datavorous/entrance-exam-dataset huggingface dataset:
Two leaves, A and B, are of different sizes but shows absolute increase in area in the unit time to give leaves A1 and B1 . Find out the absolute growth and relative growth rate of both the leaves.

Options:
A

Absolute Growth of A	Absolute Growth of B	Relative Growth Rate of A	Relative Growth Rate of B
5 cm2	10 cm2	10%	50%

B

Absolute Growth of A	Absolute Growth of B	Relative Growth Rate of A	Relative Growth Rate of B
10 cm2	5 cm2	5%	10%

C

Absolute Growth of A	Absolute Growth of B	Relative Growth Rate of A	Relative Growth Rate of B
5 cm2	5 cm2	10%	100%

D

Absolute Growth of A	Absolute Growth of B	Relative Growth Rate of A	Relative Growth Rate of B
5 cm2	5 cm2	100%	10%

The correct answer is:

Absolute Growth of A	Absolute Growth of B	Relative Growth Rate of A	Relative Growth Rate of B
5 cm2	5 cm2	100%	10%

Solution:
Growth rate is measured by increased growth per unit time. Growth in an organism or its part can occur in many different ways.
When the total growth of the living system is measured and compared in a particular unit time is referred as absolute growth rate. This growth rate is for that particular living system in that time interval and not compared with other living system.
Absolute growth = Growth in given time period = Growth (final time) - Growth (initial time)
Absolute growth of A = Growth A'-Growth A=10-5=5 cm2.
Absolute growth of B = Growth B'-Growth B=55-50=5 cm2 .
When the growth of living system measured per unit time is expressed on a common basis is referred to as relative growth rate. It is more of a comparative growth rate between initial time period and current time period.
Relative growth rate = Growth in given time/Initial growth*100
Relative growth rate A = (Final Growth - Initial Growth)/Initial Growth*
= (10-5)/5*100 = 100%
Relative growth rate B= (Final Growth - Initial Growth)/Initial Growth*
= (55-50)*100/50 = 10%

Topic: Biology, Plant - Growth and Development, NEET
Two lenses of power +10 D and -5 D are placed in contact. Where should an object be held from the lens, so as to obtain a virtual image of magnification 2?

Options:
A 5 cm
B -5 cm
C 10Â cm
D -10 cm

The correct answer is:
-10 cm

Solution:
Here, P1=+10 D, P2=-5 D
Combined power,
P=P1+P2=10 D-5 D
P=5 D
â´ Combined focal length
=1P m=100P cm=1005=20 cm
Given magnification,
m=2
vu=2
v=2u
By lens formula,
1v-1u=1f
12u-1u=120
1-22u=120
2u=20
u=-10 cm
Hence, to obtain virtual image, object should be place at a distance of 10 cm.

Topic: Physics, Ray Optics, NEET
Two light sources are coherent when

Options:
A their amplitudes are equal
B their frequencies are equal
C their wavelengths are equal
D their frequencies are equal and their phase difference is constant with time

The correct answer is:
their frequencies are equal and their phase difference is constant with time

Solution:
Coherency is defined between two waves if their phase difference is not changing. This can result in sustainable interference pattern.

Topic: Physics, Wave Optics, NEET
Two light waves of intensities I1 and I2 having same frequency pass through the same medium at a time in the same direction and interfere. The sum of the minimum and maximum intensities is

Options:
A I1+I2
B 2I1+I2
C I1 +I2
D I1 -I2

The correct answer is:
2I1+I2

Solution:
I max = ( a 1 + a 2 ) 2
I min = ( a 1 − a 2 ) 2
I max + I min = a 1 2 + a 2 2 + a 1 2 + a 2 2
=2( a 1 2 + a 2 2 )
=2( I 1 + I 2 )

Topic: Physics, Wave Optics, NEET
Two light ways having the same wavelength in vacuum are in phase initially. Then, the first ray travels a path L1 through a medium of refractive index μ1 while the second ray travels a path L2 through a medium of refractive index μ2. The two waves are then combined to observe interference. The phase difference between the two waves is

Options:
A 2πλL1μ1-L2μ2
B 2πλL2-L1
C 2πλμ2L1-μ1L2
D 2πλμ1L1-μ2L2

The correct answer is:
2πλμ1L1-μ2L2

Solution:
Optical path for first ray =μ1L1
Optical path for second ray =μ2L2
So, phase difference is given by
Δϕ=2πλ× path difference
=2πλ ×Δx
⇒ Δϕ=2πλμ1L1-μ2L2

Topic: Physics, Wave Optics, NEET
Two like charges of magnitude 1Ã10-9 coulomb and 9Ã10-9 coulomb are separated by a distance of 1 meter. The point on the line joining the charges, where the force experienced by a charge placed at that point is zero, is

Options:
A 0.25 m from the charge 1×10-9 C
B 0.75 m from the charge 9×10-9 C
C 0.25 m from the charge 1×10-9 C, 0.75 m from the charge 9×10-9 C
D at all points on the line joining the charges

The correct answer is:
0.25 m from the charge 1×10-9 C, 0.75 m from the charge 9×10-9 C

Solution:
âµ Force on charge q is zero
â´âââââF21=F23
orâ kqq1x2=kqq21-x2
orâ 1Ã10-9x2=9Ã10-91-x2
orâ 1x=31âx
orâ 1-x=3x
orâ x=0.25âm from 1Ã10-9 C
â´ âFrom 9Ã10-9C ,
distance =1-x
=1-0.25
=0.75

Topic: Physics, Electrostatics, NEET
Two linked genes a and b show 20 % recombination. The individuals of a dihybrid cross between + + / + + crossed with ab / ab shall show gametes

Options:
A + + 80 : ab : 20
B + + 50 : ab : 50
C + + 40 : ab 40 : + a 10 : + b 10
D + + 30 : ab : 30 : + a 20 : + b : 20

The correct answer is:
+ + 40 : ab 40 : + a 10 : + b 10

Solution:
Two linked genes a and b show 20 % recombination, thus linkage is 80%. The given phenotype of parents is ++/++ and ab/ab.
The parental genotype will be ++ and ab, while the recombinant type will be +a and +b.
Since % of recombination i.e. cross over frequency is 20%, the recombinant phenotype will be 10% each.
+a: 10%
+b: 10%
Similarly, since linkage is 80%, parental types will be 80% or 40% each for ++ and ab respectively.

Topic: Biology, Principles of Inheritance and Variation, NEET
Two liquid drops have diameters of 1 cm and 1.5 cm. The ratio of excess of pressure inside them is

Options:
A 1:1
B 5:3
C 2:3
D 3:2

The correct answer is:
3:2

Solution:
Excess pressure inside a liquid drop
âp=2TR
Where T is surface tension and R is radius of liquid drop.
∴ âp1âp2=R2R1=0.750.50
â¹ âp1âp2=32

Topic: Physics, Mechanical Properties of Fluids, NEET
Two liquids A and B form an ideal solution. At 300 K the vapour pressure of a solution of 1 mole of A and x moles of B is 550 mm of Hg. If the vapour pressures of pure A and B are 400 mm of Hg nd 600 mm of Hg respectively, then x is -

Options:
A 1
B 2
C 3
D 4

The correct answer is:
3

Solution:
PT=PA°XA+PB°XB
550=40011+x+600x1+x
55=401+x+60x1+x
55+55x=40+60x
x=3

Topic: Chemistry, Solutions, NEET
Two liquids A and B are at 32 oC and 24 oC. When mixed in equal masses, the temperature of mixture is found to be 28 oC. Their specific heats are in the ratio

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

The correct answer is:
1:1

Solution:
Let specific heats of liquids A and B are respectively SA and SB .
It is given that
TA=32 oC,âTB=24 oC
and Tmix=28 oC
When these liquids are mixed, then heat lost by A= heat gained by B
â mSAÎT=mSBÎTâ²
â mSATA-Tmix=mSBTmix-TB
â SA32-28=SB28-24
â SAÃ4=SBÃ4
â SA:SB=1:1

Topic: Physics, Thermal Properties of Matter, NEET
Two liquids (A) and (B) can be separated by the method of fractional distillation. The boiling point of liquid (A) is less than boiling point of liquid (B). Which of the liquids do you expect to come out first in the distillate? Explain.

Options:
A Liquid [A] will come out first because it has a higher boiling point.
B Liquid [B] will come out first because it has a lower boiling point.
C Liquid [A] will come out first because it has a lower boiling point.
D Liquid [B] will come out first because it has a higher boiling point.

The correct answer is:
Liquid [A] will come out first because it has a lower boiling point.

Solution:
Correct Option is : (C)
Liquid [A] will come out first because it has a lower boiling point.

Topic: Chemistry, General Organic Chemistry, NEET
Two liquids A and B have PAo and PBo the ratio of 1 : 3. If the ratio of number of moles of A and B are 1 : 3, the mole fraction of ‘A’ in vapour phase in equilibrium with the solution is equal to –

Options:
A 0.1
B 0.2
C 0.5
D 1.0

The correct answer is:
0.1

Solution:
PAoPBo=13
n A n B = 1 3
XA=14
XB=34
PA=PA0XA=PA0×14;PB=PB0XB=PB0×34
PA0PB0=13⇒3PA0=PB0
P=PA0XA+PA0XB=PA0×14+94PA0∵PB0=3PA0
P=104PA0XA=14PA0104PA0=110=0.1

Topic: Chemistry, Solutions, NEET
Two liquids which do not react chemically are placed in a bent tube a shown in the figure. The heights of the liquids above their surface of separation are

Options:
A directly proportional to their densities
B inversely proportional to their densities
C directly proportional to square of their densities
D equal

The correct answer is:
inversely proportional to their densities

Solution:
The pressure at the interface must be same, calculated via either tube. Since, both tube are open to the atmosphere, we must have
h1ρ1g=h2ρ2g
or h1ρ1=h2ρ2
or hρ=constant
or h∝1ρ

Topic: Physics, Mechanical Properties of Fluids, NEET
Two liquids X and Y form an ideal solution at 300 K, vapour pressure of the solution containing 1 mol of X and 3 mol of Y is 550 mmHg. At the same temperature, if 1 mol of Y is further added to this solution, vapour pressure of the solution increases by 10 mmHg. Vapour pressure (in mmHg) of X and Y in their pure states will be, respectivelyâ

Options:
A 300 and 400
B 400 and 600
C 500 and 600
D 200 and 300

The correct answer is:
400 and 600

Solution:
According to Dalton’s law of partial pressure PT=PX0Xx+Py0Xy
14Px0+34Py0=5501
15Px0+45Py0=5602
On solving Px0=400Py0=600

Topic: Chemistry, Solutions, NEET
Two liquids X and Y form an ideal solution. At 300 K, vapour pressure of the solutions containing 1 mol of X and 3 mol of Y is 550 mmHg. At the same temperature, if 1 mol of Y is further added to this solution, vapour pressure of the solution increases by 10 mmHg. Vapour pressure [in mmHg] of X and Y in their pure states will be, respectively:

Options:
A 300 and 400
B 400 and 600
C 500 and 600
D 200 and 300

The correct answer is:
400 and 600

Solution:
PT=PXoXX+PYoXY
XX = mol fraction of X
XY = mol fraction of Y
550=PXo11+3+PYo31+3
=PXo4+3PYo4
560Ã4= P X o +3 P Y o â¦(1)
560Ã5= P x o +4 P y o .....(2)
By Solving (i) and (ii)
We get, PXo=400 mmHg
PYo=600 mmHg

Topic: Chemistry, Solutions, NEET
Two lithium nuclei in a lithium vapour at room temperature do not combine to form a carbon nucleus because

Options:
A Carbon nucleus is an unstable particle
B It is not energetically favourable
C Nuclei do not come very close due to Coulombic repulsion
D Lithium nucleus is more tightly bound than a carbon nucleus

The correct answer is:
Nuclei do not come very close due to Coulombic repulsion

Solution:
Lithium nucleus and carbon nucleus are positively charge. According to coulomb law, same charge repel each other. So, nuclei do not come very close.

Topic: Physics, Nuclear Physics, NEET
Two litres of an ideal gas at a pressure of 10 atm expands isothermally at 25 °C into a vacuum until its total volume is 10 litres. Consider the expansion given in problem,for 1 mol of an ideal gas conducted reversibly.How much heat is absorbed in the expansion?

Options:
A 542.23 L atm
B 393.66 L atm
C 465.75 L atm
D 625.12 L atm

The correct answer is:
393.66 L atm

Solution:
Correct Option is : (B)
393.66 L atm

Topic: Chemistry, Thermodynamics (C), NEET
Two litres of an ideal gas at a pressure of 10 atm expands isothermally at 25 °C into a vacuum until its total volume is 10 litres. Consider this same expansion, but this time against a constant external pressure of 1 atm. How much heat is absorbed in the expansion?

Options:
A 6 litre-atm
B 10 litre-atm
C 8 litre-atm
D 12 litre-atm

The correct answer is:
8 litre-atm

Solution:
Correct Option is : (C)
8 litre-atm

Topic: Chemistry, Thermodynamics (C), NEET
Two litres of an ideal gas at a pressure of 10 atm expands isothermally at 25 °C into a vacuum until its total volume is 10 litres. How much heat is absorbed and how much work is done in the expansion?

Options:
A No work is done and no heat is absorbed
B W≠0 and Q≠0
C W≠0 but Q=0
D W=0 but Q≠0

The correct answer is:
No work is done and no heat is absorbed

Solution:
Correct Option is : (A)
No work is done and no heat is absorbed

Topic: Chemistry, Thermodynamics (C), NEET
Two long parallel conductors are placed at right angles to a metre scale at the 2 cm and 6 cm marks as shown in the figure

They carry currents of 1 A and 3 A respectively. They will produce zero magnetic field at (ignore earth's magnetic field)

Options:
A 5 cmmark
B 3 cmmark
C 1 cmmark
D 8 cmmark

The correct answer is:
3 cmmark

Solution:
Here, x is the distance from the 2 cm mark.
μ o I 1 2π x = μ o I 2 2π( 4−x )                  ( 4−x )×1=x×3                     4−x=3x                       4=4x⇒x=1

Topic: Physics, Magnetic Effects of Current, NEET
Two long parallel conductors carry currents I and 2I in the same direction. The magnetic induction at a point exactly midway between them is B. If the current in the first conductor is reversed in direction, the magnetic induction at the same point will be

Options:
A B3
B 2B
C 3B
D B2

The correct answer is:
3B

Solution:
In the first case, the two fields are in opposite directions and as a result, the resultant field is the difference of the individual fields whereas, in the next case, the two fields are in the same direction and add up to give the resultant.

Topic: Physics, Magnetic Effects of Current, NEET
Two long parallel wires are 30 cm apart carrying currents 10 A and 15 A respectively in the same direction. The force acting over a length of 5 m of the wires is

Options:
A 5Ã10-4 N , (attraction)
B 1Ã10-4N , (attraction)
C 5Ã10-4N , (repulsion)
D 1Ã10-4N , (repulsion)

The correct answer is:
5Ã10-4 N , (attraction)

Solution:
Force acting between two parallel wire
F=Î¼0i1i22Ïrl
F=2Ã10-710150.3Ã5
F=5Ã10-4 N

Topic: Physics, Magnetic Effects of Current, NEET
Two long parallel wires are at a distance 2d apart. They carry steady equal current flowing out of the plane of the paper as shown. The variation of the magnetic field along the line XX′is given by

Options:
A
B
C
D

The correct answer is:

Solution:
The magnetic field in between because of each will be in opposite direction
Bin between=Î¼0i2Ïx j^-Î¼0i2Ï (2d-x) (-j^)
=Î¼0i2Ï1x-12d-x (j^)
At x=d, Bin between=0
For x<d, Bin between =(j^)
For x>d, Bin between=(-j^)
Towards x, net magnetic field will add up and direction will be -j^.
Towards xâ², net magnetic field will add up and direction will be -j^.

Topic: Physics, Magnetic Effects of Current, NEET
Two long parallel wires are at a distance 2d apart. They carry steady equal currents flowing out of the plane of the paper, as shown. The variation of the magnetic field B along the line XX' is given by

Options:
A
B

C
D

The correct answer is:

Solution:
If the current flows out of the paper, the magnetic field at points to the right of the wire will be upwards and to the left will be downwards, as shown in the figure.
Now, let us come to the problem.
Magnetic field at C = 0
Magnetic field in region BX' will be upwards (+ve) because all points lying in this region are to the right of both the wires.
Magnetic field in region AC will be upwards (+ve), because points are closer to A, compared to B. Similarly magnetic field in region BC will be downwards (-ve).
This graph satisfies all these conditions. It is correct answer

Topic: Physics, Magnetic Effects of Current, NEET
Two long parallel wires placed 0.08 m apart, carry currents 3 A and 5 A in the same direction. What is the distance from the conductor carrying the larger current to the point where the resultant magnetic field is zero?

Options:
A 0.5 m
B 0.04Â m
C 0.05Â m
D 0.4Â m

The correct answer is:
0.05 m

Solution:
Let the distance from the conductor carrying larger current be x.
Therefore, the point at which the resultant magnetic filed will be zero, will be at,
B1=B2
Where, B1 and B2 is the magnetic field produced by two parallel wires.
â Î¼0I12Ïx=Î¼0I22Ï0.08-x
â 50.08-x=3x
â x=0.05âm

Topic: Physics, Magnetic Effects of Current, NEET
Two long straight conductors AOB and COD are perpendicular to each other and carry currents i1 and i2. The magnitude of the magnetic induction at a point P at a distance a from the point O in a direction perpendicular to the plane ABCD is

Options:
A Î¼02Ïai1+i2
B Î¼02Ïai1-i2
C μ02πai12+i22
D Î¼02Ïai1i2i1+i2

The correct answer is:
μ02πai12+i22

Solution:
Conductors AOB and COD are perpendicular to each other shown in figure.
At distance a above O,
B1=Î¼0i12Ïa and B2= Î¼0i22Ïa
B1 is perpendicular to B2.
Resultant of B1 and B2,
B=B12+B22
= Î¼02Ïai12+i22

Topic: Physics, Magnetic Properties of Matter, NEET
Two long straight conductors with current I1 and I2 are placed along X and Y axes. The equation of locus of point of zero magnetic induction is -

Options:
A y=x
B y=I2I1x
C y=I1I2x
D y=xI1I2

The correct answer is:
y=I1I2x

Solution:
BP=0
B1- B2=0
⇒B1=B2
Î¼0I12Ï y=Î¼02Ï I2x â´ y=I1I2 x

Topic: Physics, Magnetic Effects of Current, NEET
Two long wires carrying current are kept crossed (not joined at O ). The locus where magnetic field is zero is -

Options:
A I1=xyI2
B I1=yxI2
C I1=I2
D I1=-I2

The correct answer is:
I1=xyI2

Solution:
Magnetic field could be zero in 1st or 3rd quadrant.
Î¼0I12Ï x=Î¼0I22Ï y
or. I1=xyI2

Topic: Physics, Magnetic Effects of Current, NEET
Two loops P and Q are made from a uniform wire. The radii of P and Q are r1 and r2 respectively, and their moments of inertia are I1 and I2 respectively. If I2=4I1 , then r2r1 equals

Options:
A 423
B 413
C 4-23
D 4-13

The correct answer is:
413

Solution:
Moment of inertia of a loop is,
I=Mr2=2πrAρr2
Therefore I∝r3
or r∝I13
For the two loops
r2r1=I2I113
As I2=4I1
r2r1=4113

Topic: Physics, Rotational Motion, NEET
Two loud speakers are being compared. One is perceived to be 32 times louder than the other. The difference in intensity levels between the two, when measured in decibels, is

Options:
A 60
B 40
C 50
D 30

The correct answer is:
50

Solution:
âÎ²=Î²2-Î²1=10logI2I0-10logI1I0
=10logI2I1
Increasing the intensity by a factor 10 changes the sound level by 10 dB, increasing by 100 changes Î² by 20 dB.

Since, 32 = 25
âÎ²=510=50 dB

Topic: Physics, Waves and Sound, NEET
Two loudspeakers L1 and L2, driven by a common oscillator and amplifier, are arranged as shown. The frequency of the oscillator is gradually increased from zero and the detector at D records a series of maxima and minima. If the speed of sound is 330 m s-1, then the frequency at which the first maximum is observed is

Options:
A 165 Hz

B 330 Hz

C 495 Hz

D 660 Hz

The correct answer is:
330 Hz

Solution:
Let Δx = path difference
⇒ Δ x = L 2 D - L 1 D
⇒ Δ x = 40 2 + 9 2 - 40
⇒ Δ x = 41 - 40
⇒ Δ x = 1 m
For first maximum, n = 1
Δx=nλ
Δx=n V f
f= nV x
f=330

Topic: Physics, Waves and Sound, NEET
Two magnets each of magnetic moment M are placed so as to form a cross at right angle to each other. Magnetic moment of the system will be

Options:
A 2M
B M2
C M2
D M2

The correct answer is:
M2

Solution:
M→net=M→1+M→2
As M→1⊥M→2
∴ Mnet=M12+M22
Mnet=M2+M2
Mnet=M2

Topic: Physics, Magnetic Properties of Matter, NEET
Two magnets of magnetic moments 4 Am2 and 3 Am2 are joined to form a cross sign (+), then the magnetic moment of the combination is

Options:
A 4 Am2
B 1 Am2
C 7 Am2
D 5 Am2

The correct answer is:
5 Am2

Solution:
Magnetic dipole moment is a vector quantity. and they are at angle of 90Â°
Thus, they can be added as perpendicular vectors
M=M12+M22
N=32+42=5

Topic: Physics, Magnetic Properties of Matter, NEET
Two masses A and B connected with an inextensible string of length l lie on a smooth horizontal plane. A is given a velocity of v m s-1 along the ground perpendicular to line AB as shown in the figure. Find the tension in a string during their subsequence motion.

Options:
A 2mv23l

B 3mv22l
C mv22l
D 4mv23l

The correct answer is:
2mv23l

Solution:
From conservation of linear momentum
â v cm = m × 0 + 2mv m + 2m ⇒ v cm = 2v 3
position of the centre of mass
ycm=0×2m+m×l2m⇒ycm=l3
blocks will perform circular motion in the center of mass frame
velocity of A in COM frame is v A ′ = v - 2 v/3 ⇒ v A ′ = v/3
Tension will provide the required centripetal force
T = 2m v A ′ 2 ℓ / 3 ⇒ T = 2m v/3 2 ℓ / 3 ⇒ T = 2mv 2 3 ℓ

Topic: Physics, Center of Mass Momentum and Collision, NEET
Two masses A and B of 10 kg and 5 kg respectively, are connected with a string passing over a frictionless pulley fixed at the corner of a table as shown in the figure. The coefficient of friction of A with the table is 0.2, the minimum mass of C that may be placed on A to prevent it from moving is

Options:
A 15 kg

B 5 kg

C 10 kg

D 0 kg

The correct answer is:
15 kg

Solution:
Since the given system is in equilibrium, therefore, tension in the string is 5g.
When we consider the combination of A and C, then
T=μR
or T=0.2 10+mass of Cg
or 5g=0.2 10+mass of Cg
Or mass of C = 15 kg

Topic: Physics, Laws of Motion, NEET
Two masses m1=5 kg and m2=4.8 kg tied to a string are hanging over a light frictionless pulley. What is the acceleration of the masses when they are free to move g=9.8 m s-2

Options:
A 0.2 m s-2
B 9.8 m s-2
C 5 m s-2
D 4.8 m s-2

The correct answer is:
0.2 m s-2

Solution:
a=m1-m2m1+m2g=5-4.85+4.8×9.8=0.2m s-2

Topic: Physics, Laws of Motion, NEET
Two masses m1 and m2 are suspended together by a massless spring of spring constant k (Fig.). When the masses are in equilibrium, m1 is removed without disturbing the system. Find the angular frequency of osillation of m2

Options:
A ω= km2
B ω=km2-m1
C Can't be determined
D None of these

The correct answer is:
ω= km2

Solution:
When m1 is removed, only m2 is left. Therefore, angular

Frequency

ω= km2

Topic: Physics, Oscillations, NEET
Two masses m1 and m2 m1>m2 are connected by a massless flexible and inextensible string passed over a massless and frictionless pulley. The acceleration of the centre of mass is

Options:
A m1-m2m1+m22g
B m1-m2m1+m2g
C m1+m2m1-m2g
D Zero

The correct answer is:
m1-m2m1+m22g

Solution:
Acceleration of each mass=a=m1-m2m1+m2g
Now acceleration of centre of mass of the system
Acm=m1a1→+m1a2→m1+m2
As both masses move with same acceleration but in opposite direction so a1→=-a2→= a (let)
m1-m2m1+m2×m1-m2m1+m2×g
=m1-m2m1+m22×g

Topic: Physics, Laws of Motion, NEET
Two masses of 10 kg and 5 kg are suspended from a rigid support as shown in the figure. The system is pulled down with a force of 150 N attached to the lower mass. The string attached to the support breaks and the system accelerates downwards.

In case the force continues to act, the tension acting between the two masses is

Options:
A 300 N

B 200 N

C 100 N

D Zero

The correct answer is:
100 N

Solution:
After string breaks, the acceleration of block will be
a=150+15×1015=20 m s-2
150+50-T=5×20
T=100 N

Topic: Physics, Laws of Motion, NEET
Two materials having coefficients of thermal conductivity 3K and K and thickness d and 3d respectively, are joined to form a slab as shown in the figure. The temperatures of the outer surfaces are θ2 and θ1 respectively, θ2>θ1. The temperature at the interface is

Options:
A θ2+θ12
B Î¸16+5Î¸26
C Î¸13+2Î¸23
D Î¸110+9Î¸210

The correct answer is:
θ110+9θ210

Solution:
Let the temperature of the junction T Â°C .

Rate of heat flow in Rod 1 = rate of heat flow in Rod 2

3kAdÎ¸2-T=kA3dT-Î¸1

â9Î¸2-T=T-Î¸1

â10T=9Î¸2+Î¸1

âT=9Î¸2+Î¸110=Î¸110+9Î¸210

Topic: Physics, Thermal Properties of Matter, NEET
Two men with weights in the ratio 4 : 3 run up a staircase in time in the ratio 12 : 11. The ratio of the power of the first to that of second is

Options:
A 4 3
B 12 11
C 48 33
D 11 9

The correct answer is:
11 9

Solution:
Power=worktime=energytime=mght=Wâ¡hâ¡t( W = weight )
â´Pâ¡1Pâ¡2=Wâ¡1Wâ¡2Ãt2t1=43Ã1112=119 ( h is same in both cases )

Topic: Physics, Work Power Energy, NEET
Two men with weights in the ratio 4:3 run up a staircase in time in the ratio 12:11. The ratio of the power of the first to that of second is

Options:
A 4 3
B 12 11
C 48 33
D 11 9

The correct answer is:
11 9

Solution:
Power=worktime=energytime=mght=Wâ¡hâ¡t( W = weight )
â´Pâ¡1Pâ¡2=Wâ¡1Wâ¡2Ãt2t1=43Ã1112=119 ( h is same in both cases )

Topic: Physics, Work Power Energy, NEET