Unit-5 Pressure and Deformation in Solids
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Unit-5 Pressure and Deformation in Solids
Answer: Mercury's high density means atmospheric pressure supports only about 76 cm of it, unlike water's 10 m.
Answer: Strain = change in length / original length = 0.01/2 = 0.005.
Answer: Within the elastic limit the deformation is elastic and fully reversible.
Answer: P = F/A = 60/3 = 20 Pa.
Answer: h = P/(rho g) = 50,000/(1000 x 10) = 5 m.
Answer: Relative density = density of substance / density of water = 13.6.
Answer: P = h rho g = 2 x 1200 x 10 = 24,000 Pa.
Answer: At very high altitudes the air column above is thin, so atmospheric pressure is lower.
Answer: F = P x A = 1500 x 0.02 = 30 N.
Answer: x = F/k = 0.5/100 = 0.005 m.
Answer: Pascal's law: the applied pressure is transmitted equally and undiminished in all directions.
Answer: Density = mass/volume = 120/0.04 = 3000 kg/m^3.
Answer: Ice (about 920 kg/m^3) is less dense than water (1000 kg/m^3), so it floats.
Answer: k = F/x = 90/0.3 = 300 N/m.
Answer: F = k x = 25 x 0.4 = 10 N.
Answer: When upthrust exceeds the weight, the net force is upward and the balloon rises (buoyancy in air).
Answer: Relative density = 800/1000 = 0.8, so it is lighter than water.
Answer: In a spring balance the extension of the spring is proportional to the load, following Hooke's law.
Answer: Stress = Y x strain = 10^8 x 0.005 = 5 x 10^5 Pa.
Answer: h_mercury = h_water x (density_water/density_mercury) = 5 x 1000/13,600 = 0.368 m = 36.8 cm.