nuclei of A to those of $\mathrm{B}$ will be $(1 / e)^2$ after a time interval.

Options:
A $4 \lambda$
B $2 \lambda$
C $\frac{1}{2} \lambda$
D $\frac{1}{4} \lambda$

The correct answer is:
$\frac{1}{2} \lambda$

Solution:
According to the question
$\begin{aligned}
\frac{N_A}{N_B} & =\frac{1}{e^{4 \lambda t}} \\
\Rightarrow \quad \frac{1}{e^{4 \lambda t}} & =\frac{1}{e^2} \\
\Rightarrow \quad t & =\frac{1}{2 \lambda}
\end{aligned}$

Topic: Physics, Nuclear Physics, NEET, NEET 2007
Two raindrops reach the earth with their terminal velocities in the ratio 4:9. The ratio of their radii is

Options:
A 4:9
B 2:3
C 3:2
D 9:4

The correct answer is:
2:3

Solution:
Terminal speed is v=2r2g9ησ-ρ
therefore v∝r2
r1r2=v1v2
r1r2=49
r1r2=23

Topic: Physics, Mechanical Properties of Fluids, NEET
Two reactions are given below
(i) CO(g)+12O2(g)→CO2(g)
(ii) Ag2O(s)→2Ag(s)+12O2(g)
Which of the following statements is true?

Options:
A For (i) ΔH<ΔE and for (ii) ΔH>ΔE
B For (i) ΔH>ΔE and for (ii) ΔH<ΔE
C For both (i) and (ii) ΔH>ΔE
D For both (i) and (ii) ΔH<ΔE

The correct answer is:
For (i) ΔH<ΔE and for (ii) ΔH>ΔE

Solution:
As ΔH=ΔE+ΔngRT
For (i) Δng=1-32=-12
As here Δng is negative (-1/2) so ΔH<ΔE
For (ii) Δng=12-0=12
As here Δng is positive (+1/2) so ΔH>ΔE

Topic: Chemistry, Thermodynamics (C), NEET
Two rectangular blocks A and B of different metals have the same length and same area of cross-section. They are kept in such a way that their cross-sectional areas are in contact. The temperature at one end of A is 100 °C and that of B at the other end is 0 °C. If the ratio of their thermal conductivity is 1:3, then under steady-state, the temperature of the junction in contact will be

Options:
A 25 Â°C
B 50 Â°C
C 75 Â°C
D 100 Â°C

The correct answer is:
25 Â°C

Solution:
It is given that K1K2=13
Let K1=K & K2=3K
The temperature of the junction in contact
Î¸=K1Î¸1+K2Î¸2K1+K2
Î¸=1Ã100Ã3Ã01+3=1004
Î¸=25 Â°C

Topic: Physics, Thermal Properties of Matter, NEET
Two resistors of resistance, 100 Ω and 200 Ω are connected in parallel in an electrical circuit. The ratio of the thermal energy developed in 100 Ω to that in 200 Ω in a given time is

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

The correct answer is:
2:1

Solution:
Heat produced in resistor H=V2Rt

Here, the potential is constant tor parallel combination.

Here R1=100 Ω, R2=200 Ω

H1H2=R2R1=200100=2:1

Topic: Physics, Current Electricity, NEET, NEET 2022 (Phase 1)
Two resistors R and 2R are connected in series in an electric circuit. The thermal energies developed in R and 2R are in the ratio

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

The correct answer is:
1:2

Solution:
Thermal energy or heat developed is given by,
H=i2Rt
In series combination, the current flowing through all the elements is the same, so
H1H2=R1R2
H1H2=R2R
H1H2=12

Topic: Physics, Current Electricity, NEET
Two rings of mass m and 2m are connected with a massless spring and can slip over two frictionless parallel horizontal rails having separation equal to the unstretched length of the spring as shown in figure. Ring of mass m is given velocity v0 in the direction shown. Maximum length by which the spring will be stretched is

Options:
A mkv0
B 3mkv0
C 2m3kv0
D 2mkv0

The correct answer is:
2m3kv0

Solution:
Maximum expansion takes place only when both the rings move with the same speed.
Therefore we can apply work-energy theorem here. So maximum expansion in spring is obtained by,
12kxmax2=12Î¼v02 Î¼=Reduced mass
â xmax=Î¼k.v0=2m3kv0

Topic: Physics, Work Power Energy, NEET
Two rings of radius R and nR made up of same material have the ratio of moment of inertia about an axis passing through the centre as 1 : 8. The value of n is

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

The correct answer is:
2

Solution:
The ratio of moment of inertia of the rings
I1I2=M1M2R1R22=λl1λl2R1R22=2πR2πnRRnR2
λ = linear density of wire = constant
⇒ I1I2=1n3=18 given
∴ n3=8 ⇒ n=2

Topic: Physics, Rotational Motion, NEET
Two rings of same radius r and mass m are placed such that their centres are at a common point and their planes are perpendicular to each other. The moment of inertia of the system about an axis passing through the centre and perpendicular to plane of one of the rings is

Options:
A 12mr2
B mr2
C 32mr2
D 2mr2

The correct answer is:
32mr2

Solution:
From figure,
IY=IY1+IY2
⇒IY=12mr2+mr2
⇒IY=32mr2

Topic: Physics, Rotational Motion, NEET
Two rings of the same radius (r) and mass (m) are placed such that their centres are at a common point and their planes are perpendicular to each other. The moment of inertia of the system about an axis passing through the centre and perpendicular to the plane of one of the ring is

Options:
A 12mr2
B mr2
C 32mr2
D 2 mr 2

The correct answer is:
32mr2

Solution:
I1 = mr2
I2 = moment of inertia about one of the diameters
=12mr2
∴I=I1+I2=32mr2

Topic: Physics, Rotational Motion, NEET
Two rods A and B of different materials are welded together as shown in figure. Their thermal conductivities are K1 and K2. The thermal conductivity of the composite rod will be

Options:
A K1+K22
B 3K1+K22
C K1+K2
D 2K1+K2

The correct answer is:
K1+K22

Solution:
Thermal current
H=H1+H2
=K1AT1-T2d+K2AT1-T2d
KEQ 2AT1-T2d=AT1-T2d K1+K2
KEQ=K1+K22

Topic: Physics, Thermal Properties of Matter, NEET, NEET 2017
Two rods are joined to end, as shown. Both have a cross sectional area of 0.01 $\mathrm{cm}^2$. Each is 1 meter long. One rod is of copper with a resistivity of $17 \times 10^{-6}$ ohm-centimeter, the other is of iron with a resistivity of $10^{-5} \mathrm{ohm}$ centimeter. How much voltage is required to produce a current of 1 ampere in the rods?

Options:
A $0.00145 \mathrm{~V}$
B $0.0145 \mathrm{~V}$
C $1.7 \times 10^{-6} \mathrm{~V}$
D $0.117 \mathrm{~V}$

The correct answer is:
$0.117 \mathrm{~V}$

Solution:
According to the question
Length of each $\operatorname{rod} l=1 \mathrm{~m}$
Area of cross-section of cach rod $\mathrm{A}=0.01 \mathrm{~cm}^2=0.01 \times 10^{-4} \mathrm{~m}^2$
$\therefore$ If $\rho_{c u}=$ resistivity of copper and $\rho_{C W}$ is resistivity of iron then

$\begin{aligned}
& \rho_{c u}=1.7 \times 10^{-6} \Omega \mathrm{cm}=1.7 \times 10^{-8} \Omega \\
& \rho_{f e}=10^{-5} \Omega \mathrm{cm}=10^{-7} \Omega
\end{aligned}$
New Resistance of $\mathrm{Cu}$ rod $=\mathrm{R}_{c u}=$
$\begin{aligned}
& \rho_{c u} \frac{1}{\mathrm{~A}} \\
& \text { and Resistance of iron rod } \\
& \rho_{f e}=\rho_{f e} \frac{1}{\mathrm{~A}}
\end{aligned}$
Both are connected in series then equivalent resistance is
$\begin{aligned}
R_{e s}=R_{c e}+R_{F e} & =\rho_{c u} \frac{l}{\mathrm{~A}}+\rho_{f e} \frac{l}{\mathrm{~A}}=\frac{l}{\mathrm{~A}} \\
\left(\rho_{c u}+\rho_{f e}\right) & \\
V & =I R \quad\{\text { where } \mathrm{I}=1 \mathrm{~A}\} \\
& =I\left(R_{c u}+R_{f e}\right) \\
& =1\left(\rho_{c u}+\rho_{f e}\right) \\
& =\left(\rho_{c u}+\rho_{f e}\right)\left(\frac{1}{\mathrm{~A}}\right) \\
& =\left(1.7 \times 10^{-8}+10^{-7}\right) \\
& =10^{-7}(0.17+1)\left(10^6\right) \\
& =1.17 \times 10^{-1} \mathrm{~V} \\
& =0.117 \mathrm{~V}
\end{aligned}$

Topic: Physics, Current Electricity, NEET, NEET 2013 (Karnataka)