Tag: physics

Questions Related to physics

The electric field of a plane electromagnetic wave is given by
$\vec{E} = E _0 \dfrac{\hat{i} + \hat{j}}{\sqrt{2}} \cos (kz + \omega t)$
At $t = 0$, a positively charged particle is at the point $(x, y , z) = \left(0, 0 , \dfrac{\pi}{k} \right)$. If its instantaneous velocity at $(t = 0)$ is $v _0 \hat{k}$, the force acting on it due to the wave is :

physics observing space: telescopes maxwell's equations the nature of light introduction to electromagnetic waves

  1. parallel to $\hat{k}$

  2. parallel to $\dfrac{\hat{i} + \hat{j}}{\sqrt{2}}$

  3. antiparallel to $\dfrac{\hat{i} + \hat{j}}{\sqrt{2}}$

  4. zero

Show answer
Correct Option: B
Explanation:

$\vec{E} = E _0 \left(\dfrac{i + j}{\sqrt{2}}\right) \cos (kz + wt)$


$\therefore$ unit vector along electric field, $\vec{E} = \left(\dfrac{\hat{i} + \hat{j}}{\sqrt{2}} \right)$


Direction of electromagnetic wave is in (-z) direction 

$\therefore \hat{C} = -\hat{k}$  wave direction.

Let $\hat{B} $ is the unit vector along the direction of magnetic field.

$\hat{B} = \hat{C} \times \hat{E} = -\hat{k} \times \left(\dfrac{\hat{i} + \hat{j}}{\sqrt{2}} \right) = - \left(\dfrac{\hat{k} \times i + \hat{k} \times j}{\sqrt{2}} \right)$

$\hat{B} = -\left(\dfrac{\hat{j} + (-i)}{\sqrt{2}} \right)  = \left(\dfrac{\hat{i} - \hat{j}}{\sqrt{2}} \right)$

$\vec{F _e} =$ electric force on the charge particle 

$\vec{F _e} = $ unit vector of electric force $= \dfrac{q \hat{E}}{|q\hat{E}|} = \hat{E}$

$\vec{F} _e = \dfrac{\hat{i} + \hat{j}}{\sqrt{2}}$

$\vec{F} _b = $ magnetic force $= q \vec{V} \times \vec{B} = q \left(V _0 \hat{k} \times \dfrac{i - j}{\sqrt{2}}\right)$

$\vec{F} _b = q V _0 \left[\dfrac{\hat{k} \times \hat{i} - \hat{k} \times \hat{j}}{\sqrt{2}} \right] = q V _0 \left[\dfrac{\hat{j} - (-\hat{i})}{\sqrt{2}}\right]$

$\vec{F} _b = q V _0 \left(\dfrac{\hat{i} + \hat{j}}{\sqrt{2}} \right)$

$\vec{F} _{Net} = \hat{F} _e + \hat{F} _b = \dfrac{\hat{i} + \hat{j}}{\sqrt{2}} + \dfrac{\hat{i} + \hat{j}}{\sqrt{2}} = \dfrac{2}{\sqrt{2}} (\hat{i} + \hat{j})$

$\therefore \hat{F} _{Net} = $ unit vector $= \dfrac{\hat{i} + \hat{j}}{\sqrt{2}}$

Option (B) is correct.

The Schrodinger equation for a free electron of mass m and energy E written in terms of the wave function $\psi $ is $\frac{d^2\psi}{dx^2}+\frac{8 \pi ^2mE}{h^2}\psi =0$. The dimensions of the coefficient $\psi$ of in the second term must be

physics observing space: telescopes maxwell's equations the nature of light introduction to electromagnetic waves

  1. $[M^1L^1]$

  2. $[L^2]$

  3. $[L^{-2}]$

  4. $[M^1L^{-1}T^1]$

Show answer
Correct Option: C
Explanation:

By dimensional analysis the dimensions of each term in an equation must be the same. In the first term the second derivative with respect to distance x indicates the dimensions of the coefficient $\psi$ of to be $[L^{-2}]$ and hence the answer.

The frequency of vibration of a sonometer wire is directly proportional to linear density of the wire:

laws of vibrations of stretched strings sonometer and laws of transverse vibrations vibrations of stretched strings waves physics

  1. True

  2. False

Show answer
Correct Option: B
Explanation:

The frequency of vibration in a sonometer is given by $f=(1/2L) \sqrt{(T/\mu)}; \mu $ is the linear density of the material of the wire
 
Thus, frequency of vibration is inversely proportional to $\sqrt(\mu)$

The tension in a piano wire is $10 N$. The tension in a piano wire to produce a node of double frequency is

laws of vibrations of stretched strings sonometer and laws of transverse vibrations vibrations of stretched strings waves physics

  1. $20 N$

  2. $40 N$

  3. $10 N$

  4. $120 N$

Show answer
Correct Option: B
Explanation:

For frequency of oscillation of wire $n \propto \sqrt{T}$. 

Here $T$ is tension in the wire. In order to increase frequency twice, tension needs to be made $4$ times. 
So, new tension must be $4 \times 10 = 40 N$

A knife edge divides a sonometer wire in two parts which differ in length by 2 mm. The whole length of the wire is 1 meter. The two parts of the string when sounded together produce one beat per second. Then the frequency of the smaller and longer pans.in Hz,are

laws of vibrations of stretched strings sonometer and laws of transverse vibrations vibrations of stretched strings waves physics

  1. 250.5 and 249.5

  2. 249.5 and 250.5

  3. 124.5 and 125.5

  4. 125.5 and 124.5

Show answer
Correct Option: D

A sonometer wire of length $l _1$ vibrates with a frequency 250 Hz. If the length of wire is increased then 2 beats/s are heard. What is ratio of the lengths of the wire?

laws of vibrations of stretched strings sonometer and laws of transverse vibrations vibrations of stretched strings waves physics

  1. 124 : 125

  2. 250 : 313

  3. 5 : 3

  4. 41 : 57

Show answer
Correct Option: A
Explanation:

The frequency of sonometer wire is given by
$n=\dfrac{p}{2l}\sqrt{(\dfrac{T}{m})}$
or $n\propto \dfrac{1}{l}$     ...(i)
$\therefore \dfrac{n _1}{n _2}=\dfrac{l _2}{l _1}$
or $\dfrac{250}{250-2}=\dfrac{l _2}{l _1}$
or $\dfrac{l _1}{l _2}=\dfrac{248}{250}$
$=\dfrac{124}{125}=124: 125$

The tension in the sonometer wire is decreased by 4% by loosening the screws. It fundamental frequency

laws of vibrations of stretched strings sonometer and laws of transverse vibrations vibrations of stretched strings waves physics

  1. remains same

  2. increases by 2%

  3. decreases by 2%

  4. frequency becomes imaginary

Show answer
Correct Option: C
Explanation:

The frequency of vibration in a sonometer is given by $f=(1/2L) \sqrt(T/\mu)$

The fractional change in frequency with change in tension is given by $\Delta f/f = -(\Delta T/2T)$.

Thus, the frequency decreases by 2%

The correct option is (c)

The sonometer wire is vibrating in the second overtone. The length of the wire in terms of wavelength is:

laws of vibrations of stretched strings sonometer and laws of transverse vibrations vibrations of stretched strings waves physics

  1. $L=\lambda/2$

  2. $L=5\lambda/2$

  3. $L=3\lambda/2$

  4. $L=3\lambda$

Show answer
Correct Option: C
Explanation:

second overtone = third harmonic.
Thus, the length of the wire is $L=3\lambda/2$

The correct option is (c)

The frequency of vibration of a sonometer wire is inversely proportional to tension in the wire

laws of vibrations of stretched strings sonometer and laws of transverse vibrations vibrations of stretched strings waves physics

  1. True

  2. False

Show answer
Correct Option: B
Explanation:

The frequency of vibration in a sonometer is given by $f=(1/2L) \sqrt(T/\mu)$

Thus, frequency of vibration is directly proportional to $\sqrt(T)$

A wire has frequency f. Its length is doubled by stretching. Its frequency now will be:

laws of vibrations of stretched strings sonometer and laws of transverse vibrations vibrations of stretched strings waves physics

  1. $1.4\ f$

  2. $0.7\ f$

  3. $2\ f$

  4. $f$

Show answer
Correct Option: B
Explanation:

$We\quad have\quad f=\frac { 1 }{ 2l } \sqrt { \frac { T }{ m }  } \ \ \therefore f\propto \frac { 1 }{ l } \quad \quad and\quad f\propto \frac { 1 }{ \sqrt { m }  } \ \ And\quad we\quad know\quad that\quad as\quad l\quad becomes\quad double\quad mass/unit\quad length\quad becomes\quad half.\ \ \therefore f\quad becomes\quad \frac { 1 }{ \sqrt { 2 }  } times\quad the\quad orignal.\ \ Hence\quad { f }^{ ' }=0.7f$