Question

At temperature 23°C the resistance of a wire is 100 Q. The temperature at which resistance becomes 104 Q is nearly. (Given, temperature coefficient of resistance of material is 1.5 x 10-4 °C-1)

A)1027°C

B)827°C

C)290°C

D)290 K
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Answer 1

Answer:

The temperature at which resistance become 104 Q is 279.41 °C.

So the correct option is nearly (C) 290°C.

Explanation:

We know the resistivity of  a material is related to temperature as given below  

    ρ = ρ₀[ 1 + α (T - T₀)] ------------(i)

where  ρ = resistivity at T temp.      

      ρ₀ = resistivity at a  given temp T₀      

       α = temperature coefficient of resistance material

also  

ρ = $\frac{1}{R}$        

   (R = resistance of material)

Given :            R = 104Q      R₀ = 100Q                  

           T₀ = 23°C    α= 1.5 × 10⁻⁴ °C⁻¹

Calculation:$\frac{1}{R}=\frac{1}{R_{0} } [ 1+\alpha (T-T_{0} )]$

$\frac{1}{104} =\frac{1}{100} [1+ (1.5 10^{-4})(T-23)]$

$\frac{100}{104} = [ 1+ (1.5 * 10^{-4} )(T-23)]$

$\frac{-4}{104} = 1.5*10^{-4} (T- 23)$

$\frac{4 10^{4} }{104*1.5} =T-23$

$T = 256.41 +23$

$T= 279.41 ^{0} C$

Hence temperature at which resistance becomes 104Q  is nearly 290 °C.