TS EAMCET 2019 :: Physics :: General questions

• Physics - General questions

Assertion (A) Si and GaAs  are the preferred materials for solar cells

Reason (R)  Both these materials  have energy  band gaps much below the energy level corresponding to the maximum solar irradiance in the solar spectrum

The correct answer is

Answer:

Explanation:

Key idea: The energy band gap  of Si is 1.14 eV and GaAs is 1.42 eV

 Both Si and GaSa are used to make solar cells. In visible spectrum , energy varies from about 3eV on the violet side to 1.5 eV  on red side, so visible  light can make solar energy

But radiowaves, microwaves and infrared waves have less photon energy than the bandgap of Si and GaAs.

Hence, the assertion is correct but the reason is wrong. Therefore , the correct option is(a)

 For the circuit shown in the figure, calculate the resistance between points A and B

Answer:

Explanation:

Key idea (i)  Equivalent resistance in series combination,

$R_{eq}=R_{1}+R_{2}$+.....

(ii)    Equivalent resistance in parallel combination,

$\frac{1}{R_{eq}}= \frac{1}{R_{1}}+\frac{1}{R_{2}}+\frac{1}{R_{3}}+......$

As , the circuit is ,

.

Hence , the correct option is (b)

A metal plate of thickness  2mm and area $36 \pi  cm^{2}$ is slide into a parallel plate capacitor of plate spacing 6 mm and are $36 \pi cm^{2}$ . The metal plate is at a distance 3 mm from one of the plates. What is the capacitance of this arrangement?  

$\left( Let\frac{1}{4 \pi \in_{0}}=9 \times 10^{9} Nm^{2}C^{-2}\right)$

Answer:

Explanation:

When metal plate of thickness 2mm is inserted into a parallel plate capacitor , then the arrangement is shown as,

Given , thickness  of plate, t=2mm and area, A=$36 \pi cm^{2}$

The above circuit can be redrawn as,

 where, $C_{1}=\frac{\epsilon_{0}A}{d_{1}}$  and $C_{2}=\frac{\epsilon_{0}A}{d_{2}}$

 Now, $ C_{eq}= \frac{C_{1}C_{2}}{C_{1}+C_{2}}$  (In the series combination)

$\Rightarrow C_{eq}=\epsilon_{0}A\left[\frac{\frac{1}{d_{1}}\times\frac{1}{d_{2}}}{\frac{1}{d_{1}}+\frac{1}{d_{2}}}\right]=\frac{\epsilon_{0}A}{d_{1}+d_{2}}$

$\Rightarrow C_{eq}=\frac{1}{4 \pi \times 9 \times 10^{9}}\times\frac{36 \pi \times 10^{-4}}{3\times 10^{-3}+1\times 10^{-3}}$

                               $\left( \because \frac{1}{4 \pi\epsilon_{0}}={ 9 \times 10^{9} Nm^{2}/C^{2}}\right)$

 =$\frac{1}{4} \times 10^{-10}$=$25 \times 10^{-12}$ F=25pF

so, the correct option is (c)

The limit of resolution of a telescope is $2.5 \times 10^{-7}$ rad. If the  telescope is used to detect light of wavelength  500 nm coming from a star, the diameter of the objective lens used by the telescope is 

Answer:

Explanation:

Given , limit of resolution  , $\alpha= 2.5 \times 10^{-7}$ rad

 and wavelength , $\lambda$=500 nm

 As, limit of resolution of a telescope,

   $\alpha = \frac{1.22 \lambda}{a}$

 So,   $a= \frac{1.22 \lambda}{\alpha}= \frac{ 1.22 \times 500 \times 10^{-9}}{ 2.5 \times 10^{-7}}$

 $\Rightarrow$    a=2.44 m

 = 244 cm

 Hence , the correct option is (a) 

A drone fitted with siren in flying directly away from the drone operator and towards a distant building at a speed of 15 m/s . The siren produces the sound of frequency 780Hz. What is the frequency that the  operator hears in the echo refected from the building,[ speed of sound is 340 m/s]

Answer:

Explanation:

A drone fitted with a siren  as shown in the figure,

 Given, speed of drones , $v_{s}=15 m/s$ , actual  frequency , $n_{0}=780 Hz$ and speed of sound , v=340 m/s

As source  is moving towards  the building, so the apparent frequency reflected by the building is 

       $n= \frac{n_{0}v}{v-v_{s}}= \frac{780 \times 340}{340-15}=816 Hz$

 As, we know there is no change in the apparent  frequency by the reflection from the building Hence,

the echo heard by the observer is 816 Hz.

So, the correct option is (c)

• Physics - General questions