\(\sqrt{\frac{42W}{5L}}\)
\(\sqrt{\frac{6L}{42W}}\)
\(\frac{42W}{5L}\)
\(\frac{5L}{42W}\)
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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