Question

A silver ingot weighing 2.2 kg is held by a string so as to be completely immersed in a liquid of relative density 0.8. The relative density of silver is 10.5. The tension in the string in kg-w.t is-
A) 1.6
B) 1.94
C) 31
D) 5.25

Answer 1

Hint: If anybody is put under the liquid then there will be an upward force on the body which is known as buoyancy force. The relative density is the ratio of the density of the body and the density of a standard liquid.

Complete step by step answer:
As it is given that the mass of ingot is 2.2 kg the relative density of the silver is 10.5. as the density is the ratio of mass and volume therefore the volume of the silver ingot is given by,
Volume of silver is equal to,
$ \Rightarrow V = \dfrac{{2.2}}{{10.5}}$
$ \Rightarrow V = 0.2095{m^2}$
The relative density of the fluid in which the silver ingot is immersed is 0.8 so the upward buoyancy force on the silver ingot is given by,
$ \Rightarrow B = {V_{silver}} \cdot {\rho _{fluid}} \cdot g$
$ \Rightarrow B = \left( {0.2095} \right) \cdot \left( {0.8} \right) \cdot \left( {9. 8} \right)$
$ \Rightarrow B = 1.64N$
The net tension on the string is given by,
$ \Rightarrow T = Mg - B$
On substituting the corresponding values,
$ \Rightarrow T = \left( {2.2} \right) \cdot \left( {9.8} \right) - \left( {1.64} \right)$
On simplification,
$ \Rightarrow T = 19.92N$
$ \Rightarrow T = 2.03kg - wt$

Therefore, the closest option is option (B).

Note:
The upward force on the silver ingot is the buoyancy which results in the decrease in the weight of the silver ingot so the tension on the string is less than the original weight and the loss of weight is equal to the buoyancy force due to the liquid.