The figure shows 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 $= 10^{5} Pa$ )

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$3 \times 10^{7} N$

$4 \times 10^{7} N$

$2 \times 10^{7} N$

$5 \times 10^{7} N$

answer is C.

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Detailed Solution

F_b/w water & kerosene is $F = {πr}^{2} [P_{0} + h_{K} . ρ_{K} g]$
$r = 8 \tan 37 = 6 m ∴ F = π (6^{2}) [10^{5} + 10 (800) (9.8)] = 2 \times 10^{7} N$

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