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A block of silver of mass 4 kg hanging from a string is immersed in a liquid of relative density 0.72. If relative density of silver is 10, then tension in the string will be:[ take g = 10 m/s2 ]

Option: 1

37.12N


Option: 2

42N


Option: 3

73N


Option: 4

21N


Answers (1)

best_answer

 

Net force on the body -

F_{B}+F_{v}= W

\rightarrow 6\pi \eta rv+\frac{4}{3}\pi r^{3}\sigma g= \frac{4}{3}\pi r^{3}\rho g

\rightarrow 6\pi \eta rv=\frac{4}{3}\pi r^{3} g\left ( \rho -\sigma \right )

\rightarrow v_{t}=\frac{2}{9}\frac{ r^{2} \left ( \rho -\sigma \right )}{\eta }g

 

- wherein

F_{B}-Buoyant \: force

F_{v}-viscous \: force

w-weight

\rho \rightarrow density \: of \: ball

\sigma \rightarrow density \: of \: water

V_{T} =terminal \: velocity

 

 

Let \rho_{s}\rho_{L} be the density of silver and liquid. Also m and V be the mass and volume of silver block.

Tension in string = mg – bouyant force

T=\rho_{s}V_{g}- \rho_{L}V_{g}=\left ( \rho _{s}-\rho _{L} \right )V_{g}

Also V=\frac{m}{\rho_{s}}

\therefore T=\left ( \frac{\rho _{s}-\rho_{L}}{\rho_{s}} \right )mg = \frac{\left ( 10\times 0.72 \right )\times 10^{3}}{10\times 10^{3}}\times 4\times 10=37.12N

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chirag

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