Question

JEE ADVANCED PHYSICS QUESTION SEPTEMBER 2020

Answer 1

A spherical bubble inside water has radius R. take the pressure inside the bubble and the water pressure p₀. The bubble now gets compressed radially in an Adiabatic manner so that its radius becomes (R - a). for a << R the magnitude of work done in the process is given by, (4πp₀Ra²)X where X is a constant.

To find : The value of X is ...

solution : here given work done in the process is given by, W = (4πp₀R²a)X

we know, workdone , W = ∆P_avg dV

here V = surface area × thickness

= 4πR² a

also we know, for small change ∆P_avg = arithmetic mean = (0 + P)/2 = dP/2

so, W = dP/2 × 4πR²a

= 2πR²a dP.......(1)

from adiabatic process, PV^y = constant

so, dP = -γ P/V dV = -γ P₀/V × 4πR²a , putting it in equation (1),

so, W = -2πR²a × 1/2 × γP₀/4/3πR³ × 4πR²a

= (- 4πR²a²P₀)3γ/2

on comparing to given equation we get,

X = 3γ/2 = 3 × 41/30/2 = 41/20 = 2.05

Therefore the value of X is 2.05

Answer 2

Transpiration is the process of water movement through a plant and its evaporation from aerial parts, such as leaves, stems and flowers. Water is necessary for plants but only a small amount of water taken up by the roots is used for growth and metabolism. The remaining 97–99.5% is lost by transpiration and guttation.