Gujarati
10-1.Thermometry, Thermal Expansion and Calorimetry
easy

The length of a metallic rod is $5m$ at $0°C$ and becomes $  5.01\, m$, on heating upto $100°C$. The linear expansion of the metal will be

A

$2.33 \times 10^{-5} {°C^{-1}}$

B

$6.0 \times 10^{-5} {°C^{-1}}$

C

$4.0 \times 10^{-5} {°C^{-1}}$

D

$2.0 \times 10^{-5}{°C^{-1}}$

Solution

(d) $\alpha = \frac{{\Delta L}}{{{L_0} \times \Delta \theta }} = \frac{{0.01}}{{5 \times 100}} = 2 \times {10^{ – 5}}{°C^{-1}}$

Standard 11
Physics

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