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Q.

A nylon rope 3 cm in diameter has a breaking strength of 1.5 X 10⁵ N. The breaking strength of a similar rope 1.5 cm in diameter is

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a

3 x 10⁵N

b

6x10⁵N

c

0.375x10⁵N

d

0.75x10⁵N

answer is B.

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Detailed Solution

Breaking strength $\propto$ A
$∴ \frac{F_{1}}{F_{2}} = \frac{A_{1}}{A_{2}} = \frac{{πr}_{1}^{2}}{{πr}_{2}^{2}} = \frac{r_{1}^{2}}{r_{2}^{2}} = \frac{D_{1}^{2}}{D_{2}^{2}}$
So, $\frac{1 \cdot 5 \times 10^{5}}{F_{2}} = \frac{(3)^{2}}{(1 \cdot 5)^{2}} = 4$
$F_{2} = \frac{1 \cdot 5 \times 10^{5}}{4} = 0 \cdot 375 \times 10^{5} N$

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A nylon rope 3 cm in diameter has a breaking strength of 1.5 X 105 N. The breaking strength of a similar rope 1.5 cm in diameter is