TS EAMCET 2019 :: Physics :: General questions

• Physics - General questions

Albert Einstein  was  confirmed with the Nobel prize physics for his work on 

Answer:

Explanation:

Albert Einstein was mostly famous by his work on special theory of relativity, but he got a Nobel prize in physics for the explanation of photoelectric effect.

So, the correct option is (c)

In a transistor  circuit, the collector current is changed by 8.9 mA, if the emitter  current is changed to 9.0 mA . The value of current amplification  factor $\beta$ is 

Answer:

Explanation:

As,  change in the collector  current

$\triangle I_{c}$ =8.9 mA and change in emitter  current,

$\triangle I_{E}$ =9 mA

 So, the current amplification factor, 

$\beta=\frac{\triangle I_{c}}{\triangle I_{B}}=\frac{\triangle I_{c}}{\triangle I_{E}-\triangle I_{c}}$

                                    $(\because  \triangle I_{E}=\triangle I_{c}+\triangle I_{B})$

$\Rightarrow$      $\beta=\frac{8.9}{(9-8.9)}=\frac{8.9}{0.1}=89$

 Hence, the correct option is (a)

A photodiode sensor is used to measure the output of a 300W lamp kept 10 m away. The sensor has an opening of 2 cm in diameter. How many photons enter the sensor if the wavelength of the light is 660 nm and the exposure time is 100 ms.(Assume that all the energy of the lamp is given  off as light and h=$6.6 \times 10^{-34}$ Js)

Answer:

Explanation:

Given that, power of lamp , P=300 W

distance of lamp from sensor , d=10m

 radius of sensor opening , r=1cm, wavelength of  light emitted from  photodiode, $\lambda$=660 nm and exposure time, t= 100 ms=$100 \times 10^{-3}$ s

 Energy of photons , $E= \frac{hc}{\lambda}$

 Putting the given values , we get

 $\Rightarrow$         $E=\frac{6.6\times 10^{-34}\times 3\times10^{8}}{660 nm}=3 \times 10^{-19}$

The area of of exposure of lamp at a radius  of 10 m

 $A_{0}$ =$ 4 \pi r^{2}$= $4 \pi \times 10^{2} m^{2}$

Similarly, The area of sensor

A's=$\pi r^{2}$= $\pi  \times (10^{-2})^{2}$= $\pi \times 10^{-4} m^{2}$

Energy emitted by lamp in exposure time,

$E_{0}$ = power of lamp (p) x exposure time (t)

 $E_{0} =300 \times 100 \times 10^{-3}$ =3 J

So, the number of photons entering  to sensor is given by

 $N= \frac{E_{0}}{E}\times\frac{A'_{s}}{A_{0}}$

 =$\frac{30}{3\times 10^{-19}}\times\frac{\pi\times 10^{-4} }{4\pi\times 10^{2}}$

$\Rightarrow$       N=$2.5 \times 10^{13}$

 So, the correct option is (C)

Find the potential difference between a and b as shown in the below circuit

Answer:

Explanation:

The circuit can be redrawn as,

 45 V is drop across the branch one, so current through it

$I_{1}=\frac{45}{R_{eq1}}=\frac{45}{\frac{15}{4}}=12 A$

 Now, above circuit can be redrawn as,

 Since , in a series circuit, the current flows will be same,

hence, $V_{ab}=I_{ab}R_{eq}(ab)=12\left(\frac{15}{4}+4+10\right)$

$\Rightarrow$          $V_{ab}$=213 V

Hence , the correct option is (c)

A capacitor of capacitance 4$\mu$ F  is charged to a potential difference of 6 V with a battery. The battery is removed and in its place another capacitor of capacitance is 8$\mu$ F introduced and the circuit is closed. The potential difference attained by each of the capacitors in V is 

Answer:

Explanation:

The final circuit  of the series capacitor system is shown in figure,

 Since , we know that in a series capacitors system, the charge stored by both the capacitors is equal. 

Hence , Q = $C_{eq} V$

 where,  $C_{eq}=\frac{4\times8}{4+8}=\frac{8}{3}\mu F\Rightarrow Q=\frac{8}{3}\times 6=16\mu C$

Hence, $V_{1}=\frac{Q}{C_{1}}=\frac{16}{4}=4V $   and   $V_{2}=\frac{Q}{C_{2}}=\frac{16}{8}=2V $

 Hence, the potential difference,

$\triangle V= V_{1}-V_{2}$

$\Rightarrow$    $\triangle$ V= 4-2=2V

Hence , the correct option is (a) 

• Physics - General questions