Question

3.In the given figure, the potential difference 'V' hasbeen plotted against current T for a metallicconductor at two different temperatures 25°C and100°C respectively. The temperature co-efficient ofresistivity of the wire isT2=100°CT, =25°C°30960(1) 16x10-27°C(2) 8x10-27°C(3) 4x10-27°C(4) 2x10-27°C)​

Answer 1

Coefficient of resistivity at 100°C = 4 * 10^-27°C

Explanation:

Given: T2= 100°C

T1 = 25°C

Find: Temperature coefficient of the wire.

Solution:

According to Ohm's law  R = V/I

where R = resistance, V = Voltage and I = Current.

Also R  is directly proportional to temperature T, and inversely proportional to the slope.

R1 ∝ T1  

R2 ∝ T2  

So we get T2/T1 = R2/R1

100/25 = R2/R1

Therefore R2 = R1 * 4

Hence Option 3 is correct.

Coefficient of resistivity at 100°C = 4 * 10^-27°C

Answer 2

★ Question Given :

• In the given figure, the potential difference 'V' hasbeen plotted against current T for a metallicconductor at two different temperatures 25°C and100°C respectively. The temperature co-efficient ofresistivity of the wire isT2=100°CT, =25°C°30960

• ⟼ (1) 16x10-27°C

• ⟼ (2) 8x10-27°C

• ⟼ (3) 4x10-27°C

• ⟼ (4) 2x10-27°C)

★ Required Solution :

✰ Formula Used Here :

• ➥ T2 / T1 = R2 / R1

✰ Value Given to us :

• ➥ T2 = 100°C

• ➥ T1 = 25°C

✰ Putting Value in Formula :

• ➥ T2 / T1 = R2 / R1

• ➥ 100 / 25 = R2 / R1

• ➥ R2 = R1 × 4

• ➥ 4x10-27°C

★ Therefore :

• ➸ Coefficient of Resistivity at 100°C is 4x10-27°C

• ➸ Option (3) is correct

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