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 = 4²/₃ and L = 54, find d in terms of W and L.

$\sqrt{\frac{42W}{5L}}$

$\sqrt{\frac{6L}{42W}}$

$\frac{42W}{5L}$

$\frac{5L}{42W}$

Correct answer is A

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

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