Question

A block of silver of mass 4kg hanging from a string is immersed in a liquid of a relative density of 0.72. If the relative density of silver is 10, then tension in the string will be

Answer 1

Step1: Given data

A block of silver of mass 4kg hanging from a string is immersed in a liquid of a relative density of 0.72.

Relative density of silver is 10.

Step2: Relative density

1. The relative density of the substance is defined as the ratio of the density of the substance to that of the density of water.
2. The string is assumed to be massless. Therefore the tension in the string will be due to the net force acting on the silver block downwards.

Step3: Formula used

$$\begin{gathered} F_B={\rho }_{L}gV[F_B=buoyantforce,{\rho }_L=densityofliquid,g=accelerationduetogravity,v=volumeoftheobjectinsidewater.] \\ {F}_G=mg[F_G=gravitationalforce,m=mass] \end{gathered}$$Step4: Calculating net force on the block

1. Let us say there is uniform tension $T$ in the string.
2. From the below diagram, we can see that the buoyant force ($F_B$) acts in the upward direction
3. While the gravitational force ($F_G$) acts in the downward direction.
4. The net force on the block is zero i.e. there is no relative motion of the block. Hence from the above free body diagram of the silver block, the equation for net force on the body is given by, $F_G-F_B-T=0$

Step5: Calculating tension

Let us say a body of volume $V$ is completely immersed in a liquid of density ${\rho }_L$ .

Therefore the buoyant force on the body is given by, $F_B={\rho }_{L}gV$

Hence the equation of motion for the silver block is,

$$\begin{gathered} mg-{\rho }_{L}gV-T=0 \\ T=mg-{\rho }_{L}gV.............\left(i\right) \end{gathered}$$

Step6: Calculating the relative density

Let us say the density of silver be ${\rho }_s$, density of the liquid be ${\rho }_L$and density of water be ${\rho }_w$.

Therefore the relative density of silver is given by, ${\rho }_{s/w}=\frac{{\rho }_s}{{\rho }_w}.........\left(ii\right)$

And the relative density of the liquid is given by, ${\rho }_{L/w}=\frac{{\rho }_L}{{\rho }_w}.........\left(iii\right)$

The relative density of silver and the liquid is given to us. Hence dividing equation 3 by 2 we get,

$$\begin{gathered} \frac{{\rho }_{L/w}}{{\rho }_{s/w}}=\frac{{\displaystyle \frac{{\rho }_L}{{\rho }_w}}}{{\displaystyle \frac{{\rho }_s}{{\rho }_w}}} \\ \frac{0.72}{10}=\frac{{\rho }_L}{{\rho }_s} \\ \frac{{\rho }_L}{{\rho }_s}=0.072 \end{gathered}$$

Step7: Calculating tension in the string

Using equation (i)

$$\begin{gathered} T=mg-{\rho }_{L}gV \\ T=m\left(g-\frac{{\rho }_{L}g}{{\displaystyle \frac{m}{v}}}\right) \\ T=4kg\left(10-\frac{{\rho }_L\left(10\right)}{{\displaystyle {\rho }_s}}\right)\left[As,\frac{m}{v}={\rho }_s\right] \\ T=4[10-0.072(10\left)\right]\left[As,\frac{{\rho }_L}{{\rho }_s}=0.072\right] \\ T=4[9.28)] \\ T=37.12N \end{gathered}$$

Hence, the tension in the string will be 37.12 N.