Question

A tin can has a volume of \[1000{\rm{ c}}{{\rm{m}}^3}\] and a mass of \[100{\rm{ g}}\]. What mass of lead shot can it carry without sinking in water \[\left( {\rho = 1000{\rm{ }}{{{\rm{kg}}} {\left/
 {\vphantom {{{\rm{kg}}} {{{\rm{m}}^{\rm{3}}}}}} \right.
 } {{{\rm{m}}^{\rm{3}}}}}} \right)\]?
A. \[900{\rm{ g}}\]
B. \[100{\rm{ g}}\]
C. \[1000{\rm{ g}}\]
D. \[1100{\rm{ g}}\]

Answer 1

Hint: We will be using the Archimedes principle which states that a body will be floating in a liquid till its density is less than the density of the liquid in which body is floating.

 Complete step by step solution:
Volume of the tin can is \[{V_T} = 1000{\rm{ c}}{{\rm{m}}^3}\].
Mass of the tin can is \[{M_T} = 100{\rm{ g}}\].
Density of water is \[\rho = 1000{\rm{ }}{{{\rm{kg}}} {\left/
 {\vphantom {{{\rm{kg}}} {{{\rm{m}}^3}}}} \right.
 } {{{\rm{m}}^3}}}\].
From the statement of Archimedes principle we can say that the volume of tin floating on the water must be equal to the volume of the water displaced by the tin.
Volume of the water displaced by the tin.
\[{V_W} = 1000{\rm{ c}}{{\rm{m}}^3}\]
The expression for the weight of water displaced by the tin can.
\[{W_W} = \rho \cdot {V_W} \cdot g\]……(1)
Substitute \[1000{\rm{ c}}{{\rm{m}}^3}\] for \[{V_W}\], \[10{\rm{ }}{{\rm{m}} {\left/
 {\vphantom {{\rm{m}} {{{\rm{s}}^2}}}} \right.
 } {{{\rm{s}}^2}}}\] for g and \[1000{\rm{ }}{{{\rm{kg}}} {\left/
 {\vphantom {{{\rm{kg}}} {{{\rm{m}}^3}}}} \right.
 } {{{\rm{m}}^3}}}\] for \[\rho \] in equation (1).
$
{M_W} = \left( {1000{\rm{ }}{{{\rm{kg}}} {\left/
 {\vphantom {{{\rm{kg}}} {{{\rm{m}}^3}}}} \right.
 } {{{\rm{m}}^3}}}} \right)\left( {1000{\rm{ c}}{{\rm{m}}^3}} \right)\left( {10{\rm{ }}{{\rm{m}} {\left/
 {\vphantom {{\rm{m}} {{{\rm{s}}^2}}}} \right.
 } {{{\rm{s}}^2}}}} \right)\\
 = \left( {1000{\rm{ }}{{{\rm{kg}}} {\left/
 {\vphantom {{{\rm{kg}}} {{{\rm{m}}^3}}}} \right.
 } {{{\rm{m}}^3}}}} \right)\left( {1000{\rm{ c}}{{\rm{m}}^3} \times \dfrac{{{{\rm{m}}^3}}}{{{{10}^6}{\rm{ c}}{{\rm{m}}^3}}}} \right)\left( {10{\rm{ }}{{\rm{m}} {\left/
 {\vphantom {{\rm{m}} {{{\rm{s}}^2}}}} \right.
 } {{{\rm{s}}^2}}}} \right)\\
 = 10{\rm{ kg}} \cdot {{\rm{m}} {\left/
 {\vphantom {{\rm{m}} {{{\rm{s}}^2}}}} \right.
 } {{{\rm{s}}^2}}}\\
 = 10{\rm{ N}}
$

Weight of the tin is equal to the product of the mass of tin and volume of the tin.
\[{W_T} = {M_T} \cdot g\]

Substitute \[10{\rm{ }}{{\rm{m}} {\left/
 {\vphantom {{\rm{m}} {{{\rm{s}}^2}}}} \right.
 } {{{\rm{s}}^2}}}\] for g and \[100{\rm{ g}}\] for \[{M_T}\] in the above equation.
$
{W_T} = \left( {100{\rm{ g}} \times \dfrac{{{\rm{kg}}}}{{1000{\rm{ g}}}}} \right)\left( {10{\rm{ }}{{\rm{m}} {\left/
 {\vphantom {{\rm{m}} {{{\rm{s}}^2}}}} \right.
 } {{{\rm{s}}^2}}}} \right)\\
 = 1{\rm{ kg}}{{ \cdot {\rm{m}}} {\left/
 {\vphantom {{ \cdot {\rm{m}}} {{{\rm{s}}^2}}}} \right.
 } {{{\rm{s}}^2}}}\\
 = 1{\rm{ N}}
$

By analogy, we say that the upward force applied by the water is compensated by downward force of tin and force due to the lead shot acting vertically.

Upward force = Tin's weight + Load of the lead shot
Load of the lead shot = Upward force - Tin's weight ……(2)

This upward force applied by the water on the tin can is equal to the weight of the water displaced.

Substitute \[10{\rm{ N}}\]for upward force and \[1{\rm{ N}}\] in equation (2).

Load of the lead shot=10 N- 1 N
$ = 9 N$

Write the expression for mass of the lead shot that can be carried by the can without sinking.
$M_L$ = Load of the leadshot/g
Substitute \[9{\rm{ N}}\] for load of the lead shot in the above equation.
$
{M_L} = \dfrac{{{\rm{9 N}}}}{{10{\rm{ }}{{\rm{m}} {\left/
 {\vphantom {{\rm{m}} {{{\rm{s}}^2}}}} \right.
 } {{{\rm{s}}^2}}}}}\\
 = \dfrac{{{\rm{9 N}} \times \dfrac{{{\rm{kg}}{{ \cdot {\rm{m}}} {\left/
 {\vphantom {{ \cdot {\rm{m}}} {{{\rm{s}}^2}}}} \right.
 } {{{\rm{s}}^2}}}}}{{\rm{N}}}}}{{10{\rm{ }}{{\rm{m}} {\left/
 {\vphantom {{\rm{m}} {{{\rm{s}}^2}}}} \right.
 } {{{\rm{s}}^2}}}}}\\
 = \dfrac{9}{{10}}{\rm{ kg}} \times \dfrac{{{\rm{1000 g}}}}{{{\rm{kg}}}}\\
{\rm{ = 900 g}}
$

Therefore, the mass of lead shot that can be carried by the tin can is \[900{\rm{ g}}\].

So, the correct answer is “Option A”.

Note:
After calculating mass do not forget to multiply it with the acceleration due to gravity to get the value of weight. We can remember the conversion of Newton into its base units (kg, m, s) so that low homogeneity of units is followed.The volume of tin floating on the water must be equal to the volume of the water displaced by the tin.