Answer
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Hint: Specific Gravity is defined as the ratio of density of a given object to the density of water. For the cork to just float on water, it has to be in complete vertical equilibrium, that is the buoyant force has to be equal to the total weight.
Complete step by step answer:
We know that specific gravity is the ratio between the density of an object and density of water $\left( \rho \right)$. So to calculate density of load $\left( {{\rho _1}} \right)$,
Specific Gravity$ = \dfrac{{{\rho _1}}}{\rho }$
$\Rightarrow 11 = \dfrac{{{\rho _1}}}{\rho }$
Since, $\rho = 1g/cc$
$\Rightarrow {\rho _1} = 11g/cc$
Similarly the density of cork $\left( {{\rho _2}} \right)$ can be calculated and,
${\rho _2} = 0.2g/cc$
Mass of cork, ${m_2} = 10g$
So, the volume of cork, ${V_2} = \dfrac{{{m_2}}}{{{\rho _2}}}$
$\Rightarrow {V_2} = \dfrac{{10}}{{0.2}}$
$\Rightarrow {V_2} = 50c{m^3}$
Let the volume of load which is to be added be ${V_1}$, then in order to attain equilibrium, buoyant force $\left( {{F_B}} \right)$ has to be equal to the sum of weight of load and weight of cork$\left( {{W_2}} \right)$, that is
$\Rightarrow {F_B} = {W_1} + {W_2}$
Since the cork and load displace ${V_1} + {V_2}$ volume of water,
$\Rightarrow \left( {{V_1} + {V_2}} \right)\rho g = mg + {V_1}{\rho _1}g$
$\Rightarrow {V_1} + {V_2} = m + {V_1}{\rho _1}$
$\Rightarrow \left[ {\rho = 1g/cc} \right]$
$\Rightarrow {V_2} - m = {V_1}{\rho _1} - {V_1}$
$\Rightarrow {V_2} - m = {V_1}\left( {{\rho _1} - 1} \right)$
$\Rightarrow {V_1} = \dfrac{{{V_2} - m}}{{{\rho _1} - 1}}$
$\Rightarrow {V_1} = \dfrac{{50 - 10}}{{11 - 1}}$
$\Rightarrow {V_1} = \dfrac{{40}}{{10}}$
$\Rightarrow {V_1} = 4c{m^3}$
To calculate mass of the load to be added, ${m_1} = {\rho _1}{V_1}$
$\Rightarrow {m_1} = 11 \times 4$
$\Rightarrow {m_1} = 44g$
Therefore option B is correct.
Note: Buoyant force is an upward force exerted by the water displaced by the object. If the buoyant force is greater than or equal to the weight of the object, the object floats. If the buoyant force is less than the weight of the object, it sinks.
Complete step by step answer:
We know that specific gravity is the ratio between the density of an object and density of water $\left( \rho \right)$. So to calculate density of load $\left( {{\rho _1}} \right)$,
Specific Gravity$ = \dfrac{{{\rho _1}}}{\rho }$
$\Rightarrow 11 = \dfrac{{{\rho _1}}}{\rho }$
Since, $\rho = 1g/cc$
$\Rightarrow {\rho _1} = 11g/cc$
Similarly the density of cork $\left( {{\rho _2}} \right)$ can be calculated and,
${\rho _2} = 0.2g/cc$
Mass of cork, ${m_2} = 10g$
So, the volume of cork, ${V_2} = \dfrac{{{m_2}}}{{{\rho _2}}}$
$\Rightarrow {V_2} = \dfrac{{10}}{{0.2}}$
$\Rightarrow {V_2} = 50c{m^3}$
Let the volume of load which is to be added be ${V_1}$, then in order to attain equilibrium, buoyant force $\left( {{F_B}} \right)$ has to be equal to the sum of weight of load and weight of cork$\left( {{W_2}} \right)$, that is
$\Rightarrow {F_B} = {W_1} + {W_2}$
Since the cork and load displace ${V_1} + {V_2}$ volume of water,
$\Rightarrow \left( {{V_1} + {V_2}} \right)\rho g = mg + {V_1}{\rho _1}g$
$\Rightarrow {V_1} + {V_2} = m + {V_1}{\rho _1}$
$\Rightarrow \left[ {\rho = 1g/cc} \right]$
$\Rightarrow {V_2} - m = {V_1}{\rho _1} - {V_1}$
$\Rightarrow {V_2} - m = {V_1}\left( {{\rho _1} - 1} \right)$
$\Rightarrow {V_1} = \dfrac{{{V_2} - m}}{{{\rho _1} - 1}}$
$\Rightarrow {V_1} = \dfrac{{50 - 10}}{{11 - 1}}$
$\Rightarrow {V_1} = \dfrac{{40}}{{10}}$
$\Rightarrow {V_1} = 4c{m^3}$
To calculate mass of the load to be added, ${m_1} = {\rho _1}{V_1}$
$\Rightarrow {m_1} = 11 \times 4$
$\Rightarrow {m_1} = 44g$
Therefore option B is correct.
Note: Buoyant force is an upward force exerted by the water displaced by the object. If the buoyant force is greater than or equal to the weight of the object, the object floats. If the buoyant force is less than the weight of the object, it sinks.
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