TS EAMCET · Physics · Mechanical Properties of Fluids
A cubical block of wood, of length \(10 \mathrm{~cm}\), floats at the interface between oil of density \(800 \mathrm{~kg} / \mathrm{m}^3\) and water. The lower surface of the block is \(1.5 \mathrm{~cm}\) below the interface. If the depth of water is \(10 \mathrm{~cm}\) below the interface and oil is upto \(10 \mathrm{~cm}\) above the interface then the difference in pressure at the lower and the upper face of the wooden block is (Assume density of water, \(\rho=1000 \mathrm{~kg} / \mathrm{m}^3\) and acceleration of gravity, \(g=10 \mathrm{~m} / \mathrm{s}^2\) )
- A \(850 \mathrm{~Pa}\)
- B \(780 \mathrm{~Pa}\)
- C \(800 \mathrm{~Pa}\)
- D \(830 \mathrm{~Pa}\)
Answer & Solution
Correct Answer
(D) \(830 \mathrm{~Pa}\)
Step-by-step Solution
Detailed explanation
Given, length of cubical block, \(l=10 \mathrm{~cm}=0.1 \mathrm{~m}\) density of oil, \(\rho_0=800 \mathrm{~kg} / \mathrm{m}^3\) and density of water, \(\rho_w=10^3 \mathrm{~kg} / \mathrm{m}^3\) If \(V_1\) and \(V_2\) be the volume of wooden block in water and oil respectively,…
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