The rod given in the figure is made up of two different materials. Both have square cross sections of 3 mm side. The resistivity of the first material is 4 xx 10^(-3) Omega.m and it is 25 cm long while second material has resistivity of 5 xx 10^(-3) Omega.m and is of 70 cm long. What is the resistivity of rod between its ends?

Question

The rod given in the figure is made up of two different materials. Both have square cross sections of 3 mm side. The resistivity of the first material is

4 \times 10^{- 3} Ω . m

and it is 25 cm long while second material has resistivity of

5 \times 10^{- 3} Ω . m

and is of 70 cm long. What is the resistivity of rod between its ends?

Answer 1

Side of square cross section ,

a = 3 m m = 3 \times 10^{- 3} m

Cross section ,

A = a^{2} = (3 \times 10^{- 3})^{2} = 9 \times 10^{- 6} m^{2}

Resistivity of first material ,

ρ_{1} = 4 \times 10^{- 3} Ω . m

Resistivity of second material ,

ρ_{2} = 5 \times 10^{- 3} Ω . m

length of first material ,

L_{1} = 25 c m = 0.25 m

length of second material ,

L_{2} = 70 c m = 0.7 m

To find = Resistivity of rod
We can calculate the resistance due to the first material using :

R_{1} = \frac{ρ_{1} L_{1}}{A} R_{1} = \frac{4 \times 10^{- 3} \times 0.25}{9 \times 10^{- 6}} R_{1} = 111.11 Ω

Similarly , we can calculate the resistance due to second material using :

R_{2} = \frac{ρ_{2} L_{2}}{A} R_{2} = \frac{5 \times 10^{- 3} \times 0.7}{9 \times 10^{- 6}} R_{2} = 388.89 Ω

Since they are connected in series , the total resistance is :

R = R_{1} + R_{2} R = 111.11 + 388.89 R = 500 Ω

Hence, the total resistance of the rod is

500 Ω

Answers (1)

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