The weight W kg of a metal bar varies jointly as its length L meters and the square of its diameter d meters. If w = 140 when d = 4

^{2}/_{3}and L = 54, find d in terms of W and L.-
**A.**

(sqrt{frac{42W}{5L}}) -
**B.**

(sqrt{frac{6L}{42W}}) -
**C.**

(frac{42W}{5L}) -
**D.**

(frac{5L}{42W})

##### Correct Answer: Option A

##### Explanation

(Winfty LD^2\W=KLd^2\K=frac{W}{Ld^2}\=frac{140}{54}timesleft(4frac{2}{3}right)^2 \=frac{140}{54}timesleft(frac{14}{3}right)^2\=frac{140times 9}{54times 14times 14}\=frac{5}{42}\∴W=frac{5}{42Ld^2}\42W=5Ld^2\frac{42W}{5L}=d^2\d=sqrt{frac{42W}{5L}})