A right circular cylinder has a mass m,radius | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(D) The net force on the cylinder is the difference between the buoyant force $(F_b)$ acting upwards and the gravitational force $(mg)$ acting downwar

Vedclass · Accepted Answer

$\pi r^2\rho gh - mg$

Vedclass · Answer

(D) The net force on the cylinder is the difference between the buoyant force $(F_b)$ acting upwards and the gravitational force $(mg)$ acting downwards.The volume of the cylinder is $V = \pi r^2 h$.The buoyant force is given by Archimedes' principle: $F_b = \rho V g = \rho (\pi r^2 h) g = \pi r^2 \rho g h$.The gravitational force acting on the cylinder is $F_g = mg$.Therefore,the net force $F_{net}$ is:$F_{net} = F_b - F_g = \pi r^2 \rho g h - mg$.

$A$ right circular cylinder has a mass $m$,radius $r$,and a height $h$. The cylinder is completely submerged in a fluid of density $\rho$,as shown in the diagram. What is the magnitude of the net force on the cylinder?

Similar Questions

The number of subsets of $\{1, 2, 3, \ldots, 9\}$ containing at least one odd number is

If $f(x) = \begin{cases} \frac{1-\sqrt{2} \sin x}{\pi-4x} & \text{if } x \neq \frac{\pi}{4} \\ a & \text{if } x = \frac{\pi}{4} \end{cases}$ is continuous at $x = \frac{\pi}{4}$,then $a$ is equal to

Two blocks of $7 \ kg$ and $5 \ kg$ are connected by a heavy rope of mass $4 \ kg$. An upward force of $200 \ N$ is applied as shown in the diagram. The tension at the top end of the heavy rope at point $P$ is: $(g = 10 \ m/s^2)$ (in $N$)

During the depression in freezing point experiment,an equilibrium is established between the molecules of

Chloroethane reacts with $X$ to form diethyl ether. What is $X$?

Vedclass Test Series

Exam Paper Generator

Online Exam Module