Question

A metal cube of side 9 cm is immersed in water.Find the buoyant force .Will the cube float or sink. The mass of the cube is 500 gram.​

Answer 1

$\huge{\underline{\tt{Answer:-}}}$

•The bouyant force is 729 N.

•The cube would float on the water.

$\huge{\underline{\tt{Given:-}}}$

• a cube of side:- 9 cm

• Mass of the cube :- 500 gm.

$\huge{\underline{\tt{1.Concept:-}}}$

★Archimedes principle states that when an object immersed in a fluid it experiences a bouyant force which is equal to the volume of the liquid displaced by the object.

${\boxed{\bf{Bouyant\:Force= Volume\:of\:object}}}$

$\huge{\underline{\tt{1.Solution:-}}}$

Based on the concept, we know that

Bouyant force= volume of object

→ Volume of object:-

${ \tt{volume \: of \: cube = a {}^{3}}} \\ \implies { \tt{9 {}^{3} }} \\ \implies \: { \boxed{ \bf{729 \: cm {}^{3} }}}$

•°• Bouyant force= 729 cm³

$\huge{\underline{\tt{2.Concept:-}}}$

★ If the density of the object is less than water then it will float on the surface of water. Otherwise if it's density is more than that of water then it will sink.

$\huge{\underline{\tt{2.Solution:-}}}$

Based on concept,

We know that,

• density of water :-1 gm/cm³

•Density of an object:-

${ { \boxed{\tt{density = \frac{mass}{volume} }}}}$

•Density of Cube:-

$\implies{ \tt{\frac{500}{729} }} \\ \implies{ \boxed{{ \tt{0.685\: cm {}^{3} }}}} \: { \tt{ (approx.)}}$

→Which is less less than the density of water.

•°• It would float on the surface of water.