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

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 :

  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
Reveal answer Fill a bubble to check yourself
B Correct answer
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.

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

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

  1. $[M^1L^1]$
  2. $[L^2]$
  3. $[L^{-2}]$
  4. $[M^1L^{-1}T^1]$
Reveal answer Fill a bubble to check yourself
C Correct answer
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.

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

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

  1. True

  2. False
Reveal answer Fill a bubble to check yourself
B Correct answer
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 $\sqrt(\mu)$

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

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

  1. $20 N$
  2. $40 N$
  3. $10 N$
  4. $120 N$
Reveal answer Fill a bubble to check yourself
B Correct answer
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 $4 \times 10 = 40 N$

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

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

  1. 250.5 and 249.5

  2. 249.5 and 250.5

  3. 124.5 and 125.5

  4. 125.5 and 124.5
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

Using the frequency formula f = (1/2L) * sqrt(T/m), the difference in frequencies for two segments of lengths L1 and L2 is 1 Hz. Given L1 + L2 = 1m and L1 - L2 = 0.002m, we find L1 = 0.501m and L2 = 0.499m. The frequencies are proportional to 1/L, leading to 124.5 Hz and 125.5 Hz.

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

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?

  1. 124 : 125

  2. 250 : 313

  3. 5 : 3

  4. 41 : 57
Reveal answer Fill a bubble to check yourself
A Correct answer
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 $\dfrac{l _1}{l _2}=\dfrac{248}{250}$
$=\dfrac{124}{125}=124: 125$

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

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

  1. remains same

  2. increases by 2%

  3. decreases by 2%

  4. frequency becomes imaginary
Reveal answer Fill a bubble to check yourself
C Correct answer
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)

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

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

  1. $1.4\ f$
  2. $0.7\ f$
  3. $2\ f$
  4. $f$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

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