Get Answers to all your Questions

header-bg qa
Filter By

All Questions

In a fluorescent lamp choke ( a small transformer) 100 V of reverse voltage is produced when the choke current changes uniformly from 0.25 A to 0 in a duration of 0.025 ms. The self-inductance of the choke (in mH) is estimated to be _____.
Option: 1 10
Option: 2  20
Option: 3 30
Option: 4 40
 

 

 

 

V = L\frac{\text d i}{\text d t}\Rightarrow 100 = L\frac{0.25}{0.025} \\ \Rightarrow L = 10 \ \text{mH}

View Full Answer(1)
Posted by

avinash.dongre

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}

 

View Full Answer(1)
Posted by

vishal kumar

Crack CUET with india's "Best Teachers"

  • HD Video Lectures
  • Unlimited Mock Tests
  • Faculty Support
cuet_ads

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

View Full Answer(1)
Posted by

vishal kumar

 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}

 

 

View Full Answer(1)
Posted by

vishal kumar

JEE Main high-scoring chapters and topics

Study 40% syllabus and score up to 100% marks in JEE

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

View Full Answer(1)
Posted by

vishal kumar

At time t=0 magnetic field of 1000 Gauss is passing perpendicularly through the area defined by the closed loop shown in the figure. If the magnetic field reduces linearly to 500\: Gauss, in the next 5s, then induced EMF (in \mu V) in the loop is : 
Option: 1 56

Option: 3 48

Option: 5 36

Option: 7 28
 

 

 

 

 

 

Area of the given loop

\\A=Area\ of\ rectangle-Area\ of \ 2\ triangles\ with\ height\ 2cm \\ A=16\times4-2\times4=56cm^2

Magnitude of emf 

\\E=A\frac{dB}{dt}=56\times10^{-4}\times(\frac{(1000-500)\times10^{-4}}{5})\\ E=5600\times10^{-8}V=56\mu \ V

So the correct graph is given in option 1.

View Full Answer(1)
Posted by

vishal kumar

NEET 2024 Most scoring concepts

    Just Study 32% of the NEET syllabus and Score up to 100% marks


A loop ABCDEFA of straight edges has six corner points A(0,0,0), B(5,0,0),C(5,5,0), D(0,5,0), E(0,5,5) and F(0,0,5). The magnetic field in this region is \vec B=(3\hat{i}+4\hat{j})T.The quantity of flux through the loop ABCDEFA (in Wb) is:-    
Option: 1 175
Option: 2 60
Option: 3 25
Option: 4 170
 

 

 

Magnetic flux - - wherein

Magnetic flux-

The total number of magnetic lines of force passing normally through an area placed in a magnetic field is equal to
the magnetic flux linked with that area.

I.e for the below figure

Net magnetic flux through the surface is given by

\phi_B = \oint \vec{B}\cdot \vec{dA}= BA\cos \Theta

where 

\phi_B= Magnetic Flux

B = Magnetic field 

\Theta = The angle between area vector and magnetic field vector

 

 

  • Magnetic flux is a scalar quantity.
  • Unit of magnetic flux -

          It's S.I. unit is Weber (wb) or Tesla\times m^2 and its C.G.S. unit is maxwell(Mx).

      and 1 \ w b=1 \ T m^{2}  and 1 \ Mx = 10^{-8 } \ wb

  • The dimension of magnetic flux is ML^{2}T^{-2}A^{-1} 
  • if θ = 0 then \phi = BA and  Flux will be positive.
  • If  \theta =\frac{\pi }{2} then Flux will be zero (i.e \phi = 0)

            

 

 

 

 

 

 

\vec B=(3\hat{i}+4\hat{j})T

Total area vectot=area of ABCD+area of DEFA=5^2\hat{k}+5^2\hat{i}=25 (\hat{i}+\hat{k})

Total Magnetic flux=\vec{B}.\vec{A}=(3\hat{i}+4\hat{j}).(25 (\hat{i}+\hat{k}))=(75+100)wb=175 wb

 

So option 3 will be correct answer.

View Full Answer(1)
Posted by

Ritika Jonwal

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.

View Full Answer(1)
Posted by

Ritika Jonwal

Crack CUET with india's "Best Teachers"

  • HD Video Lectures
  • Unlimited Mock Tests
  • Faculty Support
cuet_ads

A LCR circuit behaves like a damped harmonic oscillator. Comparing it with a physical spring-mass damped oscillator having dampling constant 'b', the correct equivalence would be:-
 

Option: 1 L\leftrightarrow k,C\leftrightarrow b,R\leftrightarrow m


Option: 2 L\leftrightarrow m,C\leftrightarrow k,R\leftrightarrow b


Option: 3 L\leftrightarrow m,C\leftrightarrow \frac{1}{k},R\leftrightarrow b  


Option: 4 L\leftrightarrow \frac{1}{b},C\leftrightarrow \frac{1}{m},R\leftrightarrow \frac{1}{k}
 

 

 

Series LCR circuit -

Series LCR circuit-

                                  

The Figure given above shows a circuit containing a capacitor ,resistor and inductor connected in series through an alternating / sinosoidal voltage source.

As they are in series so the same amount of current will flow in all the three circuit components and for the voltage, vector sum of potential drop across each component would be equal to the applied voltage.

Let 'i' be the amount of current in the circuit at any time and VL,VC and VR the potential drop across L,C and R respectively then
                                          \begin{array}{l}{\mathrm{v}_{\mathrm{R}}=\mathrm{i} \mathrm{R} \rightarrow \text { Voltage is in phase with i }} \\ \\ {\mathrm{v}_{\mathrm{L}=\mathrm{i} \omega \mathrm{L}} \rightarrow \text { voltage is leading i by } 90^{\circ}} \\ \\ {\mathrm{v}_{\mathrm{c}}=\mathrm{i} / \mathrm{\omega} \mathrm{c} \rightarrow \text { voltage is lagging behind i by } 90^{\circ}}\end{array}\varepsilon

By all these we can draw phasor diagram as shown below -

                                

One thing should be noticed that we have assumed that VL is greater than VC which makes i lags behind V. If VC > VL then i lead V. So as per our assumption, there resultant will be (VL -VC). So, from the above phasor diagram V will represent resultant of vectors VR and (VL -VC). So the equation become - 

                                                   \begin{aligned} V &=\sqrt{V_{R}^{2}+\left(V_{L}-V_{C}\right)^{2}} \\ \\ &=i \sqrt{R^{2}+\left(X_{L}-X_{C}\right)^{2}} \\ \\ &=i \sqrt{R^{2}+\left(\omega L-\frac{1}{\omega C}\right)^{2}} \\ \\ &=i Z \\ \text { where, } & \\ Z &=\sqrt{R^{2}+\left(\omega L-\frac{1}{\omega C}\right)^{2}} \end{aligned}

Here, Z is called Impedence of this circuit.

Now come to the phase angle. The phase angle for this case is given as - 

                                            \tan \varphi=\frac{V_{L}-V_{C}}{V_{R}}=\frac{X_{L}-X_{C}}{R}=\frac{\omega L-\frac{1}{\omega C}}{R}

Now from the equation of the phase angle three cases will arise. These three cases are - 

(i) When,     \omega L \ > \ \frac{1}{\omega C}

    then, tanφ is positive i.e. φ is positive and voltage leads the current i.
(ii) When \omega L \ < \ \frac{1}{\omega C}

    then, tanφ is negative i.e. φ is negative and voltage lags behind the current i.
(iii) When  \omega L \ = \ \frac{1}{\omega C} ,

      then tanφ is zero i.e. φ is zero and voltage and current are in phase. This is called electrical resonance. 

 

 

-

 

 

For the spring-mass damping system, the governing equation is given by

m \frac{dx^2}{dt^2}+b\frac{dx}{dt}+kx=0...............(1)

For an LCR damped oscillator, the equation is 

L \frac{dQ^2}{dt^2}+R\frac{dQ}{dt}+\frac{Q}{C}=0...............(2)

Comparing 1 and 2

\\L\leftrightarrow m\\C\leftrightarrow \frac{1}{k}\\R\leftrightarrow b

So the answer is option (3)

View Full Answer(1)
Posted by

Ritika Jonwal

A long solenoid of radius R carries a time(t) - dependent current I(t)=I_ot(1-t). A ring of radius 2R is placed coaxially near its middle. During the time interval 0\leq t\leq 1, the induced current (IR) and the induced EMF (VR) in the ring changes as:-
Option: 1 Direction of I_{R} remains unchanged and V_{R} is zero at t = 0.25
Option: 2 Direction of I_{R} remains unchanged and V_{R} is zero at t = 0.5
Option: 3 At t = 0.25 direction of I_{R}  reverses and V_{R}  is maximum  
Option: 4 At t = 0.5 direction of I_{R}  reverses and V_{R}  is zero
 

 

Faraday's law of induction -

 Faraday’s First Law-

Whenever the number of magnetic lines of force (Magnetic Flux) passing through a circuit changes an emf called induced emf is produced in the circuit. The induced emf persists only as long as there is a change of flux.

Faraday’s Second Law-

The induced emf is given by the rate of change of magnetic flux linked with the circuit.

i.e  Rate of change of magnetic Flux= \varepsilon = \frac{-d\phi }{dt}

where d\phi\rightarrow \phi _{2}-\phi _{1}= change in flux

So,

B=\mu _0nI_0t(1-t)

using B_0=\mu _0nI_0

SO B=B_0t(1-t)

Now considering solenoid as ideal solenoid extended up to infinite and ring as its centre

 \phi =BA_s=B_0t(1-t) A_s

e=-\frac{d \phi}{dt} =-B_0A_s (1-2t)

Since induced emf is changing so current will also be changing 

because i=\frac{e}{R}

So, since direction of emf is changing so direction of current is also changing.

And VR will be zero at t=0.5 sec

 

So the option (4) is correct.

View Full Answer(1)
Posted by

Ritika Jonwal

JEE Main high-scoring chapters and topics

Study 40% syllabus and score up to 100% marks in JEE

filter_img