datavorous/entrance-exam-dataset huggingface dataset:
A large flat metal surface has a uniform charge density +Ï . An electron of mass m and charge Ï leaves the surface at the point A with speed u and returns to it at point B. Disregarding gravity the maximum value of AB is

Options:
A 4u2mÎµ0ÏÏ
B u2⋅ÏÎµ0mÏ
C 2Îµ0mu2ÏÏ
D u2ÏÏÎµ⋅m

The correct answer is:
2ε0mu2ρσ

Solution:
Electric field due to one surface of the large metal plate,
The projected electron's motion will be on a curved path like a projectile.
∴ AB will be maximum when θ=45°
Range ABmax=u2a
a=ρEm=ρ​σ2ε0m
=u2ρσ2ε0m=2ε0mu2ρσ

Topic: Physics, Motion In Two Dimensions, JEE Main
A large heavy box is sliding without friction down a smooth plane of inclination Î¸ . From a point P on the bottom of the box, a particle is projected inside the box. The initial speed of the particle with respect to the box is u and the direction of projection makes an angle Î± with the bottom as shown in the figure:

Find the distance along the bottom of the box between the point of projection P and the point Q where the particle lands. [Assume that the particle does not hit any other surface of the box. Neglect air resistance]

Options:
A u 2 sin 2α g
B u sin 2α g cos θ
C u 2 sin α g
D u 2 sin 2Î± gcos Î¸

The correct answer is:
u 2 sin 2Î± gcos Î¸

Solution:
u is the relative velocity of the particle with respect to the box. Resolve u.
u x is the relative velocity of the particle with respect to the box in x â direction.

u y is the relative velocity of the particle with respect to the box in y â direction.

Since there is no velocity of the box in the y â direction, therefore this is the vertical velocity of the particle with respect to ground also.

Y â direction motion (Taking relative terms w.r.t. box)

u y =+usinÎ±

a y =âgcosÎ¸

S y =0 (activity is taken till the time the particle comes back to the box)

t y =t

Using, S=ut+ 1 2 a t 2 we getâ

ââââââââââ0=( usinÎ± )tâ 1 2 gcosÎ¸Ã t 2 âââââââât=0ââorâât= 2usinÎ± gcosÎ¸

X-direction motion (Taking relative terms w.r.t. box)

u x =+ucosÎ± a x =0

Again, using âS=ut+ 1 2 a t 2 we get â

â S x =ucosÎ±Ã 2usinÎ± gcosÎ¸ = u 2 sin2Î± gcosÎ¸ [ âµââââ t x = 2ucosÎ± gcosÎ¸ ]

Topic: Physics, Motion In Two Dimensions, JEE Main
A large hollow metal sphere of radius R has a small opening at the top. Small drops of mercury, each of radius r and charged to a potential V, fall into the sphere. The potential of the sphere becomes V' after N drops fall into it. Then

Options:
A V′=V for any value of N
B V′=V for N=Rr2/3
C Vâ²=V for N=Rr1/3
D V′=V for N=Rr

The correct answer is:
V′=V for N=Rr1/3

Solution:
V=kqr
After N drop potential of hollow sphere=Vâ²
Vâ²=kNqR
Vâ²=V â kqr=kNqR
âN=Rr

Topic: Physics, Electrostatics, JEE Main
A large horizontal surface moves up and down in S.H.M. with an amplitude of \(1 \mathrm{~cm}\). If a mass of \(10 \mathrm{~kg}\) (which is placed on the surface) is to remain continuously in contact with it, the maximum frequency of S.H.M. will be

Options:
A \(5 \mathrm{~Hz}\)
B \(0.5 \mathrm{~Hz}\)
C \(1.5 \mathrm{~Hz}\)
D \(10 \mathrm{~Hz}\)

The correct answer is:
\(5 \mathrm{~Hz}\)

Solution:
Here, \(a=1 \mathrm{~cm}=0.01 \mathrm{~m}\); The mass will remain in contact with surface, if
\(m g=m \omega^2 a \text { or } \omega=\sqrt{g / a}\)
\(\begin{aligned}
& \text { or } 2 \pi v=\sqrt{g / a} \text { or } v=\frac{1}{2 \pi} \sqrt{\frac{g}{a}} \\
& v=\frac{7}{2 \times 22} \sqrt{\frac{980}{1}}=4.9 \mathrm{~s}^{-1}=5 \mathrm{~Hz}
\end{aligned}\)

Topic: Physics, Oscillations, AIIMS, AIIMS 2007
A large increase in the rate of a reaction for a rise in temperature is due to

Options:
A The decrease in the number of collisions
B The increase in the number of activated molecules
C The shortening of the mean free path
D The lowering of the activation energy

The correct answer is:
The increase in the number of activated molecules

Solution:
The increase in collision frequency brings in an increase in effective collisions and thus rate of reaction increases.

Topic: Chemistry, Chemical Kinetics, JEE Main
A large metal plate has a surface charge density of 8.85×10-6 C m-2. An electron having initial kinetic energy of 8×10-17 J is moving towards the center of the plate. If the electron stops just before reaching the plate then the initial distance between the electron and the plate is [Take ε0=8.85×10-12 C2 N m-2]

Options:
A 0.5 mm
B 0.1 mm
C 0.2 cm
D 0.02 cm

The correct answer is:
0.5 mm

Solution:
Electric field due to an infinite sheet of charge having surface density σ is E=σε0.

Now using energy conservation we can write,

gain in potential energy = loss in kinetic energy

⇒qEd=12mv2⇒1.6×10-19σε0d=8×10-17⇒d=8×10-178.85×10-121.6×10-198.85×10-6=0.5 mm

Topic: Physics, Electrostatics, JEE Main
A large number (n) of identical beads, each of mass m and radius r are strung on a thin smooth rigid horizontal rod of length L(L≫r) and are at rest at random positions. The rod is mounted between two rigid supports (see figure). If one of the beads is now given a speed v, the average force experienced by each support after a long time is (assume all collisions are elastic):

Options:
A mv2L-nr
B mv2L-2nr
C mv22L-nr
D Zero

The correct answer is:
mv2L-2nr

Solution:
Average time for 1 collision to occur on 1 wall,
=tavg=average distance travelledspeed=2L-2nrv
⇒tavg=2L-2nrv. Change in momentum in 1 collision =2mv. Average force on each support, favg=2mvtavg=2mv22L-2nr⇒favg=mv2L-2nr.

Topic: Physics, Laws of Motion, JEE Main, JEE Main 2015 (11 Apr Online)
A large number of droplets, each of radius, $r$ coalesce to form a bigger drop of radius, $R$. An engineer designs a machine so that the energy released in this process is converted into the kinetic energy of the drop. Velocity of the drop is ( $T=$ surface tension, $\rho=$ density)

Options:
A
$\left[\frac{T}{\rho}\left(\frac{1}{r}-\frac{1}{R}\right)\right]^{1 / 2}$

B
$\left[\frac{6 T}{\rho}\left(\frac{1}{r}-\frac{1}{R}\right)\right]^{1 / 2}$

C
$\left[\frac{3 T}{\rho}\left(\frac{1}{r}-\frac{1}{R}\right)\right]^{1 / 2}$

D
$\left[\frac{2 T}{\rho}\left(\frac{1}{r}-\frac{1}{R}\right)\right]^{1 / 2}$

The correct answer is:
$\left[\frac{6 T}{\rho}\left(\frac{1}{r}-\frac{1}{R}\right)\right]^{1 / 2}$

Solution:
When small droplets coelesce to form a bigger drop, energy released in this process is given by,
$$
4 \pi R^3 T\left[\frac{1}{r}-\frac{1}{R}\right]
$$
Where, $R=$ radius of big drop
$r=$ radius of small drop
$T=$ surface tension
According to question
$$
\begin{aligned}
& \frac{1}{2} m v^2=4 \pi R^3 T\left[\frac{1}{r}-\frac{1}{R}\right] \\
\Rightarrow & \frac{1}{2}\left[\frac{4}{3} \pi R^3 \rho\right] v^2=4 \pi R^3 T\left[\frac{1}{r}-\frac{1}{R}\right] \\
\Rightarrow V^2 & =\frac{6 T}{\rho}\left[\frac{1}{r}-\frac{1}{R}\right] \Rightarrow V=\left[\frac{6 T}{\rho}\left(\frac{1}{r}-\frac{1}{R}\right)\right]^{\frac{1}{2}}
\end{aligned}
$$

Topic: Physics, Mechanical Properties of Fluids, JEE Main, JEE Main 2012 (19 May Online)
A large number of liquid drops each of radius r coalesce to form a single drop of the radius R. The energy released in the process is converted into kinetic energy of the big drop so formed. The speed of the big drop is (given surface tension of the liquid T, density ρ)

Options:
A 2Tρ1r-1R
B 6Tρ1r-1R
C 4Tρ1r-1R
D Tρ1r-1R

The correct answer is:
6Tρ1r-1R

Solution:
n×43πr3=43πR3

∴ n=R3r3 .....i

KE=Ui-Uf

=n×4πr2T-4πR2T

=4πT(nr2-R2)

=4πTR2Rr-1 from i

1243πR3·ρυ2=4πTR2Rr-1

Rρυ26=TRr-1

ρ6υ2=T1r-1R

υ=6Tρ1r-1R

Topic: Physics, Mechanical Properties of Fluids, JEE Main, JEE Main 2014 (19 Apr Online)
A large number of liquid drops each of radius r coalesce to from a single drop of radius R. The energy released in the process is converted into kinetic energy of the big drop so formed. The speed of the big drop is (given, surface tension of liquid T, density ρ)

Options:
A Tρ1r-1R
B 2Tρ1r-1R
C 4Tρ1r-1R
D 6Tρ1r-1R

The correct answer is:
6Tρ1r-1R

Solution:
n43Ïr3=43ÏR3
n=R3r3
ÎU=nÂ 4Ïr2Â T-4ÏR2Â T
=4ÏTÂ nr2-R2
=4ÏTÂ R3r2r3-R2
â´ Energy released,
U=4ÏT R 3 [ 1 r â 1 R ]
If this energy is converted into kinetic energy then
1 2 m v 2 =4Ï R 3 T[ 1 r â 1 R ]
1 2 Ã[ 4 3 Ï R 3 Ï ] v 2 =4Ï R 3 T[ 1 r â 1 R ]
v= 6T Ï [ 1 r â 1 R ]

Topic: Physics, Mechanical Properties of Fluids, JEE Main
A large number of particles are placed around the origin, each at a distance $R$ from the origin. The distance of the centro of mass of the system from the origin is

Options:
A equal to $R$
B less than equal to $R$
C greater than $R$
D greater than equal to $R$

The correct answer is:
less than equal to $R$

Solution:
As large number of particles is situated at a distance $R$ from the ongin If particles are uniformly distubuted and make a circulat boundary atound the ongm, then centre of mass will be at the ongin While if the patticles are not uniformly distnbuted, then centre of mass will lie between particle and origin. This implies distance between centre of mass and ongin is always less than equal to $R$.

Topic: Physics, Center of Mass Momentum and Collision, JEE Main
A large number of water droplets each of radius ' $t$ ' combine to form a large drop of Radius ' $R$ '. If the surface tension of water is ' $T$ ' \& mechanical equivalent of heat is ' $\mathrm{J}$ ' then the rise in temperature due to this is

Options:
A $\frac{2 \mathrm{~T}}{\mathrm{rJ}}$
B $\frac{3 \mathrm{~T}}{\mathrm{RJ}}$
C $\frac{3 \mathrm{~T}}{\mathrm{~J}}\left(\frac{1}{\mathrm{r}}-\frac{1}{\mathrm{R}}\right)$
D $\frac{2 \mathrm{~T}}{\mathrm{~J}}\left(\frac{1}{\mathrm{r}}-\frac{1}{\mathrm{R}}\right)$

The correct answer is:
$\frac{3 \mathrm{~T}}{\mathrm{~J}}\left(\frac{1}{\mathrm{r}}-\frac{1}{\mathrm{R}}\right)$

Solution:
Radius of each droplet $=\mathrm{r}$
Radius of the drop $=\mathrm{R}$
As volume remains constant,
$$
\begin{aligned}
& \mathrm{n} \times \frac{4}{3} \pi \mathrm{r}^3=\frac{4}{3} \pi \mathrm{R}^3 \\
& \therefore \quad \mathrm{n}=\frac{\mathrm{R}^3}{\mathrm{r}^3}
\end{aligned}
$$
Decrease in surface area $=4 \pi r^2 n-4 \pi R^2 n$
$$
\begin{aligned}
\Delta \mathrm{A} & =4 \pi\left[\mathrm{nr}^2-\mathrm{R}^2\right] \\
& =4 \pi\left[\frac{\mathrm{R}^3}{\mathrm{r}^3} \mathrm{r}^2-\mathrm{R}^2\right] \\
& =4 \pi \mathrm{R}^3\left[\frac{1}{\mathrm{r}}-\frac{1}{\mathrm{R}}\right]
\end{aligned}
$$
Enery released $\mathrm{W}=\mathrm{T} \times \Delta \mathrm{T}$
$$
\begin{aligned}
& \text { Heat produced } \mathrm{Q}=\frac{\mathrm{W}}{\mathrm{J}} \\
& \text { H }
\end{aligned}
$$
Heat produced $\mathrm{Q}=\frac{\mathrm{W}}{\mathrm{J}}$
$$
\mathrm{Q}=\mathrm{m} \cdot \mathrm{s} \cdot \Delta \theta
$$
Put (i) and (iii) into (ii)
$$
\begin{aligned}
& \mathrm{m} \cdot \mathrm{S} \Delta \theta=\frac{4 \pi \mathrm{R}^3 \mathrm{~T}}{\mathrm{~J}}\left[\frac{1}{\mathrm{r}}-\frac{1}{\mathrm{R}}\right] \\
& \frac{4}{3} \pi \mathrm{R} \rho_{\text {water }} \mathrm{S}_{\text {water }} \Delta \theta=\frac{4 \pi \mathrm{R}^3 \mathrm{~T}}{\mathrm{~J}}\left[\frac{1}{\mathrm{r}}-\frac{1}{\mathrm{R}}\right] \\
& \Delta \mathrm{Q}=\frac{3 \mathrm{~T}}{\mathrm{~J}}\left[\frac{1}{\mathrm{r}}-\frac{1}{\mathrm{R}}\right]
\end{aligned}
$$

Topic: Physics, Mechanical Properties of Fluids, JEE Main
A large number of water droplets each of radius ' $t$ ' combine to form a large drop of Radius ' $R$ '. If the surface tension of water is ' $T$ ' \& mechanical equivalent of heat is ' $\mathrm{J}$ ' then the rise in temperature due to this is

Options:
A $\frac{2 \mathrm{~T}}{\mathrm{rJ}}$
B $\frac{3 \mathrm{~T}}{\mathrm{RJ}}$
C $\frac{3 \mathrm{~T}}{\mathrm{~J}}\left(\frac{1}{\mathrm{r}}-\frac{1}{\mathrm{R}}\right)$
D $\frac{2 \mathrm{~T}}{\mathrm{~J}}\left(\frac{1}{\mathrm{r}}-\frac{1}{\mathrm{R}}\right)$

The correct answer is:
$\frac{3 \mathrm{~T}}{\mathrm{~J}}\left(\frac{1}{\mathrm{r}}-\frac{1}{\mathrm{R}}\right)$

Solution:
Radius of each droplet $=\mathrm{r}$
Radius of the drop $=\mathrm{R}$
As volume remains constant,
$$
\begin{aligned}
& \mathrm{n} \times \frac{4}{3} \pi \mathrm{r}^3=\frac{4}{3} \pi \mathrm{R}^3 \\
& \therefore \quad \mathrm{n}=\frac{\mathrm{R}^3}{\mathrm{r}^3}
\end{aligned}
$$
Decrease in surface area $=4 \pi r^2 n-4 \pi R^2 n$
$$
\begin{aligned}
\Delta \mathrm{A} & =4 \pi\left[\mathrm{nr}^2-\mathrm{R}^2\right] \\
& =4 \pi\left[\frac{\mathrm{R}^3}{\mathrm{r}^3} \mathrm{r}^2-\mathrm{R}^2\right] \\
& =4 \pi \mathrm{R}^3\left[\frac{1}{\mathrm{r}}-\frac{1}{\mathrm{R}}\right]
\end{aligned}
$$
Enery released $\mathrm{W}=\mathrm{T} \times \Delta \mathrm{T}$
$$
\begin{aligned}
& \text { Heat produced } \mathrm{Q}=\frac{\mathrm{W}}{\mathrm{J}} \\
& \text { H }
\end{aligned}
$$
Heat produced $\mathrm{Q}=\frac{\mathrm{W}}{\mathrm{J}}$
$$
\mathrm{Q}=\mathrm{m} \cdot \mathrm{s} \cdot \Delta \theta
$$
Put (i) and (iii) into (ii)
$$
\begin{aligned}
& \mathrm{m} \cdot \mathrm{S} \Delta \theta=\frac{4 \pi \mathrm{R}^3 \mathrm{~T}}{\mathrm{~J}}\left[\frac{1}{\mathrm{r}}-\frac{1}{\mathrm{R}}\right] \\
& \frac{4}{3} \pi \mathrm{R} \rho_{\text {water }} \mathrm{S}_{\text {water }} \Delta \theta=\frac{4 \pi \mathrm{R}^3 \mathrm{~T}}{\mathrm{~J}}\left[\frac{1}{\mathrm{r}}-\frac{1}{\mathrm{R}}\right] \\
& \Delta \mathrm{Q}=\frac{3 \mathrm{~T}}{\mathrm{~J}}\left[\frac{1}{\mathrm{r}}-\frac{1}{\mathrm{R}}\right]
\end{aligned}
$$

Topic: Physics, Mechanical Properties of Fluids, MHT CET, MHT CET 2023 (13 May Shift 1)
A large number of water drops, each of radius r, combine to have a drop of radius R. If the surface tension is T and mechanical equivalent of heat is J, the rise in heat energy per unit volume will be:

Options:
A 2TrJ
B 3TrJ
C 2T J1r-1R
D 3T J1r-1R

The correct answer is:
3T J1r-1R

Solution:
n×43πr3=43πR3

∴ n1/3r=R

∴ Total change in surface energy

=n4πr2-4πR2T

⇒4πTnr2-R2

∴ Heat energy

=4πTnr2-R2J×43πR3=3 T Jnr2R3-1R

Put nr3=R3

∴ 3TJ1r-1R

Topic: Physics, Mechanical Properties of Fluids, JEE Main, JEE Main 2021 (26 Feb Shift 1)
A large proportion of oxygen is left unused in the human blood even after its uptake by the body tissues. This O2

Options:
A

Helps in releasing more O 2 to the epithelial tissues.

B

It is enough to keep oxyhaemoglobin saturation at 96%.

C

Raise the pC O 2 of blood to 75 mm of Hg.

D

Acts as a reserve during muscular exercise

The correct answer is:

Acts as a reserve during muscular exercise

Solution:
A large portion of the oxygen is left unused in the human blood even after its uptake by the body tissues. This O 2 acts as a reserve during muscular exercise. Tissues usually take up only 25% of the oxygen in the tissue and the rest 75% remain unused in the blood (venous).

Topic: Biology, Breathing and Exchange of Gases, NEET
A large sphere of mass M is fixed at one position and two identical particles each of mass m are kept on a line passing through the centre of M (see figure). The point masses are connected by a rigid massless rod of length l and this assembly is free to move along the line connecting them. All three masses interact only through their mutual gravitational interaction. When the point mass nearer to M is at a distance r=3l from M, the tension in the rod is zero for m=kM288. The value of k is

Options:

The correct answer is:
7

Solution:
Due to gravitational interaction connected masses have some acceleration.
Let both small masses are moving with acceleration 'a' towards larger mass M
Force equation for mass nearer to a larger mass
GMm3l2-Gm2l2=ma ...(i)
Force equation for mass away from a larger mass
GMm4l2+Gm2l2=ma ...(ii)
From equation (i) & (ii)
GM9l2-Gml2=GM16l2+Gml2
âM9-M16=m+m
7M144=2m
âm=7M288=kM288
âk=7

Topic: Physics, Gravitation, JEE Main
A large tank filled with water to a height h is to be emptied through a small hole at the bottom. The ratio of time taken for the level of water to fall from h to h2 and from h2 to zero is

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

The correct answer is:
2-1

Solution:
Time taken for the level to fall from H to H′t=AA02gH-H′
According to problem- the time taken for the level to fall from h to h2t1=AA02gh-h2
and similarly time taken for the level to fall from h2 to zero t2=AA02gh2-0â´Â t1t2=1-1212-0=2-1.

Topic: Physics, Mechanical Properties of Fluids, JEE Main
A large tank filled with water to a height ' $h$ ' is to be emptied through a small hole at the bottom. The ratio of time taken for the level of water to fall from $h$ to $\frac{h}{2}$ and from $\frac{h}{2}$ to zero is

Options:
A $\sqrt{2}$
B $\frac{1}{\sqrt{2}}$
C $\sqrt{2}-1$
D $\frac{1}{\sqrt{2}-1}$

The correct answer is:
$\sqrt{2}-1$

Solution:
Time taken for the level to fall from $H$ to $H^{\prime} \quad t=\frac{A}{A_0} \sqrt{\frac{2}{g}}[\sqrt{H}-\sqrt{H}]$
According to problem- the time taken for the level to fall from $h$ to $\frac{h}{2} \quad t_1=\frac{A}{A_0} \sqrt{\frac{2}{g}}\left[\sqrt{h}-\sqrt{\frac{h}{2}}\right]$
and similarly time taken for the level to fall from $\frac{h}{2}$ to zero $t_2=\frac{A}{A_0} \sqrt{\frac{2}{g}}\left[\sqrt{\frac{h}{2}}-0\right]$
$\therefore \frac{t_1}{t_2}=\frac{1-\frac{1}{\sqrt{2}}}{\frac{1}{\sqrt{2}}-0}=\sqrt{2}-1$.

Topic: Physics, Mechanical Properties of Fluids, JEE Main
A large tank filled with water to a height h is to be emptied through a small hole at the bottom. The ratio of time taken for the level of water to fall from h to h 2 and from h 2 to zero is

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

The correct answer is:
2 - 1

Solution:
Time taken for the level to fall from H to H'
t = A A 0 2 g H - H ′
According to problem the time taken for the level to fall from h to h 2
t 1 = A A 0 2 g [ h - h 2 ]
and similarly time taken for the level to fall from h 2
t 2 = A A 0 2 g [ h 2 - 0 ]
∴ t 1 t 2 = 1 - 1 2 1 2 - 0 = 2 - 1

Topic: Physics, Mechanical Properties of Fluids, JEE Main
A large temple has a depression in one wall. On the floor plan, it appears as an indentation having a spherical shape of radius 2.50 m . A worshiper stands on the centerline of the depression, 2.00 m out from its deepest point, and whispers a prayer. Where is the sound concentrated after reflection from the black wall of the depression?

Options:
A - 1 0 3 m
B - 1 0 7 m
C - 3 10 m
D - 7 10 m

The correct answer is:
- 1 0 3 m

Solution:
Wall will act as concave mirror
1 v + 1 u = 1 f ⇒ 1 v + 1 u = 2 R ⇒ 1 v + 1 - 2 = 2 - 2 . 5
1 v = 2 - 2 . 5 + 1 2 ⇒ 1 v = - 8 + 5 1 0
1 v = - 3 1 0
v = - 1 0 3 m

Topic: Physics, Ray Optics, JEE Main
A large vessel having a very small circular hole of diameter 1 mm at the bottom is filled with water. Up to what maximum height (in cm) water can be filled in the vessel so that there is no leakage? [surface tension of water =7.5×10-2 N m-1 and g=10 m s-2 ]

Options:

The correct answer is:
3

Solution:
Water will not leak out from the hole if the weight of water in the water column is supported by the force due to surface tension.
Using the relation,
h=2TrÏg
=2Ã7.5Ã10-20.5Ã10-3Ã103Ã10
=3Ã10-2 m=3 cm

Topic: Physics, Mechanical Properties of Fluids, JEE Main
A laser beam of power 27 mW has a cross-sectional area of 10 mm2. The magnitude of the maximum electric field in this electromagnetic wave is given by

[Given permittivity of space Îµ0â9Ã10-12 C2 N-1 m-2, speed of light c=3Ã108 m s-1

Options:
A 1 kV m-1
B 1.4 kV m-1
C 0.7 kV m-1
D 2 kV m-1

The correct answer is:
1.4 kV m-1

Solution:
I=PA=12Îµ0E02c

â´E0=2PÎµ0CA=2Ã27Ã10-3Ã36ÏÃ1093Ã108Ã10Ã10-6

E0=1.4 kV m-1

Topic: Physics, Electromagnetic Waves, JEE Main
A laser beam of power 27 mW has a cross-sectional area of 10 mm2. The magnitude of the maximum electric field in this electromagnetic wave is given by [Take ε0≈9×10-12 C2 N-1 m-2 and c=3×108 m s-1]

Options:
A 1 kV/m
B 1.4 kV/m
C 0.7 kV/m
D 2 kV/m

The correct answer is:
1.4 kV/m

Solution:
I=PA=12ε0E02c
∴E0=2Pε0cA=2×27×10-39×10-12×3×108×10×10-6
⇒E0=1.4 kV/m

Topic: Physics, Electromagnetic Waves, JEE Main
A laser light of wavelength 660 nm is used to weld Retina detachment. If a laser pulse of width 60 ms and power 0.5 kW is used, the approximate number of photons in the pulse are (Take Planck's Constant, h=6.62×10−34 J s)

Options:
A 1022
B 1019
C 1020
D 1018

The correct answer is:
1020

Solution:
The energy of photon is given by E=hcλ.

PowerP is given as, P=Et=nhcλt

From the above equation, the number of photons is n=Pλthc
=0.5×103×660×10-9×60×10-36.6×10-34×3×108
Hence, number of photons in the pulse are=100×1018=1020.

Topic: Physics, Dual Nature of Matter, JEE Main, JEE Main 2017 (09 Apr Online)
A layer of oil with a density 724 kg m-3 floats on water of density 1000 kg m-3. A block floats on the oil-water interface with 16th of its volume in oil and 56th of its volume in water, as shown in the figure. What is the density of the block?

Options:
A 1024 kg m-3
B 1276 kg m-3
C 776 kg m-3
D 954 kg m-3

The correct answer is:
954 kg m-3

Solution:
Given,
The density of oil Ï1=724 kg/m3
The density of water Ï2=1000 kg/m3
According to Archimedes' principle,
Upthrust = Weight of the liquid displaced
âVÏbg=VoÏ1g+VwÏ2g (Vo and Vw are volumes of oil and water displaced respectively)
VÏbg=V6724g+5V61000g, where Ïb=density of the block.
and, Vo=V6 and Vw=5V6
âVÏbg=724V6+5000V6g
âVÏbg=724V+5000V6g
âVÏbg=57246Vg
Ïb=954 kg/m3

Topic: Physics, Mechanical Properties of Fluids, JEE Main
A layer of oil with density 724 kg m-3 floats on water of density 1000 kg m-3. A block floats on the oil-water interface with 1/6 of its volume in oil and 5/6 of its volume in water, as shown in the figure. What is the density of the block?

Options:
A 1024 kg m-3
B 1276 kg m-3
C 776 kg m-3
D 954 kg m-3

The correct answer is:
954 kg m-3

Solution:
Given,
Density of oil Ï1=724 kg/m3
Density of water Ï2=1000 kg/m3
According to Archimedes' principle,
Up thrust = Weight of the liquid displaced
âVÏbg=VoÏ1g+VwÏ2g (Vo and Vw are volumes of oil and water displaced respectively)
VÏbg=V6724g+5V61000g, where Ïb=density of block.
and, Vo=V6 and Vw=5V6
âVÏbg=724V6+5000V6g
âVÏbg=724V+5000V6g
âVÏbg=57246Vg
Ïb=954 kg/m3

Topic: Physics, Mechanical Properties of Fluids, JEE Main
A LCR circuit is at resonance for a capacitor $C$, inductance $L$ and resistance $R$. Now the value of resistance is halved keeping all other parameters same. The current amplitude at resonance will be now:

Options:
A Zero
B same
C halved
D double

The correct answer is:
double

Solution:
In resonance $Z=R$
$\begin{aligned} & \mathrm{I}=\frac{\mathrm{V}}{\mathrm{R}} \\ & \mathrm{R} \rightarrow \text { halved } \\ & \Rightarrow \mathrm{I} \rightarrow 2 \mathrm{I}\end{aligned}$
I becomes doubled.

Topic: Physics, Alternating Current, JEE Main, JEE Main 2024 (08 Apr Shift 1)
A lead bullet of 10 g travelling at 300 m s-1 strikes against a block of wood comes to rest. Assuming 50% of heat is absorbed by the bullet, the increase in its temperature (Specific heat of lead = 150 J kg-1 k-1)

Options:
A 100 ℃
B 125 ℃
C 150 ℃
D 200 ℃

The correct answer is:
150 ℃

Solution:
Here, m=10 g =10-2 kg
v=300 m s-1, Î Î¸=?, C=150 J kâ g â1 ââ K â1
Q=5010012mv2=14Ã10-23002=225 J
From Q=mCÎÎ¸
Î¸ =QCm=225150Ã10-2=150 â

Topic: Physics, Thermodynamics, JEE Main
A lead bullet penetrates into a solid object and melts. Assuming that 40% of its kinetic energy is used to heat it, the initial speed of bullet is
(Given, initial temperature of the bullet =127°C,
Melting point of the bullet =327°C,
Latent heat of fusion of lead =2.5×104 J kg-1,
Specific heat capacity of lead =125 J kg K-1)

Options:
A 125 m s-1
B 500 m s-1
C 250 m s-1
D 600 m s-1

The correct answer is:
500 m s-1

Solution:
Loss of kinetic energy of the lead bullet=Heat generated to raise its temperature from 127 °C to 327 °C.

40 % loss of the kinetic energy of the lead bullet=Heat gained by the lead bullet to raise its temperature.

0.412mv2=m×SPb×∆T+m×LPb

0.4×12mv2=mSPbT2-t1+mLPb

0.2mv2=m×125×327-127+2.5×104

v2=5×125×200+2.5×104

v2=5×25000+25000

v=500 m s-1

Topic: Physics, Thermal Properties of Matter, JEE Main, JEE Main 2022 (27 Jun Shift 2)