Tag: physics

Questions Related to physics

The ratio of the drifty velocity $v _d$ and r.m.s. velocity of electrons is

physics electric current, potential difference and resistance drift velocity and mobility drift speed drift velocity & mobility

  1. $10^{-10}$

  2. $10^{-5}$

  3. $10^{-3}$

  4. $10^{-6}$

Show answer
Correct Option: A
Explanation:

We assume the electrons in a copper wire of radius 0.815 mm and carrying current 1 A at room temp 300 K.
Drift velocity , $v _d=\frac{I}{neA}=\frac{1}{(8.47\times 10^{28})\times (1.6\times 10^{-19})\times \pi\times (0.815\times 10^{-3} )^2 } \sim 3.45\times 10^{-5} m/s$
and $v _{rms}=\sqrt{\frac{3k _BT}{m}}=\sqrt{\frac{3\times (1.38\times 10^{-23})\times 300}{9.1\times 10^{-31}}}=1.17\times 10^5  m/s $
thus, $\frac{v _d}{v _{rms}}=10^{-10}$

When no current is passed through a conductor

physics electric current, potential difference and resistance drift velocity and mobility drift speed drift velocity & mobility

  1. the free electrons do not move

  2. the average speed of a free electron over a large period of time is zero

  3. the average velocity of a free electron over a large period of time is zero

  4. the average of the velocities of all the free electrons at an instant is zero

Show answer
Correct Option: C,D
Explanation:

When no current pass through the conductor i.e. when there is no electric field is applied to the conductor, each electron moves along a straight path at constant speeds and collide with the lattice ions(positive). With each collision the direction of electrons is changing randomly. The resulting path of any electron over along period of time, covering many collisions is a random sequence of straight segments. Due to this, the average number of electrons crossing any small area is nearly equal(in any direction). Thus the average displacement of electrons along any direction during any long period of time is also zero. Hence, the average velocity component in any direction is zero.

In metals, the time of relaxation of electrons

physics electric current, potential difference and resistance drift velocity and mobility drift speed drift velocity & mobility

  1. increases with increasing temperature

  2. decreases with increasing temperature

  3. does not depend on temperature

  4. changes suddenly at $400 K$

Show answer
Correct Option: B
Explanation:
Resistivity of a material is given by $\displaystyle\rho=\dfrac{m}{ne^2\tau}$ 

where $m=$mass of electron, $ n=$ electron density, $e=$ electronic charge and $\tau=$ relaxation time.

Resistance $\displaystyle R=\dfrac{\rho l}{A}=\dfrac{ml}{ne^2\tau A}$ or $\displaystyle R \propto \dfrac{1}{\tau}$

Since in metals Resistance increases with temperature so the relaxation of electrons decreases with increasing temperature.

A wire   whose cross - sectional area is increasing linearly from  its cone end to the other,  is connected  across a battery of V  volts.  Which of the  following quantities remain constant in the  wire ? 

physics electric current, potential difference and resistance drift velocity and mobility drift speed drift velocity & mobility

  1. drift speed

  2. current density

  3. electric current

  4. electric field

Show answer
Correct Option: C

A current of $5A$ passes through a copper conductor $(resistivity=1.7\times10^{-8}\Omega m$) of radius of cross-section $5mm$. Find the mobility of the charges if their drift velocity is $1.1\times 10^{-3}m/s$

physics electric current, potential difference and resistance drift velocity and mobility drift speed drift velocity & mobility

  1. $1.3m^2/Vs$

  2. $1.5m^2/Vs$

  3. $1.8m^2/Vs$

  4. $1.0m^2/Vs$

Show answer
Correct Option: D
Explanation:

$\mu=\dfrac{V _d}{E}$     $E=\rho J$


$=\dfrac{1.1\times 10^{-3}}{1.7\times 10^{-8}\times \dfrac{5}{\pi \times 25\times 10^{-6}}}$

$=\dfrac{1.1\times 10^{-3}\times \pi \times 25\times10^{-6}}{1.7 \times 10^{-8} \times5}\approx 1.01m^2/Vs$

Charge carriers, each of charge q, move along a wire of fixed length. The number density of the
charge carriers in the wire is n.
What is also required, for this wire, to determine the average drift velocity of the charge carriers in
terms of n and q?

physics electric current, potential difference and resistance drift velocity and mobility drift speed drift velocity & mobility

  1. current per unit of cross-sectional area

  2. potential difference per unit of length

  3. resistance and cross-sectional area

  4. resistivity and length

Show answer
Correct Option: A
Explanation:

We know that 

$I = nqAv _d$
$\implies v _d=  \dfrac{I}{A}\times \dfrac{1}{nq} $
we have the value of $n$ and $q$, 
So, to find the value of $v _d$ we require $\dfrac{I}{A}$ i.e. current per unit cross-sectional area.

A current I flows through a uniform wire of diameter d when the mean electron drift velocity is
V. The same current will flow through a wire of diameter d/2 made of the same material if the
mean drift velocity of the electron is:

physics electric current drift velocity and mobility drift speed drift velocity & mobility

  1. v/4

  2. v/2

  3. 2v

  4. 4v

Show answer
Correct Option: D
Explanation:

Let I be the current flow through wire. 

Let V be the drift velocity of wire which has diameter d. 
Let A be the area of wire. 
Consider a wire of same material but diameter is half
Therefore, Area =$\dfrac{A}{4}$
Since, $I=NeAV$
N is same for both wire because both of them made of same material. 
$I _{1}=I _{2}$
$NeA _{1}V _{1}=NeA _{2}V _{2}$
$AV=\dfrac{A}{4}V _2$
$V _{2}=4V$


The electric current in a wire may be calculated using the equation $I=Anvq$.
Which statement is not correct?

physics electric current, potential difference and resistance drift velocity and mobility drift speed drift velocity & mobility

  1. n is the number of charge carriers per unit volume of the wire

  2. nA is the number of charge carriers per unit length of the wire

  3. q is the charge of each charge carrier

  4. v is the velocity of each charge carrier

Show answer
Correct Option: D
Explanation:

Given that,

Electric current in the wire  , $I=Anvq$

Here $A$ is the cross-section of wire 
$n$ is the number of charge carriers per unit volume of wire
$v $ is  drift velocity of charge carriers 
$q$ is the charge of each charge carier 

Hence incorrect statement is $(D)$

Electrical, as well as gravitational affects, can be thought to be caused by fields. Which of the following is true of an electrical or gravitational field?

physics gravitational fields representing a gravitational field gravitational field circular motion and gravitation

  1. The field concept is often used to describe contact forces

  2. Gravitational or electric field does not exist in the space around an object

  3. Fields are useful for understanding forces acting through a distance

  4. There is no way to verify the existence of a force field since it is just a concept

Show answer
Correct Option: C
Explanation:

In physics, concept of field is a model used to explain the influence that a massive body or charged particle extends into the space around itself, producing a force on another massive body or charged body placed in space. Thus, concept of field is used to explain gravitational and electrostatic phenomena.

A satellite is revolving in a circular orbit at a height $'h'$ from the earth's surface (radius of earth $R$;$h\ <\ <\ R$). The minimum increase in its orbital velocity required, so that the satellite could escape from the earth's gravitational field, is close to:(Neglect the effect of atmosphere.)

physics gravitational fields representing a gravitational field gravitational field circular motion and gravitation

  1. $\sqrt{2gR}$

  2. $\sqrt{gR}$

  3. $\sqrt{gR/2}$

  4. $\sqrt{gR}\left(\sqrt{2}-1\right)$

Show answer
Correct Option: D
Explanation:

height$=h$

No, the given formula,
Orbital velocity
$v=\cfrac{GM}{R+h}=\cfrac{GM}{R}$ as $h<<R$
Velocity required to escape
$\cfrac{1}{2}mv^{\prime 2}=\cfrac{GMm}{R+h}\v^{\prime}=\cfrac{2GM}{R+h}=\cfrac{2GM}{R}(h<<R)$
$\therefore$ Increase in velocity
$v^{\prime}=v=\sqrt{\cfrac{2GM}{R}}-\sqrt{\cfrac{GM}{R}}\ \quad=\sqrt{2gR}-\sqrt{gR}=\sqrt{gR}(\sqrt2-1)$