MHT CET 2018 :: Physics :: General questions

• Physics - General questions

In the following network, the current flowing through 15$\Omega$ resistance is

Answer:

Explanation:

 The above circuit is a balanced Wheatstone bridge.

 Hence, the above circuit  is drawn  in the following manner

 From the above circuit,

18 l=24(21-l)

18 l= 50.4-24 l

 24 l+18l=50.4

 42 l=50.4

 l = $\frac{50.4}{42}$

$\therefore$   l= 1.2 A

An alternating voltage e =200 $\sqrt{2}$ $\sin $ (100 t) volt is connected to 1 $\mu$ F  capacitor through AC ammeter. The reading of ammeter is 

Answer:

Explanation:

 Given, e =200 $\sqrt{2}$ $\sin $ (100 t)   .......(i)

 Capacitance of capacitor  (C) = 1 $\mu$ F= 1 x 10^-6 F

 The standard  equation  of voltage of AC is given by

 V= V₀ $\sin \omega$t  ..........(ii)

 On comparing Eqs.(i) and (ii) , we get

 $V_{0}= 200\sqrt{2}$

 $\omega$ = 100

We know that,

 $l_{rms}= \frac{V_{rms}}{X_{c}}$

 But   $V_{rms}= \frac{V_{0}}{\sqrt{2}}$     $l_{rms}=\frac{V_{0}}{\sqrt{2}X_{c}}$

$l_{rms}=\frac{V_{0}\omega C}{\sqrt{2}}$    $\left(X_{c}=\frac{1}{\omega C}\right)$

$l_{rms}=\frac{200\times\sqrt{2}\times100\times1 \times 10^{-6}}{\sqrt{2}}$

$l_{rms}=20\times 10^{-3}A=20 mA$

A circular coil carrying current 'l' has radius ' R'  and magnetic  field at the  centre ' B'  . At what distance from the centre along the axis of the same coil, the magnetic field will be $\frac{B}{8}$ ?

Answer:

Explanation:

According to the question, 

Magnetic field at the centre of circular coil is given by 

$B_{centre}=\frac{\mu_{0}Ni}{2R}$  ...........(i)

 $B_{axis}=\frac{\mu_{0}}{4\pi}\frac{2\pi Ni R^{2}}{(x^{2}+R^{2})^{3/2}}$

$\frac{B}{8}=\frac{\mu_{0} Ni R^{2}}{2(x^{2}+R^{2})^{3/2}}$

 From Eq.(i) , we get

 $\frac{\mu_{0} Ni}{8\times 2R}=\frac{\mu_{0} Ni R^{2}}{2(x^{2}+R^{2})^{3/2}}$

 $\frac{1}{8 R}=\frac{1}{(x^{2}+R^{2})^{3/2}}$

$\frac{1}{8 R^{3}}=\frac{1}{(x^{2}+R^{2})^{3/2}}$

$8 R^{3}=(x^{2}+R^{2})^{3/2}$  (on taking cube root)

$2 R^{}=(x^{2}+R^{2})^{1/2}$  ( on taking square root)

$4 R^{2}=(x^{2}+R^{2})^{}$

 $4 R^{2}-R^{2}=x^{2}$

$3 R^{2}=x^{2}$

x= $R\sqrt{3}$

A disc has mass 'M' and radius  'R'. How much tangential force should be applied to the rim of the disc, so as to rotate with angular velocity '$\omega$ ' in time 't'?

Answer:

Explanation:

 Given, mass of disc= M

 Radius of disc= R

 we know that,

   $\tau= l\alpha$

 But , $\tau= F\times R$

 $\therefore$   $l= \frac{MR^{2}}{2}$

 and   $\alpha= \frac{\omega}{t}$

 therefore  , F x R= $\frac{MR^{2}}{2}\times\frac{\omega}{t}$

 F= $\frac{MR\omega}{2t}$

A transistor is used as a  common emitter amplifier with a load resistance 2 k $\Omega$ .The input resistance is 150 k $\Omega$ .Base current is changed by 20 $\mu$ A which results in a change in collector current by 1.5 mA. The voltage gain of the amplifier is 

Answer:

Explanation:

Given , load resistance (R_e)=2 k $\Omega$= 2 x 10³Ω

 Input  resistance (R_i)= 150 $\Omega$

 Change on collector current ($\triangle$ l_c)= 1.5 mA= 1.5 x 10^-3 A

Change in base current ($\triangle$ l_B)=20 $\mu$ A= 20 x 10^-6 A

 We know that

 Voltage gain=  $\frac{V_{o}}{V_{i}}=\frac{R_{o}\times\triangle l_{C}}{R_{i}\times\triangle l_{B}}$

= $\frac{2\times 10^{3}\times 1.5\times 10^{-3} }{150\times 20\times 10^{-6}}= \frac{3\times 10^{6}}{150\times 20}=1000$

• Physics - General questions