Shown in figure, a conical container of half-apex angle 37° filled with certain quantities of kerosene and water. The force exerted by the water on the kerosene is approximately, (Take atmospheric pressure =105 Pa )

Question

Shown in figure, a conical container of half-apex angle 37° filled with certain quantities of kerosene and water. The force exerted by the water on the kerosene is approximately, (Take atmospheric pressure =105 Pa ) (A) 3 × 107 N (B) 4 × 107 N (C) 2 × 107 N (D) 5 × 107 N

Tardigrade Admin · Accepted Answer

<img class="img-fluid question-image" alt="image" src="https://cdn.tardigrade.in/img/question/physics/5b15eac62a832353b15f57401102a605-.png" /> (r/8)= tan 37°=(3/4) r =6 m F =( P 0+ h ρ g ) π r 2 =(105+10 × 800 × 10) × π × 36 1.8 × 36 × π × 105=2 × 107

Q. Shown in figure, a conical container of half-apex angle $3 7^{\circ}$ filled with certain quantities of kerosene and water. The force exerted by the water on the kerosene is approximately,
(Take atmospheric pressure $= 1 0^{5} P a$ )

$3 \times 1 0^{7} N$

$4 \times 1 0^{7} N$

$2 \times 1 0^{7} N$

$5 \times 1 0^{7} N$

$\frac{r}{8} = tan 3 7^{\circ} = \frac{3}{4}$
$r = 6 m$
$F = (P_{0} + h ρ g) π r^{2}$
$= (1 0^{5} + 10 \times 800 \times 10) \times π \times 36$
$1.8 \times 36 \times π \times 1 0^{5} = 2 \times 1 0^{7}$

Q. Shown in figure, a conical container of half-apex angle 37∘ filled with certain quantities of kerosene and water. The force exerted by the water on the kerosene is approximately, (Take atmospheric pressure =105Pa )

3×107N

4×107N

2×107N

5×107N