Question

A light semi cylindrical gate of radius r is pivoted at its mid point holding the liquid of density q

Answer 1

Options is D (None of these)

Answer 2

Option (D) none of these

Explanation:

The correct question is,

A light semi cylindrical gate of radius R is pivoted at its midpoint O, of the diameter as shown in the figure holding liquid of density ρ. The force F required to prevent the rotation of the gate is equal to.

$A. 2 \pi R^{3} \rho g \\B. 2 \rho g R^{3} l \\c. \frac{2 R^{2} l \rho g}{3} \\D. none of these$

Given Data

Radius of the cylindrical gate = R

Density of the liquid = ρ

Find the force required to prevent the rotation of the gate.

The Pressure is related with the density of the liquid and the height of the liquid filled in the container.

The expression for pressure can be written as,

P = ρgh (Pa)    ------>(1)

Let us consider,

• Zero pressure at the base reference, where the top most point is taken as base reference.
• Radius of the semi circular ring = r
• Thickness of the semi circular ring = dr

From the above consideration,

The height can be calculated using the formula'

Height (h) = R -r  ------> (2)

The outside pressure (dP) can be calculated as,

dP =  ρgh

From equation (2)

dP = ρg(R−r)  

Force is defined as the pressure applied on a given or particular area.

Force = dP  × A

dF = ρg(R−r)×2πrdr

The counteracting force(F) required is equal to the ,

$\mathrm{\text{F}}=\int_{0}^{\text{R}} \rho \text{g(R-r)} \times 2 \pi \text{r d r}$

$\text{F}=\frac{\pi \rho \text{g R}^{3}}{3}$

The force we have got is not in the first three options , so the answer is none of these.

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