A Block Of Silver Of Mass 4kg 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

Let, \(\rho _{s},\rho _{1} \) be the density of silver and liquid.

Let, \(m_{s} and V_{s} \)be the mass and volume of the silver block.

In equilibrium, the sum of tension and buoyant force is equal to the weight of the silver block. Hence

\(T + V_{s} * \rho _{1} * g = V_{s} * \rho _{s} * g \) \(\Rightarrow T = (\rho _{s} – \rho _{1}) V_{s}g = (\rho _{s} – \rho _{1})m_{s}\rho _{s}g \) \(\Rightarrow T = \frac{\rho _{s} – \rho _{1}}{\rho _{s}}m_{s}g \) \(\Rightarrow T = \frac{10 – 0.72}{10} 4 * 10 = 37.12 N \)

Explore more such questions and answers at BYJU’S.

Was this answer helpful?


0 (0)


Choose An Option That Best Describes Your Problem

Thank you. Your Feedback will Help us Serve you better.

Leave a Comment

Your Mobile number and Email id will not be published. Required fields are marked *




Free Class