Doubt by ARYA DUTTA

Answers

Answer by Shreya

 we have to find temperature of a collective drop when initially 1gm and 2gm drops travel with velocities 10cm/s and 15cm/s respectively.applying law of conservation of linear momentum,
before coagulation = after coagulation
1 × 10 + 2 × 15 = (1 + 2) × u
=u = 40/3 cm/s
kinetic energy before coagulation of drops
= 1/2 × 1 × (10)² + 1/2 × 2 × (15)²
= 50 + 225 = 275 erg [we know, 1g.cm² = 1erg ]
kinetic energy after coagulation of drops = 1/2 (1 + 2) × (40/3)²
                                                           = 1/2 × 3 × 1600/9 = 800/3 erg
so, energy converted into heat = kinetic energy before coagulation of drops - kinetic energy after coagulation of drops
                   = 275 - 800/3 = 25/3 erg
                   = 25/3 × 10^-7 J [ we know, 1 erg = 10^-7 ]
we know, heat = msT where , m is mass , s is specific heat capacity and T is temperature.
S = 4.2 J/Kg/°C ,
m = 3g = 3 × 10^-3 kg
so, 25/3 × 10^-7 = 3 × 10^-3 × 4.2 × T 
T = 25/(9 × 4.2) = 6.6 × 10^-5 °C