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 `1^(st)` filament
`H= (V^(2))/(R_(1))t_(1) rArr R_(1)= (V^(2))/(H) t_(1)" "(i)`
For `2^(nd)` filament
`H=(V^(2))/(R_(2)) t_(1) rArr R_(2)=(V^(2))/(H) t_(2)" "(ii)`
When placed in parallel
`H=(V^(2))/(R_(p)) t_(p) rArr R_(p)=(V^(2))/(H) t_(p)" "(iii)`
From eq. (i), (ii), & (iii) we get
`(1)/(R_(p)) =(1)/(R_(1)) +(1)/(R_(2))`
`(H)/(V^(2)t_(p)) =(H)/(V^(2)t_(1)) +(H)/(V^(2)t_(2)) " for series "=t_(1)+t_(2)`
`(1)/(t_(p))=(1)/(t_(1)) +(1)/(t_(2)) rArr (1)/(t_(p))= (1)/(10) +(1)/(15) rArr t_(p)=6`
`t_(p)=6` min