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ExamsJEE MainPhysics

A container holding a liquid of density rho is accelerated with acceleration a directed up along an inclined plane of inclination alpha. Find the angle theta that the free surface of the liquid makes with the horizontal.

  1. tan⁻¹[(a*cos alpha)/(g + a*sin alpha)]
  2. tan⁻¹[(a + g*sin alpha)/(g*cos alpha)]
  3. tan⁻¹[a/(g*cos alpha)]
  4. tan⁻¹[(a - g*sin alpha)/(g(1 - cos alpha))]

Correct answer: tan⁻¹[(a*cos alpha)/(g + a*sin alpha)]

Solution

The acceleration a is along the incline, so its horizontal component is a*cos(alpha) and vertical component is a*sin(alpha) (upward). In the non-inertial frame, pseudo-forces act opposite to acceleration. The free surface tilts so its slope satisfies tan(theta) = a_horizontal / (g + a_vertical) = (a*cos alpha)/(g + a*sin alpha).

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