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A metal rope of density 6000 kgm^-3 has breaking stress 9.8 × 10⁸ Nm^–2. This rope is used to measure the depth of the sea. Then the depth of the sea that can be measured without breaking is -----

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10 × 10³ m

20 × 10³ m

30 × 10³ m

40 × 10³ m

answer is B.

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Detailed Solution

d_r = 6000kgm^-3; S = 9.8 x 10⁸ N/m²
$g^{1} = g (1 - \frac{d_{w}}{d_{r}}) = g (1 - \frac{1000}{6000}) \Rightarrow g^{,} = \frac{5 g}{6}$

$S = \frac{f}{A} = \frac{{mg}^{,}}{A}; S = \frac{{ρvg}^{,}}{A} = d_{r} \frac{{Alg}^{,}}{A}$

$l = \frac{s}{d_{r} g^{,}} = \frac{s}{d_{r} x \frac{5 g}{6}} = \frac{9.8 x 10^{8} x 6}{6000 x 5 x 9.8}$

$l = \frac{10^{8}}{10^{3} x 5} = \frac{10 x 10^{7}}{10^{3} x 81} = \frac{20 x 10^{6}}{10^{3}} = 20 x 10^{3} m$

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