datavorous/entrance-exam-dataset huggingface dataset:
Two identical conducting rods are first connected independently to two vessels, one containing water at 100 Â°C and the other containing ice at 0 Â°C. In the second case, the rods are joined end to end and connected to the same vessels. Let q 1 and q 2 gram per second be the rate of melting of ice in the two cases respectively. The ratio q 1 q 2 is

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

The correct answer is:
4 1

Solution:
dQ dt = L dm dt
or Temperature difference Thermal difference = L dm dt
or dm dt â 1 Thermal resistance
q â 1 R
In the first case, rods are in parallel and thermal resistance is R 2 while in second case rods are in series and thermal resistance is 2 R.
q 1 q 2 = 2R R / 2 = 4 1

Topic: Physics, Thermal Properties of Matter, NEET
Two identical conducting spheres carrying different charges attract each other with a force F when placed in air medium at a distance d apart. The spheres are brought into contact and then taken to their original positions. Now the two spheres repel each other with a force whose magnitude is equal to that of the initial attractive force. The ratio between initial charges on the spheres is

Options:
A –(3+8)only
B –3+8 only
C –(3+8) or (–3+8)
D -8

The correct answer is:
–(3+8) or (–3+8)

Solution:
F= kq1q2d2
After touching F'= kd2q1–q222
F = F'
q1q2 = q1–q222
q1q2 = – 3 + 8 or – 3 – 8

Topic: Physics, Electrostatics, NEET
Two identical conducting wires AOB and COD are placed at right angles to each other. The wire AOB carries an electric current I1 and COD carries a current I2. The magnetic field on a point lying at a distance of from O, in a direction perpendicular to the plane of the wires AOB and COD, will be given by

Options:
A μ02πI1+I2d12
B μ02πdI12+I2212
C μ02πd I1+I2
D μ02πd I12+I22

The correct answer is:
μ02πdI12+I2212

Solution:
The magnetic field inductions at a point P1 at a distance d from O in a direction perpendicular to the plane of the wires due to the plane of the wires due to currents through AOB and COD are perpendicular to each other is
Hence, B= B12+B22
=μ04π 2I1d2+μ04π 2I2d212
=μ02πd l12+l22

Topic: Physics, Magnetic Effects of Current, NEET
Two identical containers A and B have frictionless pistons. They contain the same volume of an ideal gas at the same temperature. The mass of the gas in A is mA and that in B is mB. The gas in each cylinder is now allowed to expand isothermally to double the initial volume. The change in the pressure in A and B, respectively, is Îp and 1.5Îp. Then

Options:
A 4mA=9mB
B 2mA=3mB
C 3mA=2mB
D 9mA=4mB

The correct answer is:
3mA=2mB

Solution:
For the gas in container A
Δ P = P A final - P A initial = n A RT 2 V - n A RT V
Δ P = n A RT 2 V ...(i)
For gas in container B
1.5 Δ P = P B final - P B initial = n B RT 2 V - n B RT V
1.5 Δ P = - n B RT 2 V ...(ii)
From Eqs. (i) and (ii), we get
n B = 1.5 n A
⇒ 2 n B = 3 n A
⇒ 3 m A = 2 m B

Topic: Physics, Thermodynamics, NEET
Two identical containers contain equal amounts of the same ideal gas at a temperature T0 and pressure P0. Now the two containers are connected through a tube having negligible volume and one container is heated to 2T0 while the other is kept at T0. The final common pressure will be

Options:
A 2P03
B 4P03
C P03
D 2P0

The correct answer is:
4P03

Solution:
P0VRT0+P0VRT0=PVRT0+PVR2T0
2P0=P+P2
2P0=3P2
P=4P03

Topic: Physics, Kinetic Theory of Gases, NEET
Two identical glass μg=32 equiconvex lenses of focal length, f are kept in contact. The space between the two lenses is filled with water μw=43. The focal length of the combination is

Options:
A f3
B f
C 4f3
D 3f4

The correct answer is:
3f4

Solution:
1f=μg-12R=1R, μg=32⇒R=f

1f1=-μw-12R=-23R=-23f.

1feq=1f+1f+1f1=1R+1R-23R

1feq=1f+1f-23f

1feq=2f-23f
1feq=43f
feq=3f4.

Topic: Physics, Ray Optics, NEET, NEET 2016 (Phase 2)
Two identical light waves having phase difference Ï propagate in the same direction. When they superpose, the intensity of the resultant wave is proportional to

Options:
A cos2⁡ϕ
B cos2⁡ϕ2
C cos2⁡ϕ3
D cos2⁡ϕ4

The correct answer is:
cos2⁡ϕ2

Solution:
For two identical light waves
IR=4Icos2⁡ϕ2
∴ IR∝cos2⁡ϕ2

Topic: Physics, Wave Optics, NEET
Two identical long conducting wires AOB and COD are placed at right angle to each other, with one above other such that ‘ O ’ is their common point for the two. The wires carry I1 and I2 currents, respectively. Point ‘ I ’ is lying at distance ‘ d ’ from ‘ O ’ along a direction perpendicular to the plane containing the wires. The magnetic field at the point ‘ P ’ will be:

Options:
A μ02πdI1I2
B μ02πdI1+I2
C μ02πdI12-I22
D μ02πdI12+I221/2

The correct answer is:
μ02πdI12+I221/2

Solution:
B→ due to wire 1 B→1=μ0i12πdi^
B→ due to wire 2 B→2=μ0i22πd-j^
Bnet=μ02πd i12+i22

Topic: Physics, Magnetic Effects of Current, NEET, NEET 2014
Two identical metal plates are given positive charges Q1 and Q2(<Q1) respectively. If they are now brought close together to form a parallel plates capacitor with capacitance C, the potential difference between them is

Options:
A Q1+Q22C
B Q1+Q2C
C Q1-Q2C
D Q1-Q22C

The correct answer is:
Q1-Q22C

Solution:
For charge Q1, the electric field is
E1=Q12Îµ0A
where A is area.
For charge Q2, the electric field is
E2=Q22Îµ0A
Resultant electric field
E=E1-E2=(Q1-Q2)2Îµ0A
The potential difference between plates when they are brought close together to form a parallel plate capacitor is
V=Ed=(Q1-Q2)2Îµ0Ad
Since C=Îµ0Ad for parallel plate capacitor.
â´ V=(Q1-Q2)2C

Topic: Physics, Electrostatics, NEET
Two identical metal plates are given positive charges Q1 and Q2(<Q1) respectively. If they are now brought close together to form a parallel plates capacitor with capacitance C, the potential difference between them is

Options:
A Q1+Q22C
B Q1+Q2C
C Q1-Q2C
D Q1-Q22C

The correct answer is:
Q1-Q22C

Solution:
For charge Q1, electric field is
E1=Q12ε0A
where A is area.
For charge Q2 electric field is
E2=Q22ε0A
Resultant electric field
E=E1-E2=(Q1-Q2)2ε0A
Potential difference between plates when they are brought close together to form a parallel plate capacitor is
V=Ed=(Q1-Q2)2ε0Ad
Since, C=ε0Ad for parallel plate capacitor.
V=(Q1-Q2)2C

Topic: Physics, Electrostatics, NEET
Two identical metal plates show the photoelectric effect by a light of wavelength Î»1 on plate 1 and Î»2 on plate 2 (where Î»1=2Î»2 ). Then the relation between kinetic energies will be

Options:
A 2K2=K1
B K1<K22
C K1>K22
D 2K1=K2

The correct answer is:
K1<K22

Solution:
According to Einstein photelectric equation
K.E=hcÎ»-Ï
hcÎ»1=KE1+Ï .... (i) (for Î»1)
hcÎ»2=KE2+Ï .... (ii) (for Î»2)
Let Î»1=Î» and Î»2=Î»12=Î»2
From Eqs. (i) and (ii),
KE2KE1=hcÎ»2-ÏhcÎ»1-Ï=hcÎ»2-ÏhcÎ»-Ï=2hcÎ»-ÏhcÎ»-Ï
=hcÎ»-ÏhcÎ»-Ï+hcÎ»hcÎ»-Ï
âµhcÎ»hcÎ»-Ï>1
â´KE2KE1=1+>1>2
Thus, KE1<KE22

Topic: Physics, Dual Nature of Matter, NEET
Two identical narrow slits S1 and S2 are illuminated by the light of a wavelength Î» from a point source P.

If, as shown in the diagram above, the light is then allowed to fall on a screen, and if n is a positive integer, the condition for destructive interference at Q is

Options:
A lâ¡ 1 - lâ¡ 2 = 2 n + 1 λ / 2
B lâ¡ 3 - lâ¡ 4 = 2 n + 1 λ / 2
C lâ¡ 1 + lâ¡ 2 - lâ¡ 3 + lâ¡ 4 = n λ
D lâ¡ 1 + lâ¡ 3 - lâ¡ 2 + lâ¡ 4 = 2 n + 1 λ / 2

The correct answer is:
lâ¡ 1 + lâ¡ 3 - lâ¡ 2 + lâ¡ 4 = 2 n + 1 λ / 2

Solution:
Path difference = lâ¡ 1 + lâ¡ 3 - lâ¡ 2 + lâ¡ 4 ...(i)
for destructive interference
Path difference = 2 n + 1 λ / 2 ...(ii)
lâ¡ 1 + lâ¡ 3 - lâ¡ 2 + lâ¡ 4 = 2 n + 1 λ / 2

Topic: Physics, Wave Optics, NEET
Two identical p-n junctions may be connected in series with a battery in three ways. The potential drops across the two p-n junctions are equal in

Options:
A Circuit-1 and circuit-2
B Circuit-2 and circuit-3
C Circuit-3 and circuit-1
D Circuit-1 only

The correct answer is:
Circuit-2 and circuit-3

Solution:
In circuit 2 both the diodes are forward biased and in circuit 3 both the diodes are reverse biased.
⇒ Potential difference across both diodes will be half of the emf of battery.

Topic: Physics, Semiconductors, NEET
Two identical parallel plate capacitors are placed in series and connected to a constant voltage source of V volt. If one of the capacitors is completely immersed in a liquid of dielectric constant K, then the potential difference between the plates of the other capacitor will change to

Options:
A KK+1V
B K+1K
C 2KK+1V
D K+12KV

The correct answer is:
KK+1V

Solution:
In series combination, charge is same.
Vâ1C
Now, V2V1=KCC=K1
But V1+V2 =V
or V2K+V2 =V or V2=KK+1V

Topic: Physics, Electrostatics, NEET
Two identical positive charges are placed on the y-axis at y=-a and y=+a. The variation of V (potential) along the x-axis is shown by the graph

Options:
A
B
C
D

The correct answer is:

Solution:
The potential at the centre is V0=2kQa
Potential on the x-axis is given by, VA=2kQa2+x2
At x=∞,V∞=0
∴ The graph will tend to 0 at x=±∞ and will be maximum at x=0

Topic: Physics, Electrostatics, NEET
Two identical radiators have a separation of d = Î» 4 where Î» is the wavelength of the waves emitted by either source. The initial phase difference between the sources is Ï / 4. Then the intensity on the screen at a distant point situated at an angle Î¸ = 3 0 â from the radiators is (here I0 is the intensity at that point due to one radiator alone)

Options:
A I0
B 2I0
C 3I0
D 4I0

The correct answer is:
2I0

Solution:
The intensity at a point on screen is given by
I = 4 I 0 cos 2 Ï / 2
where Ï is the phase difference. In this problem Ï arises (i) due to initial phase difference of π / 4 and (ii) due to path difference for the observation point situated at θ = 3 0 â . Thus
Ï = π 4 + 2 π λ d sin θ = π 4 + 2 π λ · λ 4 sin 3 0 â = π 4 + π 4 = π 2
Thus Ï 2 = π 4 and I = 4 I 0 cos 2 π / 4 = 2 I 0

Topic: Physics, Wave Optics, NEET
Two identical satellites are moving around the earth in circular orbits at heights 3R and R respectively where R is the radius of the Earth. The ratio of their kinetic energies is

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

The correct answer is:
1 : 2

Solution:
Kinetic energy of satellites in circular orbit is given as Ek=GMm2râ Ek1Ek2=r2r1=R+RR+3R=2R4R=12

Topic: Physics, Gravitation, NEET
Two identical short magnetic dipoles of magnetic moments 1.0 A m3 each, placed at a separation of 2 m with their axes perpendicular to each other. The resultant magnetic field at a point midway between the dipole is :

Options:
A 5 × 1 0 - 7 T
B 5 Ã 1 0 - 7 T
C 1 0 - 7 T

D 2 × 1 0 - 7 T

The correct answer is:
5 × 1 0 - 7 T

Solution:
B 1 = 2 Î¼ 0 4 Ï M r 3 (As the dipole is short)
= 1 0 - 7 Ã 1 Ã 2 1 3 = 2 Ã 1 0 - 7 T
B 2 = Î¼ 0 4 Ï M r 3 = 1 0 - 7 T
B net = 5 Ã 1 0 - 7 T

Topic: Physics, Magnetic Properties of Matter, NEET
Two identical simple pendulums A and B are fixed at same point. They are displaced by very small angles Î± and Î²Î²>Î± respectively and released from rest. If collisions are elastic and length of strings is l then find the time after which B reaches its initial position for the first time

Options:
A πlg
B 2πlg
C πβαlg
D 2πβαlg

The correct answer is:
2πlg

Solution:
Time period of both A and B .
2πlg
A and B collide after time T4
After first collision, B acquires amplitude of A and second collision takes place after additional time of T4+T4=T2 after second collision B acquires its own amplitude and reaches its initial position after additional time of T4
In this process time taken is =T4+T4+T4+T4
⇒ T=2πlg

Topic: Physics, Oscillations, NEET
Two identical spheres move in opposite directions with speeds v1 and v2 and pass behind an opaque screen, where they may either cross without touching (event 1) or make an elastic head-on collision (event 2) -

Options:
A we can never make out which event has occurred
B we can not make out which event has occurred only if v1=v2
C we can always make out which event has occurred
D we can make out which event has occurred only if v1=v2

The correct answer is:
we can never make out which event has occurred

Solution:
Balls are identical and velocity exchanges if elastic head on collision occurs.

Topic: Physics, Center of Mass Momentum and Collision, NEET
Two identical spheres P and Q lie on a smooth horizontal circular groove at opposite ends of a diameter. P is projected along the groove and at the end of time t, impinges on Q. If e is the coefficient of restitution, the second impact will occur after the shortest time of

Options:
A 2 π t e&ApplyFunction;
B π t e&ApplyFunction;
C t e&ApplyFunction;
D 2 t eâ¡

The correct answer is:
2 t e⁡

Solution:
Let r be the radius of the groove.
As P and Q lie at opposite ends of a diameter, therefore distance Covered = 1 2 2 π r⁡ = π r⁡
∴ Speed of P before impact, u⁡ = π r⁡ t
Speed of Q before impact = 0.
Let v and v' be velocities of P and Q respectively after impact.
∴ v'–v=e(u–0)=eu
Thus the velocity of Q relative to P after the first impact = eu
Let the next impact occur after t' seconds. This would happen when Q completes one round more than P in time t'.
Distance travelled in one round = 2πr
∴ t = Relative distance Relative velocity = 2 π r⁡ v ′ - v
t ′ = 2 π r⁡ eu = 2 u⁡ t eu⁡ = 2 t e⁡

Topic: Physics, Center of Mass Momentum and Collision, NEET
Two identical springs are connected in parallel and series as shown in the figure. If fp and fs are frequencies of arrangements, what is fsfp?

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

The correct answer is:
1:2

Solution:
In first case, springs are connected in parallel, so their equivalent spring constant
kp=k1+k2
So, frequency of this spring-block system is
fp=12Ï kpm
or fp=12Ï k1+k2m
but k1=k2=k
â´ fp=12Ï 2km ..(i)
Now in second case, springs are connected in series, so their equivalent spring constant
k=k1k2k1+k2
Hence, frequency of this arrangement is given by
fs=12Ï k1k2(k1+k2)m
or fs=12Ï k2m ...(ii)
Dividing Eq. (ii) by Eq. (i), we get
fsfp=12Ï k2m12Ï 2km=14
or fsfp=12

Topic: Physics, Oscillations, NEET
Two identical springs are connected in series and parallel as shown in the figure. If fs and fp are frequencies of arrangements, what is fsfp?

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

The correct answer is:
1:2

Solution:
In first case, springs are connected in parallel, so their equivalent spring constant
kp=k1+k2
So, frequency of this spring-block system is
fp=12Ï kpm
or fp=12Ï k1+k2m
but k1=k2=k
â´ fp=12Ï 2km ..(i)
Now in second case, springs are connected in series, so their equivalent spring constant
k=k1k2k1+k2
Hence, frequency of this arrangement is given by
fs=12Ï k1k2(k1+k2)m
or fs=12Ï k2m ...(ii)
Dividing Eq. (ii) by Eq. (i), we get
fsfp=12Ï k2m12Ï 2km=14
or fsfp=12

Topic: Physics, Oscillations, NEET
Two identical springs of spring constant k are attached to a block of mass m and to fixed supports as shown in Fig. Find the period of oscillations.

Options:
A $2\pi \sqrt{\frac{m}{k}}$
B $2\pi \sqrt{\frac{3m}{2k}}$
C $2\pi \sqrt{\frac{2m}{3k}}$
D $2\pi \sqrt{\frac{m}{2k}}$

The correct answer is:
$2\pi \sqrt{\frac{m}{2k}}$

Solution:
Correct Option is : (D)
$2\pi \sqrt{\frac{m}{2k}}$

Topic: Chemistry, Oscillations, NEET
Two identical square rods of metal are welded end to end as shown in the figure (1), 20 cal of heat flows through it in 4 min. If the rods are welded as shown in the figure (2), the same amount of heat will flow through the rods in

Options:
A 1 min

B 2 min

C 4 min

D 16 min

The correct answer is:
1 min

Solution:
Figure (1) → series combination
Figure (2) → parallel combination
Thermal resistance for each block to be R
Then
Thermal resistance for the case I R1=2R
Thermal resistance for the case II R2=R/2
Thermal resistance for case II
I 1 = Δ T R 1 I 2 = Δ T R 2
⇒ 4 I 1 = I 2
4ΔQ1Δt1=ΔQ2Δt2 ifΔQ1=Q2
then
4 Δ t 2 = Δ t 1
Δ t 2 = Δ t 1 / 4 = 4 4 = 1 min

Topic: Physics, Thermal Properties of Matter, NEET
Two identical thin isosceles prisms of refracting angle θ and refractive index μ are placed with their bases touching each other. A parallel beam, of width 2b, is incident on this system, as shown. The distance of the point of convergence from the prism is

Options:
A b(μ-1)θ
B b2(μ-1)θ
C 2b(μ-1)θ
D bθ(μ-1)

The correct answer is:
b(μ-1)θ

Solution:
δ=(μ-1)θ=bx⇒x=b(μ-1)θ

Topic: Physics, Ray Optics, NEET
Two identical thin Plano-convex glass lenses (refractive index 1.5 ) each having radius of curvature of 20 cm are placed with their convex surfaces in contact at the center. The intervening space is filled with oil of refractive index 1.7 The focal length of the combination is

Options:
A -20 cm
B -25 cm
C -50 cm
D 50 cm

The correct answer is:
-50 cm

Solution:
Len's maker's formula
1f = μ-11R1-1R2
1f1=μ-11 20 ⇒ 1.5-1 120=140
1f1=140
For lens 1f1+1f2 = 140+140=120
For concave 1f3=1.7-1 (-1R-1R )= -0.7×2R
1feq=120+ -0.7×2R
=120 - 1.420
feq= -200.4 =-50 cm

Topic: Physics, Ray Optics, NEET, NEET 2015 (Phase 1)
Two identical thin plano-convex glass lenses (refractive index 1.5) each having radius of curvature of 20 cm are placed with their convex surfaces in contact at the centre. The intervening space is filled with the oil of refractive index 1.7. The focal length of the combination is:

Options:
A -20 cm

B -25 cm

C -50 cm

D 50 cm

The correct answer is:
-50 cm

Solution:
Consider the situation shown in figure.
As 1f=1.5-11R-1∞
⇒1f=0.5R
⇒1fconcave=1.7-1-2R
⇒1fconcave=0.7×2R=-1.4R
∴1feq=0.5R-1.4R+0.5R
Also, =-0.4R=-0.420
⇒feq=-50 cm

Topic: Physics, Ray Optics, NEET
Two identical wires A and B have the same length L and carry the same current I. Wire A is bent into a circle of radius R and wire B is bent to form a square of side a. If B1 and B2 are the value of magnetic induction at the centre of the circle and at the centre of the square respectively, then the ratio B1B2 is

Options:
A π28
B π282
C π216
D π2162

The correct answer is:
π282

Solution:
As, B1=Î¼04ÏÃ2ÏIR
=Î¼04ÏÃ2ÏIÃ2ÏL ... (i)
[âµL=2ÏRâforâcircularâloop]
B2=Î¼04ÏÃIa2 [sin45o+sin45o]Ã4
Where a=L4
â´B2=Î¼0I4ÏLÃ8Ã4ÃI2+I2
=Î¼0I4ÏLÃ642 ..... (ii)
Hence, B1B2=Î¼04ÏÃ4Ï2ILÎ¼0I4ÏLÃ642
Or B1B2=Ï282

Topic: Physics, Electromagnetic Induction, NEET
Two important intrinsic controls which are important in plant growth and development are

Options:
A PGRs
B Genomic control
C Both (a) and (b)
D O2 concentration

The correct answer is:
Both (a) and (b)

Solution:
Plant growth regulators (PGRs) or phytohormones control growth and development of plants and play an important role along with genomic control. PGRs can be growth promoters or inhibitors.

Topic: Biology, Plant - Growth and Development, NEET
Two inductors each of inductance L are connected in parallel. One more inductor of value 5 mH is connected in series of this configuration then the effective inductance is 15 mH. The value of L is _____ mH.

Options:
A 10
B 5.0
C 2.5
D 20

The correct answer is:
20

Solution:
L2+5=15
L=20 mH

Topic: Physics, Alternating Current, NEET
Two infinitely long charged wires with linear densities Î» and 3Î» are placed along x and y-axis respectively. Determine the slope of the electric field at any point on the line y=3x

Options:
A 33
B 332
C 133
D 3

The correct answer is:
133

Solution:
E→=3λ2π∈0xi^+λ2π∈0x3j^
Slope= EyEx=13÷3=133

Topic: Physics, Electrostatics, NEET
Two insulated charged spheres of radii R1 and R2 having charges Q1 and Q2 respectively are connected to each other; then there is -

Options:
A no change in the energy of the system
B an increase in the energy of the system
C always a decrease in the energy of the system
D a decrease in energy of the system unless Q1R2=Q2R1

The correct answer is:
a decrease in energy of the system unless Q1R2=Q2R1

Solution:
ΔU=12C1C2C1+C2V1-V22
ΔU=0 , If V1=V2
or  kQ1R1=kQ2R2
or  Q1R2=Q2R1

Topic: Physics, Electrostatics, NEET
Two insulated charged spheres of radii R1 and R2 having charges Q1 and Q2, respectively, are connected to each other; then there is

Options:
A no change in the energy of the system
B an increase in the energy of the system
C always a decrease in the energy of the system
D a decrease in energy of the system unless q1R2=q2R1

The correct answer is:
a decrease in energy of the system unless q1R2=q2R1

Solution:
Decrease in energy of the system after connection is given by
ΔU=12C1C2C1+C2V1-V22
ΔU=0, If V1=V2
or kq1R1=kq2R2
or q1R2=q2R1

Topic: Physics, Electrostatics, NEET
Two ions having masses in the ratio 1:1 and charges 1:2 are projected into uniform magnetic field perpendicular to the field with speeds in the ratio 2:3. The ratio of the radius of circular paths along which the two particles move is

Options:
A 4:3

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

The correct answer is:
4:3

Solution:
m1 : m2=1 : 1
q1 : q2=1 : 2
v1 : v2=2 : 3
Radius of path of charged particle moving in uniform magnetic field,
r=mvBq
â´ r1r2=m1m2 v1v2 Ãq2q1=11Ã23Ã21=43
r1 : r2=4 : 3

Topic: Physics, Magnetic Effects of Current, NEET
Two isolated metallic spheres of radii R and 2R are charged such that both of these have same charge density σ . These sphere are located far away from each other and connected by a thin conducting wire, the new charge density on the bigger sphere -

Options:
A 3 σ2
B 5σ6
C 6 σ5
D 2 σ3

The correct answer is:
5σ6

Solution:
σ=Q14πR2=Q14πR2
σ=Q24π2R2=Q216πR2
QB′=Q1+Q2C1+C2C2
QB′=20πR2σ×2C3C=40πR2σ3
σ′=40πR2σ34π2R2=40πR2σ3×16πR2=5σ2×3=5σ6

Topic: Physics, Electrostatics, NEET
Two isomeric compounds Cl-CH2-CH2-CH2Cl and CH3-CH2-CHCl2 can be distinguished by the use of

Options:
A HOH/ OH −
B Tollens reagent
C Both HOH/ OH − and Tollens reagent
D HOH/ OH − , Tollens reagent and ceric ammonium nitrate

The correct answer is:
HOH/ OH − , Tollens reagent and ceric ammonium nitrate

Solution:
This will give red colour with ceric ammonium nitrate
CH 3 − CH 2 − CHCl 2 → HOH/ OH − CH 3 − CH 2 −CHO
This will give silver mirror with Tollens reagent.

Topic: Chemistry, Haloalkanes and Haloarenes, NEET
Two isomeric ketone, pentan-3-one and pentan-2-one can be distinguished by

Options:
A I2/NaOH only
B NaSO3H only
C NaCN/HCl
D Both A and B

The correct answer is:
Both A and B

Solution:
(i) This will not give iodoform test
(ii) It will not give adduct with NaSO3H due to steric hinderance

(i) This will give iodoform test
(ii) This will form adduct with NaSO3H

Topic: Chemistry, Aldehydes and Ketones, NEET
Two kingdom classification does not distinguish between

Options:
A Eukaryote and prokaryote
B Unicellular and multi-cellular organisms
C Photosynthetic (green algae) and non-photosynthetic (fungi) organisms
D All the above

The correct answer is:
All the above

Solution:
Two kingdom classification was proposed by Carolus Linneaus in which he separated living world into Plantae and Animalia. Algae, Bryophyta, Pteridophyta, Gymnosperm, Angiosperm and fungi were placed in kingdom Plantae on the basis of cell wall character.

Topic: Biology, Biological Classification, NEET
Two large conducting plates of a parallel plate capacitor are given charges Q and 3Q respectively. If the electric field in the region between the plates is E0, then the force of interaction between the plates is

Options:
A 3QE0

B 2QE03
C 32QE0
D QE02

The correct answer is:
32QE0

Solution:
The net electric field between the plates due to both the plates is

E0=σε0=QAε0

So the electric field due to each plate is

E=E02=Q2Aε0

So the force on one plate due to the other one is

F=QE=3Q×Q2Aε0=3Q22Aε0
=3QE02

Topic: Physics, Electrostatics, NEET
Two large parallel conducting plates are separated by 40 mm. The potential difference between the plates is V. Potential difference between points X and Y as indicated in the figure will be

Options:
A 1540V
B V2
C 2540V
D V

The correct answer is:
V2

Solution:
Electric field is uniform in horizontal direction.
VXY=V(25cos⁡37o)40=V2

Topic: Physics, Electrostatics, NEET