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A Laser light of wavelength 660 nm is used to weld Retina detachment.  If a Laser pulse of width 60 ms and power 0.5 kW is used the approximate number of photons in the pulse are : [Take Planck’s constant h=6.62×10−34 Js]
Option: 1  1020
Option: 2 1018
Option: 3 1022
Option: 4 1019
 

As we discussed in

E= N \frac{hc}{\lambda }

or N=\frac{E\lambda }{hC}\:=\: \frac{0.5\times10^{3}\times60\times10^{-3}\times660\times10^{-9}}{6.62\times10^{-34}\times3\times10^{8}}

=\: \frac{10^{-8}\times6.60\times10^{2}}{6.62\times10^{-26}}=10^{20}

 

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Posted by

vishal kumar

A signal of frequency 20 kHz and peak voltage of 5 Volt is used to modulate a carrier wave of frequency 1.2 MHz and peak voltage 25 Volts.  Choose the correct statement.
Option: 1  Modulation index=5, side frequency bands are at 1400 kHz and 1000 kHz
Option: 2 Modulation index=5, side frequency bands are at 21.2 kHz and 18.8 kHz
Option: 3 Modulation index=0.8, side frequency bands are at 1180 kHz and 1220 kHz
Option: 4 Modulation index=0.2, side frequency bands are at 1220 kHz and 1180 kHz  
 

\\\text{ Modulate}\ Index: \frac{5}{25}=\frac{1}{5}=0.2\\ \text{Side frequency} (1200+20) H z=1220 k H z \ and \ 1200-20=1180 k H z

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vishal kumar

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 Two wires W1 and W2 have the same radius r and respective densities ρ1 and ρ2 such that ρ2=4ρ1.  They are joined together at the point O, as shown in the figure.  The combination is used as a sonometer wire and kept under tension T. The point O is midway between the two bridges.  When a stationary wave is set up in the composite wire, the joint is found to be a node.  The ratio of the number of antinodes formed in W1 to W2 is :  
Option: 1  1 : 1
Option: 2  1 : 2
Option: 3  1 : 3
Option: 4 4 : 1
 

When the joint is found to be a node.

then the frequency is given as 

n=\frac{p}{2 l} \sqrt{\frac{T}{\pi r^{2} d}}

Where p is also equal to the number of antinodes formed  in the wire

As

\begin{aligned} &n_{1}=n_{2}\\ &T \rightarrow \text { same }\\ &r \rightarrow \text { same }\\ &l \rightarrow \text { same } \end{aligned}

So

\begin{aligned} &\frac{p_{1}}{\sqrt{d_{1}}}=\frac{p_{2}}{\sqrt{d_{2}}}\\ &\frac{p_{1}}{p_{2}}=\frac{1}{2} \end{aligned}

 

 

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Posted by

vishal kumar

A one metre long (both ends open) organ pipe is kept in a gas that has double the density of air at STP. Assuming the speed of sound in air at STP is 300m/s, the frequency difference between the fundamental and second harmonic of this pipe is ________Hz. 
Option: 1 106
Option: 2 600
Option: 3 310
Option: 4 210
 

 

 

   

 

\begin{array}{l}{\mathrm{V}=\sqrt{\frac{\mathrm{B}}{\rho}}} \\\\ {\frac{\mathrm{V}_{\text {pipe }}}{\mathrm{V}_{\text {air }}}=\frac{\sqrt{\frac{\mathrm{B}}{2 \rho}}}{\sqrt{\frac{\mathrm{B}}{ \rho}}}=\frac{1}{\sqrt{2}}} \\\\ {\mathrm{V}_{\text {pipe }}=\frac{V_{\text {air }}}{\sqrt{2}}}\end{array}

\begin{array}{l}{f_{n}=\frac{(n+1) V_{\text {pipe }}}{2 \ell}} \\ {\left.f_{1}-f_{0}=\frac{V_{\text {pipe }}}{2 \ell}=\frac{300}{2 \sqrt{2}}=106.38 =106\ \mathrm{Hz} \text { (If } \sqrt{2}=1.41\right)}\end{array}

So the answer will be 106

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vishal kumar

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If the magnetic field in a plane electromagnetic wave is given by \vec B=3\times10^{-8} sin (1.6\times10^3x + 48\times10^{10}t)\ \vec{j} \ T; then what will be the expression for the electric field?


Option: 1 \vec{E}=\left ( 9\sin \left ( 1.6\times 10^{3}x+48\times 10^{10}t \right )\widehat{k}\; V/m \right )
 
Option: 2 \vec{E}=\left ( 60\sin \left ( 1.6\times 10^{3}x+48\times 10^{10}t \right )\widehat{k}\; V/m \right )

Option: 3 \vec{E}=\left ( 3\times 10^{-8}\sin \left ( 1.6\times 10^{3}x+48\times 10^{10}t \right )\widehat{i}\; V/m \right )  

Option: 4 \vec{E}=\left ( 3\times 10^{-8}\sin \left ( 1.6\times 10^{3}x+48\times 10^{10}t \right )\widehat{j}\; V/m \right )
 

Nature of Electromagnetic Waves -

It is also seen from Maxwell’s equations that the magnitude of the electric and the magnetic fields in an electromagnetic wave are related as - B_{0}= \frac{E_o}{c}

given, \vec B=3\times10^{-8} sin (1.6\times10^3x + 48\times10^{10}t)T

\begin{aligned} \\ \left | \vec E \right |=BC=3\times10^{-8} sin (1.6\times10^3x + 48\times10^{10}t)\times (3\times10^8)\\ =9 sin (1.6\times10^3x + 48\times10^{10}t)T \end{aligned}

wave is propagating in -x direction, i.e., in - i direction.

the direction of the EMW wave is in the direction of \vec E\times \vec B.

Since B is in j direction and EMW is in -i direction. Therefore E is in (k) direction.

So Option (1) is correct.

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Posted by

Ritika Jonwal

A plane electromagnetic wave is propagating along the direction \frac{\hat{i}+\hat{j}}{\sqrt{2}}, with its polarization along the direction \hat{k}. The correct form the magnetic field of the wave would be (here B_{0} is an appropriate constant ) :
Option: 1 B_{0}\frac{\hat{i}-\hat{j}}{\sqrt{2}}\cos \left ( \omega t-k\frac{\hat{i+\hat{j}}}{\sqrt{2}} \right )      
Option: 2  B_{0}\frac{\hat{j}-\hat{i}}{\sqrt{2}}\cos \left ( \omega t+k\frac{\hat{i+\hat{j}}}{\sqrt{2}} \right )
Option: 3  B_{0}\frac{\hat{i}+\hat{j}}{\sqrt{2}}\cos \left ( \omega t-k\frac{\hat{i+\hat{j}}}{\sqrt{2}} \right )      
Option: 4 B_{0}\; \hat{k}\; \cos \left ( \omega t-k\frac{\hat{i+\hat{j}}}{\sqrt{2}} \right )
 

 

EM wave is in direction - \frac{\hat{i}+\hat{j}}{\sqrt{2}}

 

As we know that the axis of polarisation of the Em wave is same as Electric field direction that is -  \hat{k}

 

\vec{E}\times \vec{B} direction of propagation of EM wave = \frac{\hat{i}+\hat{j}}{\sqrt{2}}

 

                                                                        \Rightarrow \vec{k}\times \vec{B}=\frac{\hat{i}+\hat{j}}{\sqrt{2}} 

 

                                                          So  B is along                \Rightarrow \frac{\hat{i}-\hat{j}}{\sqrt{2}}

And the equation of the electromagnetic waves will be in terms of the  \cos (\omega t-\vec{K}\cdot \vec{r})

So by concluding the above result we can deduce that the option (1) is correct.

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Posted by

avinash.dongre

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A wire of length L and mass per unit length 6.0\times 10^{-3}kgm^{-1} is put under tension of 540 N. Two consecutive frequencies at which it resonates are : 420 Hz and 490 Hz . Then L in meters is :
Option: 1 2.1
Option: 2 8.1
Option: 3 1.1
Option: 4 5.1
 

 

 

Fundamental frequency   = 490 – 420 = 70 Hz

70=\frac{1}{2l}\sqrt{\frac{T}{\mu }}

\Rightarrow 70=\frac{1}{2l}\sqrt{\frac{540}{6\times 10^{-3} }}

\Rightarrow l=\frac{1}{2\times 70}\sqrt{90\times 10^{3}}=\frac{300}{140}

\Rightarrow l\approx 2.14m

Hence the correct option is (1).

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Posted by

avinash.dongre

Three harmonic waves having equal frequency v and same intensity I0, have phase angles 0, \frac{\pi}{4} and -\frac{\pi}{4} , respectively. When they are superimposed the intensity of the resultant wave is close to:
Option: 1 0.2I0
Option: 2  I0
Option: 3 3I0
Option: 4 5.8I0
 

 

 

 

 

 

 

A_{res} = A + \frac{A}{\sqrt2} + \frac{A}{\sqrt2} = A(1 + \sqrt2)

I = (1 + \sqrt2)^2I_0 = 5.8I_0

 

Hence the correct option is (4). 

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avinash.dongre

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A transverse wave travels on a taut steel wire with a velocity of v when tension in it is 2.06 x 104 N . When the tension is changed to t, the velocity changed to v/2. The value of T (in N ) is close to :
Option: 1 5.15 x 103 
Option: 2 10.2 x 102 
Option: 3 2.50 x 104 
Option: 4 30.5 x 104 
 

 

 

 

Velocity

\\v\propto\sqrt{T}\\\\\frac{T_1}{T_2}=(\frac{v_1}{v_2})^2\\\\\frac{T_1}{T_2}=(\frac{v}{v/2})^2=4\\\\\Rightarrow T_2=\frac{T_1}{4}=0.515\times10^4\\T_2=5.15\times10^3 N

Hence the correct option is (1)

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Posted by

vishal kumar

A plane electromagnetic wave of frequency 25 GHz is propagating in vacuum along the z-direction. At a particular point in space and time, the magnetic field is given by \overrightarrow{B} = 5 \times 10^{-8} \widehat{j} T. The corresponding electric field \overrightarrow{E} is (speed of light c = 3 x 108 ms-1)


Option: 1 -1.66 \times 10^{-16} \widehat{i} V/m

Option: 2 1.66 \times 10^{-16} \widehat{i} V/m

Option: 3 -15\widehat{i} V/m

Option: 4 15\widehat{i} V/m
 

 

\begin{array}{l}{\frac{E}{B}=c} \\ {E=B \times c} \\ {=15 \hat{i} \ \mathrm{V/m}}\end{array}

 

Hence the correct option is (4).

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Posted by

vishal kumar

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