Question

(a) How does a boat float in water? (b) A piece of steel has a volume of $12c{{m}^{3}}$, and a mass of $96g$. What is its density: (i) in $\dfrac{g}{c{{m}^{3}}}$ ? (ii) in $\dfrac{kg}{{{m}^{3}}}$ ?

Answer 1

Hint: When a boat is in water it displaces liquid and it floats the principle of floating is explained below. The density of an object is defined as the mass per unit volume. First calculate the density in $\dfrac{g}{c{{m}^{3}}}$ then convert gram to kilogram and centimetre to meter to get the density in $\dfrac{kg}{{{m}^{3}}}$
Formula used:
Density of an object is given by
$\rho =\dfrac{\text{mass of the object}}{\text{volume of the object}}$

Complete answer:
(a) The principle of floating of boat and any object is explained by Archimedes. According to Archimedes principle when an object is placed in water it will displace some amount of liquid. So the force exerted by the fluid on the object is equal to the weight of the fluid displaced by the object. This force which is exerted by liquid on the object is called buoyant force. The nature of this buoyant force is that it always pushes the object upwards against the weight of the object. The weight of the object is always directed downward. So an object will float if the buoyant force is more than its weight and will sink if the buoyant force is less than the weight of the object.

(b) Given that the mass of the piece of steel is, $m=96g$ and the volume of the steel is $V=12c{{m}^{3}}$.
So the density of the steel is given by the ratio of mass with the volume. i.e.
Density, $\rho =\dfrac{\text{mass of the object}}{\text{volume of the object}}=\dfrac{m}{V}=\dfrac{96g}{12{{m}^{3}}}=8g{{m}^{-3}}$
(i)So in $\dfrac{g}{c{{m}^{3}}}$ the density of the steel is $8\dfrac{g}{c{{m}^{3}}}$
(ii) In $\dfrac{kg}{{{m}^{3}}}$ the density will be
$\rho =8\dfrac{g}{c{{m}^{3}}}=\dfrac{8\times {{10}^{-3}}kg}{{{10}^{-6}}{{m}^{3}}}=8\times {{10}^{3}}\dfrac{kg}{{{m}^{3}}}\left( \because 1c{{m}^{3}}={{10}^{-6}}{{m}^{3}}\text{ and 1g}={{10}^{-3}}kg \right)$
So $\rho =8\times {{10}^{3}}\dfrac{kg}{{{m}^{3}}}$

Note:
The density of an object is also related to the buoyant force exerted. Keeping the volume same a denser material will have more weight. But the weight of water displaced by the object is related to its volume. So a denser object will float on water but a less dense object is not. This can be explained by taking an iron block and a wood block of the same dimension and placing them in water. The iron has more mass compared to the wood. So the gravitational force or weight will be more in case of iron and the weight in case of iron exceeds the buoyant force. So an iron piece sink where as a wood of the same dimension floats.