In Fig. 6, l and m are two parallel tangents to a circle with centre O, touching the circle at A and B respectively. Another tangent at C intersects the line at D and in at E. Prove that ZDOE=90.

Options

A. `(a^(3)R)/(2b)`
B. `(a^(3)R)/(b)`
C. `(a^(3)R)/(6b)`
D. `(a^(3)R)/(3b)`

Answer

C

Explanation

Given, `Q=at-bt^(2) therefore I=(dQ)/(dt)=a-2bt`
At `t=0, Q=0impliesI=0`
Also, `I=0` at `t=a//2b`
`therefore` Total heat produced in resistance R,
`H=underset(0)overset(a//2b)intI^(2)Rdt=Runderset(0)overset(a//2b)int(a-2bt)^(2)dt`
`=Runderset(0)overset(a//2b)int(a^(2)+4b^(2)t^(2)-4abt)dt`
`=R[a^(2)t+4b^(2)(t^(3))/(3)-4ab(t^(2))/(2)]_(0)^(a//2b)`
`=R[a^(2)xx(a)/(2b)+(4b^(2))/(3)xx(a^(3))/(8b^(3))-(4ab)/(2)xx(a^(2))/(4b^(2))]=(a^(3)R)/(6b)`