Question

The figure shows a conical container of hlf-apex angle ${37}^{\circ }$ filled with certain quantities of kerosene and water. The force exereted by the water on the kerosene is $y\times {10}^{7}N$. Find $y$. Answer should be in nearest integer.(take atmosphric pressure$={10}^{5}Pa$)

Answer 1

Step 1: Draw a labelled diagram.

Step 2: Find the force exerted by the water on the kerosene.
Formula Used:$P=\frac{F}{A}$
According to given situation,
$\frac{r}{8}=\tan {37}^{\circ }=\frac{3}{4}$
The force exerted by the water on the kerosene is,
$P=\frac{F}{A}$
Pressure at height $h$ is given as,
$P=P_0+h\rho g\therefore F=(P_0+h\rho g)\pi r^2$
Substitute the values, we get
$=({10}^5+10\times 800\times 10)\times \pi \times 36=1.8\times 36\times \pi \times {10}^5\approx 2\times {10}^{7}N$
Final answer: (2)