Tag: physics

Questions Related to physics


When electric current passes through a bucket full of water , lot of bubbling is observed. This suggests that the type of supply is

  1. AC

  2. DC

  3. any of the above two

  4. None of the above


Correct Option: B
Explanation:

When a dc current is passed through water, the electrolysis of water molecules $(H _{2}O)$ takes place i.e, water gets split into hydrogen and oxygen gases. The bubbles are these gases escaping.

How many electronic charges from $1$ 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:

Let the total electronic charge be n

$1=ne=n\times 1.6\times 10^{-19}$
$n=62.5\times 10^{17}$

The net charge inside an isolated system is

  1. conserved

  2. variable

  3. zero

  4. infinite.


Correct Option: A
Explanation:

According to law of conservation of charge, the net charge inside an isolated system is conserved. 

A charge of $-1C$ contains $n$ electrons, then $n=?$

  1. $6.25 \times 10^{-20}$

  2. $6.25 \times 10^{19}$

  3. $6.25 \times 10^{18}$

  4. $6.25 \times 10^{-19}$


Correct Option: C
Explanation:

Here, $n(-e)=Q=-1C$ or $n=1/e=\dfrac{1}{(1.6\times 10^{-19})}=6.25\times 10^{18}$

If $Q _1 = 5\mu C$, $Q _2 = 3 \mu C$ and $Q _3= -2 \mu C$, then the net charge in the system is :

  1. $-6 \mu C$

  2. $6 \mu C$

  3. $8 \mu C$

  4. $-2 \mu C$


Correct Option: B
Explanation:

The net charge of the given system is $Q=Q _1+Q _2+Q _3=(5+3-2)\mu C=6\mu C$

Charge separation can be affected by

  1. contact

  2. pressure

  3. heat

  4. charge


Correct Option: A,D
Explanation:
Charge separation can be affected by -
(a) Contact
(b) Charge
Because contact of force will be affected for the charge separation and charge accumulation will be the affected of charge separation pressure and heat also.

During charge separation, there is a

  1. creation of charge

  2. creation of energy

  3. creation of charged particles

  4. creation of neutrons


Correct Option: C
Explanation:

Charge separation is the process of exciting an electron in an atom to a higher energy level and then making it leave the atom to a hereby electron acceptor thus creating charged particle.

A metal sphere has a charge of $- 6 \mu C$. When $5 \times 10^{12} $ electrons are removed from the sphere, what would be net charge on it?

  1. $-6.8 \mu C$

  2. $6.8 \mu C$

  3. $5.2 \mu C$

  4. $-5.2 \mu C$


Correct Option: D
Explanation:

Charge removed=$-e\times n=-1.6\times 10^{-19}\times 5\times 10^{12}=-8\times 10^{-7}=-0.8\mu C$


Hence, charge on sphere=$-6-(-0.8)=-5.2\mu C$

Answer-(D)

Only the integral number of electrons which is transferred from one body to another, i.e. $\pm ne$. This is represented as

  1. Superposition of charge

  2. Quantisation of charge

  3. Conservation of charge

  4. Addition of charge


Correct Option: B
Explanation:

The transfer of electrons from one body to other leads to charging of bodies. It is found that charge on any body is always equal to integral multiple of charge on electron. This property of charge is known as Quantisation of electric charge according to which, $Q=ne$
$Q=$ total charge transferred
$n=$ number of electrons transferred
$e=$ charge on one electron
$n=1, \pm 2.....$ and so on.

A copper slab of mass $2 g$ contains $2 \times 10^{22}$ atoms. The charge on the nucleus of each atom is $29 e$. What fraction of the electrons must be removed from the sphere to give it a charge of $+2\mu C$? 

  1. $58 \times 10^{22}$

  2. $1.25 \times 10^{13}$

  3. $2.16 \times 10^{11}$

  4. $2.16 \times 10^{-11}$


Correct Option: D
Explanation:
Given a mass of copper slab $=2\, g$ contains $2\times 10^{22}$ atoms.
Charge on nucleus of each atom $=29e$
We have to find the fraction of electrons that must be removed from sphere to give it $+2\mu C$ charge.
We solve it as follows:
Total number of electrons in sphere $=29\times 2\times 10^{22}$
Number of electrons $=\dfrac qe$
$=\dfrac{2\times 10^{-6}}{1.6\times 10^{-19}}$
$=1.25\times 10^{13}$
Fraction of electrons removed $=\dfrac{1.25\times 10^{13}}{Total\; number \; of \;electrons \; in\; sphere}$
$=\dfrac{1.25\times 10^{13}}{29\times 2\times 10^{22}}$
$=2.16\times 10^{-11}$