One filament takes 10min to heat a kettle and another takes 15 min. If connected in parallel they combindly take………min to heat the same kettle:

Options

A. 6
B. 12.5
C. 25
D. 7.5

Answer

A

Explanation

For first filament
`H=(V^(2))/(R_(1)) t_(1)rArr R_(1)=(V^(2))/H t_(1).....(i)`
For second filament
`H=(V^(2))/(R_(2))t_(2)rArr R_(2)=(V^(2))/H T_(2).....(ii)`
When place din parallel
`H=(V^(2))/(R_(P)) t_(2) rArr R_(p)=(V^(2))/H t_(p)....(iii)`
From eqs. (i),(ii),(iii) we get
`1/(R_(P))=1/(R_(1))+1/(R_(2))`
`rArr H/(V^(2)t_(p))=H/(V^(2)t_(1))+H/(V^(2)t_(2))`
`rArr 1/(t_(p)) =1/10 +1/15 rArr t_(p)` =6 min