datavorous/entrance-exam-dataset huggingface dataset:
A specialized procedure in which an embryo is formed in the laboratory by injecting a sperm directly into the ovum is

Options:
A ICSI
B AI
C GIFT
D ZIFT

The correct answer is:
ICSI

Solution:
A specialized procedure in which an embryo is formed in the laboratory by injecting a sperm directly into the ovum is known as intra-cytoplasmic sperm injection (ICSI). Artificial insemination (AI) is the procedure in which the semen collected from the husband or a healthy donor is inseminated either into the vagina or the uterus. Gamete intra-fallopian tube transfer (GIFT) is a procedure in which the ovum collected from a donor female is transferred into the fallopian tube of a female who is unable to produce an ovum. Zygote intra-fallopian tube transfer (ZIFT) is a procedure in which the zygote or early embryo with up to 8 blastomeres is transferred to the fallopian tube.

Topic: Biology, Reproductive Health, NEET
A species which was introduced for ornamentation but has become a trouble-some weed in India:

Options:
A Parthenium hysterophorus
B Eichhornia crassipes
C Prosopis juliflora
D Trapa spinosa

The correct answer is:
Eichhornia crassipes

Solution:
Water hyacinth or Eichhornia crassipes was introduced in India because of its shape of leaves and beautiful flowers. But, due to its high vegetative reproductive rate, it became an invasive weed. As a result of its growth, the oxygen content was drastically reduced, causing the death of a large number of fish.

Topic: Biology, Organisms and Populations, NEET, NEET 2020 (Phase 2)
A specific hormone is responsible for causing a disease that is characterized by excessive urination. The urine analysis of such a person reveals that the urine is dilute, has high pH and absence of sugar in it. This is due to under secretion of hormone from

Options:
A The gland which is heterocrine in nature
B The gland which stores the hypothalamic hormones
C The gland situated on the dorsal side of the forebrain
D The gland which is influenced by the hypothalamic hormones

The correct answer is:
The gland which stores the hypothalamic hormones

Solution:
The hormone responsible for diabetes Insipidus in humans is vasopressin or anti-diuretic hormone (ADH). An impairment affecting synthesis or release of ADH results in a diminished ability of the kidney to conserve water leading to water loss and dehydration.
ADH is released by the posterior pituitary gland. It is responsible for the storage of hormones oxytocin and vasopressin secreted by the posterior pituitary. The disease is called diabetes insipidus.

Topic: Biology, Chemical Coordination and Integration, NEET
A specific recognition sequence identified by endonucleases to make cuts at specific positions within the DNA is:

Options:
A Degenerate primer sequence
B Okazaki sequences
C Palindromic Nucleotide sequences
D Poly(A) tail sequences

The correct answer is:
Palindromic Nucleotide sequences

Solution:
Each restriction enzyme recognizes a specific palindromic nucleotide sequence in DNA. Palindrome in DNA is a sequence of base pairs that reads the same on the two strands when the orientation of reading is kept the same. These sequences may range from 4-8 nucleotides in length.

Topic: Biology, Biotechnology - Principles and Processes, NEET, NEET 2021
A speeding motorcyclist sees traffic jam ahead of him. He slows down to 36km/hour. He finds that traffic has eased and a car moving ahead of him at 18km/hour is honking at a frequency of 1392 Hz. If the speed of sound is 343m/s, the frequency of the honk as heard by him will be:

Options:
A 1332Â Hz
B 1372Â Hz
C 1412 Hz
D 1454Â Hz

The correct answer is:
1412 Hz

Solution:
Apparent frequency heard by the observer is given by,
f'=f0V-V0V-Vs

Where,

f0 is the original frequency.

V is the velocity of sound.

V0 is the velocity of the observer.

Vs is the velocity of source
f'=1392343--10343--5=1412 Hz

Topic: Physics, Waves and Sound, NEET, NEET 2014
A sphere and a cube of same material and same total surface area are placed in the same evacuated space turn by turn after they are heated to the same temperature. Find the ratio of their initial rates of cooling in the enclosure

Options:
A π6:1
B π3:1
C π6:1
D π3:1

The correct answer is:
π6:1

Solution:
Ss=Sc⇒ms>mc
dQdt=4σAT03msDT
Given 6a2=4pr2
or ar=4π6=2π3
dQdt∝Am;
dQdtsdQdtc=4πr243πr3ρ6a2a3ρ=3r6a
dQdtsdQdtc=a2r=12
dQdtsdQdtc=a2r=122π3=π6:1

Topic: Physics, Thermal Properties of Matter, NEET
A sphere and a cube of same material and same total surface area are placed in the same evacuated space turn by turn after they are heated to the same temperature. Find the ratio of their initial rates of cooling in the enclosure.

Options:
A Ï 6 : 1
B π 3 : 1
C π 6 : 1
D π 3 : 1

The correct answer is:
Ï 6 : 1

Solution:
Rate of emission of energy = Ï T 4 S
Let m1 be the mass of sphere, C is specific heat and d Î¸ / dt , the rate of cooling.
For sphere
Ï T 4 S = m 1 C d Î¸ dt S ...(i)
Let m2 be the mass of cube, C its specific heat and d Î¸ / dt , the rate of cooling
For cube
Ï T 4 S = m 2 C d Î¸ dt C ...(ii)
From Eqs.(i) and (ii)
d Î¸ / dt s d Î¸ / dt c = m 2 m 1
= a 3 Ï 4 / 3 Ï r 2 Ï
where a is the side of cube and r is the radius of sphere, Ï is the density.
Required ratio R s R c = 3 a 3 4 Ï r 3
But since S (surface area) is the same,
6 a 2 = 4 Ï r 2
or a 2 = 2 / 3 Ï r 2
â´ R s R c = 3 2 Ï r 2 / 3 3 / 2 4 Ï r 3 = 2 Ï 2 Ï 3 4 Ï
= 2 Ï 1 2 = Ï 6

Topic: Physics, Thermal Properties of Matter, NEET
A sphere and a hollow cylinder roll without slipping down two separate inclined planes and travel the same distance at the same time. If the angle of the plane down which the sphere rolls is 30°. The angle of the other plane is

Options:
A 60°
B 53°
C 37°
D 45°

The correct answer is:
45°

Solution:
For sphere, K12=25R2, θ1=30°,
For hollow cylinder, K22=R2, θ2=?
1+25sin⁡30°=1+1sin⁡θ2
sin⁡θ2=57=0.7143
θ=45°

Topic: Physics, Rotational Motion, NEET
A sphere encloses an electric dipole with charge $\pm 3 \times 10^{-6} \mathrm{C}$. What is the total electric flux across the sphere?

Options:
A $-3 \times 10^{-6} \mathrm{~N}-\mathrm{m}^2 / \mathrm{C}$
B zero
C $3 \times 10^6 \mathrm{~N}-\mathrm{m}^2 / \mathrm{C}$
D $6 \times 10^{-6} \mathrm{~N}-\mathrm{m}^2 / \mathrm{C}$

The correct answer is:
zero

Solution:
When a sphere encloses a charged dipole,

Here, $q= \pm 3 \times 10^{-6} \mathrm{C}$
Thus, according to Gauss's law, the net electric flux across the closed surface is equal to the net charge enclosed by it divided by $\varepsilon_0$, i.e.
$\phi_{\mathrm{E}}=\frac{\mathrm{q}_{\mathrm{in}}}{\varepsilon_0}=\frac{+3 \times 10^{-6}-3 \times 10^{-6}}{\varepsilon_0}=0$
Hence, electric flux across the sphere is zero.

Topic: Physics, Electrostatics, NEET, NEET 2019 (Odisha)
A sphere is suspended by a thread of length l. The minimum horizontal velocity which has to be imparted to the sphere for it to reach the height of suspension is

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

The correct answer is:
2gl

Solution:
Kinetic energy at A=12mu2
Potential energy at B=mgh
For conservation of energy, 12mu2=mgl⁡
u=2gl⁡

Topic: Physics, Laws of Motion, NEET
A sphere is suspended by a thread of length l. The minimum horizontal velocity which has to be imparted to the sphere for it to reach the height of suspension is

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

The correct answer is:
2gl

Solution:
Kinetic energy at A = 1 2 mu 2
Potential energy at B = mgh
For conservation of energy, 1 2 mu 2 = mg lâ¡
u = 2 g lâ¡

Topic: Physics, Laws of Motion, NEET
A sphere of density d, specific heat capacity c and radius r is hung by a thermally insulating thread in an enclosure which is kept at a lower temperature than the sphere. The temperature of the sphere starts to drop at a rate which depends upon the temperature difference between the sphere and the enclosure. If the temperature difference is ΔT and surrounding temperature is T0, then rate of fall in temperature will be [Given that ΔT<<T0]

Options:
A 12σT02ΔTrdc
B 12σT03ΔTrdc
C 12σT04ΔTrdc
D 12σΔTrdcT03

The correct answer is:
12σT03ΔTrdc

Solution:
From Stefan's law
-ΔQΔt=σAT4-T04
mc-ΔTΔt=σAT0+ΔT4-T04
-ΔTΔt=σAT04mc1+ΔTT04-1
-ΔTΔt=σ4πr2T04d43πr3c1+4ΔTT0-1
as ΔT<<T0
-ΔTΔt=12σT03ΔTrdc

Topic: Physics, Thermal Properties of Matter, NEET
A sphere of mass M and radius R is released from the top of an inclined plane of inclination Î¸. The minimum coefficient of friction between the plane and the sphere so that it rolls down the plane without sliding is given by

Options:
A μ=tan⁡θ
B μ=23tan⁡θ
C μ=25tan⁡θ
D μ=27tan⁡θ

The correct answer is:
μ=27tan⁡θ

Solution:
μ≥tan⁡θ1+R2K2
⇒ μmin=tan⁡θ1+R2K2

Topic: Physics, Rotational Motion, NEET
A sphere of mass m and radius r rolls on a horizontal plane without slipping with the speed u. Now, if it rolls up vertically, the maximum height it would attain will be

Options:
A 3u24g
B 5u22g
C 7u210g
D 11u29g

The correct answer is:
7u210g

Solution:
The rolling sphere has rotational as well as translational kinetic energy.
â´ Kinetic energy =12mu2+12IÏ2
=12mu2+1225mr2Ï2
=12mu2+15mu2=710mu2 (âµu=rÏ)
From conservation of energy
ie, mgh=710mu2
or h=7u210g

Topic: Physics, Rotational Motion, NEET
A sphere of mass m is held between two smooth inclined walls. For sin37Â°=35 . The normal reaction of the wall 2 is equal to:

Options:
A mg
B mgsin⁡74o
C mgcos⁡74o
D None of these

The correct answer is:
mg

Solution:
Writing equilibrium equation along the incline of surface 1:
â N2cosâ¡54o-mgcosâ¡54o=0
â N2=mg

Topic: Physics, Laws of Motion, NEET
A sphere of mass m is rolling without sliding along positive x-axis on a rough horizontal surface of coefficient of friction μ . It elastically collides with a wall and then returns. The correct statement of frictional force (f) acting on the sphere is

Options:
A f&ApplyFunction; = μ mg i &Hat; before collision and f&ApplyFunction; = μ mg i&ApplyFunction; &Hat; after collision
B fâ¡ = 0 before collision and fâ¡ = + μ mg iâ¡ ^ after collision
C fâ¡ < μ mg iâ¡ ^ before collision and fâ¡ = μ mg iâ¡ ^ just after collision
D f&ApplyFunction; = μ mg before collision and just after collision

The correct answer is:
fâ¡ = 0 before collision and fâ¡ = + μ mg iâ¡ ^ after collision

Solution:
Before collision, pure uniform rolling therefore f = 0 After collision, point P is having net velocity along –x axis, Therefore f = + μ mg
∴ fâ¡ = 0 before collision and fâ¡ = + μ mg iâ¡ ^ after collision

Topic: Physics, Laws of Motion, NEET
A sphere of mass m moves with a velocity 2v and collides inelastically with another identical sphere of mass. After collision the first mass moves with velocity v in a direction perpendicular to the initial direction of motion. Find the speed of the second sphere after collision.

Options:
A v
B v5
C 23v
D v3

The correct answer is:
v5

Solution:
Initially first mass moves along x direction, after collision first mass moves along y direction.
Let the velocity of second mass be v→′
⇒P→i=P→f
⇒m2vi^=mvj^+mv→′
⇒v→′=2vi^-vj^
⇒v′=2v2+v2
⇒v′=v5

Topic: Physics, Center of Mass Momentum and Collision, NEET
A sphere of mass m moving with velocity v collides head-on another sphere of same mass which is at rest. The ratio of final velocity of the second sphere to the initial velocity of the first sphere is (e is the coefficient of restitution and collision is inelastic)

Options:
A e-12
B e2
C e+12
D e

The correct answer is:
e+12

Solution:
mu+0=m u 1 +m u 2
Ï 1 + Ï 2 =Ï ....(i)
e= Ï 2 â Ï 1 Ïâ0
eÏ= Ï 2 â Ï 1 ..(ii)
Add equation (i) & equation (ii)
eÏ+Ï=2 Ï 2
Ï 2 = Ï( e+1 ) 2
Ï 2 Ï = e+1 2

Topic: Physics, Center of Mass Momentum and Collision, NEET
A spherical ball A of mass 4 kg, moving along a straight line strikes another spherical ball B of mass 1 kg at rest. After the collision, A and B move with velocities v1 m sâ1 and v2 m sâ1 respectively making angles of 30Â° and 60Â° with respect to the original direction of motion of A. The ratio v1v2 will be

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

The correct answer is:
3 4

Solution:
Applying the law of conservation of linear momentum along a direction perpendicular to the direction of motion (i.e. along the y-axis), we get.
0 + 0 = 4v1 sin 30° – v2 sin 60°
4v1sin 30° = v2 sin 60°
v1v2=sin60o4sin30o=34

Topic: Physics, Center of Mass Momentum and Collision, NEET
A spherical ball is dropped in a long column of a highly viscous liquid. The curve in the graph shown, which represents the speed of the ball v as a function of time t is

Options:
A B
B C
C D
D A

The correct answer is:
B

Solution:
Viscous force increases with increase in velocity and becomes constant after attaining terminal velocity.

Topic: Physics, Mechanical Properties of Fluids, NEET, NEET 2022 (Phase 1)
A spherical ball of radius 1 cm and density 4Ã103 kg m-3 is dropped gently in a large container containing a viscous liquid of density 2Ã103 kg m-3, and Î·=0.1 N s m-2. The distance moved by the ball in t=0.1 s after it attains terminal velocity is

Options:
A 45 m up
B 49 m up
C 23 m down
D 49 m down

The correct answer is:
49 m down

Solution:
Once the body acquires the terminal velocity it will have uniform motion.
vt=29r2ηρ-σg, down
vt=409 m s-1, down
Distance travelled =409×0.1=49 m

Topic: Physics, Mechanical Properties of Fluids, NEET
A spherical black body with a radius of 12 cm radiates 450 W power at 500 K. If the radius is halved and the temperature is doubled, then the power dissipated by the sphere is

Options:
A 3600 W
B 450 W
C 900 W
D 1800 W

The correct answer is:
1800 W

Solution:
Initially, the power dissipated by the sphere is

P=σ5004×4π122

Later, the power dissipated by the sphere is

P'=σ10004×4π62

P'P=σ10004×4π62σ5004×4π122⇒P'=1800 W

Topic: Physics, Thermal Properties of Matter, NEET
A spherical black body with a radius of 12 cm radiates 450 W power at 500 K. If the radius were halved and the temperature doubled, the power radiated in watt would be

Options:
A 225
B 450
C 1000
D 1800

The correct answer is:
1800

Solution:
By Stefan's law rate of heat radiation from a black body =ÏAT4
â Rate of power loss
râR2 T4
r1r2=R12 T14R22 T24
=4Ã116
450r2=14
r2=1800 watt

Topic: Physics, Thermal Properties of Matter, NEET, NEET 2017
A spherical black body with a radius of 12 cm radiates 450 W power at 500 K. If the radius were halved and the temperature doubled, the power radiated in watts would be

Options:
A 225
B 450
C 900
D 1800

The correct answer is:
1800

Solution:
The energy radiated per second by a black body is given by Stefan's law,
E t = Ï T 4 Ã A
where A is the surface area of the black body
E t = Ï T 4 Ã 4 Ï r 2
Since black body is a sphere, A = 4 Ï r 2
Case (i)
E t = 4 5 0 , T = 5 0 0 K , r = 0.12 m
â´ 4 5 0 = 4 Ï Ï 5 0 0 4 0.12 2 ...(i)
Case (ii)
E t = ? , T = 1 0 0 0 K , r = 0.06 m ...(ii)
Dividing Eq. (ii) by Eq. (i), we get
E / t 4 5 0 = 1 0 0 0 4 0.06 2 5 0 0 4 0.12 2 = 2 4 2 2 = 4
â Et=450Ã4=1800 W

Topic: Physics, Thermal Properties of Matter, NEET
A spherical conductor of radius 10 cm has a charge of \(3.2 \times 10^{-7} \mathrm{C}\) distributed uniformly. What is the magnitude of electric field at a point 15 cm from the centre of the sphere?
\(\left(\frac{1}{4 \pi \varepsilon_0}=9 \times 10^9 \mathrm{Nm}^2 \mathrm{C}^{-2}\right)\)

Options:
A 1.28×105 N C-1
B 1.28×106 N C-1
C 1.28×107 N C-1
D 1.28×104 N C-1

The correct answer is:
1.28×105 N C-1

Solution:
E=kQr2=9×109×3.2×10−715×10−22

E=1.28×105 N C-1

Topic: Physics, Electrostatics, NEET, NEET 2020 (Phase 1)
A spherical liquid drop is placed on a horizontal plane. A small disturbance causes the volume of the drop to oscillate. The time period of oscillation T of the liquid drop depends on the radius (r) of the drop, density Ï and surface tension S of the liquid. Which among the following will be a possible expression for the time period? where k is a dimensionless constant.

Options:
A kρrS
B kρ2rS
C kρr3S
D kρr3S2

The correct answer is:
kρr3S

Solution:
According to the question,
time period, TâraÏbSc
T=kraÏbSc ..... (i)
Thus, putting dimension, we get
[ T ]= [ L ] a [ M L â3 ] b [ M T â2 ] c
T=Mb+câLa-3bâT-2c
Equating the dimensions of both sides, we get
b+c=0, a-3b=0
and -2c=1
â´ c=-12
â´ b=12
and a=3b=32
Putting these value of a, b and c into Equation. (i), we get
T=kâr32Ï12âS-12
=kÏSr3

Topic: Physics, Units and Dimensions, NEET
A spherical planet has a mass $M_p$ and diameter $D_p$. A particle of mass $m$ falling freely near the surface of this planet will experience an acceleration due to gravity, equal to

Options:
A $4 G M_p / D_p^2$
B $G M_p m / D_p^2$
C $G M_p m / D_p^2$
D $4 G M_p m / D_p^2$

The correct answer is:
$4 G M_p / D_p^2$

Solution:
Force,
$F=\frac{G M_e m}{R^2}$
Here, $F=\frac{G M_P m}{\left(D_P / 2\right)^2}$
$\begin{aligned}
& F=\frac{4 G M_P m}{D_P^2} \\
& F=m a
\end{aligned}$
Acceleration due to gravity
$a=\frac{F}{m}=\frac{4 G M_P}{D_P^2}$

Topic: Physics, Gravitation, NEET, NEET 2012 (Screening)
A spherical shell of radius R is rolling down an incline of inclination θ without slipping. Find the minimum value of the coefficient of friction.

Options:
A 27tan⁡θ
B 25tan⁡θ
C 23tan⁡θ
D none of these

The correct answer is:
25tan⁡θ

Solution:
Ffr=Iα
⇒ μmgcosθ=23ma
⇒ μmgcosθ=23mgsinθ1+23
⇒ μ=25tanθ

Topic: Physics, Rotational Motion, NEET
A spherical uniform planet is rotating about its axis. The velocity of a point on its equator is V. Due to the rotation of the planet about its axis the acceleration due to gravity g at equator is 12 of g at poles. The escape velocity of a particle on the planet in terms of V from the pole of the planet is

Options:
A Ve=2V
B Ve=V
C Ve=V/2
D Ve=3V

The correct answer is:
Ve=2V

Solution:
g e = g p - R Ï 2 â g 2 = g - R Ï 2
R Ï 2 = g 2 â R 2 Ï 2 = g R 2 â V 2 = g R 2 ...(1)
The escape velocity, V e = 2 g R ...(2)
From (1) and (2)
V e = 2 Ã 2 V 2 â V e = 2 V

Topic: Physics, Gravitation, NEET
A spherically symmetric gravitational system of particles has a mass density ρ=ρ0 for r≤R and ρ=0 for r>R , where ρ0 is a constant. A test mass can undergo circular motion under the influence of the gravitational field of the particles. Which figure represents its speed v as a function of distance r0<r<∞ from the center of the system?

Options:
A
B
C
D

The correct answer is:

Solution:
If M is the total mass of the system of particles, then similar to a satellite revolving around the earth, the orbital velocity of the test mass is given by
v=GMr
For râ¤R, v=Gâ43Ïr3Ï0r
i.e. vâr2 or vâr
i.e. v increases linearly with r up to r=R
Thus (1) and (4) are wrong but (2) and (3) satisfy this.
â´ââFor r>R, v= GM r âor vâ 1 r
Thus is correctly shown in figure (2). Figure (3) is wrong because for r>R,âv is shown as constant.

Topic: Physics, Gravitation, NEET
A spring balance A and a beam balance B are used to weigh the same object at a place on the equator and a place on the pole of the earth. Then

Options:
A Readings of A and B will be different at both the places
B Readings of A will be different but those of B will be same at both the places
C Readings of A will be same but those of B will be different at both the places
D Readings of both A and B are same at both the places

The correct answer is:
Readings of A will be different but those of B will be same at both the places

Solution:
Spring balance measures spring force hence readings of A will be different as g at equator is smaller than g at pole.
Beam balance measures torque hence B will be affected equally due to variation of g at two places.
So readings of B will be same.

Topic: Physics, Gravitation, NEET
A spring balance is attached to the ceiling of a lift. A man hangs his bag on the spring and the spring reads 49 N, when the lift is stationary. If the lift moves downward with an acceleration of 5 m s-2, the reading of the spring balance will be

Options:
A 49 N
B 24 N
C 74 N
D 15 N

The correct answer is:
24 N

Solution:
When the lift is stationary w=mg
⇒ 49=m×9.8
m=5 kg
When the lift is moving downward with an acceleration
R=m9.8-a=59.8-5=24 N

Topic: Physics, Laws of Motion, NEET
A spring having with a spring constant 1200 N m–1 is mounted on a horizontal table as shown in Fig. 13.19. A mass of 3 kg is attached to the free end of the spring. Let us take the position of mass when the spring is unstretched as x = 0, and the direction from left to right as the positive direction of x-axis.Give x as a function of time t for the oscillating mass if at the moment we start the stopwatch (t = 0), the mass is at the mean position,

Options:
A x=sin20t
B x=2sin40t
C x=2sin20t
D x=sin40t

The correct answer is:
x=2sin20t

Solution:
Correct Option is : (C)
x=2sin20t

Topic: Chemistry, Oscillations, NEET
A spring having with a spring constant 1200 N m–1 is mounted on a horizontal table as shown in Fig. 13.19. A mass of 3 kg is attached to the free end of the spring. Let us take the position of mass when the spring is unstretched as x = 0, and the direction from left to right as the positive direction of x-axis.Give x as a function of time t for the oscillating mass if at the moment we start the stopwatch (t = 0), the mass is at the maximum stretched position

Options:
A x=sin20t
B x=cos40t
C x=2cos20t
D x=sin40t

The correct answer is:
x=2cos20t

Solution:
Correct Option is : (C)
x=2cos20t

Topic: Chemistry, Oscillations, NEET
A spring having with a spring constant 1200 N m–1 is mounted on a horizontal table as shown in Fig. A mass of 3 kg is attached to the free end of the spring. The mass is then pulled sideways to a distance of 2.0 cm and released. Determine the frequency of oscillations

Options:
A 1.6 per second
B 3.2 per second
C 4.8 per second
D 6.4 per second

The correct answer is:
3.2 per second

Solution:
Correct Option is : (B)
3.2 per second

Topic: Chemistry, Oscillations, NEET
A spring is stretched by 5 cm by a force 10 N. The time period of the oscillations when a mass of 2 kg is suspended by it is :

Options:
A 0.0628 s
B 6.28 s
C 0.628 s
D 3.14 s

The correct answer is:
0.628 s

Solution:
Spring force, F=kx

Spring constant, k=Fx=100.05=200 Nm

Time period, T=2πmk

T=2π2200=0.2π=0.628 s

Topic: Physics, Oscillations, NEET, NEET 2021
A spring is stretched by applying a load to its free end. The strain produced in the spring is

Options:
A Volumetric
B Shear
C Longitudinal and shear
D Longitudinal

The correct answer is:
Longitudinal and shear

Solution:
When the load is applied at the lower end of the spring, the shape and the length of the spring changes. The change in length corresponds to longitudinal strain and change in shape corresponds to shearing strain.

Topic: Physics, Mechanical Properties of Solids, NEET
A spring of force constant 800 N m-1 has an extension of 5 cm. The work done in extending it from 5 cm to 15 cm is

Options:
A 16 J

B 8 J

C 32 J

D 24 J

The correct answer is:
8 J

Solution:
W⁡ = 1 2 k x 2 2 - x 1 2
W⁡ = 1 2 × 8 0 0 1 5 × 1 0 - 2 2 - 5 × 1 0 - 2 2
W= 400 × 200 × 10–4 = 8 J

Topic: Physics, Work Power Energy, NEET
A spring of force constant k extends by a length x on loading. If T is the tension in the spring then the energy stored in the spring is

Options:
A T22k
B T22k2
C 2kT2
D 2T2k

The correct answer is:
T22k

Solution:
Energy stored in a spring is given by
U=12kx2 ..........(1)
Where k is the spring constant
Now, applying Hooke's law, we have
Tension, T=kx (in magnitude)
⇒ extension x=Tk
Substituting the value in equation (1), we get
U=12T2k .

Topic: Physics, Work Power Energy, NEET
A spring of force constant k is cut into lengths of ratio 1 : 2 : 3. They are connected in series and the new force constant is k′. If now they are connected in parallel and force constant is k′′, then k' : k'' is

Options:
A 1 : 6
B 1 : 9
C 1 : 11
D 1 : 14

The correct answer is:
1 : 11

Solution:
Spring constant ∝1length
k∝1l,
Spring constant for the length ratios are k1=6k
k2=3k
k3=2k
In series connection of springs,
1k′=16k+13k+12k
1k′=66k
k′=k.
In parallel connection of springs,
k'′=6k+3k+2k
k′′=11k
k′k′′=111
i.e., k′ :k′′=1 :11

Topic: Physics, Laws of Motion, NEET, NEET 2017