datavorous/entrance-exam-dataset huggingface dataset:
A siren emitting a sound of frequency 800 Hz moves away from an observer towards a cliff at a speed of 15 m s-1. Then, the frequency of sound that the observer hears in the echo reflected from the cliff is
(Take velocity of sound in air =330 m s-1)

Options:
A 765  Hz
B 800  Hz
C 838ââHz
D 885  Hz

The correct answer is:
838  Hz

Solution:
Given
fÎ¿=800 Hz
Vsource=15 m sec-1
f a = f 0 ( V+ V 0 Vâ V s )â
âfa=330330-15Ã800=330315Ã800
fa=838 Hz

Topic: Physics, Waves and Sound, NEET
A skater of mass m standing on ice throws a stone of mass M= 2 m with a velocity of v= 5 m/s in horizontal direction. If the coefficient of friction between the skater and the ice is 0.5 , find the distance over which the skater will move. (take g=10 ms-2)

Options:
A 20 m
B 15 m
C 10 m
D 5 m

The correct answer is:
10 m

Solution:
By momentum conservation, 2mÃ5=mÃV'
So, V'=10m/sec
Let skater stops after travelling distance s.
So, 02=102-2ÃÎ¼ÃgÃs
s=10 m

Topic: Physics, Center of Mass Momentum and Collision, NEET
A slab consists of portions of different materials of the same thickness and length connected in series. The materials have conductivities K1 and K2. The equivalent thermal conductivity of the slab is

Options:
A K1+K2
B K1+K2
C 2K1 K2K1+K2
D K1K2K1+K2

The correct answer is:
2K1 K2K1+K2

Solution:
As two portions of the slab are connected in series, therefore
K=∑xi∑xiK1=d+ddK1+dK2
=2dd1K1+1K2=2K1 K2K1+K2

Topic: Physics, Thermal Properties of Matter, NEET
A slab consists of two parallel layers of copper and brass of the same thickness and having thermal conductivities in the ratio 1:4. If the free face of brass is at 100 âC and that of copper at 0 âC, the temperature of the interface is

Options:
A 80 °C
B 20 °C
C 60 °C
D 40 °C

The correct answer is:
80 °C

Solution:
We know that dQdt=KAθ2-θ1d
As slabs are in series, therefore
⇒K1θ1–θC=K2θC–θ2
⇒4K100–θ=Kθ–0
⇒ 400-4θ=θ
⇒ θ=80 °C

Topic: Physics, Thermal Properties of Matter, NEET
A slab of a glass of refractive index 1.5 and thickness 3 cm is placed with the faces perpendicular to the principal axes of a concave mirror. If the radius of curvature of the mirror is 10 cm, the distance at which an object must be placed from the mirror so that image coincides with the object is

Options:
A 9 cm

B 10 cm

C 11 cm

D 12 cm

The correct answer is:
11 cm

Solution:
μ = 1.5 = 3 / 2
Δ x = t 1 - 1 μ = 3 1 - 1 3 / 2 = 3 1 3
Δ x = 1 cm
object distance = 10 + 1 = 11 cm

Topic: Physics, Ray Optics, NEET
A slide under a microscope shows the following features:

(i) Unicellularity (ii) Well defined nucleus

(iii) Biflagellate-one flagellum lying longitudinally and the other transversely

What would you identify it as?

Options:
A Protozoan
B Bacterium
C Euglenoid
D Dinoflagellate

The correct answer is:
Dinoflagellate

Solution:
Dinoflagellates are unicellular protists, also known as marine planktons which make the sea appear like red tides due to its rapid multiplication. They have characteristics of both plants and animals and are organisms bearing two dissimilar flagella, one lying longitudinally and other transversely. They are considered to be algae but show a presence of well- defined nucleus which makes its placement controversial.

Topic: Biology, Biological Classification, NEET
A slit of width a is illuminated by white light. The first minimum for red light λ=6500 Å will fall at θ=30o , when a is

Options:
A 3250ââ«
B 6.5Ã10-4Â cm
C 1.3 Î¼m
D 2.6Ã10-4Â cm

The correct answer is:
1.3 Î¼m

Solution:
sinâ¡Î¸=Î»a
âsinâ¡30o=6500a AÂ°
â´ 12=6500aâ«
â´a=2Ã6500 â«=2Ã6500Ã10-10 m
â´a=1.3Ã10-6 m = 1.3 Î¼m

Topic: Physics, Wave Optics, NEET
A slit of width a is illuminated by white light. The first minimum for red light Î»=6500 â« will fall at Î¸=30o, when a is

Options:
A 3250 Å
B 6.5×10-4 cm
C 1.3 μm
D 2.6×10-4 cm

The correct answer is:
1.3 μm

Solution:
sin⁡θ=λa
⇒sin⁡30o=6500aÅ
∴ 12=6500aÅ
⇒a=2×6500 Å
⇒a=2×6500×10-10 m
∴ ⇒ a=1.3×10-6 m

Topic: Physics, Wave Optics, NEET
A slit of width a is illuminated by white light. The first minimum for red light Î»=6500 â« will fall at Î¸=30Â°, when a is

Options:
A 3250 â«
B 6.5Ã10-4Â cm
C 1.3 Î¼m
D 2.6Ã10-4Â cm

The correct answer is:
1.3 Î¼m

Solution:
sinâ¡Î¸=Î»a
â sinâ¡30Â°=6500aâ«
â´ 12=6500aâ«
â´ a=2Ã 6500 â«=2Ã6500Ã10-10 m
â´ a=1.3Ã10-6 m

Topic: Physics, Wave Optics, NEET
A small aquatic plant was put in each of the petri dishes X and Y containing different culture solutions. After six weeks, the plant in dish X had the same number of leaves as earlier, and all leaves were small and yellowish. The plant in dish Y had more leaves of normal size and dark green colour. Which of the following set of elements would be missing in the culture of dish X?

Options:
A Magnesium, Phosphorus, Nitrogen
B Potassium, Sulphur, Calcium
C Iron, Manganese, Boron
D Copper, Zinc, Chlorine

The correct answer is:
Magnesium, Phosphorus, Nitrogen

Solution:
The types of deficiency symptoms shown in plants include chlorosis, necrosis, stunted plant growth, premature fall of leaves and buds and inhibition of cell division. Chlorosis is the loss of chlorophyll leading to yellowing in leaves. This symptom is caused by the deficiency of elements N, K, Mg, S, Fe, Mn, Zn and Mo.

Topic: Biology, Mineral Nutrition, NEET
A small ball B of mass m is suspended with a light inelastic string of length L from a block A of same mass m which can move on a smooth horizontal surface as shown in the figure. The ball is displaced by the angle Î¸ from equilibrium position & then released.

Maximum velocity of block during subsequent motion of the system after release of ball is

Options:
A gL1-cosθ1/2
B 2gL1-cosθ1/2
C gLcosθ1/2
D Insufficient information to decide

The correct answer is:
gL1-cosθ1/2

Solution:
KE gain by system = mg L 1 - cos θ
half of that will be shared by block
12mv2=12mg L1-cosθ ⇒ v=gL1-cosθ

Topic: Physics, Center of Mass Momentum and Collision, NEET
A small ball B of mass m is suspended with light inelastic string of length L from a block A of same mass m which can move on smooth horizontal surface as shown in the figure. The ball is displaced by angle θ from equilibrium position & then released.

The displacement of block when ball reaches the equilibrium position is

Options:
A L sin θ 2
B L sin θ
C L
D None of these

The correct answer is:
L sin θ 2

Solution:
Δ x cm = 0
0 = - L sin θ - x m + mx
L sin θ - x = x
2 x = L sin θ
x = L sin θ / 2

Topic: Physics, Center of Mass Momentum and Collision, NEET
A small ball of density Ï is immersed in a liquid of density Ï >Ï to a depth h and released. The height above the surface of water up to which the ball will jump is

Options:
A σρ-1h
B ρσ-1h
C ρσ+1h
D σρ+1h

The correct answer is:
σρ-1h

Solution:
Let V be the volume of the ball.
New upward force =Vσg-Vρg
Net upward acceleration, a=Vσg-VρgVρ=σ-ρgρ
Velocity at the surface =2 (σ-ρ)ghρ
If ha is the height in air to which the ball rises, then
0-2 σ-ρghρ=2-gha
∴ ha= (σ-ρ)ghgρ=σρ-1h

Topic: Physics, Mechanical Properties of Fluids, NEET
A small ball of mass m is released from rest from the position shown. All contact surfaces are smooth. The speed of the ball when it reaches its lowest position is

Options:
A 2gR3
B 4gR3
C gR
D 2gR

The correct answer is:
4gR3

Solution:
Let the speed of ball be v and speed of the wedge be v2m

using conservation of momentum

0=mv+2mv2m

v2m=-v2

Now, using work-energy theorem on the system,

mgR=12mv2+12(2m)×v22
⇒ v=4gR3

Topic: Physics, Center of Mass Momentum and Collision, NEET
A small ball of mass m is released from rest from the position shown. All contact surfaces are smooth. The speed of the ball when it reaches its lowest position is

Options:
A 2 g R3
B 4 g R3
C g R
D 2 g R

The correct answer is:
4 g R3

Solution:
Let the speed of ball be v and speed of the wedge be v2m

using conservation of momentum

0=mv+2mv2m

v2m=-v2

Now, using work-energy theorem on the system,

mgR=12mv2+12(2m)×v22
⇒ v=4gR3

Topic: Physics, Center of Mass Momentum and Collision, NEET
A small bead of mass M slides on a smooth wire that is bent in a circle of radius R. It is released at the top of the circular part of the wire (point A in the figure) with a negligibly small velocity. Find the height H where the bead will reverse direction.

Options:
A R
B 2R
C 3R2
D 5R2

The correct answer is:
2R

Solution:
Apply conservation of energy at two different positions of the object.
According to energy conservation between points A and B
mg( 2R )+ 1 2 m ( 0 ) 2 =mgH+ 1 2 m ( 0 ) 2
â mg2R=mgH
â 2R=H
â H=2R

Topic: Physics, Work Power Energy, NEET
A small block is shot into each of the four tracks as shown below. Each of the tracks risks to the same height. The speed with which the block enters the track is the same in all cases. At the highest point of the track, the normal reaction is maximum in

Options:
A
B
C
D

The correct answer is:

Solution:
At the highest point let v' is the speed
mv'2R=mg+N â N=mv'2R-mg â R is min for first case so Nmax

Topic: Physics, Laws of Motion, NEET
A small block of mass M moves with velocity 5 m s-1 towards an another block of same mass M placed at a distance of 2 m on a rough horizontal surface. Coefficient of friction between the block and ground is 0.25. The collision between the two blocks is elastic, the separation between the blocks, when both of them come to rest, is (g=10 m s-2)

Options:
A 3 m

B 4 m

C 2 m

D 1.5 m

The correct answer is:
3 m

Solution:
Retardation due to friction
a=μg=0.25(10)
=2.5 m s-2
Collision is elastic, i.e. after collision first block comes to rest and the second block acquires the velocity of the first block. Or we can understand it in this manner that second block is permanently at rest while only the first block moves. Distance travelled by it will be
s=v22a=522(2.5)=5 m
∴ The final separation will be (5-2) =3 m

Topic: Physics, Center of Mass Momentum and Collision, NEET
A small block of super dense material has mass 2Ã1024 kg. It is at a height h<<R . It falls towards the earth. Find its speed when it is at a height h2

Options:
A 2gh3
B 3gh4
C 3gh5
D gh2

The correct answer is:
3gh4

Solution:
m=2Ã1024 kg,

Let the mass of the earth be M, then m=M3
From momentum conservation, MV1=M3V2
âV1=V23

From energy conservation
12MV12+12M3V22=M3gh2
â´ V2=3gh4

Topic: Physics, Gravitation, NEET
A small block of superdense material has mass 2×1024 kg. It is at a height h<<R. It falls towards the earth. Find its speed when it is at a height h2.

Options:
A 2gh3
B 3gh4
C 3gh5
D gh2

The correct answer is:
3gh4

Solution:
Mass of earth is, M=6×1024 kg
Mass of superdense material m=2×1024 kg=M3
From momentum conservation, MV1=M3V2
From energy conservation V1=V23
12MV12+12M3V22=GMM3h2+R-GMM3h+R
∴ v2=3gh4

Topic: Physics, Gravitation, NEET
A small block slides down on a smooth inclined plane, starting from rest at time t=0. Let Sn be the distance travelled by the block in the interval t=n-1 to t=n. Then, the ratio snsn+1 is:

Options:
A 2n-12n
B 2n+12n-1
C 2n-12n+1
D 2n2n-1

The correct answer is:
2n-12n+1

Solution:
As we know that the displacement covered by particle in nthsecond is given by,

∵Snth=u+2n-12a

but u=0

∴sn=2n-12a …1

sn+1=2(n+1)-12a …(2)

Now,
equation (1) divide by equation (2)

snsn+1=2n-12n+1

Topic: Physics, Motion In One Dimension, NEET, NEET 2021
A small body is dropped from a rising balloon. A person A stands on the ground, while another person B is on the balloon. Choose the correct statement: Immediately, after the body is released.

Options:
A A and B, both feel that the body is coming (going) down.
B A and B, both feel that the body is going up
C A feels that the body is coming down, while B feels that the body is going up
D A feels that the body is going up, while B feels that the body is going down.

The correct answer is:
A feels that the body is going up, while B feels that the body is going down.

Solution:
When the body is dropped from the balloon, it also acquires the upward velocity of the balloon. So with respect to a person on the ground, the ball appears to be going up. But a person in the balloon is also going up. So with respect to him, the velocity of the body will be zero and he will then see the body to be coming down.

Topic: Physics, Motion In One Dimension, NEET
a. Small, circular non-histone DNA also present in mitochondria and chloroplast.
b. The extension of plasma membranes mesosomes into the cell can be present in the form of tubules, lamellae and vesicles.
c. The quasi fluid nature of lipid enables flip flop movement of proteins at a specific place.
d. Components of the endomembrane system associated with secretion, packaging of material and capable of digestion of several macro organic compounds.
e. Palade particles forming a peptide bond between amino acids is a membrane-bounded organelle.

From given statements identify the incorrect ones.

Options:
A aÂ andÂ c
B bÂ andÂ d
C b,Â c,Â dÂ andÂ e
D c and e

The correct answer is:
c and e

Solution:
The Palade particles that form a peptide bond between amino acids are Ribosomes and are non-membranous organelle. Mitochondria and plastid are semi-autonomous organelles as they have circular DNA molecules similar to bacteria and partially depend on the nucleus. Mesosomes are specialized membraneous structures formed by an extension of the plasma membrane into the cell in the form of vesicles, tubules, and lamellae. Fluid mosaic model is proposed by Singer and Nicolson in 1972, and according to it, the lipids in membrane show the quasi-fluid nature that enables the lateral movement of proteins.

Topic: Biology, Cell - The Unit of Life, NEET
A small coin is placed at a distance r from the centre of a gramophone record. The rotational speed of the record is gradually increased. If the coefficient of friction between the coin and the record is Î¼, the minimum angular frequency of the record, for which the coin will fly off, is given by

Options:
A 2μgr
B μg2r
C μgr
D 2μgr

The correct answer is:
μgr

Solution:
The coin performs circular motion in a circle of radius r. It will fly off when centrifugal force is greater than the centripetal force provided by friction

Hence, at the moment when the coin is about to fly off,

Î¼N=mÏ2r

or Î¼mg=mÏ2r

Ï=Î¼gr

Topic: Physics, Laws of Motion, NEET
A small coin is resting on the bottom of a beaker filled with liquid. A ray of light from the coin travels upto the surface of the liquid and moves along its surface. How fast is the light travelling in the liquid?

Options:
A $2.4 \times 10^8 \mathrm{~m} / \mathrm{s}$
B $3.0 \times 10^8 \mathrm{~m} / \mathrm{s}$
C $1.2 \times 10^8 \mathrm{~m} / \mathrm{s}$
D $1.8 \times 10^8 \mathrm{~m} / \mathrm{s}$

The correct answer is:
$1.8 \times 10^8 \mathrm{~m} / \mathrm{s}$

Solution:
$\because \quad \frac{1}{\mu}=\sin \theta_c=\frac{3}{5}$
$\begin{aligned}
& \Rightarrow \quad \mu=\frac{5}{3} \\
& \text { And } v=\frac{c}{\mu}=\frac{3 \times 10^8}{5 / 3} \\
& =\frac{9}{5} \times 10^8=1.8 \times 10^8 \mathrm{~ms}^{-1}
\end{aligned}$

Topic: Physics, Ray Optics, NEET, NEET 2007
A small coin of mass 40 g is placed on the horizontal surface of a rotating disc. The disc starts from rest and is given a constant angular acceleration Î±=2 rad s-2. The coefficient of static friction between the coin and the disc is Î¼ s = 3 / 4 and the coefficient of kinetic friction is Î¼ k = 0.5. The coin is placed at a distance r=1 m from the centre of the disc. The magnitude of the resultant force on the coin exerted by the disc just before it starts slipping on the disc is :

Options:
A 0.2 N
B 0.3 N
C 0.4 N
D 0.5 N

The correct answer is:
0.5 N

Solution:
The friction force on coin just before coin is to slip will be : f = Î¼ s mg
Normal reaction on the coin ; N = mg
The resultant reaction by disk to the coin is
= N 2 + f 2 = mg 2 + Î¼ s mg 2 = mg 1 + Î¼ s 2
= 4 0 Ã 1 0 - 3 Ã 1 0 Ã 1 + 9 1 6 = 0.5 N

Topic: Physics, Laws of Motion, NEET
A small conducting sphere of radius r is lying concentrically inside a bigger hollow conducting sphere of radius R. The bigger and smaller spheres are insulated from each other and are charged to Q and qQ>q respectively. The potential difference between the sphere will be

Options:
A 14πε0qr-qR
B 14πε0qr+qR
C 14πε0qr-QR
D 14πε0qr+QR

The correct answer is:
14πε0qr-qR

Solution:
First we calculate total potential on sphere A and B
VA=kqr+kQR
VB=kqR+kQR
â´ VA-VB=kqr-kqR

Topic: Physics, Electrostatics, NEET
A small hole of area cross-section 2 mm2 is present near the bottom of a fully filled open tank of height 2 m. Taking g=10 m s-2, the rate of flow of water through the open hole would be nearly

Options:
A 12.6×10-6 m3 s-1
B 8.9×10-6 m3 s-1
C 2.23×10-6 m3 s-1
D 6.4×10-6 m3 s-1

The correct answer is:
12.6×10-6 m3 s-1

Solution:
Velocity of efflux v=2gh
v=2×10×2
=2×10
Rate of flow = area ×v
=2×10-6×2×10
=4×3.14×10-6
≅12.6×10-6

Topic: Physics, Mechanical Properties of Fluids, NEET, NEET 2019 (All India)
A small magnet of dipole moment M is kept on the arm of a deflection magnetometer set in tanâ¡A position at a distance of 0.2 m. If the deflection is 60Â°, find the value of M BH=0.4Ã10-4 T

Options:
A 2.77 A m2
B 8 A m2
C 0.2 A m2
D 0.9 A m2

The correct answer is:
2.77 A m2

Solution:
∵ B=BHtan⁡θ
Or μ04π .2Mr3=0.4×10-4tan⁡60°
∴ μ0M2π 0.23=0.4×10-4tan⁡60°
M=0.4×4π×0.23×10-4×32× μ0
=0.00277×10-410-7
∴ M=2.77 A m2

Topic: Physics, Magnetic Properties of Matter, NEET
A small mass executes SHM around a point O with amplitude A & time period T. Its displacement from O at time T/8 after passing through O is

Options:
A A8
B A22
C A2
D A2

The correct answer is:
A2

Solution:
Equation of S.H.M.
x=Asin⁡ωt ; so when t=T/8
x=Asin⁡ω×T/8=Asin⁡2πT×T8
=Asin⁡π4=A2

Topic: Physics, Oscillations, NEET
A small, metal wire loop is dragged across the gap between the poles of a magnet in 0.4 s. If the change in magnetic flux for the wire is 8Ã10-4 Wb, then the emf induced in the wire is

Options:
A 8Ã10-3Â V
B 6Ã10-3Â V
C 4Ã10-3Â V
D 2Ã10-3 V

The correct answer is:
2Ã10-3 V

Solution:
e=dϕdt
= 8×10-40.4=2×10-3 V

Topic: Physics, Electromagnetic Induction, NEET
A small object of mass of 100 g moves in a circular path. At a given instant velocity of the object is 10i^ m s-1 and acceleration is (20i^+10j^) m s-2. At this instant of time, the rate of change of kinetic energy of the object is

Options:
A 200 kg m2s-2
B 300 kg m2 s-2
C 10000 kg m2 s-2
D 20 kg m2 s-2

The correct answer is:
20 kg m2 s-2

Solution:
Given, the mass of the object m=100 g
=100Ã10-3 kg
Velocity of object v=10i^ m s-1
Acceleration of object a=(20i^+10j^) m s-2
We know that,
d(KE)dt=F.v=ma.v [âµK.E.=12mv2]
=(100Ã10-3)(20i^+10j^).(10i^)
=100Ã10-3Ã200
=100Ã2001000=20 kg m2 s-3

Topic: Physics, Work Power Energy, NEET
A small object of uniform density rolls up a curved surface with an initial velocity $v^{\prime}$. It reaches up to a maximum height of $\frac{3 v^2}{4 g}$ with respect to the initial position. The object is

Options:
A ring
B solid sphere
C hollow sphere
D disc

The correct answer is:
disc

Solution:
As $v=\sqrt{\frac{2 g h}{1+\frac{k^2}{r^2}}}$
Given $h=\frac{3 v^2}{4 g}$
$v^2=\frac{2 g h}{1+\frac{k^2}{r^2}}=\frac{2 g 3 v^2}{4 g\left(1+\frac{k^2}{r^2}\right)}=\frac{6 g v^2}{4 g\left(1+\frac{u^2}{v^2}\right)}$
$1=\frac{3}{2\left(1+\frac{k^2}{v^2}\right)}$
or $1+\frac{k^2}{r^2}=\frac{3}{2}$ or $\frac{k^2}{r^2}=\frac{3}{2}-1=\frac{1}{2}$
$k^2=\frac{1}{2} r^2$ (Equation of disc)
Hence, the object is disc.

Topic: Physics, Rotational Motion, NEET, NEET 2013 (All India)
A small planet is revolving around a very massive star in a circular orbit of radius r with a period of revolution T. If the gravitational force between the planet and the star is proportional to r-5/2, then T will be proportional to

Options:
A r3/2
B r5/3
C r7/4
D r3

The correct answer is:
r7/4

Solution:
Gravitational force provides the necessary centripetal force required by the given planet for circular motion. So
mv 2 r ∝ r - 5 / 2
⇒ mv 2 r = kr - 5 / 2 , k = constant
∴ v = kr - 3 / 2 m
T = 2 π r v = 2 π r × m k r - 3 / 2
i.e., T ∝ r 7 / 4 .

Topic: Physics, Gravitation, NEET
A small protein that is attached to the outer surface of the inner membrane of mitochondria and acts as a mobile carrier for electron transfer in oxidative phosphorylation is

Options:
A Ubiquinone that receives electron from complex I only
B Cytochrome c1 between complex III and complex IV
C Cytochrome c between complex III and complex IV
D Plastocyanin

The correct answer is:
Cytochrome c between complex III and complex IV

Solution:
The metabolic pathway through which the electron passes from one carrier to another, is called the electron transport system (ETS). This pathway is present in the inner mitochondrial membrane. Electrons from NADH (produced in the mitochondria matrix) are oxidized by an NADH dehydrogenase (Complex I) and are transferred to ubiquinone (Complex II). The reduced ubiquinone (ubiquinol) is then oxidised with the transfer of electrons to cytochrome c via cytochrome bc1 complex (complex III). Cytochrome c is a small protein attached to the outer surface of the inner membrane and acts as a mobile carrier for transfer of electrons between complex III and IV. Complex IV refers to cytochrome c oxidase complex containing cytochromes a and a3, with two copper centres. When the electrons pass from one carrier to another via complex I to IV in the electron transport chain, they are coupled to ATP synthase (complex V). This coupling is necessary for the production of ATP from ADP and inorganic phosphate.

Topic: Biology, Respiration in Plants, NEET
A small signal voltage Vt=V0sin⁡ωt is applied across an ideal capacitor C:

Options:
A Current I(t), lags voltage V(t) by 90o
B Over a full cycle the capacitor C does not consume any energy from the voltage source
C Current I(t) is in phase with voltage V(t).
D Current I(t) leads voltage V(t) by 180o

The correct answer is:
Over a full cycle the capacitor C does not consume any energy from the voltage source

Solution:
In capacitor current leads the voltage by ϕ=π2 phase. Average power dissipated in AC circuit is given by

Pav=VrmsIrmscosϕ

In capacitor ϕ=π2 so cosπ2 is zero.

Topic: Physics, Alternating Current, NEET, NEET 2016 (Phase 1)
A small sphere is given vertical velocity of magnitude v0=5m/s and it swings in a vertical plane about the end of the massless string. If maximum tension string can withstand is equal to twice of the weight of sphere, then angle Î¸ with vertical at which string will break is

Options:
A cos-1⁡23
B cos-114
C 60°
D 30°

The correct answer is:
cos-114

Solution:
FBD of body at the position shown in diagram
TâmgcosÎ¸=mv2l .....(i)
Applying work energy theorem
âmglcosâ¡Î¸=12mv2-12mv02 ....(ii)
T=2mg ......(iii)
By solving â´ cosâ¡Î¸=14

Topic: Physics, Laws of Motion, NEET
A small sphere of mass m is suspended by a thread of length l . It is raised up to the height of suspension with thread fully stretched and released. Then the maximum tension in the thread will be

Options:
A mg
B 2mg
C 3mg
D 6mg

The correct answer is:
3mg

Solution:
The tension in string is maximum at the lowest position of the ball.
∴Tmax-mg=mv2l
∴Tmax=mg+mv2l ...(i)
The loss in potential energy = gain in kinetic energy
Or mgl=12mv2
∴v=2gl ...(ii)
∵mv2l=ml2gl=2mg ...(iii)
From equation (i), (ii) and (iii)
Tmax=mg+2mg=3mg

Topic: Physics, Work Power Energy, NEET
A small sphere of radius r falls from rest in a viscous liquid. As a result, heat is produced due to viscous force. The rate of production of heat when the sphere attains terminal velocity, is proportional to

Options:
A r5
B r2
C r3
D r4

The correct answer is:
r5

Solution:
Rate of heat produced = power of viscous force
terminal velocity V T = 2 r 2 ( ρ−σ )g 9η
⇒ V T ∝ r 2
dQ dt =6πηr  V T × V T
⇒ dQ dt ∝ r 5

Topic: Physics, Mechanical Properties of Fluids, NEET, NEET 2018
A small square loop of wire of side l is placed inside a large square loop of side L Lâ«l. If the loops are coplanar and their centres coincide, the mutual induction of the system is directly proportional to

Options:
A lL
B l2L
C lL2
D l2L2

The correct answer is:
l2L

Solution:
M=Ïi=BAi
M=4ÃÎ¼oi4ÏÎ¹iÃ22l2i
=4ÃBone sideiâl2L as Aâl2 and Bâ1L

Topic: Physics, Electromagnetic Induction, NEET