Heater of an electric kettle is made of a wire of length L and diameter d. It takes 4 minutes to raise the temperature of 0.5 kg water by 40K. This heater is replaced by a new heater having two wires of the same material, each of length L and diameter 2d. The way these wires are connected is given in the options. How much time in minutes will it take to raise the temperature of the same amount of water by 40 K?

Options

A. 4 if wires are in parallel
B. 2 if wires are in series
C. 1 if wires are in series
D. 0.5 if wires are in parallel

Explanation

Heat required to increase the temperature of water by 40 K.
`Delta Q=mc Delta T=(e^2/R)t implies Delta Q=0.5 times c times 40=(e^2/R) times 4m` in, where
The resistance of wire of length L and diameter 2d `R_1=(pL)/(pi (2d)^2//4)=R/4`
In resistance are in series `R_(eq)=R/4+R/4=R/2`, Heat required `Delta Q=(e^2/R)4=e^2/((R//4)) times t`
Which gives t = 2min If resistances are in parallel, `R_(eq)=(R/4 times R/4)/(R/4+R/4)=R/8`
Heat required `Delta Q=(e^2/R)4=e^2/((R//8)) times t` which gives t=0.5 min