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: Dec 16, 2025
- 2 cm
- 6 cm
- 4.5 cm
- 2.25 cm
Hide Solution
Verified By Collegedunia
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
$0- v ^{2}=2 \times h \left(\frac{-3}{2} g \right) X$
$\Rightarrow 2 g \times 9=3\, gh$
$\therefore h =6\, cm$
Was this answer helpful?
0
0
Top Questions on Pressure
- The volume contraction of a solid copper cube of edge length 10 cm, when subjected to a hydraulic pressure of \( 7 \times 10^6 \) Pa, would be ________________________ mm\(^3\). (Given bulk modulus of copper = \( 1.4 \times 10^{11} \) N m\(^{-2}\))
- 1 torr = ________
- KEAM - 2025
- Physics
- Pressure
- The volume contraction of a solid copper cube of edge length 10 cm, when subjected to a hydraulic pressure of \( 7 \times 10^6 \) Pa, would be _____ mm\(^3\). (Given bulk modulus of copper = \( 1.4 \times 10^{11} \) N m\(^{-2}\))
- Mercury is filled in a tube of radius \(2 \, \text{cm}\) up to a height of \(30 \, \text{cm}\). The force exerted by mercury on the bottom of the tube is ___N.
(Given, atmospheric pressure = \(10^5 \, \text{N/m}^2\), density of mercury = \(1.36 \times 10^4 \, \text{kg/m}^3\), \(g = 10 \, \text{m/s}^2\), \(\pi = \frac{22}{7}\)) - In a container, 1 g of hydrogen and 1 g of oxygen are taken. Find the ratio of hydrogen pressure to oxygen pressure.
View More Questions
Questions Asked in BITSAT exam
- A simple pendulum has a time period of 2 s on Earth's surface. What is its time period at a height equal to the Earth's radius (R)? (Acceleration due to gravity at height h is \( g_h = \frac{g}{(1 + h/R)^2} \)).
- BITSAT - 2025
- Mechanics
- What is the dot product of the vectors \( \mathbf{a} = (2, 3, 1) \) and \( \mathbf{b} = (1, -1, 4) \)?
- BITSAT - 2025
- Vector Algebra
For the reaction \( A + B \to C \), the rate law is found to be \( \text{rate} = k[A]^2[B] \). If the concentration of \( A \) is doubled and \( B \) is halved, by what factor does the rate change?
- BITSAT - 2025
- Chemical Kinetics
- A coil of resistance 10 \( \Omega \) is connected to a battery of 12 V. If the current flowing through the coil is 2 A, what is the power dissipated in the coil?
- BITSAT - 2025
- Resistance
- A particle moves along the x-axis under a force $ F(x) = 6x^2 $ N. The work done by this force in moving the particle from $ x = 1 \, \text{m} $ to $ x = 2 \, \text{m} $ is:
- BITSAT - 2025
- work, energy and power
View More Questions
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.