Question

A piece of aluminum with mass $1.00kg$ and density $2700kg/m^3$ is suspended in air from a string and then completely immersed in a container of water. Calculate the tension in the string $\left(a\right)$ before and $\left(b\right)$ after the metal is immersed. Take, $g=10m/s^2$, density of water $={10}^{3}kg{m}^{-3}$.

• A. $10N,2.3N$
• B. $2.3N,10N$
• C. $5N,1.15N$
• D. None of these

Answer 1

The correct option is A $10N,2.3N$

Given, mass of place of aluminium, $m=1kg$;

density of aluminium, ${\rho }_s=2700kg/m^3$;

density of water, ${\rho }_w=1000kg{m}^{-3}$

As shown in figure, we have to find tension in the string in each case.

(a) When the aluminium piece is suspended in air

Tension, $T=mg$ [Ref fig. (a)]

or $T=1\times 10=10N$

(b) When the aluminium piece is completely immersed in water,

$mg=T+F_B$ [Ref fig.(b)]

Where, $F_B=$ buoyant force acting on aluminium

$T=$ tension in the string

$\therefore T=mg-F_B={\rho }_s{V}_{s}g-{\rho }_L{V}_{L}g$

$=({\rho }_s{V}_s-{\rho }_L{V}_L)g$

$=({\rho }_s{V}_s-{\rho }_L{V}_s)g$

$={\rho }_s{V}_s(1-\frac{{\rho }_L}{{\rho }_s})g=\left({\rho }_s{V}_{s}g\right)(1-\frac{{\rho }_L}{{\rho }_s})$

$=mg\left(1-\frac{{\rho }_L}{{\rho }_s}\right)$

or $T^{\prime }=1\times 10(1-\frac{1000}{2700})=10(1-\frac{10}{27})$

or $T^{\prime }=10\left(\frac{27-10}{27}\right)=10\left(\frac{17}{27}\right)$

$=6.296N$.

Hence, the tension in the string before and after the metal is immersed are $10N$ and $6.296N$ respectively.