If the coil of a heater is cut to raduce its length by 20% then effect on its power will be

Options

A. 0.125
B. 0.21
C. 0.25
D. 0.42

Answer

C

Explanation

Let V be the voltage rating and R the initial resistance of heater coil. Power of heater coil can be written as follows:
`P_(1) = (V^(2))/(R )` …(i)
We know that resistance is directly proportional to the length of conductor and hence on reducing length of coil by 20%, its resistance should decrease proportionally. New resistance will become 0.8R. New power can be written as follows
`P_(2) = (V^(2))/(0.8R)` ...(ii)
`P_(2) -P_(1) = (V^(2))/(R )[(1)/(0.8) -1] = (V^(2))/(4R)`
`(P_(2) - P_(1))/(P_(1)) xx 100 = 25`% Hence power will increase by 25%. So, option (c ) is correct