Question

Two wires a and b with lengths in the ratio of 3: 1, diameters in the ratio of 1:2 and resistivity in the ratio of 1:20 are joined in parallel with a source of e.M.F. 2v. Ratio of the rates of heat production of wire a with respect to wire b is

Answer 1

R1/R2 = 3/5

So H1/H2 = 5/3

Answer 2

Answer:

The wire's heat-production ratio is 5:3 (length=3:1=L1/L2).

Explanation:

Diameter=1:2=d₁/d₂

Resistivity=1:20=P₁/P2

2 volts electromotive force

Q₁=V²t/R₁

Q₂=V²t/R₂

Q₁/Q₂ heat production=R₂/R₁=(L₂/L₁)ₓ(P₂/P₁)ₓ(A₁/A₂)

         =(L₂/L₁)ₓ(P₂/P₁)ₓ(d₁/d₂)

         =1/3ₓ20ₓ1/4

          =5/3

Answer=5:3

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