It is possible to take a high quality photograph of a very fast moving object by illuminating the object for quite a small fraction of a second. You may have come across photographs of a bullet penetrating a banana or an apple in many text books or magazines. This is called 'stop action' photography because the fast moving object travels a very short distance during the time of illumination. Harold Edgerton, the inventor of stroboscope, was a pioneer of this kind of photography. A normal photographic plate works properly if it receives an energy of 4J during the exposure. To release this energy in a very small fraction of time, huge amount of power is required. Such huge power can not be generated directly from a battery because of its high internal resistance. To produce such power a capacitor is used. The time in which a capacitor discharges can be very short. Although, theoretically it would take a long time for a capacitor to discharge completely, it discharges almost completely in about 10 time constant. Consider the following situation A capacitor of 200muF storing 4J energy is made to discharge through a flash light in 2ms. This setup is used to take the picture of a bullet moving at a speed of `100m//s` Assume that the flash light acts as a resistor and there is no other resistance in the circuit.
What is the order of energy delivered to the flash light in `0.2` ms(approx).

Options

A. `0.4` J
B. `1.83` J
C. `2.74` J
D. `3.45` J

Answer

D

Explanation

`Q=Q_(0)e^(-t//RC) RC=timeconstant=2m//s`
After one time constant `Q=0.37Q_(0)`
`(U_(c))_(f)=Q^(2)/(2C)=(0.37Q_(0))^(2)/(2C)=0.37xx0.37Q^(2)/(2C)`
`(U_(c))_(f)=(0.37xx0.37)(U_(C))`
rest energy is lost to resister`=3.45` J