The charge flowing through a resistance `R` varies with time `t as Q = at - bt^(2)`. The total heat produced in `R` is

Options

A. `(alpha^(2)R)/(2beta)`
B. `(alpha^(3)R)/(6beta)`
C. `(alpha^(3)R)/(beta)`
D. None

Answer

B

Explanation

As `q=alpha t -beta t^(2)`
`i=(dq)/(dt)= alpha-2beta t`
Current flow from t = 0 to `t=(alpha)/(2beta)`
`rArr H= int_(0)^(t)i^(2)Rdt` on solving we get
`H=(alpha^(3)R)/(6beta)`