Two rods are joined end to end, as shown. Both have a cross-sectional area of 0.01 cm2. Each is 1 - 6 meter long. One rod is of copper with a resistivity of 1.7 × 10 ohm-centimeter, the other is of iron - 5 with a resistivity of 10 ohm-centimeter. How much voltage is required to produce a current of 1 ampere in the rods? (a) 0.00145 V (b) 0.0145 V (c) 1.7 × 10 - 6 V (d) 0.117 V

Answer: (d). R = ρ l/A with l = 100 cm, A = 0.01 cm², so l/A = 10⁴ cm⁻¹. RCu = 1.7×10⁻⁶×10⁴ = 0.017 Ω, RFe = 10⁻⁵×10⁴ = 0.1 Ω. In series: R = 0.117 Ω. V = IR = 1×0.117 = 0.117 V. The answer is (d).

Two rods are joined end to end, as shown. Both have a cross-sectional… | NEET 2013

Q58NEET 2013Medium

Ohm's Law, Resistivity & Temperature

Two rods are joined end to end, as shown. Both have a cross-sectional area of 0.01 cm2. Each is 1 $-$ 6

meter long. One rod is of copper with a resistivity of 1.7 $\times$ 10 ohm-centimeter, the other is of iron $-$ 5

with a resistivity of 10 ohm-centimeter.

How much voltage is required to produce a current of 1 ampere in the rods?

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Answer: (d)

$R = \rho l/A$ with $l = 100$ cm, $A = 0.01\ \text{cm}^2$ , so $l/A = 10^4\ \text{cm}^{-1}$ .
$R_{Cu} = 1.7\times10^{-6}\times10^4 = 0.017\ \Omega$ , $R_{Fe} = 10^{-5}\times10^4 = 0.1\ \Omega$ .
In series: $R = 0.117\ \Omega$ .
$V = IR = 1\times0.117 = 0.117$ V. The answer is (d).

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