Question

What mass of lead will weigh as much as 8 g of iron, when both are immersed in water? (Given specific gravities of iron and lead area 8 and 11, respectively).
A. $1.1$g
B. $2.2$g
C. $5.5$ g
D. $7.7$ g

Answer 1

Hint
When solids are weighed in a liquid medium like water, the measured apparent weight is different from the actual weight of the solid. This is because of an upwards buoyant force exerted on the solid by the liquid.
$\Rightarrow F = V\rho g$, where F is the buoyancy force exerted by a liquid of density $\rho $ on volume V, and g is the acceleration due to gravity.

Complete step by step answer
We are provided with the specific densities of iron and lead, and are asked to find the mass of lead which will be equivalent to the mass of iron in water. The data given to us is:
Specific gravity of iron $\dfrac{{{\rho _i}}}{{{\rho _w}}} = 8$
Specific gravity of lead $\dfrac{{{\rho _l}}}{{{\rho _w}}} = 11$
Mass of iron ${m_i} = 8$ g
We know that the apparent weight of a solid in water is given as:
$\Rightarrow mg - F$
Where m is the actual mass of the solid, and F is the buoyant force exerted by the liquid. We know that the apparent weight of lead is equal to the apparent weight of iron:
$\Rightarrow {m_l}g - {F_l} = {m_i}g - {F_i}$ [Eq. 1]
We also know that the buoyant force is given as:
$\Rightarrow F = V\rho g$
For iron, this will be equal to:
$\Rightarrow {F_i} = {V_i}{\rho _w}g = {V_i}{\rho _w}g \times \dfrac{{{\rho _i}}}{{{\rho _i}}}$ [Multiplying and dividing by the density of iron]
$\Rightarrow {F_i} = {V_i}{\rho _i}g \times \dfrac{{{\rho _w}}}{{{\rho _i}}} = {m_i}g\dfrac{{{\rho _w}}}{{{\rho _i}}}$ [As mass is equal to volume times the density]
Similarly, the buoyant force for lead will be equal to:
$\Rightarrow {F_l} = {V_l}{\rho _l}g \times \dfrac{{{\rho _w}}}{{{\rho _l}}} = {m_l}g\dfrac{{{\rho _w}}}{{{\rho _l}}}$
Putting this in the Eq. 1 gives us:
$\Rightarrow {m_l}g - {m_l}g\dfrac{{{\rho _w}}}{{{\rho _l}}} = {m_i}g - {m_i}g\dfrac{{{\rho _w}}}{{{\rho _i}}}$
Taking out the common terms:
$\Rightarrow {m_l}g\left( {1 - \dfrac{{{\rho _w}}}{{{\rho _l}}}} \right) = {m_i}g\left( {1 - \dfrac{{{\rho _w}}}{{{\rho _i}}}} \right)$
Now, we substitute the known values in this equation:
$\Rightarrow {m_l}\left( {1 - \dfrac{1}{{11}}} \right) = {m_i}\left( {1 - \dfrac{1}{8}} \right)$
Solving for the mass of lead:
$\Rightarrow {m_l}\dfrac{{10}}{{11}} = 8 \times \dfrac{7}{8} = 7$
This finally gives us:
$\Rightarrow {m_l} = 7 \times \dfrac{{11}}{{10}} = 7.7$ g
Hence, the correct answer is option (D).

Note
Specific gravity is the same as relative density of a substance. It is defined as the ratio of the density of a substance to the density of a reference material. It is nearly always measured with respect to water, and is helpful in determining the extent to which the substance’s density varies as compared to the same volume of water.