Question

1) A METAL ROD AB OF LENGTH 80cm AND MASS 12gm CAN BE BALANCED AT A DISTANCE OF 35cm FROM THE END A. CALCULATE THE POSITION OF THE SUPPORT FOR THE SAME ROD FOR IT TO BE IN EQUILIBRIUM IF A WEIGHT OF 40gf IS SUSPENDED ON IT AT A DISTANCE OF 30cm FROM THE END B​

Answer 1

Answer:

Let a be the cross-sectional area of the rod and k be the thermal conductivity of the rod's material. The rates of heat flow from the point p towards the end A and the end B, are given by

d(Qa)/dt = kA(Tp-Ta)/λx

= 400ka/λx

d(Qb)/dt = kA(Tp-Tb)/(10x-λx)

= 300ka/(10x-λx)

Let the rate of melting of ice be dmi/dt

and the rate of evaporation of water be dmw/dt. The rates of heat required for the ice and the water are

dQA/dt=(dm/dt)Li,

dQB/dt=(dmw/dt)Lw,

where Li=80 cal/gram and Lw=540 cal/g are the latent heats of fusion and evaporation, respectively. Use

dmi/dt=dmw/dt and simplify to get,

(400/λ)(10-λ/300) = 80/540

which gives,

λ=9.