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A spring balance reads $200 \,gF$ when carrying a lump of lead in air. If the lead is now immersed with half of its volume in brine solution, what will be the new reading of the spring balance? specific gravity of lead and brine are $11.4$ and $1.1$ respectively ........... $gF$
$190.4$
$180.4$
$210$
$170.4$
Solution
(a)
The weight of lump is given as,
$W _1= mg$
The volume of a lump is given as,
$V =\frac{ m }{\rho_1}$
The weight of lump in brine solution is given as,
$W _2= mg -\frac{\rho g V }{2}$
$= mg \left(1-\frac{\rho}{2 \rho_1}\right)$
The new readings of the spring balance is given as,
$W^{\prime}=m g\left(1-\frac{\rho}{2 p_1}\right)$
$=200 \times\left(1-\frac{1.1}{2 \times 11.4}\right)$
$=190.35\,gF$
Thus, the new reading of the spring balance is $190.35\,gF$.
