datavorous/entrance-exam-dataset huggingface dataset:
A man weighs $80 \mathrm{~kg}$. He stands on a weighing scale in a lift which is moving upwards with a uniform acceleration of $5 \mathrm{~m} / \mathrm{s}^2$. What would be the reading on the scale? $\left(g=10 \mathrm{~m} / \mathrm{s}^2\right)$

Options:
A zero
B $400 \mathrm{~N}$
C $800 \mathrm{~N}$
D $1200 \mathrm{~N}$

The correct answer is:
$1200 \mathrm{~N}$

Solution:
Reading of weighing scale
$$
\begin{aligned}
& =m(g+a) \\
& =80(10+5)=1200 \mathrm{~N}
\end{aligned}
$$

Topic: Physics, Laws of Motion, NEET, NEET 2003
A man whose father was colour blind marries a woman who had a colour blind mother and normal father. What percentage of male children of this couple will be colour blind?

Options:
A 25%
B 0%
C 50%
D 75%

The correct answer is:
50%

Solution:
Colour Blidness : It is a sex-linked recessive disorder due to defect in either red or green cone of eye resulting in failure to discriminate between red and green colour. This defect is due to mutation in certain genes present in the X chromosome. It occurs in about 8 per cent of males and only about 0.4 per cent of females. This is because the genes that lead to red-green colour blindness are on the X chromosome. Males have only one X chromosome and females have two.

A man whose father was colour blind (XCY) has a zero per cent chance of being colour blind. a woman who had a colour blind mother (XCXC) and normal father (XY) is not herself colour blind because the gene is recessive. That means the man will be normal (XY) and women will be carrier (XCX). The cross between them is as:

Thus, 50% percentage of male children of this couple will be colour blind.

Topic: Biology, Principles of Inheritance and Variation, NEET, NEET 2014
A man with a certain disease marries a normal woman. They have eight children (3 daughters and 5 sons). All the daughters suffer from their father's disease but none of the sons is affected. What will be the mode of inheritance of this disease?

Options:
A Sex-linked recessive
B Sex-linked dominant
C Autosomal dominant
D Sex-limited recessive

The correct answer is:
Sex-linked dominant

Solution:
The daughters receive one X chromosome from the father and one X chromosome from the mother. Since all the daughters suffer from their father's disease, the X chromosome from the father must be carrying a dominant trait.

Topic: Biology, Principles of Inheritance and Variation, NEET
A man with blood group 'A' marries a woman with blood group 'B'. What are all the possible blood groups of their offsprings?

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

The correct answer is:
A, B, AB and O

Solution:

Topic: Biology, Principles of Inheritance and Variation, NEET, NEET 2015 (Phase 1)
A man with blood group A marries AB blood group woman. Which type of progeny indicate that the man is not homozygous?

Options:
A AB
B B
C A
D More than one option is correct

The correct answer is:
B

Solution:
Man with blood group A can be of genotype â I A I O or â I A I A . If the genotype of man would have been homozygous, then the progeny formed would be:
If the genotype of man would have been heterozygous â I A I O then the progeny formed would be:
Thus, it is only progeny with B blood group that can be considered to be sure than man is not homozygous.

Topic: Biology, Principles of Inheritance and Variation, NEET
A man with genotype EEFfGg can produce a P number of genetically different sperms and women with genotype LLMmNn can produce a Q number of genetically different eggs. Determine the values of P and Q.

Options:
A P - 4, Q - 4
B P - 4, Q - 8
C P - 8, Q - 4
D P - 8, Q -8

The correct answer is:
P - 4, Q - 4

Solution:
A man with genotype EEFfGg can produce 4 number of genetically different sperms as the individual is heterozygous for two gene locus. A women with genotype LLMmNn can produce 4 number of genetically different eggs as she is heterozygous for two gene locus. Number of gametes formed is determined by 2n where n is number of heterozygous gene.

Topic: Biology, Principles of Inheritance and Variation, NEET
A man with genotype EEFfGgHH can produce a (P) number of genetically different sperms and a woman with genotype LLMmNnOo can produce (Q) number of genetically different eggs. Determine the values of (P) and (Q).

Options:
A PÂ -Â 4,Â QÂ -Â 4
B P - 4, Q - 8
C PÂ -Â 8,Â QÂ -Â 4
D PÂ -Â 8,Â QÂ -8

The correct answer is:
P - 4, Q - 8

Solution:
A man with genotype EEFfGgHH can produce 4 number of genetically different sperms as the individual is heterozygous for two gene locus. Woman with genotype LLMmNnOo can produce 8 number of genetically different eggs as she is heterozygous for three gene locus.
The number of gametes formed can be calculated by the formula (2)n, where n = no. of heterozygous genes.

Topic: Biology, Principles of Inheritance and Variation, NEET
A manometer connected to a closed tap reads 3.5Ã105 N m-2. When the valve is opened, the reading of manometer falls to 3.0Ã105 N m-2, then the velocity of flow of water is

Options:
A 100 m s-1
B 10 m s-1
C 1 m s-1
D 1010 m s-1

The correct answer is:
10 m s-1

Solution:
Bernoulli's theorem for unit mass of liquid is
PÏ+12v2= constant
As the liquid starts flowing, its pressure energy decreases
12v2=ÎPÏ
â 12v2=3.5Ã105-3Ã105103
â v2=2Ã0.5Ã105103
â v2=100
â v=10 m/s

Topic: Physics, Mechanical Properties of Fluids, NEET
A man's near point is 0.5 m and far point is 3 m. Power of spectacle lenses required for

(i) reading purposes

(ii) seeing distant objects, respectively are

Options:
A -2 D and + 3 D
B +2 D and - 3 D
C +2 D and -0.33 D
D -2 D and +0.33 D

The correct answer is:
+2 D and -0.33 D

Solution:
For reading purpose
u=-25 cm,v=-50 cm, f=?
∴1f=1v-1u=150+125=150
P=100f=+2 D
For distinct vision,
f′=distance of far point =-3 m
P = 1f′=-13 D =-0.33 D

Topic: Physics, Ray Optics, NEET
A man's near point is 0.5 m and the far point is 3 m. Power of spectacle lenses required for
(i) reading purposes
(ii) seeing distant objects, respectively are

Options:
A -2 D and + 3 D
B +2 D and - 3 D
C +2 D and -0.33 D
D -2 D and +0.33 D

The correct answer is:
+2 D and -0.33 D

Solution:
For reading purpose
u=-25 cm,v=-50 cm, f=?
∴1f=1v-1u=150+125=150
P=100f=+2 D
For distinct vision,
f′=distance of far point =-3 m
P = 1f′=-13 D =-0.33 D

Topic: Physics, Ray Optics, NEET
A marble block of mass 2 kg lying on ice when given a velocity of 6 m s-1 is stopped by friction in 10 s . Then the coefficient of friction is

Options:
A 0.02
B 0.03
C 0.06
D 0.01

The correct answer is:
0.06

Solution:
Let coefficient of friction be Î¼, and then retardation will be Î¼âg .
From equation of motion, v=u+at
â 0=6-Î¼ gÃ10
â Î¼=6100=0.06

Topic: Physics, Laws of Motion, NEET
a marble block of mass 2 kg lying on the ice when given a velocity of 6 m sâ1 is stopped by frictional force in 10 s. Then the coefficient of friction is
(take g=10 m s-2)

Options:
A 0.01
B 0.02
C 0.03
D 0.06

The correct answer is:
0.06

Solution:
v=u+at

v=uâÎ¼gtââââââââââââââââââââââââââââ ( since ââÎ¼mg=ma )

0=6âÎ¼Ã10Ã10

or 100Î¼=6

or Î¼=6100=0.06

Topic: Physics, Laws of Motion, NEET
A marine cartilaginous fish that can produce electric current is:

Options:
A Pristis
B Torpedo
C Trygon
D Scoliodon

The correct answer is:
Torpedo

Solution:
Torpedo (electric ray) possesses modified muscles acting as electric organs. These marine cartilaginous fish produces 8-220 volt electric charge or current though their lateral muscle plates innervated by cranial nerves.

Topic: Biology, Animal Kingdom, NEET, NEET 2014
A marriage between a colourblind man and a homozygous normal woman produces _____.

Options:
A all carrier daughter and normal sons
B 50% carrier daughters, 50% normal daughters
C 50% colourblind sons, 50% normal sons
D all carrier offsprings

The correct answer is:
all carrier daughter and normal sons

Solution:
Colour blindness is a X-linked recessive disorder. In it, the affected individual has is not able to distinguish colour especially read and green. X-lined disorders are inherent disorders due to mutation in the X-chromosome which causes the phenotype to be always expressed in male and in female only in homozygous condition. The daughter will receive one X chromosome from father (the chromosome with colour blindness gene) and other X (normal chromosome) from mother. Therefore, all daughters born will be carriers.
The son will receive Y chromosome from father and a X chromosome from mother. Thus, all sons will be normal.

Topic: Biology, Principles of Inheritance and Variation, NEET
A mass 3m, initially at rest at the origin explodes into three fragments of equal mass. Two of the fragments have speed v each and move perpendicular to each other. The third fragment will move with a speed

Options:
A v2
B v2
C v
D v2

The correct answer is:
v2

Solution:
From law of conservation of momentum
net moment of the of the first two particles moving with velocities v is = 2  v
From the conservation of linear momentum momentum of the third particle must be 12  v
to make resultant of momentum due to all three particles zero

Topic: Physics, Center of Mass Momentum and Collision, NEET
A mass attached to one end of a string crosses topmost point on a vertical circle with critical speed. Its centripetal acceleration when string becomes horizontal will be ( g= gravitational acceleration)

Options:
A g
B 3g
C 4g
D 6g

The correct answer is:
3g

Solution:
Critical velocity at top most point v= rg
â´ Velocity when string become horizontal
v'= Ï 2 +2gr
= 3gr
Centripetal acceleration = Ï 2 r = 3rg r =3g

Topic: Physics, Laws of Motion, NEET
A mass is suspended from a vertical spring which is executing S.H.M. of frequency 5 Hz. The spring is unstretched at the highest point of oscillation. The maximum speed of the mass is [acceleration due to gravity g=10 m s-2 ]

Options:
A 2π m s-1
B Ï m s-1
C 12π m s-1
D 1π m s-1

The correct answer is:
1π m s-1

Solution:
T=2π m k
n= 1 2π k m
25= 1 4 π 2 k m
k=100 π 2 m
kA=mg
A= mg k
V max =ωA
= 2π T A
=2πnA
= 2π×5×mg k
Vmax=10π×m×10100π2m=1π m s-1

Topic: Physics, Oscillations, NEET
A mass is suspended separately by two different springs in successive order, then the time period is $T_1$ and $T_2$, respectively. If it is connected by both spring as shown in figure, then the time period $T_0$, so the correct reaction is:

Options:
A $T_0^2=T_1^2+T_2^2$
B $T_0^{-2}=T_1^{-2}+T_2^{-2}$
C $T_0^{-1}=\mathrm{T}_1^{-1}+\mathrm{T}_2^{-1}$
D $T_0=T_1+T_2$

The correct answer is:
$T_0^{-2}=T_1^{-2}+T_2^{-2}$

Solution:
Time period of spring $T=2 \pi \sqrt{\frac{m}{\mathrm{~K}}}$
$\begin{aligned}
& \therefore K=4 \pi^2 \frac{m}{T^2} \\
& \Rightarrow K \propto \frac{1}{T^2}
\end{aligned}$
For parallel combination of spring
$K=K_1+K_2$
$\begin{aligned}
& \therefore \frac{1}{T_0^2}=\frac{1}{T_1^2}+\frac{1}{T_2^2} \\
& \Rightarrow T_0^{-2}=T_1^{-2}+T_2^{-2}
\end{aligned}$

Topic: Physics, Oscillations, NEET, NEET 2002
A mass is whirled in a circular path with constant angular velocity and its angular momentum is L. If the string is now halved keeping the angular velocity the same, the angular momentum is

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

The correct answer is:
L 4

Solution:
L = mr2 ω. For given m and ω, L ∝ r2. If r is halved, the angular momentum L becomes one-fourth.

Topic: Physics, Rotational Motion, NEET
A mass m is attached to a thin wire and whirled in a vertical circle. The wire is most likely to break, when

Options:
A the mass is at the highest point
B the wire is horizontal
C the mass is at the lowest point
D inclined at an angle of 60° from vertical

The correct answer is:
the mass is at the lowest point

Solution:
Tension is maximum at lowest point most likely string will break at lowest point.

Topic: Physics, Laws of Motion, NEET, NEET 2019 (All India)
A mass M is lowered with the help of a string by a distance h at a constant acceleration g2. The work done by the string will be

Options:
A Mgh2
B -Mgh2
C 3Mgh2
D -3Mgh2

The correct answer is:
-Mgh2

Solution:
Tension in the string
T=Mg-a=Mg-g2=Mg2
W= Force × displacement
=-Mgh2

Topic: Physics, Work Power Energy, NEET
A mass m is raised from a distance 2R from the surface of the earth to 3R . Work done to do so against gravity will -

Options:
A mgR10
B mgR11
C mgR12
D mgR14

The correct answer is:
mgR12

Solution:
Ui=- GMm2R+R
Uf=- GMmR+3R
W=ÎU=Uf- Uf
=GMm12R=GMR2 Ã mR12
=mgR/12

Topic: Physics, Gravitation, NEET
A mass M is split into two parts m and (M-m), which are then separated by a certain distance. The ratio m:M which maximizes the gravitational force between the parts is

Options:
A 1:4

B 1:2

C 4:1

D 2:1

The correct answer is:
1:2

Solution:
F=Gm(M-m)x2; For maxima,
dFdm=Gx2M-2m=0
or mM=12

Topic: Physics, Gravitation, NEET
A mass M is suspended by a rope from rigid support at A as shown in the figure. Another rope is tied at the end B and it is pulled horizontally with a force F. If the rope AB makes an angle Î¸ with the vertical, then the tension in the string AB is

Options:
A Fsin⁡θ
B Fsinθ
C Fcos⁡θ
D Fcosθ

The correct answer is:
Fsinθ

Solution:
Tsin⁡90o=Fsin⁡π-θ=Mgsin⁡π2+θ
Or T=Fsinθ=Mgcosθ

Topic: Physics, Laws of Motion, NEET
A mass M is suspended from a light spring. An additional mass m added, displaces the spring further by a distance X . Now, the combined mass will oscillate on the spring with time period

Options:
A T=2πmg(M+m)X
B T=2π(M+m)Xmg
C T=π2mg(M+m)X
D T=2π(M+m)mgX

The correct answer is:
T=2π(M+m)Xmg

Solution:
As an additional mass m is added, the spring is further stretched by a distance X , therefore

kX=mg

⇒k=mgX

The time period of the spring-mass system when both the masses m+M are suspended from the spring,
T=2π(M+m)k

⇒T=2πm+MXmg

Topic: Physics, Oscillations, NEET
A mass M is suspended from a spring of negligible mass. The spring is pulled a little and then released so that the mass executes SHM of time period T. If the mass is increased by m, the time period becomes 5T/3. Then the ratio of m/M is

Options:
A 3 / 5
B 25 / 9
C 16 / 9
D 5 / 3

The correct answer is:
16 / 9

Solution:
Initially, T=2πM/k
Finally, 5T3=2πM+mk
∴ 53×2πMk=2πM+mk
or 259Mk=M+mk
or 9 m+9 M=25M
or mM=169 .

Topic: Physics, Oscillations, NEET
A mass m moves in a circle on a smooth horizontal plane with velocity v0 at a radius R0 . The mass is attached to a string which passes through a smooth hole in plane as shown.

The tension in the string is increased gradually and finally m moves in a circle of radius R02. The final value of the kinetic energy is:

Options:
A mv02
B 14mv02
C 2 mv02
D 12mv02

The correct answer is:
2 mv02

Solution:
From conservation of angular momentum.
\(\begin{aligned}
& m v_0 R_0=m v^{\prime}\left(\frac{R_0}{2}\right) \\
& \Rightarrow v^{\prime}=2 v_0
\end{aligned}\)
Hence, final \(K E=\frac{1}{2} m v^{\prime 2}=\frac{1}{2} m\left(2 v_0\right)^2\)
Final \(KE=2 m v_0^2\)

Topic: Physics, Rotational Motion, NEET, NEET 2015 (Phase 1)
A mass m moves with a velocity v and collides inelastically with another identical mass. After collision the first mass moves with velocity v/3 in a direction perpendicular to the initial direction of motion. Find the speed of the 2nd mass after collision

Options:
A 2 3 ν
B ν 3
C ν
D 3 ν

The correct answer is:
2 3 ν

Solution:
let v1= speed of second mass after collision Momentum is conserved
Along X-axis, mâ¡ ν 1 cosâ¡ θ = mâ¡ ν
Along Y-axis, mâ¡ ν 1 sinâ¡ θ = mâ¡ ν 3
From (i) and (ii)
∴ ( mâ¡ ν 1 cosâ¡ θ ) 2 + ( mâ¡ ν 1 sinâ¡ θ ) 2 = ( mâ¡ ν ) 2 + ( mâ¡ ν 3 ) 2
or mâ¡ 2 ν 1 2 = 4 mâ¡ 2 ν 2 3
or ν 1 = 2 3 ν

Topic: Physics, Center of Mass Momentum and Collision, NEET
A mass m moving horizontally (along the x-axis) with velocity v collides and sticks to the mass of 3m moving vertically upward (along the y-axis) with velocity 2v. The final velocity of the combination is

Options:
A 14vi^+32v j^
B 13vi^+23v j^
C 23vi^+13v j^
D 32vi^+14v j^

The correct answer is:
14vi^+32v j^

Solution:
As the two masses stick together after collision, hence it is inelastic collision. Therefore, only momentum is conserved.
â´mvi^+3m2vj^=4m vâ
vâ=v4i^+64vj^=v4i^+32vj^

Topic: Physics, Laws of Motion, NEET
A mass M of 100ââkg is suspended with the use of strings A,B and C as shown in the figure, where W is the vertical wall and R is a rigid horizontal rod. The tension in the string B is

Options:
A 100g N
B Zero
C 1 0 0 2 g N
D 1 0 0 2 g N

The correct answer is:
100g N

Solution:
Let T be the tension in the string C . Hence

T cos45Â°=Mg
Tâsinââ45Â°= tension in B
Hence, tension in B=Mg=100gâN

Topic: Physics, Laws of Motion, NEET
A mass of 1kg carrying a charge of 2 C is accelerated through a potential of 1 V. The velocity acquired by it is

Options:
A 2 m s-1
B 2 m s-1
C 12 m s-1
D 12 m s-1

The correct answer is:
2 m s-1

Solution:
Given, m=1 kg, q=2C, V=1 V
qV=12 mv2
â 2Ã1=12Ã1Ãv2
v=2 m s-1

Topic: Physics, Electrostatics, NEET
A mass of 1 kg is acted upon by a single force Fâ=(4i^+4j^) N . Due to force, mass is displaced from (0,0) to 1 m,1 m . If initially, the speed of the particle was 2 m s-1 , its final speed should approximately be

Options:
A 6 m s-1
B 4.5 m s-1
C 8 m s-1
D 7.2 m s-1

The correct answer is:
4.5 m s-1

Solution:
Mass is displaced from (0, 0) to (1, 1). Therefore displacement is,
Sâ=1-0i^+1-0j^
Sâ=i^+j^ m
Given force
Fâ=4i^+4j^ N
â´ Work done by a constant force,
W=Fâ. Sâ
W= 4i^+4j^.i^+j^
W=4+4=8 J
From work-energy theorem
Î K=W
12mÎ½22-Î½12=W
121Î½22-22=8
Î½22=20
Î½2=4.5 m s-1

Topic: Physics, Motion In Two Dimensions, NEET
A mass of 1 kg is suspended by a thread. It isâ
(i) lifted up with an acceleration 4.9âmâsâ2,
(ii) lowered with an acceleration 4.9âmâsâ2.
The ratio of the tensions in (i) and (ii) is

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

The correct answer is:
3:1

Solution:
In case (i) we have â

In case (ii),

â´ââ T 1 T 2 = 14.7 4.9 = 3 1

Topic: Physics, Laws of Motion, NEET
A mass of 2.0 kg is put on a flat pan attached to a vertical spring fixed on the ground as shown in the figure. The mass of the spring and the pan is negligible. When pressed slightly and released the mass executes slightly and released the mass executes a simple harmonic motion. The spring constant is 200 N m-1. What should be the minimum amplitude of the motion, so that the mass gets detached from the pan? (Take g=10 ms-2 )

Options:
A 8 cm

B 10 cm

C Any value less than 12 cm
D 4 cm

The correct answer is:
10 cm

Solution:
Let the minimum amplitude of SHM is a.
Restoring force on spring
F=ka
Restoring force is balanced by weight mg of block. For mass to execute simple harmonic motion of amplitude a.
ka=mg
or a=mgk
Here, m=2 kg, k=200 N m-1, g=10 m s-2
â´ a=2Ã10200=10100 m
= 10100Ã100 cm=10 cm
Hence, minimum amplitude of the motion should be 10 cm, so that the mass gets detached from the pan.

Topic: Physics, Oscillations, NEET
A mass of 2kg is suspended with thread AB (Fig). Thread CD of the same type is attached to the other end of 2 kg mass. Lower thread is pulled gradually, harder and harder in the downward direction so as to apply force on AB. Which of the threads will break

Options:
A AB
B CD
C Both
D None

The correct answer is:
AB

Solution:
Correct Option is : (A)
AB

Topic: Chemistry, Laws of Motion, NEET
A mass of $0.5 \mathrm{~kg}$ moving with a speed of $1.5 \mathrm{~m} / \mathrm{s}$ on a horizontal smooth surface, collides with a nearly weightless spring of force constant $k=50 \mathrm{~N} / \mathrm{m}$. The maximum compression of the spring would be:

Options:
A $0.15 \mathrm{~m}$
B $0.12 \mathrm{~m}$
C $1.5 \mathrm{~m}$
D $0.5 \mathrm{~m}$

The correct answer is:
$0.15 \mathrm{~m}$

Solution:
According to the question
$$
\begin{aligned}
\frac{1}{2} m v^2 & =\frac{1}{2} k x^2 \\
\Rightarrow \quad x & =v \sqrt{\frac{m}{k}}=1.5 \sqrt{\frac{0.5}{50}} \\
& =0.15 \mathrm{~m}
\end{aligned}
$$

Topic: Physics, Work Power Energy, NEET, NEET 2004
A mass of 6Ã1024 kg is to be compressed in a sphere in such a way that the escape velocity from the sphere is 3Ã108. What should be the radius of the sphere?

Options:
A 9 km
B 9 m
C 9 mc
D 9 mm

The correct answer is:
9 mm

Solution:
Let escape velocity be ve, then kinetic energy is
=12mve2 ….i
and escape energy=+GMemRe …ii
Equating Eqs. (i) and (ii), we get
1 2 m e v⁡ e⁡ 2 = G⁡ M⁡ e⁡ m⁡ R⁡ e⁡
⟹ ve=2GMeRe
⟹ R=2GMeve2
Given,
G⁡=6.67×10-11N m2 kg-2
R=2×6.67×10-11×6×1024(3×108)2
R=8.89×10-3
R≈9×10-3 m=9 mm

Topic: Physics, Gravitation, NEET
A mass of $2.0 \mathrm{~kg}$ is put on a flat pan attached to a vertical spring fixed on the ground as shown in the figure. The mass of the spring and the pan is negligible. When pressed slightly and released the mass executes a simple harmonic motion. The spring constant is $200 \mathrm{~N} / \mathrm{m}$. What should be the minimum amplitude of the motion so that the mass gets detached from the pan (take $g=10 \mathrm{~m} / \mathrm{s}^2$ ).

Options:
A $10.0 \mathrm{~cm}$
B any value less than $12.0 \mathrm{~cm}$
C $4.0 \mathrm{~cm}$
D $8.0 \mathrm{~cm}$

The correct answer is:
$10.0 \mathrm{~cm}$

Solution:
For given condition
$\begin{aligned}
m g & =m \omega^2 a=K a \\
\Rightarrow \quad a & =\frac{m g}{k}=\frac{2 \times 10}{200} \\
& =0.1=10 \mathrm{~cm}
\end{aligned}$

Topic: Physics, Oscillations, NEET, NEET 2007
A massless spring of length l and spring constant k is placed vertically on a table. A ball of mass m is just kept on top of the spring. The maximum velocity of the ball is

Options:
A gmk
B g2mk
C 2gmk
D g2mk
E gm2k

The correct answer is:
gmk

Solution:
If x- displacement of free end of spring, then
at equilibrium
mg=kx
or x=mgk
Time period of oscillation, T=2πmk
⇒ω= Angular frequency
=2πT=km
Maximum velocity of oscillating mass
=Vmax=Aω
=mgk×km=gmk

Topic: Physics, Oscillations, NEET
A massless string going over a clamped pulley of mass m supports a block of mass M as shown in the figure. The force on the pulley by the clamp is given

Options:
A 2 Mg
B 2 mg
C gM+m2+m2
D g M + m 2 + M 2

The correct answer is:
g M + m 2 + M 2

Solution:
R = M 2 g 2 + m + M 2 g 2 ⇒ R = M 2 + m + M 2 g

Topic: Physics, Laws of Motion, NEET
A mature corn plant absorbs almost _____ of water in ____.

Options:
A 5 litres, in a day
B 1 litre, one hour
C 5 litres in a half day
D 3 litres, a day

The correct answer is:
3 litres, a day

Solution:
A plant water relationship is concerned with how plants control hydration of cells, including the collection of water from the soil and its transport within the plant. Terrestrial plants absorb a large quantity of water but most of it is lost through transpiration. A mature corn plant absorbs 3 litres of water per day and also some amount is lost in the air through transpiration.

Topic: Biology, Transport in Plants, NEET
A mature sieve tube differs from a vessel in

Options:
A Lacking a functional nucleus
B Absence of lignified walls
C Being nearly dead
D Lacking cytoplasm

The correct answer is:
Absence of lignified walls

Solution:
A mature sieve tube lacks lignified walls. Sieve tube in flowering plants has elongated living cells (sieve-tube elements) of the phloem, the nuclei of which are fragmented and disappeared, and the transverse end walls are pierced by sieve-like groups of pores (sieve plates). They are the conduits of food (mostly sugar) transport.

Topic: Biology, Anatomy of Flowering Plants, NEET
A mature sieve tube differs from a vessel in

Options:
A lacking a functional nucleus.
B absence of lignified walls.
C being nearly dead.
D lacking cytoplasm.

The correct answer is:
absence of lignified walls.

Solution:
The vessel is a long cylindrical tube-like structure made up of many cells called vessel members, each with lignified walls and a large central cavity. The vessel cells are also devoid of protoplasm. Vessel members are interconnected through perforations in their common walls. The presence of vessels is a characteristic feature of angiosperms.
Sieve tube elements are also long, tube-like structures, arranged longitudinally and are associated with the companion cells. Their end walls are perforated in a sieve-like manner to form the sieve plates. A mature sieve element possesses a peripheral cytoplasm and a large vacuole but lacks a nucleus.

Topic: Biology, Anatomy of Flowering Plants, NEET