datavorous/entrance-exam-dataset huggingface dataset:
A spherical capacitor has outer sphere of radius \(5 \mathrm{~cm}\) and inner sphere of radius \(2 \mathrm{~cm}\). When the inner sphere is earthed, its capacity is \(C_1\) and when the outer sphere is earthed its capacity is \(C_2\). Then \(\frac{C_1}{C_2}\) is

Options:
A \(\frac{5}{2}\)
B \(\frac{2}{5}\)
C \(\frac{7}{3}\)
D \(\frac{3}{7}\)

The correct answer is:
\(\frac{5}{2}\)

Solution:
Key Idea If a spherical capacitor has outer radius \(R_2\) and inner radius \(R_1\). Then the capacitance,
(a) when outer shell is earthed,
\(C=4 \pi \varepsilon_0 \frac{R_1 R_2}{R_2-R_1}\)
(b) When inner shell is earthed,
\(C^{\prime}=C+4 \pi \varepsilon_0 R_2\)
Given, radius of outer sphere of capacitor, \(R_1=2 \mathrm{~cm}\) and Radius of inner sphere of capacitor, \(R_2=5 \mathrm{~cm}\)
\(\begin{aligned}
\therefore \quad C_1 & =4 \pi \varepsilon_0\left[\frac{R_1 R_2}{R_2-R_1}+R_2\right]=4 \pi \varepsilon_0\left[\frac{2 \times 5}{5-2}+5\right] \\
C_1 & =4 \pi \varepsilon_0 \frac{25}{3} \text { Farad }
\end{aligned}\)
and \(C_2=4 \pi \varepsilon_0 \frac{R_1 R_2}{R_2-R_1}=4 \pi \varepsilon_0 \frac{2 \times 5}{5-2}=4 \pi \varepsilon_0 \frac{10}{3}\)
\(\frac{C_1}{C_2}=\frac{\frac{25}{3}}{\frac{10}{3}}=\frac{25}{10}=\frac{5}{2}\)
Hence, the correct option is (a).

Topic: Physics, Capacitance, JEE Main
A spherical cavity is made in a lead sphere of radius R such that its surface touches the outside surface of lead sphere and passes through the centre. The shift in the centre of mass of the lead sphere as a result of this hollowing, is

Options:
A R 7
B R 14
C R 2
D R

The correct answer is:
R 14

Solution:
Let centre of mass of lead sphere after hollowing be at point O2, where OO2=x
Mass of spherical hollow m=43ÏR22M42ÏR3=M8 and
x=OO1=R2
â´x=MÃ0-M8ÃR2M-M8=MR167M8=-R14
â´ shift = R 1 4

Topic: Physics, Center of Mass Momentum and Collision, JEE Main
A spherical conducting shell of inner radius ' $r_1$ ' and outer radius ' $r_2$ ' has a charge ' $Q$ '. A charge $-q$ is placed at the center of the shell. The surface charge density on the inner and outer surface of the shell will be

Options:
A $\frac{\mathrm{q}}{4 \pi \mathrm{r}_1^2}$ and $\frac{\mathrm{Q}-\mathrm{q}}{4 \pi \mathrm{r}_2^2}$
B $\frac{\mathrm{q}}{4 \pi \mathrm{r}_1^2}$ and $\frac{\mathrm{Q}}{4 \pi \mathrm{r}_2^2}$
C $\frac{-\mathrm{q}}{4 \pi \mathrm{r}_1^2}$ and $\frac{\mathrm{Q}+\mathrm{q}}{4 \pi \mathrm{r}_2^2}$
D zero and $\frac{\mathrm{Q}-\mathrm{q}}{4 \pi \mathrm{r}_2^2}$

The correct answer is:
$\frac{\mathrm{q}}{4 \pi \mathrm{r}_1^2}$ and $\frac{\mathrm{Q}-\mathrm{q}}{4 \pi \mathrm{r}_2^2}$

Solution:
Due to charge $-q$ at the center of the shell, a charge $q$ will be induced on the inner surface and $-q$ on the outer surface. The charge on outer surface will become $Q-q$. Hence surface charge densities will be $\frac{\mathrm{q}}{4 \pi \mathrm{r}_1^2}$ and $\frac{\mathrm{Q}-\mathrm{q}}{4 \pi \mathrm{r}_2^2}$

Topic: Physics, Electrostatics, JEE Main
A spherical drop of capacitance $1 \mu \mathrm{F}$ is broken into eight drops of equal radius, then the capacitance of each small drop is

Options:
A $\frac{1}{2} \mu \mathrm{F}$
B $\frac{1}{4} \mu \mathrm{F}$
C $\frac{1}{8} \mu \mathrm{F}$
D $8 \mu \mathrm{F}$

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

Solution:
As, the volume remains constant, so
$\begin{array}{ll} & \frac{4}{3} \pi R^{3}=8\left(\frac{4}{3} \pi r^{3}\right) \\ \Rightarrow \quad & R=2 r \text { or } r=\frac{R}{2} \\ \text { Capacitance of spherical capacitor, } \\ \Rightarrow \quad C & =4 \pi \varepsilon_{0} R \\ \Rightarrow \quad & C R \\ \therefore \quad & \frac{C_{1}}{C_{2}}=\frac{R}{r}=\frac{2 r}{r}=2 \quad \Rightarrow \quad C_{2}=\frac{C_{1}}{2}=\frac{1}{2} \mu \mathrm{F}\end{array}$

Topic: Physics, Capacitance, COMEDK, COMEDK 2013
A spherical drop of liquid carrying charge, $Q$ has potential $V_0$ at its surface. If two drops of same charge and radius combine to form a single spherical drop, then the potential at the surface of new drop is (Assume, $V=0$ at infinity.)

Options:
A $2^{\frac{1}{3}} V_0$
B $4^{\frac{1}{3}} V_0$
C $6^{\frac{1}{3}} V_0$
D $2^{\frac{-1}{3}} V_0$

The correct answer is:
$4^{\frac{1}{3}} V_0$

Solution:
Given,
charge of spherical drop of liquid $=Q$
and potential at its surface $=V_0$
If $r$ be the radius of drop, $V_0=\frac{1}{4 \pi \varepsilon_0} \frac{Q}{r}$
If $R$ be the radius of big drop when two drops of radius $r$ are combined, then volume of big drop $=$ volume of both small drops
i.e.,
$$
\begin{aligned}
\frac{4}{3} \pi R^3 & =\frac{4}{3} \pi r^3+\frac{4}{3} \pi r^3 \\
R^3 & =2 r^3 \Rightarrow R=2^{1 / 3} r
\end{aligned}
$$
$\therefore$ charge on the big drop, $Q^{\prime}=Q+Q=2 Q$
$\therefore$ Potential at the surface of big (new) drop,
$$
\begin{aligned}
V^{\prime} & =\frac{1}{4 \pi \varepsilon_0} \frac{Q^{\prime}}{R}=\frac{1}{4 \pi \varepsilon_0} \frac{2 Q}{2^{1 / 3} r} \\
& =\frac{1}{4 \pi \varepsilon_0} \frac{Q}{r} \cdot 2^{2 / 3}=V_0 \cdot 4^{1 / 3}=4^{1 / 3} \cdot V_0
\end{aligned}
$$

Topic: Physics, Electrostatics, JEE Main
A spherical drop of liquid splits into 1000 identical spherical drops. If ui is the surface energy of the original drop and uf is the total surface energy of the resulting drops, the (ignoring evaporation), ufui=10x. Then value of x is ______:

Options:

The correct answer is:
1

Solution:
Let R be the radius of bigger drop and r be the radius of smaller drop.

Here, volume will remain conserved, so VR=Vr.

⇒43πR3=1000×43πr3

⇒R=10r

As surface energy is stated as u=SA, here, S is surface tension and A is surface area.

Thus, ui=S4πR2 and uf=S4πr2×1000

The ratio is ufui=1000r2R2

⇒ufui=101

Hence, the value of x=1.

Topic: Physics, Mechanical Properties of Fluids, JEE Main, JEE Main 2023 (25 Jan Shift 2)
A spherical drop of water has radius 1 mm If the surface tension of water is 70Ã10â3 N m-1. Difference of pressures between inside and outside of the spherical drop is

Options:
A 35 N m−2
B 70 N m−2
C 140 N m−2
D Zero

The correct answer is:
140 N m−2

Solution:
ΔP=2TR=2×70×10−31×10−3=140 N​ m-2

Topic: Physics, Mechanical Properties of Fluids, JEE Main
A spherical gas balloon of radius 16 meter subtends an angle 60° at the eye of the observer A while the angle of elevation of its center from the eye of A is 75°. Then the height (in meter) of the top most point of the balloon from the level of the observer's eye is :

Options:
A 8(2+23+2)
B 8(6+2+2)
C 8(2+2+3)
D 8(6-2+2)

The correct answer is:
8(6+2+2)

Solution:
Assume, O is the centre of sphere

and P,Q are the point of contact of tangents from A

Let T be top most point of balloon & R be foot of perpendicular from O to ground.

From triangle OAP, sin30°=16OA

⇒OA=16cosec30°=32

From triangle ARO, sin75°=OROA

⇒OR=OAsin75°

⇒32(3+1)22=86+82

So the height (in meter) of the top most point of the balloon from the level of the observer's eye is OR+OT

=8(6+2+2)

Topic: Mathematics, Heights and Distances, JEE Main, JEE Main 2021 (25 Jul Shift 1)
A spherical hole is made in a solid sphere of radius R. The mass of the sphere before hollowing was M. The gravitational field at the centre of the hole due to the remaining mass is -

Options:
A zero
B GM8R2
C GM2R2
D GMR2

The correct answer is:
GM2R2

Solution:
By the principle of superposition of fields
E→=E1→+E2→
Here, E→=net field at the centre of hole due to entire mass
E1→=field due to remaining mass
and E2→=field due to mass in hole=0
∴ E1→=E→=GMR3.r
where r=R2
∴ E→=GM2R2

Topic: Physics, Gravitation, NEET
A spherical hole is made in a solid sphere of radius R. The mass of the sphere before hollowing was M. The gravitational field at the centre of the hole due to the remaining mass is

Options:
A Zero
B GM8R2
C GM2R2
D GMR2

The correct answer is:
GM2R2

Solution:
By the principle of superposition of fields
E â = E â 1 + E â 2
Here, E â = net field at the centre of hole due to entire mass
E â 1 = field due to remaining mass
and E â 2 = field due to mass in hole = 0
â´ E 1 = E = GM R 3 . r where r = R 2
â´ E = GM 2 R 2

Topic: Physics, Gravitation, JEE Main
A spherical iron ball $10 \mathrm{~cm}$ in radius is coated with a layer of ice of uniform thickness than melts at a rate of $50 \mathrm{~cm}^3 / \mathrm{min}$. When the thickness of ice is $5 \mathrm{~cm}$, then the rate at which the thickness of ice decreases, is

Options:
A
$\frac{1}{36 \pi} \mathrm{cm} / \mathrm{min}$

B
$\frac{1}{18 \pi} \mathrm{cm} / \mathrm{min}$

C
$\frac{1}{54 \pi} \mathrm{cm} / \mathrm{min}$

D
$\frac{5}{6 \pi} \mathrm{cm} / \min$

The correct answer is:
$\frac{1}{18 \pi} \mathrm{cm} / \mathrm{min}$

Solution:
$$
\begin{aligned}
& \frac{d v}{d t}=50 \\
& 4 \pi r^2 \frac{d r}{d t}=50 \\
& \Rightarrow \frac{d r}{d t}=\frac{50}{4 \pi(15)^2} \quad \text { where } r=15 \\
& =\frac{1}{16 \pi}
\end{aligned}
$$

Topic: Mathematics, Application of Derivatives, JEE Main, JEE Main 2005
A spherical iron ball of 10cm radius is coated with a layer of ice of uniform thickness that melts at a rate of 50cm3/min . When the thickness of ice is 5cm , then the rate (in cm/min .) at which of the thickness of ice decreases, is:

Options:
A 56π
B 154π
C 136π
D 118π

The correct answer is:
118π

Solution:
Let thickness =x cm
Total volume V=43π10+x3
dVdt=4π10+x2dxdt…..(i)
Given dVdt=50 cm3/min
At x=5 cm
50=4π10+52dxdt
dxdt=118π cm/min

Topic: Mathematics, Application of Derivatives, JEE Main, JEE Main 2020 (09 Jan Shift 1)
A spherical iron ball of \( 10 \mathrm{~cm} \) radius is coated with a layer of ice of uniform thickness that melts at a rate of \( 50 \mathrm{~cm}^{3} / \mathrm{min} \).
When the thickness of ice is \( 5 \mathrm{~cm} \), then the rate (in \( \mathrm{cm} / \mathrm{min} \).) at which of the thickness of ice decreases, is:

Options:
A \( \frac{5}{6 \pi} \)
B \( \frac{1}{3 \pi} \)
C \( \frac{1}{36 \pi} \)
D \( \frac{1}{18 \pi} \)

The correct answer is:
\( \frac{1}{18 \pi} \)

Solution:
Let thickness =x cm
Total volume V=43π10+x3
dVdt=4π10+x2dxdt…..(i)
Given dVdt=50 cm3/min
At x=5 cm
50=4π10+52dxdt
dxdt=118π cm/min

Topic: Mathematics, Application of Derivatives, JEE Main
A spherical iron ball of radius 10 cm is coated with a layer of ice of uniform thickness that melts at a rate of 50 cm3/min. When the thickness of the ice is 5 cm, then the rate at which the thickness (in cm/min) of the ice decreases, is :

Options:
A 19π
B 136π
C 118π
D 56π

The correct answer is:
118π

Solution:
We know that the volume of a sphere of radius r is 43πr3.

Given the radius of the spherical iron ball is R=10 cm and the thickness of the ice is x cm.

Volume of ice: V=43πR+x3-43πR3

⇒V=43π10+x3-43π103

Differentiating with respect to t, we get

dVdt=4π10+x2dxdt-0

Given dVdt=50 cm3/min

⇒50=4π10+x2dxdt

⇒dxdt=504π10+x2

⇒dxdtx=5=504π152=118π cm/min.

Topic: Mathematics, Application of Derivatives, JEE Main, JEE Main 2019 (10 Apr Shift 2)
A spherical liquid drop is placed on a horizontal plane. A small disturbance causes the volume of the drop to oscillate. The time period of oscillation T of the liquid drop depends on the radius (r) of the drop, density Ï and surface tension S of the liquid. Which among the following will be a possible expression for the time period? where k is a dimensionless constant.

Options:
A kρrS
B kρ2rS
C kρr3S
D kρr3S2

The correct answer is:
kρr3S

Solution:
According to the question,
time period, TâraÏbSc
T=kraÏbSc ..... (i)
Thus, putting dimension, we get
[ T ]= [ L ] a [ M L â3 ] b [ M T â2 ] c
T=Mb+câLa-3bâT-2c
Equating the dimensions of both sides, we get
b+c=0, a-3b=0
and -2c=1
â´ c=-12
â´ b=12
and a=3b=32
Putting these value of a, b and c into Equation. (i), we get
T=kâr32Ï12âS-12
=kÏSr3

Topic: Physics, Units and Dimensions, JEE Main
A spherical mirror is placed 10 cm below the level of water. A point object is placed in air 30 cm above the water surface on the axis of the mirror such that two images seen by an observer above the water surface coincide. The images are formed by partial reflection at the water surface and due to emerging light after reflection from the mirror. Find the focal length of the mirror. ( Î¼water=43 )

Options:
A 1003 cm
B 75 cm
C 1108 cm
D 50 cm

The correct answer is:
75 cm

Solution:
For partial reflection, the final image should be 30 cm below water surface i.e. 20 cm below the mirror.

For the image 30 cm below surface, its actual depth should be h=3043=40cm below surface i.e. 30 cm below the mirror
For mirror, u=-10+3043

âu=-50 cm and v=30 cm

1f=-150+130=-3+510=2150

or f=75cm

convex mirror of focal length 75 cm.

(diagram drawn was just for reference actually it should be a convex mirror but it was not known so we assumed concave but at final calculation, we found focal length +ve so it must be a convex mirror)

Topic: Physics, Ray Optics, JEE Main
A spherical planet far out in space has a mass $M_0$ and diameter $D_0$. A particle of mass $m$ falling freely near the surface of this planet will experience an acceleration due to gravity which is equal to

Options:
A $G M_0 / D_0^2$
B $4 m G M_0 / D_0^2$
C $4 G M_0 / D_0^2$
D $G m M_0 / D_0^2$

The correct answer is:
$4 G M_0 / D_0^2$

Solution:
$g=\frac{G M}{R^2}=\frac{G M_0}{\left(D_0 / 2\right)^2}=\frac{4 G M_0}{D_0^2}$

Topic: Physics, Gravitation, JEE Main
A spherical planet has a mass $M_p$ and diameter $D_p$. A particle of mass $m$ falling freely near the surface of this planet will experience an acceleration due to gravity, equal to

Options:
A $4 G M_p / D_p^2$
B $G M_p m / D_p^2$
C $G M_p m / D_p^2$
D $4 G M_p m / D_p^2$

The correct answer is:
$4 G M_p / D_p^2$

Solution:
Force,
$F=\frac{G M_e m}{R^2}$
Here, $F=\frac{G M_P m}{\left(D_P / 2\right)^2}$
$\begin{aligned}
& F=\frac{4 G M_P m}{D_P^2} \\
& F=m a
\end{aligned}$
Acceleration due to gravity
$a=\frac{F}{m}=\frac{4 G M_P}{D_P^2}$

Topic: Physics, Gravitation, JEE Main
A spherical portion $A$ of radius $R$ is removed from a solid sphere $B$ of radius $2 R$ such that the centre of the removed portion is same as the centre of the sphere $B$. The ratio of the moments of inertia of the remaining spherical shell and the solid sphere B about the axes passing through their diameters is

Options:
A $31: 32$
B $7: 8$
C $15: 16$
D $4: 7$

The correct answer is:
$31: 32$

Solution:
Angular acceleration, $\alpha=\frac{8}{2}=4 \mathrm{rev} / \mathrm{s}^2$
$\begin{aligned} & \omega_0=0 \\ & \theta=\omega_0 t+\frac{1}{2} \alpha t^2 \\ & =0+\frac{1}{2} \times 4 \times 5^2 \\ & =50 \times 2 \pi \text { radian, } N=\frac{50 \times 2 \pi}{2 \pi}=50\end{aligned}$

Topic: Physics, Rotational Motion, AP EAMCET, AP EAMCET 2023 (17 May Shift 2)
A spherical portion has been removed from a solid sphere having a charge distributed uniformly in its volume as shown in the figure. The electric field inside the emptied space is

Options:
A zero everywhere
B non-zero and uniform
C non-uniform
D zero only at its centre

The correct answer is:
non-zero and uniform

Solution:
Inside the cavity, field at any point is uniform and non-zero.

Topic: Physics, Electrostatics, JEE Main
A spherical rubber balloon carries a charge, uniformly distributed over the size, the total electric flux wrface. As the balloon coming out of the surface

Options:
A decreases.
B remains unchanged.
C increases.
D becomes zero.

The correct answer is:
remains unchanged.

Solution:
remains unchanged

Topic: Physics, Electrostatics, MHT CET, MHT CET 2020 (16 Oct Shift 2)
A spherical rubber balloon carries a charge, uniformly distributed over the surface. As the balloon is blown up and increases in size, the total electric flux coming out of the surface

Options:
A Becomes zero
B Increases
C Decreases
D Remains unchanged

The correct answer is:
Remains unchanged

Solution:
Total electric flux through the surface $=\frac{\mathrm{q}}{\varepsilon_0}$.
As charge remains unchanged when size of balloon increases, electric flux through the surface remains unchanged.

Topic: Physics, Electrostatics, MHT CET, MHT CET 2022 (07 Aug Shift 1)
A spherical soap bubble of radius 3 cm is formed inside another spherical soap bubble of radius 6 cm. If the internal pressure of the smaller bubble of radius 3 cm in the above system is equal to the internal pressure of the another single soap bubble of radius r cm. The value of r is _____

Options:

The correct answer is:
2

Solution:
Pressure inside a bubble is given by, P0+4Tr.

Pressure inside the bubble of smaller radius will be,

P2-P1=P0+4T6+4T3

Pressure inside the bubble with radius r, P0+4Tr

Now,

⇒P0+4T6+4T3=P0+4Tr

⇒1r=16+13

⇒r=2 cm

Topic: Physics, Mechanical Properties of Fluids, JEE Main, JEE Main 2022 (27 Jul Shift 2)
A spherical solid ball of volume $V$ is made of a material of density $\rho_1$. It is falling through a liquid of density $\rho_2\left(\rho_2 < \rho_1\right)$. Assume that the liquid applies a viscous force on the ball that is proportional to the square of its speed v, i.e., $F_{\text {viscous }}=-k v^2(k>0)$. The terminal speed of the ball is

Options:
A
$\sqrt{\frac{\operatorname{Vg}\left(\rho_1-\rho_2\right)}{\mathrm{k}}}$

B
$\frac{\mathrm{Vg} \rho_1}{\mathrm{k}}$

C
$\sqrt{\frac{V g \rho_1}{\mathrm{k}}}$

D
$\frac{\operatorname{Vg}\left(\rho_1-\rho_2\right)}{\mathrm{k}}$

The correct answer is:
$\sqrt{\frac{\operatorname{Vg}\left(\rho_1-\rho_2\right)}{\mathrm{k}}}$

Solution:
$$
\begin{aligned}
& \rho_1 V g-\rho_2 V g=k v_T^2 \\
& \Rightarrow \quad V_T=\sqrt{\frac{V g\left(\rho_1-\rho_2\right)}{k}}
\end{aligned}
$$

Topic: Physics, Mechanical Properties of Fluids, JEE Main, JEE Main 2008
A spherical solid ball of volume $V$ is made up of material of density $\rho$. It is falling through a liquid of density $\sigma(\sigma<\rho)$. Assume that the liquid applies a viscous force on the ball that is proportional to square of the terminal speed $v_{\mathrm{T}}, F=-K v_{\mathrm{T}}^2, \forall(K>0)$, then the terminal speed of the ball is ( $g=$ acceleration due to gravity)

Options:
A $\left[\frac{V g \rho}{K}\right]^{\frac{1}{2}}$
B $\left[\frac{V g(\rho-\sigma)}{K}\right]^{\frac{1}{2}}$
C $\frac{V g(\rho-\sigma)}{K}$
D $\frac{V g \rho}{K}$

The correct answer is:
$\frac{V g(\rho-\sigma)}{K}$

Solution:
The condition for terminal speed $v_{\mathrm{T}}$ is given by,
weight $(W)=$ buoyant force $(f)+$ viscous force $(F)$
We know, $W=\rho V g, f=\sigma V g$ and $F=K v_{\mathrm{T}}^2$
$\therefore \rho V g=\sigma V g+K v_{\mathrm{T}}^2$
$v_{\mathrm{T}}=\sqrt{\frac{(\rho-\sigma) V g}{K}}$

Topic: Physics, Mechanical Properties of Fluids, JEE Main
A spherical uniform planet is rotating about its axis. The velocity of a point on its equator is 7.5 km s-1. Due to the rotation of the planet about its axis, the acceleration due to gravity g at equator is 12 of g at poles. What is the escape velocity (in km s-1) of a particle on the planet from the pole of the planet?

Options:

The correct answer is:
15

Solution:
ge=gp-Rω2⇒g2=g-Rω2
Rω2=g2⇒R2ω2=gR2⇒ V2=gR2 ...(1)
The escape velocity,Ve=2gR ...(2)
From (1) and (2)
Ve=2×2V2⇒Ve=2V

Ve=15 km s-1

Topic: Physics, Gravitation, JEE Main
A spherical uniform planet is rotating about its axis. The velocity of a point on its equator is V. Due to the rotation of the planet about its axis the acceleration due to gravity g at equator is 12 of g at poles. The escape velocity of a particle on the planet in terms of V from the pole of the planet is

Options:
A Ve=2V
B Ve=V
C Ve=V/2
D Ve=3V

The correct answer is:
Ve=2V

Solution:
g e = g p - R Ï 2 â g 2 = g - R Ï 2
R Ï 2 = g 2 â R 2 Ï 2 = g R 2 â V 2 = g R 2 ...(1)
The escape velocity, V e = 2 g R ...(2)
From (1) and (2)
V e = 2 Ã 2 V 2 â V e = 2 V #

Topic: Physics, Gravitation, JEE Main
A spherical volume contains a uniformly distributed charge of density $1.0 \times 10^{-6} \mathrm{C} / \mathrm{m}^3$.
Find the electrical field (in N/C) at a point inside the volume at a distance $1 \mathrm{~mm}$ from the centre. $\left(\right.$ Let $\left.\frac{1}{4 \pi \epsilon_0}=9 \times 10^9 \mathrm{Nm}^2 \mathrm{C}^{-2}\right)$

Options:
A $\frac{8}{\pi}$
B $6 \pi$
C $\frac{\pi}{6}$
D $12 \pi$

The correct answer is:
$12 \pi$

Solution:
Given, the volume charge density of sphere,
$\begin{aligned} \rho_v & =1 \times 10^{-6} \mathrm{C}^2 \mathrm{~m}^3, \\ \frac{1}{4 \pi \varepsilon_0} & =9 \times 10^9 \mathrm{Nm}^2 \mathrm{C}^{-2} \\ r & =1 \mathrm{~mm}=10^{-3} \mathrm{~m}\end{aligned}$
Charge on sphere,
$q=$ charge density $\times$ volume
$=\rho_v \times \frac{4}{3} \pi r^3 \quad\left(\because V=\frac{4}{3} \pi r^3\right)$
$\therefore$ Electric field intensity at a distance $r$,
$\begin{aligned} & E=\frac{1}{4 \pi \varepsilon_0} \cdot \frac{q}{r^2} \\ & E=\frac{1}{4 \pi \varepsilon_0} \cdot \frac{\rho_v \times \frac{4}{3} \pi r^3}{r^2}=\frac{1}{4 \pi \varepsilon_0} \cdot \frac{\rho_v \times 4 \pi r}{3}\end{aligned}$
Putting the given values, we get
$\begin{aligned} & =9 \times 10^9 \times \frac{1 \times 10^{-6} \times 4 \times 10^{-3} \times \pi}{3} \\ E & =12 \pi\end{aligned}$
Hence, the electric field inside the sphere at a distance $1 \mathrm{~mm}$ from the centre is $12 \pi \mathrm{N} / \mathrm{C}$.

Topic: Physics, Electrostatics, JEE Main
A spherically symmetric charge distribution is characterised by a charge density having the following variations:
$\rho(r)=\rho_o\left(1-\frac{r}{R}\right)$ for $r < R$
$\rho(\mathrm{r})=0$ for $r \geq \mathrm{R}$
Where $r$ is the distance from the centre of the charge distribution $\rho_{\mathrm{o}}$ is a constant. The electric field at an internal point $(r < R)$ is:

Options:
A
$\frac{\rho_{\mathrm{o}}}{4 \varepsilon_{\mathrm{o}}}\left(\frac{\mathrm{r}}{3}-\frac{\mathrm{r}^2}{4 \mathrm{R}}\right)$

B
$\frac{\rho_0}{\varepsilon_0}\left(\frac{r}{3}-\frac{r^2}{4 R}\right)$

C
$\frac{\rho_{\mathrm{o}}}{3 \varepsilon_{\mathrm{o}}}\left(\frac{\mathrm{r}}{3}-\frac{\mathrm{r}^2}{4 \mathrm{R}}\right)$

D
$\frac{\rho_{\mathrm{o}}}{12 \varepsilon_{\mathrm{o}}}\left(\frac{\mathrm{r}}{3}-\frac{\mathrm{r}^2}{4 \mathrm{R}}\right)$

The correct answer is:
$\frac{\rho_0}{\varepsilon_0}\left(\frac{r}{3}-\frac{r^2}{4 R}\right)$

Solution:
Let us consider a spherical shell of radius $x$ and thickness $\mathrm{dx}$.
Charge on this shell
$$
d q=\rho .4 \pi x^2 d x=\rho_0\left(1-\frac{x}{R}\right) \cdot 4 \pi x^2 d x
$$
$\therefore$ Total charge in the spherical region from centre to $r(r < R)$ is
$$
\begin{aligned}
&q=\int d q=4 \pi \rho_0 \int_0^r\left(1-\frac{x}{R}\right) x^2 d x \\
&=4 \pi \rho_0\left[\frac{x^3}{3}-\frac{x^4}{4 R}\right]_0^r \\
&=4 \pi \rho_0\left[\frac{r^3}{3}-\frac{r^4}{4 R}\right] \\
&=4 \pi \rho_0 r^3\left[\frac{1}{3}-\frac{r}{4 R}\right]
\end{aligned}
$$
$\therefore$ Electric field at $\mathrm{r}, \mathrm{E}=\frac{1}{4 \pi \varepsilon_0} \cdot \frac{\mathrm{q}}{\mathrm{r}^2}$
$$
\begin{aligned}
&=\frac{1}{4 \pi \varepsilon_0} \cdot \frac{4 \pi \rho_0 r^3}{r^2}\left[\frac{1}{3}-\frac{r}{4 R}\right] \\
&=\frac{\rho_0}{\varepsilon_0}\left[\frac{r}{3}-\frac{r^2}{4 R}\right]
\end{aligned}
$$

Topic: Physics, Electrostatics, JEE Main