A brass rod of length $2 ext{ m}$ is heated from $20^\circ ext{C}$ to $100^\circ ext{C}$. If $\alpha = 2 imes 10^{-

Question

A brass rod of length $2	ext{ m}$ is heated from $20^\circ	ext{C}$ to $100^\circ	ext{C}$. If $\alpha = 2 	imes 10^{-5} 	ext{ K}^{-1}$, what is the increase in length?

Accepted Answer

Increase in length $\Delta L = L \alpha \Delta T = 2 	imes (2 	imes 10^{-5}) 	imes (100 - 20) = 4 	imes 10^{-5} 	imes 80 = 320 	imes 10^{-5} 	ext{ m} = 3.2 	imes 10^{-3} 	ext{ m} = 3.2 	ext{ mm}$. Hence, option A is correct.

Thermal Properties of Matter

❓ Sibling Questions in this Chapter