Unit-5 Pressure and Deformation in Solids
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Unit-5 Pressure and Deformation in Solids
Answer: The weight is the same, but the smaller contact area gives a larger pressure.
Answer: P = F/A = 25/5 = 5 Pa.
Answer: A wider strap spreads the load over a larger area, reducing pressure on the shoulders.
Answer: P = F/A = 120/0.6 = 200 Pa.
Answer: F = P x A = 250 x 4 = 1000 N.
Answer: Atmospheric pressure pushes on the mercury surface in the reservoir and supports the column.
Answer: P = h rho g = 10 x 1000 x 10 = 100,000 Pa.
Answer: P = h rho g depends only on depth, density and g, not on the container's shape or area.
Answer: A = F/P = 500/250 = 2 m^2.
Answer: F2 = F1 x A2/A1 = 50 x (2/0.02) = 5000 N.
Answer: Hydraulic lifts, presses and brakes work on Pascal's law of transmission of pressure in liquids.
Answer: Upthrust = V rho g, so it depends on the submerged volume and the liquid's density.
Answer: A hydrometer floats in the liquid, and its sinking depth depends on the liquid's density (buoyancy).
Answer: Filling or emptying its tanks changes the submarine's average density and hence the upthrust balance.
Answer: Upthrust = V rho g = 0.05 x 1000 x 10 = 500 N.
Answer: The hollow hull encloses a large volume of air, so the ship's average density is less than water's.
Answer: Density = mass/volume = 2500/0.5 = 5000 kg/m^3.
Answer: The ship floats because the upthrust equals the weight of the displaced water, which balances its own weight.
Answer: k = F/x = 40/0.08 = 500 N/m.
Answer: Y = stress/strain = 200/0.002 = 100,000 Pa.