Question:
A ball whose density is $0.4\times 10^3 kg/m^3$ falls into water from a height of $9\, cm$. To what depth does the ball sink ?
A ball whose density is $0.4\times 10^3 kg/m^3$ falls into water from a height of $9\, cm$. To what depth does the ball sink ?
Updated On: Jan 30, 2025
- 2 cm
- 6 cm
- 4.5 cm
- 2.25 cm
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The Correct Option is B
Solution and Explanation
The velocity of ball before entering the water surface
$v =\overline{2 gh }=\overline{2 g } X$
When ball enters into water, due to upthrust of water, the velocity of ball decreases (or retarded)
The retardationm
$a =\frac{\text { apparent weight }}{\text { mass of ball }}$
$=\frac{ v (\rho-\sigma) g }{ V \rho}=\frac{(\rho-\sigma}{=} \frac{(\rho-\sigma) g }{\rho}$
$=\left(\frac{(0.4-1)}{0.4}\right) g =-\overline{2} g$
If $h$ be the depth upto which ball $\sin x$, then
$0- v ^{2}=2 \times h \left(\frac{-3}{2} g \right) X$
$\Rightarrow 2 g \times 9=3\, gh$
$\therefore h =6\, cm$
$v =\overline{2 gh }=\overline{2 g } X$
When ball enters into water, due to upthrust of water, the velocity of ball decreases (or retarded)
The retardationm
$a =\frac{\text { apparent weight }}{\text { mass of ball }}$
$=\frac{ v (\rho-\sigma) g }{ V \rho}=\frac{(\rho-\sigma}{=} \frac{(\rho-\sigma) g }{\rho}$
$=\left(\frac{(0.4-1)}{0.4}\right) g =-\overline{2} g$
If $h$ be the depth upto which ball $\sin x$, then
$0- v ^{2}=2 \times h \left(\frac{-3}{2} g \right) X$
$\Rightarrow 2 g \times 9=3\, gh$
$\therefore h =6\, cm$
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Concepts Used:
Pressure
Pressure is defined as the force applied perpendicular to the surface of an object per unit area over which that force is distributed.
Everyday examples of pressure are:
- The working of the vacuum cleaner is an example of pressure. The fan inside the vacuum creates a low-pressure region which makes it easy to suck the dust particles inside the vacuum.
- Using a knife for cutting is another example of pressure. The area exposed from the knife is small but the pressure is high enough to cut the vegetables and fruits.
Formula:
When a force of ‘F’ Newton is applied perpendicularly to a surface area ‘A’, then the pressure exerted on the surface by the force is equal to the ratio of F to A. The formula for pressure (P) is:
P = F / A
Units of Pressure:
The SI unit of pressure is the pascal (Pa)
A pascal can be defined as a force of one newton applied over a surface area of a one-meter square.