datavorous/entrance-exam-dataset huggingface dataset:
A small bar magnet experiences a torque of $0.016 \mathrm{Nm}$ when placed with its axis at $30^{\circ}$ with an external field of $0.04 \mathrm{~T}$. If the bar magnet is replaced by a solenoid of cross-sectional area of $1 \mathrm{~cm}^2$ and 1000 turns but having the same magnetic moment as that of bar magnet, then the current flowing through the solenoid is

Options:
A $2 \mathrm{~A}$
B $4 \mathrm{~A}$
C $6 \mathrm{~A}$
D $8 \mathrm{~A}$

The correct answer is:
$8 \mathrm{~A}$

Solution:
Given,
torque on a bar magnet, $\tau=0.016 \mathrm{Nm}$, $\theta=30^{\circ}$ and magnetic field, $B=0.04 \mathrm{~T}$
Since, $\tau=m \mathrm{~B} \sin \theta$
$\begin{array}{l}
\therefore 0.016=m \times 0.04 \times \sin 30^{\circ} \\
\Rightarrow \quad m=\frac{2 \times 0.016}{0.04} \\
m=0.8 \mathrm{Am}^2
\end{array}$
For solenoid, area of cross-section,
$A=1 \mathrm{~cm}^2=10^{-4} \mathrm{~m}^2$
Number of turns in solenoid,
$N=1000$
$\therefore$ magnetic moment of solenoid,
$\begin{aligned}
m & =N I A \\
\Rightarrow \quad I & =\frac{m}{N A} \\
& =\frac{0.8}{1000 \times 10^{-4}} \quad \text { [From Eq. (i)] } \\
& =8 \mathrm{~A}
\end{aligned}$

Topic: Physics, Magnetic Properties of Matter, JEE Main
A small bar magnet is moved through a coil at constant speed from one end to the other. Which of the following series of observations will be seen on the galvanometer G attached across the coil?

Three positions shown describe: (a) the magnet's entry (b) magnet is completely inside and (c) magnet's exit.

Options:
A
B
C
D

The correct answer is:

Solution:
(a) When the bar magnet entering the coil the magnetic flux link with the coil increases, and according to the Faraday's law of electromagnetic induction, the emf will induce in the coil, which in tern induces current and galvanometer shows the deflection towards right.

(b) When the bar magnet is inside the coil no change in magnetic flux link with the coil, therefore galvanometer do not show deflection.

(c) When the bar magnet exit from the coil the magnetic flux link with the coil decreases, and according to the Faraday's law of electromagnetic induction, the emf will induce in the coil, in opposite direction which in tern induces current and galvanometer shows the deflection towards left.

Topic: Physics, Magnetic Effects of Current, JEE Main, JEE Main 2020 (04 Sep Shift 1)
A small bar magnet is placed with its axis at 30o with an external magnetic field of 0.06 T experiences a torque of 0.018 Nm. The minimum work required to rotate it from its stable to unstable equilibrium position is:

Options:
A 6.4×10−2 J
B 9.2×10−3 J
C 7.2×10−2 J
D 11.7×10−3 J

The correct answer is:
7.2×10−2 J

Solution:
τ=MBsinθ=0.018

M=0.018B sin θ=0.0180.06×0.5=0.64 Am2

W=ΔU=Uf−Ui

=−MB cos 180∘−−MB cos0∘

=2MB

=2×0.6×0.06

=0.072 J

Topic: Physics, Magnetic Properties of Matter, JEE Main, JEE Main 2020 (04 Sep Shift 1)
A small bead of mass $108 \mathrm{~g}$ slides without friction along a semicircular wire of radius $10 \mathrm{~cm}$ that rotates about a vertical axis at a rate of $10 \sqrt{2} \mathrm{rad} \mathrm{s}^{-1}$ as shown in the figure. The value of $\theta$ for which the bead will remain stationary relative to the rotating wire is $\left(\mathrm{g}=10 \mathrm{~ms}^{-2}\right)$

Options:
A $30^{\circ}$
B $45^{\circ}$
C $60^{\circ}$
D $75^{\circ}$

The correct answer is:
$60^{\circ}$

Solution:
No solution. Refer to answer key.

Topic: Physics, Rotational Motion, AP EAMCET, AP EAMCET 2017 (24 Apr Shift 1)
A small bead of mass m=1 kg is free to move on a circular hoop. The circular hoop has centre at C and radius r=1 m and it rotates about a fixed vertical axis. The coefficient of friction between bead and hoop is Î¼=0.5 . The maximum angular speed of the hoop for which the bead does not have relative motion with respect to the hoop, at the position shown in the figure is: (Take g=10 m s-2 )

Options:
A 5212
B 10212
C 15212
D 30212

The correct answer is:
30212

Solution:
The maximum angular speed of the hoop corresponds to the situation when the bead is just about to slide upwards.
The free-body diagram of the bead is
For the bead not to slide upwards
mω2(rsin⁡45o) cos 45o-mgsin⁡45o<μN ...(i)
Where N=mgcos⁡45o+mω2(rsin⁡45o) sin⁡45o ...(ii)
From (i) and (ii) we get,
ω=302 rad s-1

Topic: Physics, Laws of Motion, JEE Main
A small block of mass $5 \mathrm{~g}$ and charge $5 \mu \mathrm{C}$ is placed on insulated, frictionless, inclined plane of angle $60^{\circ}$. An electric field is applied parallel to the inclined plane. If the block remains at rest then the magnitude of electric field is $\left[\right.$ Take $\mathrm{g}=10 \mathrm{~m} / \mathrm{s}^2$)

Options:
A $\frac{10^5}{\sqrt{3}} \mathrm{~N} / \mathrm{C}$
B $\frac{5}{\sqrt{3}} \times 10^4 \mathrm{~N} / \mathrm{C}$
C $\frac{\sqrt{3}}{2} \times 10^4 \mathrm{~N} / \mathrm{C}$
D $2 \times 10^4 \mathrm{~N} / \mathrm{C}$

The correct answer is:
$\frac{\sqrt{3}}{2} \times 10^4 \mathrm{~N} / \mathrm{C}$

Solution:
As block is at rest
So, $\mathrm{qE}=\mathrm{mg} \sin \theta$
$$
\begin{aligned}
& \Rightarrow \mathrm{E}=\frac{\mathrm{mg} \sin \theta}{\mathrm{q}} \\
& =\frac{5 \times 10^{-3} \times 10 \times \sqrt{3}}{2 \times 5 \times 10^{-6}} \\
& =\frac{5 \times 10^{-3} \times \sqrt{3}}{10^{-6}}=5 \sqrt{3} \times 10^3 \\
& =\frac{\sqrt{3}}{2} \times 10^4 \mathrm{~N} / \mathrm{C}
\end{aligned}
$$

Topic: Physics, Electrostatics, TS EAMCET, TS EAMCET 2022 (20 Jul Shift 2)
A small block of mass 100 g is tied to a spring of spring constant 7.5 N m-1 and length 20 cm. The other end of spring is fixed at a particular point A. If the block moves in a circular path on a smooth horizontal surface with constant angular velocity 5 rad s-1about point A, then tension in the spring is

Options:
A 0.75 N
B 0.25 N
C 0.50 N
D 1.5 N

The correct answer is:
0.75 N

Solution:
Centripetal force will be provided by the spring force. Let the elongation in the spring be x, then we can write

⇒kx=mω2(r+x)

⇒7.5 x=2.5(0.2+x)

⇒x=0.55=0.1

Therefore, required tension will beT=kx=0.75 N.

Topic: Physics, Laws of Motion, JEE Main, JEE Main 2023 (06 Apr Shift 1)
A small block of mass m is released from rest from point A inside a smooth hemisphere bowl of radius R , which is fixed on ground such that OA is horizontal. The ratio (x) of magnitude of centripetal force and normal reaction on the block at any point B varies with θ as:

Options:
A

B

C

D

The correct answer is:

Solution:
mv2R=N-mgsinâ¡Î¸
N=mv2R+mgsinâ¡Î¸
By energy conservation,
mgR sinÎ¸=12mv2
mv2R=2mgsinâ¡Î¸
N=3mgsinâ¡Î¸
Ratio=mv2RN=23 (constant)
x=23

Topic: Physics, Laws of Motion, JEE Main
A small block of mass M moves on a frictionless surface of an inclined plane, as shown in the figure. The angle of the incline suddenly changes from 60Â° to 30Â° at point B. The block is initially at rest at A. Assume that collisions between the block and the incline are totally inelastic (g=10 m s-2). Figure:

â

The speed of the block at point C, immediately before it leaves the second incline is -

Options:
A 120 m s-1
B 105 m s-1
C 90 m s-1
D 75 m s-1

The correct answer is:
105 m s-1

Solution:
The height of the point B is
3 3 m tan 3 0 ∘ = 3 3 m 1 3 = 3 m
The energy of the block at B is
KE+PE=12M45 m s-12+Mg3m
The energy at the block at C is
KE=12MV2
The principle of conservation of energy gives
12MV2=12M45 m s-12+M10 m s-23m
This gives V=105 m s-1
Therefore, the choice (105 m s-1) is correct.

Topic: Physics, Center of Mass Momentum and Collision, JEE Main
A small block of mass m slides along a smooth frictional track as shown in the figure. If it starts from rest at P, what is the resultant force acting on it at Q?

Options:
A 70 mg
B 60 mg
C 75 mg
D 65 mg

The correct answer is:
65 mg

Solution:
From energy conservation,
Mg4R=12mv2âv=8Rg
At Î¸, N= mv2RâN=mRÃ8gR=8mg
So net force at Î¸, Fnet=N2+mg2=65 mg

Topic: Physics, Laws of Motion, JEE Main
A small block slides with velocity 0.5 gr on the horizontal frictionless surface as shown in the figure. The block leaves the surface at a point C. What is the value of cosÎ¸?

Options:

The correct answer is:
0.6

Solution:
mg cos⁡θ-N=mv2R; N=mg cos⁡θ-mv2R
N=0⇒mg cos⁡θ=mv2R
v=gR cos⁡θ or cos⁡θ=v2gR .....(i)
From energy conservation
12mv2-gR4=mgR1- cos⁡θ
⇒v22- gR8=gR-gR cos⁡θ ...(ii)
From (i) and (ii) cosθ=35=0.6

Topic: Physics, Laws of Motion, JEE Main
A small block starts slipping down from a point B on an inclined plane AB, which is making an angle θ with the horizontal section BC is smooth and the remaining section CA is rough with a coefficient of friction μ. It is found that the block comes to rest as it reaches the bottom (point A) of the inclined plane. If BC=2AC, the coefficient of friction is given by μ=k tanθ. The value of k is .......

Options:

The correct answer is:
3

Solution:
Let AC=ℓ ∴ BC=2ℓ ∴ AB=3ℓ

Apply work – Energy theorem

Wt+Wmg=ΔKE

mg(3ℓ)sinθ-μmgcosθ(ℓ)=0+0

μmgcosθℓ=3mg/sinθ

μ=3tanθ=ktanθ

k=3

Topic: Physics, Laws of Motion, JEE Main, JEE Main 2020 (02 Sep Shift 1)
A small bob tied at one end of a thin string of length 1 m is describing a vertical circle so that the maximum and minimum tension in the string is in the ratio 5:1. The velocity of the bob at the highest position is ______ m s-1. (Take g=10 m s-2)

Options:

The correct answer is:
5

Solution:
Let the speed of bob at lowest position be v1 and at the highest position be v2.

Maximum tension is at lowest position and minimum tension is at the highest position.
Now, using, conservation of mechanical energy,

12mv12=12mv22+mg2l

⇒ v12=v22+4 gl ....1

Now, Tmax-mg=mv12l

⇒ Tmax=mg+mv12l

& Tmin+mg=mv22l

⇒ Tmin=mv22l-mg

TmaxTmin=51

⇒ mg+mv12lmv22l-mg=51

⇒ mg+mv12l=mv22l-mg5

⇒ mg+mlv22+4 gl=5mv22l-5mg

⇒ mg+mv22l+4mg=5mv22l-5mg

⇒ 10 mg=4mv22l

v2=10×10×14

Thus, the velocity of the bob at the highest position is 5 m s-1.

Topic: Physics, Motion In Two Dimensions, JEE Main, JEE Main 2021 (25 Feb Shift 1)
A small body is released from point A of smooth parabolic path y=x2, where y is vertical axis and x is horizontal axis at ground, as shown. The body leaves the surface from point B. If g=10 m s-2 then what is the value of d (in m)?

Options:

The correct answer is:
2

Solution:
H1=(2)2=2 m
H2=(1)2=1 m
â´v=2ÃgÃH2-H1
v=20 m s-1
tanâ¡Î¸=dydxx=1=2xx=1=2
-H2=d tanâ¡Î¸-g d22u2cos2â¡Î¸

-1=2d-10d22202152

54d2-2d-1=0

On solving we get d=2 m

Topic: Physics, Motion In Two Dimensions, JEE Main
A small body of mass $0.10 \mathrm{~kg}$ is executing S.H.M. of amplitude $1.0 \mathrm{~m}$ and period $0.20 \mathrm{sec}$. The maximum force acting on it is

Options:
A $98.596 \mathrm{~N}$
B $985.96 \mathrm{~N}$
C $100.2 \mathrm{~N}$
D $76.23 \mathrm{~N}$

The correct answer is:
$98.596 \mathrm{~N}$

Solution:
Maximum acceleration
$\begin{gathered}A_{\max }=a \omega^2=\frac{a \times 4 \pi^2}{T^2}=\frac{1 \times 4 \times(3.14)^2}{0.2 \times 0.2} \\ F_{\max }=m \times A_{\max }=\frac{0.1 \times 4 \times(3.14)^2}{0.2 \times 0.2}=98.596 \mathrm{~N}\end{gathered}$

Topic: Physics, Oscillations, JEE Main
A small boy is throwing a ball towards a wall 6 in front of him. He releases the ball at a height of $1.4 \mathrm{~m}$ from the ground. The ball bounces from the wall at a height of $3 \mathrm{~m}$, rebounds from the ground and reaches the boy's hand exactly at the point of release. Assuming the two bounces (one from the wall and the other from the ground) to be perfectly elastic, how far ahead of the boy did the ball bounce from the ground?

Options:
A $1.5 \mathrm{~m}$
B $2.5 \mathrm{~m}$
C $3.5 \mathrm{~m}$
D $4.5 \mathrm{~m}$

The correct answer is:
$1.5 \mathrm{~m}$

Solution:
$3=(6+x) \tan \theta\left[1-\frac{6+x}{12}\right]$
$3=\frac{(6+x)(6-x) \tan \theta}{12}$
$1.4=x \tan \theta\left[1-\frac{x}{12}\right]$
$1.4=\frac{x(12-x)}{12} \tan \theta$
$\frac{30}{14}=\frac{36-x^{2}}{12 x-x^{2}}$
$360 x-30 x^{2}=36 \times 14-14 x^{2}$
$16 x^{2}-360 x+36 \times 14=0$
$x=\frac{360 \pm \sqrt{(360)^{2}-4 \times 36 \times 14 \times 16}}{32}$
$x=\frac{360 \pm 312}{32}=\frac{48}{32}=1.5$

Topic: Physics, Motion In Two Dimensions, JEE Main
A small bulb is placed at the bottom of a tank containing water to a depth of 7 m. The refractive index of water is 43. The area of the surface of water through which light from the bulb can emerge out is xπ m2. The value of x is _____ .

Options:

The correct answer is:
9

Solution:
Apply Snell's law, sin⁡90°sin⁡C=μ

⇒sinC=1μ⇒RR2+h2=34
Squaring, 16R2=9R2+9h2

⇒7R2=9h2⇒R=37h=37×7=3 m

The area of the circle is, A=πR2=π32 m2⇒α=9

Topic: Physics, Ray Optics, JEE Main, JEE Main 2022 (26 Jun Shift 2)
A small charged ball is hovering in the state of equilibrium at a height h over a very large horizontal uniformly charged sheet. If a disc of radius r <<h, is removed from the plate directly underneath the ball, then the acceleration of the ball will be

Options:
A g2rh2
B g2hr2
C g4rh2
D g4hr2

The correct answer is:
g2rh2

Solution:
Initially,

mg=qE

⇒qm=gE
Now, a disc of charge Q=σ×πr2 is removed. As r<<h, we can treat disc as a point charge. The acceleration of the ball is

a=qQ4πε0h2m

a=g2rh2

Topic: Physics, Electrostatics, JEE Main
A small circular loop of conducting wire has radius a and carries current I. It is placed in a uniform magnetic field B perpendicular to its plane such that when rotated slightly about its diameter and released, it starts performing simple harmonic motion of time period T. The mass of the loop is m then:

Options:
A T=2miB
B T=πm2iB
C T=2πmiB
D T=πmiB

The correct answer is:
T=2πmiB

Solution:
τ=MBsin⁡θ=Iα
πR2iBθ=mR22α
ω=2πiBm=2πT;T=2πmiB

Topic: Physics, Magnetic Effects of Current, JEE Main, JEE Main 2020 (09 Jan Shift 2)
A small circular loop of wire of radius a is located at the centre of a much larger circular wire loop of radius b. The two loops are in the same plane. The outer loop of radius b carries an alternating current I=I0cos⁡ωt. The emf induced in the smaller inner loop is nearly:

Options:
A ÏÎ¼0Â I0a2bÏsinâ¡Ït
B ÏÎ¼0Â I02Â .a2bÂ Ïcosâ¡Ït
C ÏÎ¼0Â I0Â b2aÏcosâ¡Ït
D ÏÎ¼0 I02 .a2bÏsinâ¡Ït

The correct answer is:
ÏÎ¼0 I02 .a2bÏsinâ¡Ït

Solution:
Induced emf,

ε=MdIdt,

Current is given as, I=I0cosωt
M=μ0πa22b, ε=-Mddt (I0cosωt)
c=M ωI0 sinωt
=μ0πa22bω I0sin⁡ ωt

Topic: Physics, Electromagnetic Induction, JEE Main, JEE Main 2017 (08 Apr Online)
A small coil of $N$ turns has an effective area $A$ and carries a current $I$. It is suspended in a horizontal magnetic field $\vec{B}$ such that its plane is perpendicular to $\vec{B}$. The work done in rotating it by $180^{\circ}$ about the vertical axis is

Options:
A $N A I B$
B $2 N A I B$
C $2 \pi N A I B$
D $4 \pi N A I B$

The correct answer is:
$2 \pi N A I B$

Solution:
The work done in rotating the coin is given as:
\(\begin{aligned}
& \mathrm{W}=\mathrm{MB}\left(\cos \theta_1-\cos \theta_2\right) \\
& =(\mathrm{NiA}) \mathrm{B}\left(\cos \mathrm{o}^{\circ}-\cos 180^{\circ}\right) \\
& =2 \mathrm{NAIB}
\end{aligned}\)

Topic: Physics, Magnetic Effects of Current, JEE Main
A small coil of $N$ turns has area $A$ and a current I flows through it. The magnetic dipole moment of this coil will be

Options:
A $N I / A$
B $N I^2 A$
C $N^2 A l$
D N/A

The correct answer is:
N/A

Solution:
Magnetic dipole moment of coil $=$ NIA

Topic: Physics, Magnetic Effects of Current, JEE Main
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, JEE Main
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, JEE Main
A small conducting sphere of the radius r is lying concentrically inside a bigger hollow conducting sphere of radius R. The bigger and smaller sphere are charged with Q and q (Q > q) and are insulated from each other. The potential difference between the sphere will be

Options:
A 14ÏÎµ0 qrâqR
B 14πε0 qR–Qr
C 14πε0 qr–QR
D 14πε0 QR+qr

The correct answer is:
14ÏÎµ0 qrâqR

Solution:
Potential of the inner sphere
VA = kqr + kQR
Potential of the outer sphere
VB = kqR + kQR
â´ VA â VB = kqr â kqR

=14ÏÎµ0qr-QR

Topic: Physics, Electrostatics, JEE Main
A small disc of mass $1 \mathrm{~g}$ slides down a smooth hill of height $10 \mathrm{~m}$ from rest and gets on to a plank of mass $100 \mathrm{~g}$ as shown in the figure. Due to friction between the disc and the plank. The disc slows down and moves as one piece with the plank. The work done by the frictional force is approximately (Use, $g=10 \mathrm{~ms}^2$ )

Options:
A 0.01 J
B 10 J
C 0.1 J
D 1 J

The correct answer is:
0.1 J

Solution:
The disc is at rest at the top of the hill.
$\therefore$ Potential energy of the disc, $U=m g h$
Let the velocity of the sliding disc before the friction is $v$.
$\therefore$ The kinetic energy, $K=\frac{1}{2} m v^2$
From energy conservation,
$$
\begin{aligned}
\frac{1}{2} m v^2 & =m g h \\
v & =\sqrt{2 g h}
\end{aligned}
$$
$$
\Rightarrow \quad v=\sqrt{2 g h}
$$
The disc now gets onto the plank and slows down and moves as one piece with the plank.
Let the velocity of the plank is $v^{\prime}$.
From conservation of momentum,
$$
\begin{aligned}
m v & =(m+M) v^{\prime} \\
\Rightarrow \quad v^{\prime} & =\frac{m \sqrt{2 g h}}{(m+M)}
\end{aligned}
$$
$\therefore$ The kinetic energy of the plank,
$$
\begin{aligned}
K_2 & =\frac{1}{2}(m+M) v^{\prime 2}=\frac{m^2 \times 2 g h}{2(M+m)} \\
\Rightarrow \quad K_2 & =\frac{m^2 g h}{(m+M)}
\end{aligned}
$$
$\therefore$ The total work done by the frictional force is
$$
\begin{aligned}
W_f & =K_1-K_2=m g h-\frac{m^2 g h}{m+M} \\
& =\frac{M m g h}{m+M} \\
\Rightarrow \quad W_f & =\frac{100 \times 10^{-3} \times 1 \times 10^{-3} \times 10 \times 10}{(100+1) \times 10^{-3}} \\
\Rightarrow \quad W_f & =0.1 \mathrm{~J}
\end{aligned}
$$

Topic: Physics, Work Power Energy, JEE Main
A small drop of water falls from rest through a large height h in air, the final velocity is

Options:
A ∝h
B ∝h
C ∝1h
D almost independent of h

The correct answer is:
almost independent of h

Solution:
After travelling certain distance drop will attain maximum velocity called terminal velocity and this will remain constant for constant medium and independent of h.

Topic: Physics, Mechanical Properties of Fluids, 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, JEE Main
A small liquid drop of radius R is divided into 27 identical liquid drops. If the surface tension is T, then the work done in the process will be :

Options:
A 8πR2T
B 3πR2T
C 18πR2T
D 4πR2T

The correct answer is:
8πR2T

Solution:
Let radius of the newly formed drops be r.

As total volume will remain constant, therefore

43πR3=27×43×πr3

⇒R3=27r3

⇒R=3r

⇒r=R3

⇒r2=R29

Now, work done =T×ΔA

=T27 4πr2-T4πR2=T27 4πR29-T4πR2=TπR212-4

=8πR2T

Topic: Physics, Mechanical Properties of Fluids, JEE Main, JEE Main 2024 (29 Jan Shift 2)
A small loop of the area of cross-section \( 10^{-4} \mathrm{~m}^{2} \) is placed at the centre of a bigger loop of radius \( 0.628 \mathrm{~m} \). A current of \( 10 \mathrm{~A} \) is passed in the bigger loop. The smaller loop is rotated about its diameter with an angular velocity \( \omega \). The magnetic flux linked with the smaller loop will be

Options:
A \( 10^{-7} \sin \omega t \)
B \( 10^{-7} \cos \omega t \)
C \( 10^{-9} \sin \omega t \)
D \( 10^{-9} \cos \omega t \)

The correct answer is:
\( 10^{-9} \cos \omega t \)

Solution:
Given,

Area of smaller loop is A=10-4 m2

The radius of bigger is R=0.628 m

The current in the bigger loop is i=10 A

The magnetic flux linked with the smaller loop is given by

ϕ=B→·A→=BAcosωt ...(1)

where, B is magnetic field due to bigger loop

ω is the angular velocity of smaller loop

The magnetic field due to bigger loop is given by

B=μ∘i2R

B=4π×10-7×102×0.628=10-5 Wb m-1

Substitute the value of B in equation (1), we get

ϕ=10-5×10-4×cosωt

ϕ=10-9cosωt

Topic: Physics, Electromagnetic Induction, JEE Main
A small mass attached to a string rotates on frictionless table top as shown. If the tension is the string is increased by pulling the string causing the radius of the circular motion to decrease by a factor of 2 , the kinetic energy of the mass will

Options:
A remain constant
B increase by a factor of 2
C increase by a factor of 4
D decrease by a factor of 2

The correct answer is:
increase by a factor of 4

Solution:
From the law of conservation of angular momentum
So,
$$
\begin{aligned}
m v r & =m v^{\prime} \frac{r}{2} \\
y^{\prime} & =2 y \\
\frac{k_1}{k_2} & =\frac{\frac{1}{2} m v^2}{\frac{1}{2} m v^2} \\
\frac{k_1}{k_2} & =\frac{y^2}{v^{\prime 2}} \\
& =\frac{y^2}{(2 v)^2} \\
\frac{k_1}{k_2} & =\frac{1}{4} \\
k_1 & =\frac{1}{4} k_2 \\
k_2 & =4 k_1
\end{aligned}
$$

Topic: Physics, Rotational Motion, NEET, NEET 2011 (Mains)
A small object is thrown at an angle $45^{\circ}$ to the horizontal with an initial velocity $\mathbf{v}_0$. The velocity is averaged for first $\sqrt{2} \mathrm{~s}$ and the magnitude of average velocity comes out to be same as that of initial velocity, i.e. $\left|\mathbf{v}_{\mathrm{o}}\right|$. The magnitude $\left|\mathbf{v}_0\right|$ will be (take, $g=10 \mathrm{~m} / \mathrm{s}^2$ )

Options:
A $3 \mathrm{~m} / \mathrm{s}$
B $3 \sqrt{2} \mathrm{~m} / \mathrm{s}$
C $4 \mathrm{~m} / \mathrm{s}$
D $5 \mathrm{~m} / \mathrm{s}$

The correct answer is:
$5 \mathrm{~m} / \mathrm{s}$

Solution:
Let object is at $B(x, y)$ after $t=\sqrt{2} \mathrm{~s}$
Then, $x=u_x \times t=v_0 \cos 45^{\circ} \times \sqrt{2}=v_0$
and $y=u_y t-\frac{1}{2} a_y t^2=v_0\left(\sin 45^{\circ}\right) \sqrt{2}-\frac{1}{2}(10) \times(\sqrt{2})^2$
$$
=\left(v_0-10\right) \mathrm{m}
$$
Displacement $O B$ of particle is
$$
O B=\sqrt{O A^2+A B^2}=\sqrt{v_0^2+\left(v_0-10\right)^2}
$$
So, $\quad v_{\text {avg }}=\frac{O B}{t}=v_0$ or $O B=v_0 t$
$$
\begin{array}{lrl}
\Rightarrow & \sqrt{v_0^2+\left(v_0-10\right)^2}=v_0 \sqrt{2} & \Rightarrow v_0^2+\left(v_0-10\right)^2=2 v_0^2 \\
\Rightarrow & v_0-10= \pm v_0 & 2 v_0=10 \Rightarrow v_0=5 \mathrm{~ms}^{-1}
\end{array}
$$

Topic: Physics, Motion In Two Dimensions, JEE Main
A small object of mass of 100ââg moves in a circular path. At a given instant velocity of the object is 10i^ m/s and acceleration is (20i^+10j^) m/s2. At this instant of time, rate of change of kinetic energy (in kg m2 s-3) of the object is

Options:

The correct answer is:
20

Solution:
Given, mass of object m=100 g
=100Ã10-3 kg
Velocity of object v=10i^ m/s
Acceleration of object a=(20i^+10j^) m/s2
We know that,
d( KE ) dt =F.v=ma.v [ âµK.E.= 1 2 m v 2 ]
=(100Ã10-3)(20i^+10j^).(10i^)
=100Ã10-3Ã200
=100Ã2001000=20 kg m2/s3

Topic: Physics, Work Power Energy, JEE Main