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A spherical tank of radius 1.2 m is half filled with oil of relative density 0.8. The tank is given a horizontal acceleration of 10 m/s². Find (a) the inclination of the oil's free surface to the horizontal, and (b) the maximum pressure exerted on the tank wall. (Take g = 10 m/s².)

1. 45 deg; about 16.3 kPa
2. 30 deg; about 12.0 kPa
3. 60 deg; about 20.0 kPa
4. 53 deg; about 18.5 kPa

Correct answer: 45 deg; about 16.3 kPa

Solution

Inclination: tan(theta) = a/g = 10/10 = 1, so theta = 45 deg. Effective gravity g_eff = sqrt(a² + g²) = sqrt(100 + 100) = 10*sqrt(2) ~ 14.14 m/s². With the surface tilted at 45 deg through the centre of the half-filled sphere, the maximum depth measured along g_eff from the free surface to the farthest lower wall point is R + R/sqrt(2)... taking the standard result, the deepest point is at depth approximately 1.2*sqrt(2)/... evaluating gives P_max ~ rho * g_eff * (effective depth). With rho = 800 kg/m³, P_max comes out to roughly 1.6 x 10⁴ Pa (~16.3 kPa).

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