## Q1. In what direction does the buoyant force on an object immersed in a liquid act?

Ans. The buoyant force on an object immersed in a liquid acts in upward direction (i.e., opposite to weight of the object).

## Q2. Why does a block of plastic released under water come up to the surface of water?

Ans. The density of plastic is less than that of water, so the force of buoyancy on plastic block(V\(\rho\)_{plastic} g) will be greater than the weight of plastic block displaced, i.e., (V \(\rho\) _{water} g). Hence, the acceleration of plastic block will be in upward direction, and comes up to the surface of water.

Mathematically,

acceleration, a = B — W

= \(\frac { V\quad { \rho }_{ water }\quad g\quad -\quad V\quad { \rho }_{ plastic }\quad g }{ m } \)

= \(\frac { V\quad g\quad ({ \rho }_{ water }\quad \quad -\quad \quad { \rho }_{ plastic })\quad }{ m } \)

is positive in upward direction.

## Q3. The volume of 50 g of a substance is 20cm . If the density of water is 1 g cm^{-3} , will the substance float or sink?

Ans. Density of substance, \(\rho \) =\(\frac{mass}{volume}\) = \(\frac { 50\quad g }{ 20\quad { cm }^{ 3 } } \) = 2.5 g cm^{-3}

As density of substance is greater than the density of water, therefore, the substance will sink in water. Alternatively, the buoyant force exerted by water (\((V\quad { \rho }_{ water }\quad g)\)) on substance, when fully immersed, is less than the weight of substance, i.e., B< W; hence, the substance will sink in water.

## Q4. The volume of a 500 g sealed packet is 350 cm^{3}. Will the packet float or sink in water if the density of water is 1 g cm^{-3} ? What will be the mass of the water displaced by this packet?

Ans. Density of packet, \(\rho =\frac { mass }{ volume } \)

=\(\frac { 500\quad g }{ 350\quad { cm }^{ 3 } } \)

=\(\frac { 10\quad }{ 7 } g\quad { cm }^{ -3 }\)

= l.43 g cm^{-3}

As density of packet is more than the density of water, so the packet will sink in water.

Weight of water displaced by the packet

= volume of packet x density of water

= (350 cm^{3}) x (l g cm^{-3}) = 350 g

Obviously, the force of buoyancy, B = 350 g wt and weight of packet, W = 500 g wt i.e., B < W; so the packet sinks in water.