Tag: physics

Questions Related to physics

The principle of conservation of linear momentum can be strictly applied during a collision between two particles provided the time of impact is

collisions in one dimension collisions work, energy and power mechanics physics

  1. Extremely small

  2. Moderately small

  3. Extremely large

  4. Depends on a particular case

Show answer
Correct Option: B
Explanation:

The principle of conservation of linear momentum can be strictly applied during a collision between two particles provided the time of impact is  $Moderately\,\,small.$

Hence,
option $(B)$ is correct answer.

Two thin wire rings each having a radius R are placed at a distance d apart with their axes coinciding . The charges on the two rings are + q and -q . The potential difference between the centres of the two rings is

potential energy of a system of charges potential energy of various configurations electrostatic potential and capacitance electrostatics physics

  1. $\frac {{ Q.R }\quad} {\quad { 4\pi }{ \varepsilon } _{ 0 }{ d }^{ 2 }}$

  2. $\frac { Q }{ 2\pi { \varepsilon } _{ 0 } } [\frac { 1 }{ R } -\frac { 1 }{ \sqrt { { R }^{ 2 }+{ d }^{ 2 } } } ]$

  3. $\frac { Q }{ { 4\pi \varepsilon } _{ 0 } } [\frac { 1 }{ R } -\frac { 1 }{ \sqrt { { { R }^{ 2 } }+{ { d }^{ 2 } } } } ]$

  4. 0

Show answer
Correct Option: A
Explanation:

Two thin wire ring each having a radius $=R$

distance $=d$
two rings are $+q$ and $-q$.
$PD=\dfrac { 1 }{ 4\pi { \epsilon  } _{ 0 } } \dfrac { QR }{ { d }^{ 2 } } $
Now in this question, we get the solution.

what is the potential difference between two points, if 2J of work must be done to move a 4 mC charge from one point to another is:

potential energy of a system of charges potential energy of various configurations electrostatic potential and capacitance electrostatics physics

  1. 50 V

  2. 500 V

  3. 5 V

  4. 5000 V

Show answer
Correct Option: B
Explanation:

Answer is B.

The total work done = energy transferred.
so, we might see the equation energy = voltage x charge, E = V * Q, written as, 
work = voltage x charge, W = V * Q.
In this case, the charge is 4 mC, that is, 0.004 C and work done is 2 J.
Therefore, V=W/Q = 2/0.004 = 500 V.
Hence, the potential difference between two points if 2 J of work must be done to move a 4 mC charge from one point to another is 500 V.

A point charge q is rotated along a circle in the electric field generated by another point charge Q. The work done by the electric field on the rotating charge in one complete revolution is_______

potential energy of a system of charges potential energy of various configurations electrostatic potential and capacitance electrostatics physics

  1. zero

  2. positive

  3. negative

  4. zero if the charge Q is at the center and nonzero otherwise.

Show answer
Correct Option: A
Explanation:

The net displacement round one complete circle is 0.

So, the work done is 0.

$100J$ of work is done when $2 \mu C$ charge is moved in an electric field between two points. The p.d. between the points is

potential energy of a system of charges potential energy of various configurations electrostatic potential and capacitance electrostatics physics

  1. $2\times10^{-4}V$

  2. $2\times10^{-8}V$

  3. $2\times10^{-6}V$

  4. $5\times10^{7}V$

Show answer
Correct Option: D
Explanation:

We know work done $=q\Delta V $
Where $\Delta V $ is change in potential $V _2-V _1$
$100=\Delta V\times 2\times 10^{-6}$
$50\times 10^6=\Delta V$
$\therefore p\cdot d= 5\times 10^7 V$

A hollow metal sphere of radius $5\ cm$ is charged such that the potential on its surface is $10$ volts. The potential of the centre of the sphere is:

potential energy of a system of charges potential energy of various configurations electrostatic potential and capacitance electrostatics physics

  1. zero

  2. $10$ volts

  3. Same as at a point $5\ cm$ away from the surface.

  4. Same as at a point $25\ cm$ away from the surface.

Show answer
Correct Option: B
Explanation:
For hollow spherical conductor -
$\Rightarrow$ Potential on its surface $=$ Potential on center $= 10\;volts.$
Because, the electric field inside the spherical shell is zero, due to which the potential inside the shell is constant and will be equal to the potential at the surface.
Hence, the answer is $10\;volts.$

A spherical shell of radius $R _1$ with uniform charge $q$ is expanded to a radius $R _2$. Find the work performed by the electric forces during the shell expansion from $R _1$ to radius $R _2$.

potential energy of a system of charges potential energy of various configurations electrostatic potential and capacitance electrostatics physics

  1. $\dfrac{q^2}{2\pi in _0}\left(\dfrac{1}{R _1} - \dfrac{1}{R _2}\right)$

  2. $\dfrac{q^2}{3\pi in _0}\left(\dfrac{1}{R _1} - \dfrac{1}{R _2}\right)$

  3. $\dfrac{q^2}{5\pi in _0}\left(\dfrac{1}{R _1} - \dfrac{1}{R _2}\right)$

  4. $\dfrac{q^2}{8\pi in _0}\left(\dfrac{1}{R _1} - \dfrac{1}{R _2}\right)$

Show answer
Correct Option: D
Explanation:
Spherical shell of radius $={ R } _{ 1 }$
charge $=q$
work done $=W=q\left( { V } _{ B }-{ V } _{ A } \right) $
${ V } _{ A }=\dfrac { { K } _{ q } }{ { R } _{ 1 } } $
${ V } _{ B }=\dfrac { { K } _{ q } }{ { R } _{ 2 } } $
$W=q\left( \dfrac { { K } _{ q } }{ { R } _{ 1 } } -\dfrac { { K } _{ q } }{ { R } _{ 2 } }  \right) $
     $=\dfrac { { q }^{ 2 } }{ 8\pi { \epsilon  } _{ 0 }{ n } _{ 0 } } \left( \dfrac { 1 }{ { R } _{ 1 } } -\dfrac { 1 }{ { R } _{ 2 } }  \right) $    (Proved).

Two insulated charged spheres of radii ${R} _{1}$ and ${R} _{2}$ having charges ${Q} _{1}$ and ${Q} _{2}$ respectively are connected to each other, then there is:

potential energy of a system of charges potential energy of various configurations electrostatic potential and capacitance electrostatics physics

  1. no change in the energy of the sytem

  2. an increase in the energy of the system

  3. always a decrease in the energy of the system

  4. a decrease in energy of the system unless ${q} _{1}{R} _{2}={q} _{2}{R} _{1}$

Show answer
Correct Option: D
Explanation:

Two insulated charged spheres of radii $ R _1,R _2$having charges $q _1,q _2$respectively are brought in contact with each other.

Charges will flow across the point of contact from where they are connected until their potential at their surfaces became same. 
Energy is decreased because system changes to get stability.
Potential at their surfaces are same,
So,
$V _1=V _2$
$\dfrac{Kq _1}{R _1}=\dfrac{Kq _2}{R _2}$
$q _1R _2=q _2R _1$
Hence Proved.

Two small spheres have mass ${m} _{1}$ and ${m} _{2}$ and hanging from massless insulating threads of lengths ${l} _{1}$ and ${l} _{2}$. Two spheres carry charges ${q} _{1}$ and ${q} _{2}$ respectively. The spheres hang such that they are on the same horizontal level and the threads are inclined to the vertical at angle ${\theta} _{1}$ and ${\theta} _{2}$ respectively. If $F _1 = F _2$, then:

potential energy of a system of charges potential energy of various configurations electrostatic potential and capacitance electrostatics physics

  1. ${ \theta } _{ 1 }={ \theta } _{ 2 }$

  2. ${ M } _{ 1 }={ M } _{ 2 }$

  3. $\cfrac { l _{ 1 } }{ \tan { { \theta } _{ 1 } } } =\cfrac { l _{ 2 } }{ \tan { { \theta } _{ 2 } } } \quad $

  4. $\cfrac { q _{ 1 } }{ \tan { { \theta } _{ 1 } } } =\cfrac { q _{ 2 } }{ \tan { { \theta } _{ 2 } } } $

Show answer
Correct Option: B
Explanation:

For sphere $1$

In equilibrium, from figure.
$\begin{array}{l} { T _{ 1 } }\cos { \theta _{ 1 } } ={ M _{ 1 } }g; \ { T _{ 1 } }\sin { \theta _{ 1 } } ={ F _{ 1 } } \ \therefore \tan { \theta _{ 1 } } =\dfrac { { { F _{ 1 } } } }{ { { M _{ 1 } }g } } . \end{array}$
For sphere $2$
In equilibrium, from figure.
$\begin{array}{l} { T _{ 2 } }\cos { \theta _{ 2 } } ={ M _{ 2 } }g; \ { T _{ 2 } }\sin { \theta _{ 2 } } ={ F _{ 2 } } \ \therefore \tan { \theta _{ 2 } } =\dfrac { { { F _{ 2 } } } }{ { { M _{ 2 } }g } } . \end{array}$
Force of repulsion between two charges are same
$\therefore F _1=F _2$
$\theta _1=\theta _2$ only if $\dfrac{{{F _1}}}{{{M _1}g}} = \dfrac{{{F _2}}}{{{M _2}g}}.$
But $F _1=F _2$, then $M _1=M _2$.

Two particles $X$ and $Y$ having equal charges after being accelerated thorough the same potential difference enter a region of uniform magnetic field and describe circular paths of radius $R _1$ and $R _2$ respectively. the ratio of mass of $X$ to that of $Y$ is

potential energy of a system of charges potential energy of various configurations electrostatic potential and capacitance electrostatics physics

  1. $\sqrt { R _1{/R _2} }$

  2. $R _2{/R _1}$

  3. $(R _1{/R _2})^2$

  4. $R _1{/R _2}$

Show answer
Correct Option: C
Explanation:

Given that,

The radius of particle X$={{R} _{1}}$

The radius of particle Y$={{R} _{2}}$


We know that,

Work done of each particle = $qV$


Now, suppose the particle starts from rest, and final kinetic energy is

For, X particle,

$\dfrac{1}{2}{{m} _{1}}v _{1}^{2}=qV$

For, Y particle

$\dfrac{1}{2}{{m} _{2}}v _{2}^{2}=qV$


Now,

${{m} _{1}}v _{1}^{2}={{m} _{2}}v _{2}^{2}.....(I)$

Now, from the magnetic force is

$ {{F} _{{{m} _{1}}}}=q{{v} _{1}}B $

$ {{F} _{{{m} _{2}}}}=q{{v} _{2}}B $


Now, the magnetic force is equal to the centripetal force is

For, X

$ \dfrac{{{m} _{1}}v _{1}^{2}}{{{R} _{1}}}=qB{{v} _{1}} $

$ {{m} _{1}}{{v} _{1}}=qB{{R} _{1}} $

For, Y

$ \dfrac{{{m} _{2}}v _{2}^{2}}{{{R} _{2}}}=qB{{v} _{2}} $

$ {{m} _{2}}{{v} _{2}}=qB{{R} _{2}} $


Now, putting the value in equation (I)

$ {{m} _{1}}{{\left( \dfrac{qB{{R} _{1}}}{{{m} _{1}}} \right)}^{2}}={{m} _{2}}{{\left( \dfrac{qB{{R} _{2}}}{{{m} _{2}}} \right)}^{2}} $

$ \dfrac{R _{1}^{2}}{{{m} _{1}}}=\dfrac{R _{2}^{2}}{{{m} _{2}}} $

$ \dfrac{{{m} _{1}}}{{{m} _{2}}}={{\left( \dfrac{{{R} _{1}}}{{{R} _{2}}} \right)}^{2}} $

 Hence, the ratio of the mass is ${{\left( \dfrac{{{R} _{1}}}{{{R} _{2}}} \right)}^{2}}$