Tag: physics

Questions Related to physics

A thinwalled, spherical conducting shell S of radius R is given charge Q. The same amountof charge is also placed at its centre C. Which of the following statements are correct?

applications of gauss's law coulomb's law physics

  1. On the outer surface of S, the charge density is $\displaystyle \frac {Q} {2 \pi R^2} $

  2. The electric field is zero at all points inside S

  3. At a point just outside S, the electric field is double the field at a point just inside S

  4. At any point inside S, the electric field is inversely proportional to the square of its distance from C

Show answer
Correct Option: A,C,D
Explanation:

because of charge Q at center there will be induced charge -Q at inner surface of sphere. hence charge density $\dfrac{2Q}{4\pi r^2} = \dfrac{Q}{2\pi r^2}$

because of 2Q charge outside the electric field is double that of inside.
At any point inside S, the electric field is inversely proportional to the square of its distance from C

Two sphere's are isolated from each other. They each have an identical net positive charge and have the same radius, however, one sphere is solid and insulating, while the other is a hollow conducting sphere whose charge is uniformly distributed.
For which sphere is the electric field the greatest distance $x$ from the center of the spheres?
Assume $x$ is less than the radius of the spheres.

applications of gauss's law coulomb's law physics

  1. The conducting hollow sphere has a greater E-field.

  2. The insulating solid sphere has a greater E field.

  3. Both spheres have the same E field.

  4. Neither sphere would cause there to be an Electric field.

Show answer
Correct Option: B
Explanation:

The charge in a conducting hollow sphere resides on the surface only. Hence on applying Gauss Law on a spherical surface enclosed in the sphere, we get electric field to be zero inside it.

For a solid insulating sphere with uniform charge distribution, finite electric field exists in the sphere.
Hence correct answer is option B.

As one penetrates through uniformly charged conducting sphere, what happens to the electric field strength:

applications of gauss's law coulomb's law physics

  1. decreases inversely as the square of the distance

  2. decreases inversely as the distance

  3. becomes zero

  4. increases inversely as the square of distance

Show answer
Correct Option: C
Explanation:

Electric field strength inside the uniform charged sphere is zero.

$\therefore$  As one penetrates through uniformly charged sphere, electric field strength inside the sphere becomes zero. 

The magnitude of the electric field on the surface of a sphere of radius $r$ having a uniform surface charge density $\sigma$ is

applications of gauss's law coulomb's law physics

  1. $\sigma / \epsilon _{0}$

  2. $\sigma / 2\epsilon _{0}$

  3. $\sigma / \epsilon _{0}r$

  4. $\sigma / 2\epsilon _{0}r$

Show answer
Correct Option: A
Explanation:
The magnitude of the electric field on the surface of radius $=r$
Charge density $=6$
Then, $E=\dfrac { 6 }{ { \epsilon  } _{ 0 } } $
The electric field is independent of the surface radius.

Consider a thin spherical shell of radius $R$ consisting of uniform surface charge density $\sigma$. The electric field at a point of distance $x$ from its centre and outside the shell is

applications of gauss's law coulomb's law physics

  1. inversely proportional to $\sigma$

  2. directly proportional to ${x}^{2}$

  3. directly proportional to $R$

  4. inversely proportional to ${x}^{2}$

Show answer
Correct Option: D
Explanation:
For a thin uniformly charged spherical shell, the field points outside the shell at a distance $x$ from the centre is
$E=\cfrac { 1 }{ 4\pi { \varepsilon  } _{ 0 } } \cfrac { Q }{ { x }^{ 2 } } $
If the radius of the sphere is $R,Q=\sigma 4\pi { R }^{ 2 }$
$\therefore E=\cfrac { 1 }{ 4\pi { \varepsilon  } _{ 0 } } \cfrac { \sigma 4\pi { R }^{ 2 } }{ { x }^{ 2 } } =\cfrac { \sigma { R }^{ 2 } }{ { { \varepsilon  } _{ 0 }x }^{ 2 } } $
This is inversely proportional to square of the distance from the centre. It is as if the whole charge is concentrated at the centre

Two charged spheres having radii a and b are joined with a wire then the ratio of electric field $\dfrac{E _a}{E _b}$ on their surface is?

applications of gauss's law coulomb's law physics

  1. a/b

  2. b/a

  3. ba

  4. None of these

Show answer
Correct Option: B
Explanation:

When the two spheres are connected by a wire, then both of them acquire the same potential say $V$.


We also know that the electric field on the surface of a sphere $E=\dfrac{Q}{4\pi\epsilon _o r^2}$
and potential on the surface is given by $V=\dfrac{Q}{4\pi\epsilon _or}$

$\implies E=\dfrac{V}{r}$

Here, V is constant , hence  $E\propto \dfrac{1}{r}$

$\implies \dfrac{E _a}{E _b}=\dfrac{b}{a}$

Charges $Q _1$ and $Q _2$ are placed inside and outside respectively of an uncharged conducting shell. Their seperation is r.

applications of gauss's law coulomb's law physics

  1. The force on $Q _1$ is zero.

  2. The force on $Q _1$ is $\displaystyle k \frac{Q _1 Q _2}{r^2}$

  3. The force on $Q _2$ is $\displaystyle k \frac{Q _1 Q _2}{r^2}$

  4. The force on $Q _2$ is zero.

Show answer
Correct Option: A,C
Explanation:

As the electric field inside the conducting shell is zero , so the force on the inner charge, $Q _1$ will be zero.
The electric field at outside charge $Q _2$ due to $Q _1$ is $E=k\frac{Q _1}{r^2}$
Force on $Q _2$ is $F=Q _2E=k\frac{Q _1Q _2}{r^2}$

A block released from rest from the top of a smooth inclined plane of angle $\theta _1$ reaches the bottom in time $t _1$. The same block released from rest from the top of another smooth inclined plane of angle $\theta _2$, reaches the bottom in time $t _2$. If the two inclined planes have the same height, the relation between $t _1$ and $t _2$ is 

friction laws of motion physics

  1. $\cfrac{t _2}{t _1} = \left(\cfrac{sin \theta _1}{sin \theta _2}\right)^{1/2}$

  2. $\cfrac{t _2}{t _1} = 1$

  3. $\cfrac{t _2}{t _1} = \left(\cfrac{sin \theta _1}{sin \theta _2}\right)$

  4. $\cfrac{t _2}{t _1} = \left(\cfrac{sin^2 \theta _1}{sin^2 \theta _2}\right)$

Show answer
Correct Option: C
Explanation:
From the figure
we see that
down the incline acceleration us $a=g\sin \theta$
distance $s=\dfrac {h}{\sin \theta _1}$

using $s=ut+\dfrac {1}{2} at^2 \ ;\ u=0$ (initially rest)

gives $t=\sqrt {\dfrac {2s}{a}}$

For $\theta _1 \quad t _1=\sqrt {\dfrac {2\ h}{\sin \theta _1 \times g\sin \theta _1}}=\dfrac {1}{\sin \theta _1} \sqrt {\dfrac {2\ h}{g}}$

so for $O _2, \ t _2=\dfrac {1}{\sin \theta _2}\sqrt {\dfrac {2\ h}{g}}$

$\Rightarrow \ \dfrac {t _2}{t _1}=\dfrac {\sin \theta _1}{\sin \theta _2}$

A coin placed on a rotating turn table just slips if it is at a distance of $40\  cm$ from the centre if the angular velocity of the turntable is doubled, it will just slip at a distance of:

friction laws of motion physics

  1. $10\ cm$

  2. $20\ cm$

  3. $40\ cm$

  4. $80\ cm$

Show answer
Correct Option: A
Explanation:

It slips when.
$\mu mg = m{\omega ^2}r$
$ \Rightarrow \mu g = {\omega ^2}r$
$ \Rightarrow \mu g = {\omega ^2}40 - \left( 1 \right)$
$Now,\,\omega  = 2\omega $
$ \Rightarrow \mu g = {\left( {2\omega } \right)^2}r \Rightarrow 4{\omega ^2}r - \left( 2 \right)$
$\therefore {\omega ^2} \times 40 = 4{\omega ^2}r$
$ \Rightarrow r = 10cm$

The coefficient of friction between two surfaces is 0.2. The angle of friction is 

friction laws of motion physics

  1. sin$^{-1}$(0.2)

  2. cos $^{-1}$(0.2)

  3. tan$^{-1}$(0.1)

  4. cot$^{-1}$(5)

Show answer
Correct Option: D
Explanation:
The correct option is D

We have,

 The coefficient of friction is $0.2$

Since we know that,

The coefficient of friction $= tan \theta$  Where $\theta $  is the angle of friction.

$\dfrac{1}{5}=tan \theta$

$\theta=tan^{-1}\dfrac{1}{5}$

$=cot^{-1}5$