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The tension in a massless cable connected to an iron ball of 50 kg when it is submerged in sea water is close to $(ρ_{i r o n} = 8 \times 10^{3} k g / m^{3} a n d ρ_{s e a w a t e r} = 1000 k g / m^{3}, g = 10 m / s^{2})$

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438 N

876 N

950 N

933 N

answer is C.

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Detailed Solution

$T = m g - F_{B u o y a n c y}$

$T = m g - σ v g$

$= m g [1 - \frac{σ}{ρ}]$

$= 50 \times 10 [1 - \frac{1000}{8 \times 1000}]$

$T = 500 [\frac{7}{8}] = 437.5 N \approx 438 N$

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