datavorous/entrance-exam-dataset huggingface dataset:
A thin liquid film formed between a U-shaped wire and a light slider supports a weight of $1.5 \times 10^{-2} \mathrm{~N}$ (see figure). The length of the slider is $30 \mathrm{~cm}$ and its weight negligible. The surface tension of the liquid film is

Options:
A
$0.0125 \mathrm{~Nm}^{-1}$

B
$0.1 \mathrm{~Nm}^{-1}$

C
$0.05 \mathrm{~Nm}^{-1}$

D
$0.025 \mathrm{~Nm}^{-1}$

The correct answer is:
$0.025 \mathrm{~Nm}^{-1}$

Solution:
The force of surface tension acting on the slider balances the force due to the weight.
$\begin{array}{ll}
\Rightarrow & \mathrm{F}=2 \mathrm{~T} \ell=\mathrm{w} \\
\Rightarrow & 2 \mathrm{~T}(0.3)=1.5 \times 10^{-2} \\
\Rightarrow & \mathrm{T}=2.5 \times 10^{-2} \mathrm{~N} / \mathrm{m}
\end{array}$

Topic: Physics, Mechanical Properties of Fluids, JEE Main, JEE Main 2012 (Offline)
A thin metal wire of density Ï floats on water surface horizontally. If it is NOT to sink in water then maximum radius of wire is proportional to ( T= surface tension of water, g= gravitational acceleration)

Options:
A 2Tπρg
B πρgT
C Tπρg
D πρgT

The correct answer is:
2Tπρg

Solution:
Mg=2Tl
π r 2 lρg=2Tl
r 2 = 2T ÏÏg
r= 2T ÏÏg

Topic: Physics, Mechanical Properties of Fluids, JEE Main
A thin non conducting disc of mass M= 2 kg, charge Q=2Ã10-2 C and radius R=16 m is placed on a frictionless horizontal plane with its centre at the origin of the coordinate system. A non-uniform, radial magnetic field Bâ=B0 r^ exists in space, where B0=10 T and r^ is a unit vector in the radially outward direction. The disc is set in motion with an angular velocity Ï=xÃ102 rad s-1, about an axis passing through its centre and perpendicular to its plane, as shown in the figure. At what value of x, the disc will lift off from the surface.

Options:

The correct answer is:
9

Solution:
The force an any small part of the disc is in the vertically upward direction
dF=QπR22πr drωr B0
dF=2QωB0R2 r2 dr
F=23QωB0R=Mg
⇒ω=3Mg2QB0R=3×2×102×2×10-2×10×16=9×102 rad s-1

Topic: Physics, Magnetic Effects of Current, JEE Main
A thin piece of thermal conductor of constant thermal conductivity insulated on the lateral sides connects two reservoirs which are maintained at temperatures $T_{1}$ and $T_{2}$ as shown. Assuming that the system is in steady state, which of the following plots best represents the dependence of the rate of change of entropy of the ratio of temperatures $\mathrm{T}_{1} / \mathrm{T}_{2}$

Options:
A
B
C
D

The correct answer is:

Solution:
$\begin{aligned} \mathrm{ds} &=\frac{-\mathrm{Qdt}}{\mathrm{T}_{1}}+\frac{\mathrm{Qdt}}{\mathrm{T}_{2}} \\ \frac{\mathrm{ds}}{\mathrm{dt}} &=-\mathrm{Q}\left[\frac{1}{\mathrm{~T}_{1}}-\frac{1}{\mathrm{~T}_{2}}\right] \\ &=\frac{-\left(\mathrm{T}_{1}-\mathrm{T}_{2}\right)}{\mathrm{R}}\left[\frac{\mathrm{T}_{2}-\mathrm{T}_{1}}{\mathrm{~T}_{1} \mathrm{~T}_{2}}\right] \\ &=+\left(\frac{\mathrm{T}_{1}^{2}-\mathrm{T}_{2}^{2}}{\mathrm{~T}_{1} \mathrm{~T}_{2}}\right) \frac{1}{\mathrm{R}} \\ \frac{\mathrm{ds}}{\mathrm{dt}} &=\left(\frac{\mathrm{T}_{1}}{\mathrm{~T}_{2}}-\frac{\mathrm{T}_{2}}{\mathrm{~T}_{1}}\right) \frac{1}{\mathrm{R}} \\ &=\left(\mathrm{x}-\frac{1}{\mathrm{x}}\right) \frac{1}{\mathrm{R}} \end{aligned}$

Topic: Physics, Thermodynamics, KVPY, KVPY 2017 (5 Nov SB/SX)
A thin plank of mass $m$ and length $I$ is pivoted at one end and it is held stationary in horizontal position by means of a light thread as shown in the figure then find out the force on the pivot.

Options:
A $9 \mathrm{mg} / 4$
B $2 \mathrm{mg} / 3$
C $\mathrm{mg} / 2$
D $\mathrm{mg} / 4$

The correct answer is:
$\mathrm{mg} / 2$

Solution:
Free body diagram of the plank is shown in figure.
$\therefore$ Plank is in equilibrium condition
So $F_{\text {net }} \& T_{\text {net }}$ on the plank is zero
(i) from $F$
$\begin{aligned}
& F_{\text {net }}=O \\
& \Rightarrow F_{\text {net } \times}=0 \\
& N_1=O
\end{aligned}$
$\begin{aligned}
& \text { Now } \mathrm{F}_{\text {net, }}=\mathrm{O} \\
& \quad \Rightarrow \mathrm{N}_2+\mathrm{T}=\mathrm{mg} \ldots \text { (i) } \\
& \text { from } \tau_{\text {net }}=\mathrm{O} \\
& \Rightarrow \tau_{\text {net }} \text { about point } \mathrm{A} \text { is zero } \\
& \text { so } \mathrm{N}_2 \cdot \ell=\mathrm{mg} \cdot \ell / 2 \\
& \quad \Rightarrow \quad \mathrm{N}_2=\frac{\mathrm{mg}}{2}
\end{aligned}$

Topic: Physics, Rotational Motion, JEE Main
A thin Plano-convex glass lens (Î¼= 1.5) has its plane surface silvered and R is the radius of curvature of curved part, then which of the following ray diagram is true for an object placed at O?
(O is at a distance of 2R from the lens)

Options:
A
B
C
D None of the above

The correct answer is:

Solution:
For Refraction at first surface 1.5v-1u=1.5-1R
u=-2R,
v=â
Therefore, rays strike the mirror normally. So, by principle of reversibility, O and I coincide.

Topic: Physics, Ray Optics, JEE Main
A thin plano-convex lens acts like a concave mirror of focal length 0.2 m, when silvered on its plane surface. The refractive index of the material of the lens is 1.5. The radius of curvature of the convex surface of the lens will be

Options:
A 0.1 m

B 0.2 m

C 0.4 m

D 0.8 m

The correct answer is:
0.2 m

Solution:
F=-0.2m, f m =â
Î¼ = 1.5 fl=RÎ¼-1, 1fF=1fm-2flâ¡
â F=-flâ¡2
âF=-R2Î¼-1
â´ -0.2=-R21.5-1
â´ 0.2=R1
â´ R=0.2 m

Topic: Physics, Ray Optics, JEE Main
A thin plano-convex lens fits exactly into a plano-concave lens with their plane surface parallel to each other as shown in the figure. The radius of curvature of the curved surface \( R=30 \mathrm{~cm} \). The lens is made of a different material having refractive index \( \mu_{1}=\frac{3}{2} \) and \( \mu_{2}=\frac{5}{4} \) as shown in the figure. If the plane surface of the plano-convex lens is silvered, then calculate the equivalent focal length of this system (in \( \mathrm{cm} \) ).

Options:

The correct answer is:
60.00

Solution:
(i) Peq=2Pl1+2Pl2+Pm ...1
For plano-concave lens : 1fl1=32-11∞-130
fl1=-60 cm⇒Pl1=100fl1⇒Pl1=-53 D
For plano-convex lens : â1fl2=54-1130-1∞ ⇒ 1fl2=1120
fl2=+120 cm⇒Pl2=100fl2⇒Pl2=+56 D
For plane mirror : Pm=0sincefm=∞
Effective power of the combination is
Peq=2Pl1+2Pl2+Pm
Peq=-2×53+2×56+0
⇒Peq=-106 D ∵Peq=-1feq
⇒ feq=60 cm

Topic: Physics, Ray Optics, JEE Main
A thin plano-convex lens fits exactly into a plano-concave lens with their plane surface parallel to each other as shown in the figure. The radius of curvature of the curved surface R=30 cm. The lens is made of a different material having refractive index Î¼ 1 = 3 2 and Î¼ 2 = 5 4 as shown in the figure.
If the plane surface of the plano-convex lens is silvered, then calculate the equivalent focal length of this system in cm.

Options:

The correct answer is:
60

Solution:
(i) Peq = 2 Pâ1 + 2 Pââ2 + Pm ...(1)
For plano-convex lens : 1 f ℓ 1 = 3 2 - 1 1 ∞ - 1 3 0
f ℓ 1 = - 6 0 cm ⇒ P ℓ 1 = 1 0 0 f ℓ 1 ⇒ P ℓ 1 = - 5 3 D
For plano-convex lens : â 1 f ℓ 2 = 5 4 - 1 1 30 - 1 ∞ ⇒ 1 f ℓ 2 = 1 120
f ℓ 2 = 1 2 0 cm ⇒ P ℓ 2 =100 f ℓ 2 ⇒ P ℓ 2 = 5 6 D
For plane mirror : P m = 0 since f m = ∞ ⇒ P eq = 2 P ℓ 1 + 2 P ℓ 2 + P m
P eq = - 1 0 3 + 1 0 6 + 0
⇒ P eq = - 1 0 6 D ∵Peq=-1Feq
⇒ Feq=60 cm

Topic: Physics, Ray Optics, JEE Main
A thin plastic sheet of refractive index 1.6 is used to cover one of the slits, among the two slits, in a double-slit arrangement. The central point on the screen is now occupied by what would have been the 7th bright fringe before the thin plastic sheet was used. If the wavelength of light is 600 nm, the thickness (in Î¼m) of the thin plastic sheet is

Options:
A 7
B 4
C 8
D 6

The correct answer is:
7

Solution:
The change in path =μ-1t. This path length change the position of bright fringe upto seventh bright fringe.
t( μ−1 )=7λ
t= 7( 600× 10 −9 ) ( 1.6−1 )
=7×10−6 m ⇒7 μm

Topic: Physics, Wave Optics, JEE Main
A thin prism of angle $15^{\circ}$ made of glass of refractive index $\mu_1=1.5$ is combined with another prism of glass of refractive index $\mu_2=1.75$. The combination of the prism produces dispersion without deviation. The angle of the second prism should be

Options:
A $7^{\circ}$
B $10^{\circ}$
C $12^{\circ}$
D $5^{\circ}$

The correct answer is:
$10^{\circ}$

Solution:
For without deviation
$$
\begin{aligned}
\frac{A}{A^{\prime}} & =\frac{\mu^{\prime}-1}{\mu-1} \\
\frac{15^{\circ}}{A^{\prime}} & =\frac{1.75-1}{1.5-1} \\
\frac{15}{A^{\prime}} & =\frac{0.75}{0.50} \\
A^{\prime} & =\frac{0.50 \times 15}{0.75}=10^{\circ}
\end{aligned}
$$

Topic: Physics, Ray Optics, NEET, NEET 2011 (Mains)
A thin prism of angle 5Â° is placed at a distance of 10 cm from the object. The distance of the image from the object is (given Î¼ of prism=1.5)

Options:
A π8 cm
B π12 cm
C 5Ï36 cm
D π7 cm

The correct answer is:
5Ï36 cm

Solution:
δ=μ-1A
δ=1.5-1×5°=2.5°
=2.5×π180=π72
δ≈tanδ=d10
π72=d10⇒d=5π36 cm

Topic: Physics, Ray Optics, JEE Main
A thin prism of angle 5° made of glass of refractive index μ1=1.5 is combined with another prism of glass of refractive index μ2=1.75. The combination of the prisms produces dispersion without deviation. The angle of the second prism should be

Options:
A 5°
B 7°
C 10°3
D 1.2°

The correct answer is:
10°3

Solution:
For dispersion without deviation

δ1+δ2=0

μ1-1A⁡1+μ2-1A⁡2=0

A⁡2=-μ1-1A⁡1μ2-1

Substituting the given values, we get

A⁡2=-1·5-15°1·75-1=-10°3

The negative sign shows that the two prisms must be joined in opposition.

Topic: Physics, Ray Optics, JEE Main
A thin prism P1 with an angle 6° and made of glass of refractive index 1.54 is combined with another prism P2 made from glass of refractive index 1.72 to produce dispersion without average deviation. The angle of prism P2 is :

Options:
A 6°
B 1.3°
C 7.8°
D 4.5°

The correct answer is:
4.5°

Solution:
For no average deviation,

δ1+δ2=0⇒Aμ1-1-Bμ2-1=0

⇒ 6° (1.54 – 1) = θ(1.72 – 1)

⇒θ=6°×0.540.72

=4.5°

Topic: Physics, Ray Optics, JEE Main, JEE Main 2023 (30 Jan Shift 2)
A thin rectangular magnet suspended freely has a period of oscillation equal to $T$. Now it is broken into two equal halves (each having half of the original length) and one piece is made to oscillate freely in the same field. If its period of oscillation is $T^{\prime}$, the ratio $\frac{T^{\prime}}{T}$ is

Options:
A
$\frac{1}{2 \sqrt{2}}$

B
$\frac{1}{2}$

C
2

D
$\frac{1}{4}$

The correct answer is:
$\frac{1}{2}$

Solution:
$\mathrm{T} \propto \sqrt{1}$
$\frac{T_1}{T_2}=\sqrt{\frac{I_1}{I_2}} ; \quad \mathrm{I} \propto \mathrm{L}^2 \quad\left[\Rightarrow \frac{\mathrm{I}_1}{\mathrm{I}_2}=\frac{\mathrm{L}_1^2}{\mathrm{~L}_2^2}=\frac{\left(2 \mathrm{~L}_2\right)^2}{\mathrm{~L}_2^2}=4\right] \Rightarrow \frac{\mathrm{T}_1}{\mathrm{~T}_2}=\sqrt{4}=2$
$\Rightarrow \mathrm{T}_2=\frac{\mathrm{T}_1}{2} \Rightarrow \frac{\mathrm{T}_2}{\mathrm{~T}_1}=\frac{1}{2}$

Topic: Physics, Magnetic Properties of Matter, JEE Main, JEE Main 2003
A thin ring, a disk and an annular cylinder, of same mass $M$, are released from a point $3.6 \mathrm{~m}$ from the ground up an inclined plane of $30^{\circ}$ degree inclination. The ring and the disk have the same radius $R$. Times taken by the ring and disk to reach the ground are in the ratio,

Options:
A $\sqrt{2}: \sqrt{1.5}$
B $\sqrt{1.4}: \sqrt{1.5}$
C $\sqrt{1.5}: \sqrt{1.5}$
D $\sqrt{1.5}: \sqrt{1.6}$

The correct answer is:
$\sqrt{2}: \sqrt{1.5}$

Solution:
$t=\frac{1}{\sin } \theta \sqrt{\frac{2 h}{g}\left[1+\frac{K^2}{R^2}\right]} \Rightarrow t \propto\left(1+\frac{K^2}{R^2}\right)\left|s o, t_1: t_2=\sqrt{2}: \sqrt{1.5}\right|$

Topic: Physics, Rotational Motion, JEE Main
A thin ring of 10 cm radius carries a uniformly distributed charge. The ring rotates at a constant angular speed of 40π rads-1 about its axis, perpendicular to its plane. Is the magnetic field its centre is 3.8×10-9T , then the charge carried by the ring is close to μ0=4π×10-7N/A2.

Options:
A 4×10-5 C
B 3×10-5 C
C 2×10-6 C
D 7×10-6 C

The correct answer is:
3×10-5 C

Solution:
i=qt=q2π/ω=qω2π
B=μ0i2R=μ0qω4πR
4π×10-7×q×40π4π×(0.1)=3.8×10-9
q×400π=3.8×10-2
q=3×10-5 C

Topic: Physics, Magnetic Effects of Current, JEE Main, JEE Main 2019 (12 Apr Shift 1)
A thin rod $A B$ is held horizontally so that it can freely rotate in a vertical plane about the end $A$ as shown in the figure. The potential energy of the rod when it hangs vertically is taken to be zero. The end $B$ of the rod is released from rest from a horizontal position. At the instant the rod makes an angle $\theta$ with the horizontal

Options:
A the speed of end $B$ is proportional to $\sqrt{\sin \theta}$
B the potential energy is proportional to $(1-\cos \theta)$
C the angular acceleration is proportional to $\cos \theta$
D the torque about $A$ remains the same as its initial value

The correct answers are:
the speed of end $B$ is proportional to $\sqrt{\sin \theta}$, the angular acceleration is proportional to $\cos \theta$

Solution:
Loss in potential energy = gain in kinetic energy

$\Rightarrow \quad m g \frac{L}{2} \sin \theta=\frac{1}{2} I \omega^{2}$
So $\omega \propto \sqrt{\sin \theta}$
and $\quad v \propto \sqrt{\sin \theta} \quad\left(\because K=\frac{1}{2} \mathrm{H} \omega^{2}\right)$
The speed of end $B$ is proportional to $\sqrt{\sin \theta}$
and
$$
U=m g h=m g \frac{L}{2}(1-\sin \theta) \quad(\because \tau=l \alpha)
$$
$\Rightarrow \quad m g \frac{L}{2} \cos \theta=\frac{m I^{2}}{3} \times \boldsymbol{\alpha}$
$$
\alpha \propto \cos \theta
$$
The angular acceleration is proportional to cos $\theta$

Topic: Physics, Rotational Motion, WBJEE, WBJEE 2014
A thin rod having length $L_0$ at $0^{\circ} \mathrm{C}$ and coefficient of linear expansion $\alpha$ has its two ends maintained at temperatures $\theta_1$ and $\theta_2$, respectively. Find its new length.

Options:

None

Solution:
As the temperature of rod varies from $\theta_1$ to $\theta_2$ from one end to another, so mean temperature of root at $C$ point.
Let us consider the diagram.
$$
\theta=\left(\frac{\theta_1+\theta_2}{2}\right)
$$
$\begin{array}{rrr}A & C & B \\ \theta_1 \bigcirc & \theta & \Theta^\theta\end{array}$
Let temperature varies linearly in the rod from its one end to other end. Let $\theta$ be the temperature of the mid-point $(C)$ of the rod. As steady state, so rate of flow of heat from $A$ to $C$ and $C$ to $B$ are equal. $\left(\theta_1>\theta>\theta_2\right)$
Then rate of flow of heat,
$$
\left(\frac{d Q}{d t}\right)=\frac{K A\left(\theta_1-\theta\right)}{\left(L_0 / 2\right)}=\frac{K A\left(\theta-\theta_2\right)}{\left(L_0 / 2\right)}
$$
where, $K$ is coefficient of thermal conductivity of the rod.
$$
\Rightarrow \theta_1-\theta=\theta-\theta_2 \Rightarrow \theta=\frac{\theta_1+\theta_2}{2}
$$
We know that,
$$
L=L_0(1+\alpha \theta)=L_0\left[1+\alpha\left(\frac{\theta_1+\theta_2}{2}\right)\right]
$$

Topic: Physics, Thermal Properties of Matter
A thin rod $\mathrm{MN}$, free to rotate in the vertical plane about the fixed end $\mathrm{N}$, is held horizontal. When the end $\mathrm{M}$ is released the speed of this end, when the rod makes an angle $\alpha$ with the horizontal, will be proportional to: (see figure)

Options:
A
$\sqrt{\cos \alpha}$

B
$\cos \alpha$

C
$\sin \alpha$

D
$\sqrt{\sin \alpha}$

The correct answer is:
$\sqrt{\cos \alpha}$

Solution:
When the rod makes an angle $\alpha$
$$
\begin{aligned}
&\text { Displacement of centre of mass }=\frac{l}{2} \cos \alpha \\
&m g \frac{l}{2} \cos \alpha=\frac{l}{2} I \omega^2 \\
&m g \frac{l}{2} \cos \alpha=\frac{m l^2}{6} \omega^2 \quad(\because \quad \text { M.I. of thin }
\end{aligned}
$$
uniform rod about an axis passing through its centre of mass and perpendicular to the
$$
\begin{aligned}
&\left.\operatorname{rod} I=\frac{m l^2}{12}\right) \\
&\Rightarrow \omega=\sqrt{\frac{3 g \cos \alpha}{l}}
\end{aligned}
$$
Speed of end $=\omega \times l=\sqrt{3 g \cos \alpha l}$
i.e., Speed of end, $\omega \propto \sqrt{\cos \alpha}$

Topic: Physics, Rotational Motion, JEE Main, JEE Main 2018 (15 Apr Shift 2 Online)
A thin rod of length $L$ and mass $M$ is bent at its midpoint into two halves so that the angle between them is $90^{\circ}$. The moment of inertia of the bent rod about an axis passing through the bending point and perpendicular to the plane defined by the two halves of the rod is

Options:
A $\frac{\sqrt{2} M L^2}{24}$
B $\frac{M L^2}{24}$
C $\frac{M L^2}{12}$
D $\frac{M L^2}{6}$

The correct answer is:
$\frac{M L^2}{12}$

Solution:
Distribution of masses about axis of rotation remain unchanged whether it is straight or bend.
$$
I=\frac{M L^2}{12}
$$

Topic: Physics, Rotational Motion, JEE Main
A thin rod of length $L$ and mass $M$ is bent at its midpoint into two halves so that the angle between them is $90^{\circ}$. The moment of inertia of the bent rod about an axis passing through the bending point and perpendicular to the plane defined by the two halves of the rod is

Options:
A $\frac{\sqrt{2} M L^2}{24}$
B $\frac{M L^2}{24}$
C $\frac{M L^2}{12}$
D $\frac{M L^2}{6}$

The correct answer is:
$\frac{M L^2}{12}$

Solution:
Distribution of masses about axis of rotation remain unchanged whether it is straight or bend.
$$
I=\frac{M L^2}{12}
$$

Topic: Physics, Rotational Motion, NEET, NEET 2008 (Mains)
A thin rod of length ' $L$ ' is lying along the $x$-axis with its ends at $x=0$ and $x=L$. Its linear density (mass/length) varies with $x$ ask $\left(\frac{x}{L}\right)^n$, where $n$ can be zero or any positive number. If the position $x_{\mathrm{CM}}$ of the centre of mass of the rod is plotted against ' $n$ ', which of the following graphs best approximates the dependence of $x_{\mathrm{CM}}$ on $n$ ?

Options:
A

B

C

D

The correct answer is:

Solution:
$$
\begin{aligned}
& x_{c m}=\frac{\int d m x}{\int d m}=\frac{\int \lambda d x \cdot x}{\int d m}=\frac{\int k\left(\frac{x}{L}\right)^n \cdot x d x}{\int k\left(\frac{x}{L}\right)^n d x}=\left[\frac{\frac{k x^{n+2}}{(n+2) L^n}}{\frac{k x^{n+1}}{(n+1) L^n}}\right]_0^L=\left[\frac{x(n+1)}{n+2}\right]_0^L \\
& x_{c m}=\frac{L}{2}, \frac{2 L}{3}, \frac{3 L}{4}, \frac{4 L}{5}, \frac{5 L}{6}, \ldots
\end{aligned}
$$

Topic: Physics, Center of Mass Momentum and Collision, JEE Main
A thin rod of length ' $L$ ' is lying along the $x$-axis with its ends at $x=0$ and $x=L$. Its linear density (mass/length) varies with $x$ ask $\left(\frac{x}{L}\right)^n$, where $n$ can be zero or any positive number. If the position $x_{\mathrm{CM}}$ of the centre of mass of the rod is plotted against ' $n$ ', which of the following graphs best approximates the dependence of $x_{\mathrm{CM}}$ on $n$ ?

Options:
A

B

C

D

The correct answer is:

Solution:
$$
\begin{aligned}
& x_{c m}=\frac{\int d m x}{\int d m}=\frac{\int \lambda d x \cdot x}{\int d m}=\frac{\int k\left(\frac{x}{L}\right)^n \cdot x d x}{\int k\left(\frac{x}{L}\right)^n d x}=\left[\frac{\frac{k x^{n+2}}{(n+2) L^n}}{\frac{k x^{n+1}}{(n+1) L^n}}\right]_0^L=\left[\frac{x(n+1)}{n+2}\right]_0^L \\
& x_{c m}=\frac{L}{2}, \frac{2 L}{3}, \frac{3 L}{4}, \frac{4 L}{5}, \frac{5 L}{6}, \ldots
\end{aligned}
$$

Topic: Physics, Center of Mass Momentum and Collision, JEE Main, JEE Main 2008
A thin rod of mass 0.9 kg and length 1 m is suspended, at rest, from one end so that it can freely oscillate in the vertical plane. A particle of move 0.1 kg moving in a straight line with velocity 80 m s-1 hits the rod at its bottom most point and sticks to it (see figure). The angular speed (in rad s-1) of the rod immediately after the collision will be …………

Options:

The correct answer is:
20

Solution:
Given,

Mass of the rod, M=0.9 kg

Mass of particle, m=0.1 kg

Length of the rod, l=1 m

Velocity of particle, v=80 m s-1

No external torque acting on the system of rod and particle, so angular momentum of system about the suspension is conserved, i.e.; Li=Lf

⇒mvl=Iω ...(1)

Where, I is the momentum of inertia of rod and particle system, ω is the angular velocity of system just after the collision.

⇒mvl=Ml23+ml2ω ...(2)

Substitute the values in equation (2)

⇒0.1×80×1=0.9×(1)23+0.1×(1)2×ω

ω=20 rad s-1

Topic: Physics, Center of Mass Momentum and Collision, JEE Main, JEE Main 2020 (05 Sep Shift 2)
A thin rod of negligible mass and area of cross-section 4Ã10-6 m2, suspended vertically from one end, has a length of 0.5 m at 100 Â°C. The rod is cooled to 0 Â°C but prevented from contracting by attaching a mass at the lower end. The value of this mass is
(Given, coefficient of linear expansion is 10-5 Â°C-1, Young's modulus is Y=1011 N m-2 and g=10 m s-2)

Options:

The correct answer is:
40

Solution:
The change in length due to decrease in temperature,
Δl1=LαΔθ=0·510-50-100
Δl1=-0·5×10-3 m ... (i)
Negative sign implies that length is decreasing. Now let M be the mass attached to the lower end. Then, change in length due to suspension of load is
Δl2=MgLAY=M100.54×10-61011
Δl2=1.25×10-5M ... (ii)
Δl1+Δl2=0
⇒1.25×10-5M=0.5×10-3 (from equations (i) and (ii))
⇒M=0.5×10-31.25×10-5 kg
⇒M=40 kg

Topic: Physics, Mechanical Properties of Solids, JEE Main
A thin semi-circular ring of radius $r$ has a positive charge $q$ distributed uniformly over it. The net field $\vec{E}$ at the centre $O$ is

Options:
A
$\frac{q}{4 \pi^2 \varepsilon_0 r^2} \hat{j}$

B
$-\frac{q}{4 \pi^2 \varepsilon_0 r^2} \hat{j}$

C
$-\frac{q}{2 \pi^2 \varepsilon_0 r^2} \hat{j}$

D
$\frac{\mathrm{q}}{2 \pi^2 \varepsilon_0 \mathrm{r}^2} \hat{\mathrm{j}}$

The correct answer is:
$-\frac{q}{2 \pi^2 \varepsilon_0 r^2} \hat{j}$

Solution:
Linear charge density $\lambda=\left(\frac{q}{\pi r}\right)$
$E=\int d E \sin \theta(-\hat{\mathrm{j}})=\int \frac{\mathrm{K} \cdot d q}{\mathrm{r}^2} \sin \theta(-\hat{\mathrm{j}})$
$\mathrm{E}=\frac{\mathrm{K}}{\mathrm{r}^2} \int \frac{\mathrm{qr}}{\pi r} \mathrm{~d} \theta \sin \theta(-\hat{\mathrm{j}})$
$=\frac{\mathrm{K}}{\mathrm{r}^2} \frac{\mathrm{q}}{\pi} \int_0^\pi \sin \theta(-\hat{\mathrm{j}})$
$=\frac{\mathrm{q}}{2 \pi^2 \varepsilon_0 \mathrm{r}^2}(-\hat{\mathrm{j}})$

Topic: Physics, Electrostatics, JEE Main
A thin semi-circular ring of radius $r$ has a positive charge $q$ distributed uniformly over it. The net field $\vec{E}$ at the centre $O$ is

Options:
A
$\frac{q}{4 \pi^2 \varepsilon_0 r^2} \hat{j}$

B
$-\frac{q}{4 \pi^2 \varepsilon_0 r^2} \hat{j}$

C
$-\frac{q}{2 \pi^2 \varepsilon_0 r^2} \hat{j}$

D
$\frac{\mathrm{q}}{2 \pi^2 \varepsilon_0 \mathrm{r}^2} \hat{\mathrm{j}}$

The correct answer is:
$-\frac{q}{2 \pi^2 \varepsilon_0 r^2} \hat{j}$

Solution:
Linear charge density $\lambda=\left(\frac{q}{\pi r}\right)$
$E=\int d E \sin \theta(-\hat{\mathrm{j}})=\int \frac{\mathrm{K} \cdot d q}{\mathrm{r}^2} \sin \theta(-\hat{\mathrm{j}})$
$\mathrm{E}=\frac{\mathrm{K}}{\mathrm{r}^2} \int \frac{\mathrm{qr}}{\pi r} \mathrm{~d} \theta \sin \theta(-\hat{\mathrm{j}})$
$=\frac{\mathrm{K}}{\mathrm{r}^2} \frac{\mathrm{q}}{\pi} \int_0^\pi \sin \theta(-\hat{\mathrm{j}})$
$=\frac{\mathrm{q}}{2 \pi^2 \varepsilon_0 \mathrm{r}^2}(-\hat{\mathrm{j}})$

Topic: Physics, Electrostatics, JEE Main, JEE Main 2010
A thin spherical conducting shell of radius $\mathrm{R}$ has a charge $\mathrm{q}$. Another charge $\mathrm{Q}$ is placed at the centre of the shell. The electrostatic potential at a point $\mathrm{P}$ a distance $\frac{\mathrm{R}}{2}$ from the centre of the shell is

Options:
A
$\frac{2 Q}{4 \pi \varepsilon_0 R}$

B
$\frac{2 \mathrm{Q}}{4 \pi \varepsilon_0 \mathrm{R}}-\frac{2 \mathrm{q}}{4 \pi \varepsilon_0 \mathrm{R}}$

C
$\frac{2 Q}{4 \pi \varepsilon_0 R}+\frac{q}{4 \pi \varepsilon_0 R}$

D
$\frac{(\mathrm{q}+\mathrm{Q}) 2}{4 \pi \varepsilon_0 \mathrm{R}}$

The correct answer is:
$\frac{2 Q}{4 \pi \varepsilon_0 R}+\frac{q}{4 \pi \varepsilon_0 R}$

Solution:
Potential due to spherical shell, $\mathrm{v}_1=\frac{\mathrm{q}}{4 \pi \varepsilon_0 \mathrm{R}}$. Potential difference due to charge at the centre
$\mathrm{V}_2=\frac{2 \mathrm{Q}}{4 \pi \varepsilon_0 r} ; \mathrm{V}=\mathrm{V}_1+\mathrm{V}_2=\frac{2 \mathrm{Q}}{4 \pi \varepsilon_0 \mathrm{R}}+\frac{\mathrm{q}}{4 \pi \varepsilon_0 \mathrm{R}}$

Topic: Physics, Electrostatics, JEE Main, JEE Main 2003
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