MHT CET 2017 :: Physics :: General questions

• Physics - General questions

The magnetic flux near the axis and inside  the air-core solenoid of length 60 cm carrying current 'l' is 1.57 x 10^-6 Wb. its magnetic moment will be (cross-sectional area  of a solenoid is very small as compared to its length, $\mu_{0}=4\pi \times 10^{-7}$ SI unit)

Answer:

Explanation:

Given, L= 60cm, $\phi$= 1.57 x 10^-6Wb

 Magnetic induction inside the solenoid is 

$B=\frac{\mu_{0}Nl}{L}$  .......(i)

 Magnetic flux, $\phi$  =BA

     $=\frac{\mu_{0}Nl.A}{L}$     [ $\because$  from Eq.(i)]

 Magnetic  moment = NlA $ =\frac{\phi L}{\mu_{0}}$

 Substituting the given values, we get

$M=\frac{1.57 \times 10^{-6}\times0.6} {4 \times 3.14 \times 10^{-7}}=0.75 A$

In communication system, the process of superimposing a low  frequency signal on a high frequency wave is known as

Answer:

Explanation:

Modulation is the phenomenon of superimposing the low-frequency signal on a high-frequency wave.

Attenuation refers to the loss of strength of a signal during its propagation through a communication channel. Demodulation  is the reverse process of modulation 

Repeater amplifier the signal received from the transmitter and retransmits it to the receiver.

Two coils P and Q  are kept near each other. When no current flows through coil P and current increases in coil  Q at the rate 10 A/s, the emf in coil P is 15 mV. When coil Q carries no current and current of 1.8 A flows through the coil P, the magnetic flux linked with  the coil Q is 

Answer:

Explanation:

Given,  $\frac{dl_{\theta}}{dt}=10 A/s$

 $e_{p} = 15 mA$

 = 15 x 10^-3 V, l_p =1.8 A

 Induced emf in coil  P is given as,

$|e_{p}| = M.\frac{dl_{Q}}{dt}$

 Substituting the given values , we get

15 x 10^-3  =M x 10

 $\Rightarrow$   M= 15 x 10^-4 H

 magnetic fluxed with linked coil Q is givne as,

$\phi _{Q}=Ml_{p}=15  \times 10^{-4} \times 1.8$

 = $27.0 \times 10^{-4}$

 =2.7 $\times 10^{-3}$= 2.7 mWb

The observer is moving with velocity 'v₀' towards the stationary source of sound and then after crossing moves away from the source with velocity 'v₀' . Assume that the medium through which the sound waves travel is at rest.If v is the velocity of sound and n is the frequency emitted by the source, then the difference  between apparent frequencies heard by the observer is 

Answer:

Explanation:

When the observer is moving towards the source and source is stationary , then the apparent  frequency is given as,

   $n'=n\left(\frac{v+v_{0}}{v}\right)$  ..........(i)

When the observer is moving away from the source and the source is stationary , then the apparent frequency is given as,

$n''=-n\left(\frac{v-v_{0}}{v}\right)$    .......(ii)

 From Eqs .(i) and (ii) , we get

$n'- n''=\frac{n}{v}(v'+v_{0}-v+v_{0})$

 = $\frac {2nv_{0}}{v}$

A particle performs linear SHM at a particular instant, velocity of the particle is 'u' and acceleration is '$\alpha$' while at another instant velocity is 'v' and acceleration is '$\beta$  ( 0 <$\alpha$ < $\beta$ ). The distance between the two positions is 

Answer:

Explanation:

Let  the distance be p when velocity is u and acceleration $\alpha$.

 Let the distance  q when velocity is v and acceleration $\beta$.

 If $\omega$  is the angular frequency  , then 

$ \alpha=\omega^{2} p $   ans $\beta=\omega^{2} q$

 $\therefore$      $\alpha +\beta $= $\omega ^{2} (p+q)$      ........(i)

 Also,    $u^{2}=\omega ^{2} A^{2}-\omega ^{2} p^{2}$

and    $v^{2}=\omega ^{2} A^{2}-\omega ^{2} q^{2}$

 $\Rightarrow$   $ v^{2}-u^{2}= \omega ^{2}(p^{2}-q^{2})$

 $ v^{2}-u^{2}= \omega ^{2}(p^{}-q^{})(p+q)$    .........(ii)

 by Eqs .(i) and (ii), we get

$v^{2}-u^{2}= (p^{}-q^{})(\alpha+\beta)$

 $\therefore$   $  p-q= \frac{v^{2}-u^{2}}{\alpha+\beta}$  or   $q-p= \frac{u^{2}-v^{2}}{\alpha+\beta}$

• Physics - General questions