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A liquid does not wet the sides of a solid, if the angle of contact is
The compressibility of water is 4 × 10-5 per unit atmospheric pressure. The decrease in volume of 100 cm3 of water under a pressure of 100 atmosphere will be
\(K=\frac{1}{B}=\frac{\Delta V/V}{P}\)Here, P = 100 atm,
K = 4 × 10-5 and V = 100 cm3.
Hence, ΔV = 0.4 cm3
A beaker containing a liquid of density ρ moves up with an acceleration a. The pressure due to the liquid at a depth h below the free surface of the liquid is
When a beaker containing a liquid of density ρ moves up with an acceleration a, it will work as a lift moving upward with acceleration a. The effective acceleration due to gravity in lift = (a + g)
∴ Pressure of liquid of height h = h ρ(a + g)
An ice-berg floating partly immersed in sea water of density 1.03 g/cm3. The density of ice is 0.92 g/cm3. The fraction of the total volume of the iceberg above the level of sea water is
Let v be the volume of the ice-berg outside the seawater and V be the total volume of ice-berg. Then as per question
0.92 V= 1.03 (V – v) or v/V = 1 – 0.92/1.03
= 11/103
∴(v/V) × 100 = 11 × 100 / 103 ≅ 11%
The density of ice is x gram/litre and that of water is y gram/litre. What is the change in volume when m gram of ice melts?
Volume of m g of ice = m/x and volume of m g of water = m/y. So change in volume
\(=\frac{m}{y}-\frac{m}{x}=m\left ( \frac{1}{y}-\frac{1}{x} \right )\)
The excess of pressure inside a soap bubble is twice the excess pressure inside a second soap bubble. The volume of the first bubble is n times the volume of the second where n is
Given,\(\frac{4T}{r_{1}}=2\times \frac{4T}{r_{2}}\: r_{2}=2r_{1}\)
\(\frac{4}{3}\pi r_{1}^{3}=n\times \frac{4}{3}\pi r_{2}^{3}=n\times \frac{4}{3}\pi (2r_{1})^{3}\: or\: n=\frac{1}{8}=0.125\)