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The arm of a teapot is 10 cm long and inclined at an angle of 60^o to the vertical. The center of the arm base is 2 cm above the base of the beaker. Water is poured into the beaker such that half the arm is filled with it. What will be the pressure at the base of the beaker if the atmospheric pressure is 101.3 kPa?

(a) 101.3

(b) 101.5

(c) 101.7

(d) 101.9

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The correct answer is (c) 101.7

Explanation: Total height of the water in the beaker = 2 + ^1⁄2 *  10 cos 60^o cm = 4:5 cm. Pressure at the base of the beaker = 101.3 + 10^3 *  9.81 * 0.045 Pa = 101.3 + 0.44 kPa = 101.74 kPa.

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