Ncert Exercises & Solutions

11.3 The electrical resistance in ohms of a certain thermometer varies with temperature according to the approximate law:

$R = R_{o} [1 + α (T - T_{o})]$

The resistance is 101.6 Ω at the triple-point of water 273.16 K, and 165.5 Ω at the normal melting point of lead (600.5 K). What is the temperature when the resistance is 123.4 Ω?

At the triple point of water, T $_{0}$ = 273.15 K

The resistance, R $_{0}$ = 101.6 Ω

At normal melting point of lead, T = 600.5 K

The resistance, R = 165.5 Ω

Then,

$R = R_{0} [1 + α (T - T_{0})] 165.5 = 101.6 [1 + α (600.5 - 273.15)] ∴ α = 1.92 \times 10^{- 3} K^{- 1}$

For resistance, R $_{1}$ = 123.4 Ω

$123.4 = 101.6 [1 + 1.92 \times 10^{- 3} (T - 273.15)] 1.214 = 1 + 1.92 \times 10^{- 3} (T - 273.15) \frac{0.214}{1.92 \times 10^{- 3}} = T - 273.15 ∴ T = 384.61 K$

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