TS EAMCET 2018 :: Physics :: General questions

• Physics - General questions

For a television network,  $5 \times 10^{5}$  channels are granted. If the central frequency of the microwave link is 25 GHz and the allotted bandwidth for each channel is 2KHz, then how much percentage of the link is used for the network?

Answer:

Explanation:

 Let x% of the link be used for network.

 Given , central frequency =25 GHz

 = $25 \times 10^{6} Hz$

 Bandwidth of channels = x% of 25 GHz

 = $\frac{x}{100} \times 25 \times 10^{9}=25 x \times 10^{7}$

 Number of channel= Total bandwidth/ Bandwidth needed per channel

 $5 \times 10^{5}=\frac{25 x \times 10^{7}}{2 \times 10^{3}}$

 $10 \times 10^{8}=25x \times 10^{7}$

 25x=100

$\Rightarrow$    x=4%

 In the diode-based rectifier  circuit  given below, If $V_{s}-V_{m}\sin \omega t$ and the diode is ideal, then the average value of $V_{L}$ is 

Answer:

Explanation:

 Given , AC voltage , $V_{s}=V_{m}\sin  \omega t$

and $V_{m}$= maximum value of voltage 

In the given circuit diode will only conduct current in forward bias, so output across $R_{L}$ will be

 Average Voltage , $V_{av}=\frac{V_{m}}{\pi}$

average current $I_{av}= \frac{V_{av}}{R}$

 Total resistance  , $R= R_{S}+R_{L}$

 $I_{av}=\frac{V_{m}}{(R_{S}+R_{L})\pi}$

Now, voltage across $R_{L}$

 $V_{L}=I_{av}.R_{L}$

$V_{L}$= $\frac{R_{L}}{(R_{L}+R_{S})}\frac{V_{m}}{\pi}$

 A half-spherical glass lens with a refractive index 1.5 is placed in a liquid with a refractive index of 1.3 (see the following figure) . The radius of the half spherical lens is 10 cm. A parallel  beam of light travelling  in the liquid  is refracted by the glass  lens, Then the absolute  value of the position  of the  image from the centre of the glass lens will be 

Answer:

Explanation:

 Focal length of the lines

 $\frac{1}{f}=\left(\frac{n_{2}}{n_{1}}-1\right)\left(\frac{1}{R_{1}}-\frac{1}{R_{2}}\right)$

 Here, refractive index for galss lens, $n_{2}=1.5$

 Refractive index for water , $n_{1}=1.3$

 Radius of curvature  , $R_{1}=10 cm$

                                  $R_{2}= \infty$

 Putting these values , we get

  $\frac{1}{f}=\left(\frac{1.5}{1.3}-1\right)\left(\frac{1}{10}-\frac{1}{\infty}\right)$

 $\frac{1}{f}=\frac{0.15}{10}$

 $\Rightarrow$    f=65 cm

 Now parallel beam of light will converge  at the focus of lens, so image position v=65 cm.

 The position vector of a particle  moving in a plane is given by $r= a \cos \omega t \hat{i}+b \sin  \omega t \hat{j}$,  where $\hat{i}$  and $\hat{j}$  are the unit  vectors  along the rectangular axes X and Y ;a,b and $\omega$  are constants  and t is time . The acceleration  of the particle is directed along the vector 

Answer:

Explanation:

Given  , position vector ,

     $r= a \cos \omega t \hat{i}+b \sin  \omega t \hat{j}$ ........(i)

 on differentiating both sides w.r.t t, we get

 velocity  , $\frac{dx}{dt}=v=-a \omega \sin \omega t \hat{i}+b \omega \cos \omega t \hat{j}$

 Again , on differentiating w.r.t t, we get

 acceleration  $\frac{dv}{dt}=a=-a \omega^{2} \cos  \omega t \hat{i}-b \omega^{2} \sin \omega t \hat{j}$

                      $a=-\omega^{2}(a \cos \omega t \hat{i}+b \sin \omega t \hat{j})$

Form Eq. (i) , we get  $a= \omega^{2}(r)= -\omega^{2}r$

 Hence, acceleration is along -r

Consider 'a' vehicle moving with a velocity 54 km/h. at a distance of 400m, from the traffic light brakes are applied. The acceleration of the vehicle, after the application of brakes, is $-0.3 m/s^{2}$. The vehicle's  position relative to the traffic light is 

Answer:

Explanation:

Given, initial velocity,

$u=54 km/h=54 \times \frac{5}{18}$m/s

 u=15 m/s

 Distance of signal from vehicle , d=400 m

 acceleration , a= -0.3m/s²

 Now, distance covered  by vehicle  , when it stops  is given by $v^{2}=u^{2}+2as$

  $0=(0.5)^{2}-2 \times 0.3 \times s \Rightarrow s= \frac{225}{0.6}=375m$

 Relative position =400-375=25m

• Physics - General questions