Question

A vessel containing water is suspended from a spring balance. The spring balance reads $15kg$. When a piece of iron of mass $1.8kg$, suspended by a string is immersed with half of its volume inside the water the reading of the spring balance is (relative density of iron$=7.8$)

• A. $15kg$
• B. $16.8kg$
• C. $15.115kg$
• D. $16.756kg$

Answer 1

The correct option is C $15.115kg$

Density is iron $\rho =R.D\times \text{density of water}=7.8\times {10}^{3}kg/m^3$

Volume of the iron piece, $V=\frac{M}{\rho }=\frac{1.8}{7.8\times {10}^3}=0.23\times {10}^{-3}{m}^3$

Since half iron piece is immersed in water, so volume of water displaced by iron piece $V^{\prime }=\frac{V}{2}=\frac{0.23\times {10}^{-3}}{2}{m}^3$

Upthrust on iron piece =$V^{\prime }{\rho }_{w}g=\frac{0.23\times {10}^{-3}}{2}\times {10}^3\times 9.8N=4.9\times 0.23N=\frac{4.9\times 0.23}{9.8}kgf=0.115kgf$

Total upward reaction $=15+0.115=15.115kgf$ Reading on the spring balance = total upward reaction =$15.115kgf$