COMEDK · Physics · 11. Mechanical Properties of Fluids
A spherical metal ball of density '\(\rho\)' and radius '\(r\)' is immersed in a liquid of density '\(\sigma\)'. When an electric field is applied in the upward direction the metal ball remains just suspended in the liquid. Then the expression for the charge on the metal ball is :
- A \(q=\dfrac{\left[4 \pi r^2(\rho-\sigma) g\right]}{E}\)
- B \(q=\dfrac{\left[4 \pi r^2 \rho g\right]}{\sigma E}\)
- C \(q=\dfrac{\left[\dfrac{4}{3} \pi r^3 \rho g\right]}{\sigma E}\)
- D \(q=\dfrac{\left[4 \pi r^3(\rho-\sigma) g\right]}{3 E}\)
Answer & Solution
Correct Answer
(D) \(q=\dfrac{\left[4 \pi r^3(\rho-\sigma) g\right]}{3 E}\)
Step-by-step Solution
Detailed explanation
For the metal ball to remain suspended in the liquid, the net force acting on it must be zero. The forces acting on the ball are the gravitational force acting downwards, the buoyant force acting upwards, and the electric force acting upwards. The gravitational force is…
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