Question

A Ping-Pong ball is held submerged in a bucket of water by a string attached to the bucket's bottom.

Salt is now added to the water in the bucket, increasing the density of the liquid. What happens to the tension in the string ?

	The tension does not change.
	The tension increases.
	The tension decreases.

What happens to the tension in the string if the Ping-Pong ball is replaced by a smaller spherical object of equal weight?

	The tension does not change.
	The tension increases.
	The tension decreases.

Answer 1

Concepts and reason

The concepts required to solve the given question is tension in the string.

Initially, write an expression for the tension in terms of the volume, density of liquid, and the density of solid. Later, compare the tension on basis of the density of matter. Finally, compare the tension on basis of the weight of the spherical ball.

Fundamentals

The expression for the tension in terms of the volume, density of liquid, and the density of solid is as follows:

$T = V\left( {{\rho _{\rm{l}}} - {\rho _{\rm{s}}}} \right)$

Here, V is the volume, ${\rho _{\rm{l}}}$ is the density of liquid, and ${\rho _{\rm{s}}}$ is the density of solid.

(A)

The tension in the equation $T = V\left( {{\rho _{\rm{l}}} - {\rho _{\rm{s}}}} \right)$ is directly proportional to the volume, density of liquid, and the density of solid. Thus, as the density of liquid increases, the tension increases. Thus, the tension in the string neither decrease nor remains same.

Form the equation $T = V\left( {{\rho _{\rm{l}}} - {\rho _{\rm{s}}}} \right)$ , it is clear that the tension increases with increase in the density of the liquid. After adding salt to water, the density of liquid increases. Hence, the tension in the string also increases.

(B)

The weight of the spherical balls is same but the volume is decreased, Therefore, in the second terms of the equation $T = V\left( {{\rho _{\rm{l}}} - {\rho _{\rm{s}}}} \right)$ , the density of the solid remains constant, but, the density of the liquid decreases. Therefore, the net tension in the string neither increases not remains same.

The weight of the spherical balls is same but the volume is decreased, Therefore, in the second terms of the equation $T = V\left( {{\rho _{\rm{l}}} - {\rho _{\rm{s}}}} \right)$ , the density of the solid remains constant, but, the density of the liquid decreases. Therefore, the net tension decreases.

Ans: Part A

The tension in the string increases.

Part B

The net tension in the string decreases.