Tag: physics

Questions Related to physics

Which of the following charges is/are impossible?

  1. $4.8 \times 10^{-18}C$

  2. $5.8 \times 10^{-18}C$

  3. $12.8 \times 10^{-18}C$

  4. $20.8 \times 10^{-18}C$


Correct Option: B
Explanation:
From $Q=ne$,
where $Q =$ total charge
$n=$ number of electrons
$e=$ charge on one electron 
So, only that charge is possible for which charge is whole multiple of $1.6\times 10^{-19}$ (i.e, charge on one electron).
We can see that multiplying $1.6\times 10^{-19}$ with $30$ gives $A$
Multiplying $1.6\times 10^{-19}$ with $80$ gives $C$
 Multiplying $1.6\times 10^{-19}$ with $130$ gives $D$
We can see that a, c and d are multiple of charge $1.6\times 10^{-19}\:C$.
Only $5.8\times 10^{-18}\: C$ given in B) option is not a multiple of $1.6\times 10^{-19}\:C$. 

Charge can be a multiple of $e$, i.e. $ne$. Which of the following can be a value of $n$?

  1. 27

  2. 1.5

  3. 30.5

  4. 0.99999


Correct Option: A
Explanation:

Charge has to be an integral multiple of electronic charge e.

If a conductor has $10^8$ number of electrons , then the total charge of the conductor is:

  1. $+1.6 \times 10^{-19}$ C

  2. $-1.6 \times 10^{-19}$ C

  3. $+1.6 \times 10^{-11}$ C

  4. $-1.6 \times 10^{-11}$ C


Correct Option: D
Explanation:

Any charge exists in discrete lumps or packets of a certain minimum charge e where e is the charge of an electron. According to quantization of charge, the charge on a body can be only an integral multiple of charge on the electron i.e., $q = ne$ where n = 1, 2, 3 and $e=1.6\times { 10 }^{ -19 }$ C.

So, for the given case, $q=10^8 \times (-1.6 \times 10^{-19})$ C$=-1.6 \times 10^{-11}$ C.

When a glass rod is rubbed against silk.

  1. Glass rod loses electrons and becomes positively charged

  2. Silk loses electrons and becomes positively charged

  3. Glass rod losses electrons and becomes negatively charged

  4. Silk loses electrons and becomes negatively charged


Correct Option: A
Explanation:

 Due to rubbing, heat is generated which supplies energy for removal of electrons. As the work function of the glass rod is smaller than the silk cloth, it easily loses electrons to the silk cloth.

How many electronic charges form 1 coulomb?

  1. $9.1\times 10^{-31}$

  2. $1.6\times 10^{18}$

  3. $62.5\times 10^{17}$

  4. $1.76\times 10^{11}$


Correct Option: C
Explanation:

$q=ne$
$\therefore n=\dfrac {q}{e}$
Given $q=1C$
$e=-1.6\times 10^{-19}C$
$n=\dfrac {1C}{1.6\times 10^{-19}}$
$6.25\times 10^{18}$ or $62.5\times 10^{17}$

A spherical metal shell $A$ of radius ${R} _{A}$ and a solid metal sphere $B$ of radius ${R} _{8}\left( <{ R } _{ A } \right)$ are kept far apart and each is given charge $+Q$. Now they are connected by a thin metal wire. Then 

  1. ${ E } _{ A }^{ inside }=0$

  2. $\quad { Q } _{ A }>{ Q } _{ B }$

  3. $\dfrac { { \sigma } _{ A } }{ { \sigma } _{ B } } =\dfrac { { R } _{ B } }{ R _{ A } }$

  4. ${ E } _{ A }^{ on\quad surface }<{ E } _{ B }^{ on\quad surface }$


Correct Option: A,B,C,D
Explanation:
Electric field inside a spherical metallic shell with charge on surface $=0$
$\therefore (a)$ is correct
On connecting Both with wise
Electric potential will be equal say $V$
$\therefore \dfrac{1}{4\pi Co}\dfrac{Q _A}{R _A}=\dfrac{1}{4\pi Co}\dfrac{Q _B}{R _B}=V$
as $R _A> R _B\therefore Q _A > Q _B$ Hence $(b)$ is correct
as $\dfrac{\sigma _A}{\sigma _B}=\dfrac{Q _B}{4\pi R _{B}^{2}}=\dfrac{R^{2}B}{R^{2}A}\times \dfrac{4\pi Co R _{A}V}{4\pi Co R _{A}R}$
$\dfrac{\sigma A}{\sigma B}=\dfrac{R _B}{R _A}$             $(C)$ is correct
Also $E _{A}=\dfrac{\sigma _A}{\sigma _B}=\dfrac{R _B}{R _A}<1\therefore E _A < E _B$
Hence $(d)$ is correct

Electron accelerated by potential $V$ are diffracted from a crystal. If $d=1 A$ and $i = 30^\circ $. $V$ should be about  $h = 6.6 \times {10^{ - 24}}Js\,{m _e} = 9.1 \times {10^{ - 33}}kg.e = 1.6 \times {10^{ - 19}}C$

  1. $2000 V$

  2. $50 V$

  3. $500 V$

  4. $1000 V$


Correct Option: B
Explanation:

$d=1A$

$i={ 30 }^{ o  }$
$\theta ={ 60 }^{ o  }$
$h=6.6\times { 10 }^{ -34 }Js$
${ m } _{ e }=9.1\times { 10 }^{ -31 }kg$
$e=1.6\times { 10 }^{ -19 }c$
$n\lambda =2dsin\theta $
$\lambda =\dfrac { 2\times 1A\times sin{ 60 }^{ o  } }{ 1 } $
$\lambda =\sqrt { 3A } $
$\sqrt { V } =\dfrac { 1.27\times { 10 }^{ -10 } }{ \sqrt { 3 } \times { 10 }^{ -10 } } =50.18volts$

All free electric charges can be 
($e=$ single unit of charge i.e. the magnitude of charge on electron )

  1. odd multiples of $e$

  2. fractional multiples of $e$

  3. even multiples of $e$

  4. negative multiples of $e$


Correct Option: A,C,D
Explanation:

Charges are acquired by either gain or loss of electrons .

And electron transfer can occur only in form of integers, fraction of electron can't be shared.

Hence, charge on a body can be positive or negative integral multiple of $e$.

Answer-(A),(C),(D)

Charge $q$ on a body in terms of transfer of electrons can be expressed as
( $n=$ net number of electrons transacted)

  1. $ q = ne$

  2. $ q = \dfrac{e}{n}$

  3. $ q = - \dfrac{e}{n}$

  4. None of these


Correct Option: A
Explanation:

Charge on each electron$=e$


If  n be the number of electrons transferred, then charge $q=ne$.

Answer-(B)

In particular system of unit, if the unit of mass becomes twice and that of time becomes half, then $8$ Joules will be written as _______ units of work

  1. $16$

  2. $1$

  3. $4$

  4. $64$


Correct Option: D
Explanation:
The unit of work is $kg\ m^2/s^2$.

$8Joule=\dfrac { 8M{ L }^{ 2 } }{ { T }^{ 2 } }$

$ =\dfrac { 8(1kg){ (1m) }^{ 2 } }{ { (1s) }^{ 2 } } $

On doubling the unit of mass and unit of time is halved:
$=\cfrac { 8(2\quad unit\quad of\quad mass){ (1m) }^{ 2 } }{ { \left( \frac { 1 }{ 2 } unit\quad of\quad time \right)  }^{ 2 } }$

$ =\dfrac { 8\times 2 }{ { \left( \frac { 1 }{ 2 }  \right)  }^{ 2 } } units\quad of\quad work\\ =64\ units\quad of\quad work.$