datavorous/entrance-exam-dataset huggingface dataset:
A thin spherical shell of radius $R$ has charge $Q$ spread uniformly over its surface. Which of the following graphs most closely represents the electric field $E(r)$ produced by the shell in the range $0 \leq$ $r < \infty$, where $r$ is the distance from the centre of the shell?

Options:
A

B

C

D

The correct answer is:

Solution:

Topic: Physics, Electrostatics, JEE Main, JEE Main 2008
A thin spherical shell of radius R is cut into two equal halves and then each of its halves is charged to the same uniform surface charge density Ï. If F is the force exerted on each half to keep them at equilibrium, then F is proportional to

Options:
A 1 ε 0 σ 2 R
B 1 ε 0 σ 2 R
C 1 ε 0 σ 2 R 2
D 1 ε 0 σ 2 R 2

The correct answer is:
1 ε 0 σ 2 R 2

Solution:
Electrical force per unit area = 1 2 ε 0 E 2 = 1 2 ε 0 σ ε 0 2 = σ 2 2 ε 0
Projected area = π R 2
∴ Net electrical force = σ 2 2 ε 0 π R 2
In equilibrium, this force should be equal to the applied force.
∴ F = π σ 2 R 2 2 ε 0
or F ∝ σ 2 R 2 ε 0

Topic: Physics, Magnetic Properties of Matter, JEE Main
A thin strip 10 cm long is on a U shaped wire of negligible resistance and it is connected to a spring of spring constant 0.5 N m-1 (see figure). The assembly is kept in a uniform magnetic field of 0.1 T. If the strip is pulled from its equilibrium position and released, the number of oscillations it performs before its amplitude decreases by a factor of e is N . If the mass of the strip is 50 grams, its resistance 10Ω and air drag negligible, N will be close to:

Options:
A 1000
B 5000
C 10000
D 50000

The correct answer is:
5000

Solution:
Equation of motion for slider:
-md2xdt2=kx+ilB
md2xdt2+kx+BlvRlB=0 ⇒md2xdt2+B2l2R.dxdt+kx=0
Comparing with md2xdt2+b.dxdt+kx=0
m=m ;b=B2l2R;k=k
We know: A=A0e-b2m×t=A0e-B2l22mR×t
For amplitude to be A0/e,
A0e-B2l22mR×t=Ae⇒B2l22mR×t=1
t=2mRB2l2=10,000 sec
Time period = 2πmk=2 s
Number of oscillations = 10,0002=5,000

Topic: Physics, Electromagnetic Induction, JEE Main, JEE Main 2019 (08 Apr Shift 1)
A thin transparent sheet is placed in front of Young's double-slit. The fringe-width will

Options:
A Increase
B Decrease
C Remain same
D Become non-uniform

The correct answer is:
Remain same

Solution:
By placing a sheet optical path length changes by μ - 1 t , using that justify that by placing a sheet fringe width remains same only entire wave pattern shift's by a distance Δy = μ - 1 tD d

Topic: Physics, Wave Optics, NEET
A thin tube sealed at both ends is $100 \mathrm{~cm}$ long. It lies horizontally, the middle $20 \mathrm{~cm}$ containing mercury and two equal ends containing air at standard atmospheric pressure. If the tube is now turned to a vertical position, by what amount will the mercury be displaced ?
(Given : cross-section of the tube can be assumed to be uniform)

Options:
A
$2.95 \mathrm{~cm}$

B
$5.18 \mathrm{~cm}$

C
$8.65 \mathrm{~cm}$

D
$0.0 \mathrm{~cm}$

The correct answer is:
$5.18 \mathrm{~cm}$

Solution:
No Solution Available

Topic: Physics, Mechanical Properties of Fluids, JEE Main, JEE Main 2013 (23 Apr Online)
A thin uniform annular disc of mass M, has an outer radius of 4R and an inner radius of 3R as shown in the figure. The work required to take a unit mass from point P on its axis to infinity is

Options:
A 2 GM 7 R 4 2 - 5
B -2GM7R(42-5)
C GM 4 R
D 2 GM 5 R 2 - 1

The correct answer is:
2 GM 7 R 4 2 - 5

Solution:
W = Δ U = U f - U i = U ∞ - U P
= - U P = - m V P = - V P as m = 1
Upon integration, potential at point P will be obtained as given below
Let dM be the mass of small ring as shown
dM = M π 4 R 2 - π 3 R 2 2 π r dr = 2 Mr dr 7 R 2
dV P = - G d M 1 6 R 2 + r 2
= - 2 GM 7 R 2 ∫ 3 R 4 R r 1 6 R 2 + r 2 dr
=-2GM7R(42-5)
∴ W = + 2 GM 7 R 4 2 - 5

Topic: Physics, Gravitation, JEE Main
A thin uniform bar of length $\mathrm{L}$ and mass $8 \mathrm{~m}$ lies on a smooth horizontal table. Two point masses $\mathrm{m}$ and $2 \mathrm{~m}$ moving in the same horizontal plane from opposite sides of the bar with speeds $2 \mathrm{v}$ and $v$ respectively. The masses stick to the bar after collision at a distance $\frac{\mathrm{L}}{3}$ and $\frac{\mathrm{L}}{6}$ respectively from the centre of the bar. If the bar starts rotating about its center of mass as a result of collision, the angular speed of the bar will be:

Options:
A
$\frac{\mathrm{v}}{6 \mathrm{~L}}$

B
$\frac{6 \mathrm{v}}{5 \mathrm{~L}}$

C
$\frac{3 \mathrm{v}}{5 \mathrm{~L}}$

D
$\frac{\mathrm{v}}{5 \mathrm{~L}}$

The correct answer is:
$\frac{\mathrm{v}}{6 \mathrm{~L}}$

Solution:
No Solution Available

Topic: Physics, Rotational Motion, JEE Main, JEE Main 2018 (15 Apr Shift 2 Online)
A thin uniform equilateral plate rests in a vertical plane with one of its vertex A on a rough horizontal floor and another vertex B on a smooth vertical wall. If the coefficient of friction Î¼=13, then the least angle Î¸ its base AB can make with the horizontal surface is

Options:
A θ=cot-1μ+13
B θ=tan-1μ+13
C θ=tan-12μ+13
D Î¸=cot-12Î¼+13

The correct answer is:
θ=cot-12μ+13

Solution:
Let us assume that the side-length of the triangle is a, then

l=a3

N1=Mg

f1=Î¼N1=N2

N2=Î¼Mg

Balancing the torque about point A

MgÃlcos30Â°+Î¸-N2ÃasinÎ¸=0

âMgÃa3cos30Â°+Î¸-Î¼MgÃasinÎ¸=0

1332cosÎ¸-12sinÎ¸=Î¼sinÎ¸

On solving further we get

Î¸=cot-12Î¼+13

Topic: Physics, Rotational Motion, JEE Main
A thin uniform rod of length 2 m, cross sectional area A and density d is rotated about an axis passing through the centre and perpendicular to its length with angular velocity ω. If value of ω in terms of its rotational kinetic energy E is αEAd, then the value of α is ______.

Options:

The correct answer is:
3

Solution:
Rotational kinetic energy is given by E=12Iω2, here, I is moment of inertia of rod about an axis through its centre and perpendicular to it.

So, E=12ml212ω2

⇒E=12dAl312ω2 ∵mass=density×volume

⇒E=dA(2)324ω2

⇒ω=3EdA

Thus, the value of α=3.

Topic: Physics, Rotational Motion, JEE Main, JEE Main 2023 (30 Jan Shift 1)
A thin uniform rod of length $\ell$ and mass $m$ is swinging freely about a horizontal axis passing through its end. Its maximum angular speed is $\omega$. Its centre of mass rises to a maximum height of

Options:
A
$\frac{1}{3} \frac{\ell^2 \omega^2}{g}$

B
$\frac{1}{6} \frac{\ell \omega}{\mathrm{g}}$

C
$\frac{1}{2} \frac{\ell^2 \omega^2}{g}$

D
$\frac{1}{6} \frac{\ell^2 \omega^2}{\mathrm{~g}}$

The correct answer is:
$\frac{1}{6} \frac{\ell^2 \omega^2}{\mathrm{~g}}$

Solution:
$$
\begin{aligned}
& \text { } \\
& \text { T.E }=T . E_{\mathrm{f}} \\
& \frac{1}{2} I \omega^2=m g h \\
& \frac{1}{2} \times \frac{1}{3} m \ell^2 \omega^2=m g h \Rightarrow h=\frac{1}{6} \frac{\ell^2 \omega^2}{\mathrm{~g}}
\end{aligned}
$$

Topic: Physics, Rotational Motion, JEE Main, JEE Main 2009
A thin uniform rod,pivoted at O, is rotating in the horizontal plane with constant angular speed Ï, as shown in the figure. At time t = 0, a small insect starts from O and moves with constant speed v,with respect to the rod towards other end. It reaches the end of the rod at t = T and stops.The angular speed of the system remains Ï throughout. The magnitude of the torque Ï â about O,as a function of time is best represented by which plot?

Options:
A
B
C
D

The correct answer is:

Solution:
angular momentum
J=Iw
z=dJdt=wdIDt
[I=Irod+Iinsect
=C+mr2âc+m(vt)2]
Ï = Ï dI dt = Ï d dt C + mv 2 t 2
= m Ï v 2 2 t
So, the graph between Ï and t is straight line.

Topic: Physics, Rotational Motion, JEE Main
A thin uniform tube is bent into a circle of radius r in the vertical plane. Equal volumes of two immiscible liquids, Whose densities are ρ1 and ρ2 (ρ1>ρ2 ), fill half the circle. The angle θ between the radius vector passing through the common interface and the vertical is:

Options:
A Î¸=tan-1â¡Ï2Ï1+Ï2Ï1-Ï2
B Î¸=tan-1â¡Ï1-Ï2Ï1+Ï2
C Î¸=tan-1â¡Ï2Ï2Ï1
D Î¸=tan-1â¡ÏÏ1Ï2

The correct answer is:
Î¸=tan-1â¡Ï1-Ï2Ï1+Ï2

Solution:
Pressure at the point A is the same on both sides of the tube,
ρ 1 gr(cosθ−sinθ)= ρ 2 gr(sinθ+cosθ)
⇒ρ1ρ2=sinθ+cosθcosθ-sinθ=tanθ+11-tanθ
⇒ρ1-ρ1tanθ=ρ2+ρ2tanθ
⇒ρ1+ρ2tanθ=ρ1-ρ2
θ=tan-1⁡ρ1-ρ2ρ1+ρ2

Topic: Physics, Mechanical Properties of Fluids, JEE Main, JEE Main 2018 (15 Apr)
A thin uniform tube is bent into a circle of radius $r$ in the virtical plane. Equal volumes of two immiscible liquids, whose densities are $\rho_1$ and $\rho_2\left(\rho_1>\rho_2\right)$ fill half the circle. The angle $\theta$ between the radius vector passing through the common interface and the vertical is

Options:
A
$\theta=\tan ^{-1}\left[\frac{\pi}{2}\left(\frac{\rho_1-\rho_2}{\rho_1+\rho_2}\right)\right]$

B
$\theta=\tan ^{-1} \frac{\pi}{2}\left(\frac{\rho_1+\rho_2}{\rho_1-\rho_2}\right)$

C
$\theta=\tan ^{-1} \pi\left(\frac{\rho_1}{\rho_2}\right)$

D
$\theta=\tan ^{-1} \frac{\pi}{2}\left(\frac{\rho_2}{\rho_1}\right)$

The correct answer is:
$\theta=\tan ^{-1}\left[\frac{\pi}{2}\left(\frac{\rho_1-\rho_2}{\rho_1+\rho_2}\right)\right]$

Solution:
Pressure at interface A must be same from both the sides to be in equilibrium.
$\begin{aligned} \therefore \quad &(\mathrm{R} \cos \theta+\mathrm{R} \sin \theta) \rho_2 \mathrm{~g} \\ &=(\mathrm{R} \cos \theta-\mathrm{R} \sin \theta) \rho_1 \mathrm{~g} \end{aligned}$
$\Rightarrow \quad \frac{\rho_1}{\rho_2}=\frac{\cos \theta+\sin \theta}{\cos \theta-\sin \theta}=\frac{1+\tan \theta}{1-\tan \theta}$
$\Rightarrow \rho_1-\rho_1 \tan \theta=\rho_2+\rho_2 \tan \theta$
$\Rightarrow \quad\left(\rho_1+\rho_2\right) \tan \theta=\rho_1-\rho_2$
$\therefore \quad \theta=\tan ^{-1}\left(\frac{\rho_1-\rho_2}{\rho_1+\rho_2}\right)$, soclosest answer is (a)

Topic: Physics, Mechanical Properties of Fluids, JEE Main, JEE Main 2018 (15 Apr Shift 1 Online)
A thin wiform circular disc of mass $M$ and radius $R$ is rotating in a horizontal plane about an axis passing through its centre and perpendicular to its plane with an angular velocity $\omega$. Another disc of same thickness and radius but of mass $\frac{1}{8} M$ is placed gently on the first disc coaxially. The angular velocity of the system is now

Options:
A $\frac{8}{9} w$
B $\frac{5}{9} w$
C $\frac{1}{3} w$
D $\frac{2}{9} w$

The correct answer is:
$\frac{8}{9} w$

Solution:
Given, mass of first circular disc $\left(m_1\right)=M$
Radius of first circular disc $\left(r_1\right)=R$
Mass of second circular disc $\left(m_2\right)=\frac{1}{8} M$
For first disc,
$$
\begin{aligned}
& I_1=\frac{M R^2}{2} \\
& \Rightarrow \quad L_1=\frac{M R^2 \omega_1}{2} \\
&
\end{aligned}
$$
For first and second disc system,
$$
I_2=\frac{M R^2}{2}+\frac{M}{8} \cdot \frac{R^2}{2}
$$
or,
$$
\begin{aligned}
& I_2=\frac{9 M R^2}{8 \times 2} \\
& L_2=\frac{9 M R^2}{8 \times 2} \omega_2
\end{aligned}
$$
Now from conservation of angular momentum,
$$
\begin{aligned}
L_1 & =L_2 \\
\Rightarrow \quad \frac{M R^2 \omega_1}{2} & =\frac{9 M R^2}{8 \times 2} \omega_2 \Rightarrow \omega_1=\frac{9}{8} \omega_2
\end{aligned}
$$
Given,
$$
\begin{aligned}
& \omega_2=\omega \\
& \omega_2=\frac{8}{9} \omega_1=\frac{8}{9} \omega
\end{aligned}
$$

Topic: Physics, Rotational Motion, JEE Main
A thin wire of length $L$ and uniform linear mass density $\rho$ |is bent into a circular loop with centre at 0 as shown. The moment of inertia of the loop about an axis $\mathrm{XY}$ is

Options:
A $\frac{\rho L^3}{8 \pi^2}$
B $\frac{\rho L^3}{16 \pi^2}$
C $\frac{5 \rho L^3}{8 \pi^2}$
D $\frac{3 \rho L^3}{8 \pi^2}$

The correct answer is:
$\frac{3 \rho L^3}{8 \pi^2}$

Solution:
The moment of inertia of a thin hoop about its diameter is \(\frac{1}{2} \mathrm{MR}^2\) here, \(\mathrm{M}=\mathrm{LP}\) also, we have
\(\begin{aligned}
& 2 \pi \mathrm{R}=\mathrm{L} \\
& \Rightarrow \mathrm{R}=\frac{\mathrm{L}}{2 \pi}
\end{aligned}\)
So, we have
\(\mathrm{I}=\frac{1}{2} \mathrm{MR}^2=\frac{1}{2} \mathrm{LP}\left(\frac{\mathrm{L}}{2 \pi}\right)^2=\frac{\mathrm{L}^3 \mathrm{P}}{8 \pi^2}\)
Now using parallel axis theorem we have
\(\begin{aligned}
& \mathrm{L}_{\mathrm{xx}^1}=\mathrm{I}_{\mathrm{cm}}+\mathrm{MR}^2=\frac{\mathrm{L}^3 \mathrm{P}}{8 \pi^2}+\mathrm{LP}\left(\frac{\mathrm{L}}{2 \pi}\right)^2 \\
& =\frac{\mathrm{L}^3 \mathrm{P}}{8 \pi^2}+\frac{\mathrm{L}^3 \mathrm{P}}{4 \pi^2}=\frac{3 \mathrm{~L}^3 \mathrm{P}}{8 \pi^2}
\end{aligned}\)

Topic: Physics, Rotational Motion, JEE Main
A thin wire of length of $99 \mathrm{~cm}$ is fixed at both ends as shown in the figure. The wire is kept under a tension and is divided into three segments of lengths $l_1, l_2$ and $l_3$ as shown in figure. When the wire is made to vibrate, the segments vibrate respectively with their fundamental frequencies in the ratio $1: 2: 3$. Then, the lengths $l_1, l_2$ and $l_3$ of the segments respectively are (in cm)

Options:
A $27,54,18$
B $18,27,54$
C $54,27,18$
D $27,9,14$

The correct answer is:
$54,27,18$

Solution:
Ratio of fundamental frequency
$$
\left(n_1: n_2: n_3\right)=1: 2: 3
$$
$$
\begin{aligned}
l_1: l_2: l_3 & =\frac{1}{n_1}: \frac{1}{n_2}: \frac{1}{n_3} \\
& =\frac{1}{1}: \frac{1}{2}: \frac{1}{3}=6: 3: 2 \\
l_1 & =\frac{6 \times 99}{11}=54 \\
l_2 & =\frac{3 \times 99}{11}=27 \\
l_3 & =\frac{2 \times 99}{11}=18
\end{aligned}
$$

Topic: Physics, Waves and Sound, JEE Main
A thread is tied slightly loose to a wire frame as in figure and the frame is dipped into a soap solution and taken out. The frame is completely covered with the film. When the portion $A$ punctured with a pin, the thread.

Options:
A (a) Becomes concave toward $A$
B Becomes convex towards $A$
C Remains in the initial position
D Either (a) or (b) depending on the size of $A$ w.r.t. $B$

The correct answer is:
(a) Becomes concave toward $A$

Solution:
Because film tries to cover minimum surface area

Topic: Physics, Mechanical Properties of Fluids, JEE Main
A thread is tied slightly loose to a wire frame as in figure and the frame is dipped into a soap solution and taken out. The frame is completely covered with the film. When the portion $A$ is punctured with a pin, the thread

Options:
A becomes concave towards $A$
B becomes convex towards $A$
C remains in the initial position
D Either (a) or (b) depending on the size of $A$ w.r.t. $B$

The correct answer is:
becomes concave towards $A$

Solution:
The potential energy due to surface tension is given by the formula $U=S \times A$
where, $S$ is the surface tension and $A$ is the area of film. Since after puncturing, the film will try to minimise its potential energy and hence the surface area. Thus, the thread will become concave towards A.

Topic: Physics, Mechanical Properties of Fluids, COMEDK, COMEDK 2019
A thread is tied slightly loose to a wire frame as in figure and the frame is dropped into a soap solution and taken out. The frame is completely covered with the film. When the portion \(A\) is punctured with a pin, the thread

Options:
A becomes concave towards \(A\)
B becomes convex towards \(A\)
C either (~a) or (~b) depending on the size of \(A\) with respect to \(B\)
D remains in the initial position.

The correct answer is:
either (~a) or (~b) depending on the size of \(A\) with respect to \(B\)

Solution:
No Solution Available

Topic: Physics, Mechanical Properties of Fluids, AIIMS, AIIMS 2008
A thread is tied slightly loose to a wire frame as shown in figure and the frame is dipped into a soap solution and taken out. The frame is completely covered with the film. When the portion $A$ is punctured with a pin, the thread

Options:
A becomes concave towards $A$
B becomes convex towards $5 A$
C Either (a) or (b) depending on the size of $A$ with respect to $B$
D remains in the initial position

The correct answer is:
becomes concave towards $A$

Solution:
When portion $A$ is punctured, the soap solution comes out of the hole. So, due to surface tension, the solution in part $B$ tries to pull it toward itself. Thus, the thread becomes concave towards $A$.

Topic: Physics, Mechanical Properties of Fluids, COMEDK, COMEDK 2013
A tightly wound solenoid of radius âaâ and length âlâ has n turns per unit length. It carries an electric current i. Magnetic field at a distance l4 from one of the end (inside the solenoid on its axis) is B=Î¼0ni5+3K for l=4a . Then find the value of K.

Options:

The correct answer is:
40

Solution:
Hint: B=Î¼0ni2 sinÎ¸1+sinÎ¸2
â B=Î¼0ni212+310
â B=Î¼0ni5+340
â´ K=40

Topic: Physics, Magnetic Effects of Current, JEE Main
A time dependent force F=6t acts on a particle of mass 1 kg. If the particle starts from the rest, the work done by the force during the first 1 sec will be:

Options:
A 18 J
B 4.5 J
C 22 J
D 9 J

The correct answer is:
4.5 J

Solution:
m=1 kg
F=6t
∴ a=Fm=6t
dvdt=6t
∴ ∫0v0dv=∫016t·dt
∴ v0=6×122=3 m s-1
Δw=kf-ki=12×1×9-0=4.5 J

Topic: Physics, Work Power Energy, JEE Main, JEE Main 2017 (02 Apr)
A tiny spherical oil drop carrying a net charge q is balanced in still air with vertical uniform electric field of strength 81Ï7Ã105 V m-1.When the field is switched off, the drop is observed to fall with terminal velocity 2Ã10-3ms-1. Given g=9.8 ms-2, viscosity of the air =1.8Ã10-5Ns m-2 and the density of oil =900 kg m-3, the magnitude of q is

Options:
A 1.6×10-19 C
B 3.2×10-19 C
C 4.8×10-19 C
D 8.0×10-19 C

The correct answer is:
8.0×10-19 C

Solution:
qE =mg ...(i)
6πη rv=mg
43πr3 ρg=mg ...(ii)
∴r=3mg4πρg1/3 ...(iii)
Substituting the value of r in Eq. (ii), we get
6πηv3mg4πρg1/3=mg
or 6πηv3 3mg4πρg =mg3
Again substituting mg=qE, we get
qE2=34πρg6πηv3
Or qE=34πρg1/2 6πηv3/2
∴ q=1E34πρg126πηv3/2
Substituting the values, we get
q=781π×10534π×900×9.8×216π3
×1.8×10-5×2×10-33 =8.0×10-19 C

Topic: Physics, Electromagnetic Waves, JEE Main
A toroid of $n$ turns, mean radius $R$ and cross-sectional radius a carries current $I$. It is placed on a horizontal table taken as $x y$-plane. Its magnetic moment $m$

Options:
A is non-zero and points in the z-direction by symmetry
B points along the axis of the toroid $(m=m \phi)$
C is zero, otherwise there would be a field falling as $\frac{1}{r^3}$ at large distances outside the toroid
D is pointing radially outwards

The correct answer is:
is zero, otherwise there would be a field falling as $\frac{1}{r^3}$ at large distances outside the toroid

Solution:
We know that a toroid is a form of solenoid which is bend in the form of a circular coil. The magnetic field due to the toroid remains inside the circular coil of the toroid and it remains in the form of the concentric circles.
The magnetic field outside the toroid is zero and it varies as inversely proportional to the cube of distance of the point.
i.e.$\left(M \propto \frac{1}{r^3}\right)$

Topic: Physics, Magnetic Effects of Current, AP EAMCET, AP EAMCET 2021 (24 Aug Shift 2)
A torque meter is calibrated to reference standards of mass, length and time each with 5% accuracy. After calibration, the measured torque with this torque meter will have net accuracy of

Options:
A 15%
B 25%
C 75%
D 5%

The correct answer is:
25%

Solution:
Dimensional formula for Torque

τ=ML2 T-2

Now the percentage error in torque can be calculated by

τ%=1×∆MM+2×∆LL+2×∆TT×100⇒τ%=25%

Topic: Physics, Mathematics in Physics, JEE Main, JEE Main 2022 (27 Jul Shift 1)
A total of 6 Boys and 6 girls are to sit in a row alternatively and in a circle. Let m be the number of arrangements in the row and n be the number of arrangements in the circle. If k=m10n, then the value of k is

Options:

The correct answer is:
1.2

Solution:
Linear permutations with boy in the first place are of the form B1G1B2G2B3G3B4G4B5G5B6G6 and the number of such arrangements is 6!Ã6!.
The number of linear permutations with girl in the first place is 6!Ã6!
So, the number of arrangements in row =m=2Ã6!Ã6!
(In circular permutations, start with a place which can be filled by a boy or a girl and after that the arrangements become linear)

Placing the boys first and then arranging the girls in 6 gaps, the number of such circular arrangements is 5!Ã6!ân=5!Ã6!
âk=m10n=2Ã6!Ã6!10Ã5!Ã6!=1210=1.2

Topic: Mathematics, Permutation Combination, JEE Main
A total of $48 \mathrm{~J}$ heat is given to one mole of helium kept in a cylinder. The temperature of helium increases by $2^{\circ} \mathrm{C}$. The work done by the gas is:
Given, $\mathrm{R}=8.3 \mathrm{~J} \mathrm{~K}^{-1} \mathrm{~mol}^{-1}$.

Options:
A $24.9 \mathrm{~J}$
B $72.9 \mathrm{~J}$
C $48 \mathrm{~J}$
D $23.1 \mathrm{~J}$

The correct answer is:
$23.1 \mathrm{~J}$

Solution:
$1^{\text {st }}$ law of thermodynamics
$\begin{aligned}
& \Delta \mathrm{Q}=\Delta \mathrm{U}+\mathrm{W} \\
& \Rightarrow+48=\mathrm{nC}_{\mathrm{v}} \Delta \mathrm{T}+\mathrm{W} \\
& \Rightarrow 48=(1)\left(\frac{3 \mathrm{R}}{2}\right)(2)+\mathrm{W} \\
& \Rightarrow \mathrm{W}=48-3 \times \mathrm{R} \\
& \Rightarrow \mathrm{W}=48-3 \times(8.3) \\
& \Rightarrow \mathrm{W}=23.1 \text { Joule }
\end{aligned}$

Topic: Physics, Thermodynamics, JEE Main, JEE Main 2024 (06 Apr Shift 2)
A totally reflecting small plane mirror placed horizontally faces a parallel beam of light as shown in the figure. The mass of the mirror is 200 g. Assume that there is no absorption in the lens and that 20% of the light emitted by the source goes through the lens. The power of the source needs to support the weight of the mirror is P Ã 108 watt, where P is [take g = 10 m s-2]

Options:

The correct answer is:
15

Solution:
mg=2PC
P'=200×10-3×10×108×32
P‘ = 300 × 106 = 300 MW
P=P'0.2=1500×106=15×108 W

Topic: Physics, Electromagnetic Waves, JEE Main
A tower AB leans towards west making an angle Î± with the vertical. The angular elevation of B, the top most point of the tower, is 60o as observed from a point C due east of A at a distance of 10 ft from A. If the angular elevation of B from a point D due east of C at a distance of 20 ft from C is 45o, then the value of 2tanâ¡Î± is equal to

Options:
A 3+1
B 3+13-1
C 3-1
D 3-13+1

The correct answer is:
3-1

Solution:
By m-n theorem, we have

30cotâ¡60o=10cotâ¡45o-20cotÏ2+Î±
⇒103=10+20tanâ¡Î±
⇒2tanâ¡Î±=3-1

Topic: Mathematics, Heights and Distances, JEE Main
A tower leans towards west making an angle α with the vertical. The angular elevation of B, the top most point of the tower, is 75o as observed from a point C due east of A at a distance of 20 units. If the angular elevation of B from a point due east of C at a distance of 20 units from C is 45o, then tan⁡α is equal to

Options:
A 3+12
B 3+13-1
C 3-23
D 3-12

The correct answer is:
3-23

Solution:
By m–n theorem, we have
40cot⁡75o=20cot⁡45o-20cot⁡π2+α
⇒402-3=20+20tan⁡α
⇒tan⁡α=3-23

Topic: Mathematics, Heights and Distances, JEE Main
A tower PQ stands on a horizontal ground with base Q on the ground. The point R divides the tower in two parts such that QR=15m. If from a point A on the ground the angle of elevation of R is 60° and the part PR of the tower subtends an angle of 15° at A, then the height of the tower is

Options:
A 523+3m
B 53+3m
C 103+1m
D 1023+1m

The correct answer is:
523+3m

Solution:
Plotting the diagram of the given values we get,

From the right triangle RQS,

tan60°=15a

a=153=53

In right triangle PQS,

tan75°=ha

⇒h53=2+3

⇒h=103+15

⇒h=523+3

Topic: Mathematics, Heights and Distances, JEE Main, JEE Main 2022 (25 Jul Shift 1)
A tower stands at the centre of a circular park. $A$ and $B$ are two points on the boundary of the park such that $A B(=a)$ subtends an angle of $60^{\circ}$ at the foot of the tower, and the angle of elevation of the top of the tower from $A$ or $B$ is $30^{\circ}$. The height of the tower is

Options:
A
$\frac{2 a}{\sqrt{3}}$

B
$2 a \sqrt{3}$

C
$\frac{a}{\sqrt{3}}$

D
$a \sqrt{3}$

The correct answer is:
$\frac{a}{\sqrt{3}}$

Solution:
$\triangle \mathrm{O A B}$ is equilateral
$\therefore \mathrm{OA}=\mathrm{OB}=\mathrm{AB}=\mathrm{a}$
Now $\tan 30^{\circ}=\frac{h}{a}$
$\therefore h=\frac{a}{\sqrt{3}} \text {. }$

Topic: Mathematics, Heights and Distances, JEE Main, JEE Main 2007
A tower subtends an angle 75o at a point on the same level as the foot of the tower and at another point, 10 meters above the first, the angle of depression of the foot of the tower is 15o. The height of the tower is (in meters)

Options:
A 103+12
B 103-12
C 102+32
D None of these

The correct answer is:
102+32

Solution:
From the figure, in right-angled ∆ABD,
ABBD=tan⁡75o
∴hx=tan⁡75o⇒h=xtan⁡75o……..i
In right-angled ∆BCE,

BEEC=tan⁡15o
⇒10x=tan⁡15o∴x=10tan⁡15o……ii
From i and ii, we get,
∴h=10tan⁡15o×tan⁡75o

=102+32 meters

Topic: Mathematics, Heights and Distances, JEE Main