TS EAMCET 2019 :: Physics :: General questions

• Physics - General questions

 A message signal is used to modulate a carrier signal of frequency  5 MHz and peak voltage of 40 V . In the process, two side-bands are produced separated by 40 kHz. If the modulation  index is 0.75, then the peak voltage  and frequency of the messages  signal, respectively are

Answer:

Explanation:

 Given, frequency of carrier signal , f=5 MHz  and peak voltage , $V_{c}$= 40V

 Modulation index , $\mu$ =0.75

  $\therefore$           $\mu=\frac{V_{m}}{V_{c}}$          [$\because   V_{m}$   is peak voltage of message signal]

                    $0.75 =\frac{V_{m}}{40}$

   $\Rightarrow$           $V_{m}=40  \times  0.75$

 $V_{m}=30V$

 Since, a difference of frequencies of two sidebands is equal to the bandwidth  (2 fm).

 i.e, Band width =40 kHz

   $2f_{m}=40kHz \Rightarrow f_{m}=20kHz$

 Hence, the peak voltage and frequency will be 30V and 20 kHz

 A person applies a sine wave and square wave to an AND gate as shown in the figure (i) and (ii) . Assuming that both the voltages are applied in phase , the person observers the output  at E and F on (i) and (ii) , respectively. [Assume minimum voltage of 5V is equivalent to logic (i)]

Answer:

Explanation:

(i) Sine wave , 50 Hz, 2V

  Square wave  107 Hz,6 V

 (ii)  Sine wave, 100 Hz , 8 V

  Square wave, 100 Hz, 6 V

 In figure(i)  , sine wave and square wave both have different frequencies, respectively, 50Hz and 107 Hz . Hence , these voltages are not processed by AND gate, therefore, no output  is obtained  from gate (i)  . In figure (ii)  , the sine wave and square wave both have same frequency of 100 Hz  , hence , these voltage are processed by AND gate. Since , logic gate(AND gate) gives output  in two-level ( 0 and 1)  only at same frequency of input. Therefore, the output  of gate (ii) will be the pulsed  wave of 100 Hz

 Consider a toroid with rectangular cross-section, of inner radius a,  outer radius b  and height h, carrying n number of turns. Then  the self-inductance of the toroidal  coil when  current I passing through the toroid is 

Answer:

Explanation:

 Given, a toroid with a rectangular cross-section of inner radius a  and outer radius b.

 Height of the solenoid=h

Magnetic field inside a rectangular toroid is given by

$B=\frac{\mu_{0}n^{}I}{2 \pi r}$   ................(i)

Using the infinitesimal cross-sectional area element,

  dx=h dr

 $\therefore$    Flux passing through  the cross-section of toroid 

$\phi=\int B.dx =\int_{a}^{b} \frac{\mu_{0}n^{}I}{2 \pi r}.(h dr)$

 $\phi= \frac{\mu_{0}n^{}Ih}{2 \pi }\int_{a}^{b} \frac{1}{r}. dr$

 $\Rightarrow$       $\phi= \frac{\mu_{0}n^{}Ih}{2 \pi }[\log r]_{a}^{b}$

 $\Rightarrow$       $\phi= \frac{\mu_{0}n^{}Ih}{2 \pi }  (\log b-\log a)$

                               $\phi= \frac{\mu_{0}n^{}Ih}{2 \pi }ln\left(\frac{b}{a}\right)$

 Now, self inductance  of rectangular toroid.

   $L= \frac{n \phi}{I}$

 Putting  the value  of $\phi_{}$ , we get

L=  $\frac{\mu_{0}n^{2}h}{2 \pi}ln\left(\frac{b}{a}\right)$

Two  particles carrying equal charges move parallel  to each other  with the speed 150 km/s. If $F_{1}$ and $F_{2}$  are magnetic  and electric  forces between  two charged  particles  then 

$\frac{|F_{1}|}{|F_{2}|}$ is    $\left( Let \mu_{0}\epsilon_{0}=\frac{1}{9 \times 10^{16}} s^{2}/m^{2}\right)$

Answer:

Explanation:

  Given, two charge particles having same charge, q move with same speed

 i.e,  $v_{1}=v_{2}=150 km/s =1.5 \times 10^{5}  m/s$

 Electric force between two moving charge  particle is same as they are static,

 i.e,    $|F_{2}|=\frac{1}{4 \pi \epsilon_{0}}.\frac{q^{2}}{r^{2}}$ ..........(i)

 Magnetic force between two moving charge particles is given by

$F_{1}=\frac{\mu_{0}}{4 \pi }.\frac{q_{1}.q_{2} v_{1} v_{2}}{r^{2}}$

                                                                         $[\because q_{1}=q_{2}=q$ and $v_{1}=v_{2}=v]$

 $|F_{1}|=\frac{\mu_{0}}{4 \pi}\frac{q^{2} v^{2}}{r^{2}}$.........(ii)

 From   Eq.(i) and (ii) , we get

$\therefore$    $\frac{|F_{1}|}{|F_{2}|}=\frac{\frac{\mu_{0}}{4 \pi}\frac{q^{2} v^{2}}{r^{2}}}{\frac{1}{4 \pi\epsilon_{0}}\frac{q^{2} }{r^{2}}}= \epsilon_{0}\mu_{0}v^{2}$

      $=\frac{1}{9 \times 10^{16}}\times (1.5 \times 10^{5})^{2}=2.5 \times 10^{-7}$

 Hence, $\frac{F_{1}}{F_{2}}$ is $2.5 \times 10^{-7}$

 Four identical metal plates are located in the air at equal distance d from each other . The area of each plate in S. If the outermost plates are connected by a  conducting  wire as shown in the figure the  capacitance between points A and B will be

Answer:

Explanation:

The equivalent arrangement of given combination  is shown in the following figure,

We know that the capaciance , $C=\frac{\epsilon_{0}S}{d}$

 where, S is the area of each plate and d is distance between two plates.

 So, equivalent capacitance  between points A and B,

 $ C_{AB}= \frac{ C \times C}{C+C}+C  $

     $=\frac{C}{2}+C=\frac{3C}{2}=\frac{3}{2}\frac{\epsilon_{0}S}{d}$

• Physics - General questions