Ncert Exercises & Solutions

11.8 A hole is drilled in a copper sheet. The diameter of the hole is 4.24 cm at 27.0 °C. What is the change in the diameter of the hole when the sheet is heated to 227 °C? Coefficient of linear expansion of copper = 1.70 × $10^{- 5} K^{- 1}$ .

Initial temperature, $T_{1}$ = 27.0°C

The diameter of the hole at T $_{1}$ , d $_{1}$ = 4.24 cm

Final temperature, T $_{2}$ = 227°C

The diameter of the hole at T $_{2}$ = d $_{2}$

Co-efficient of linear expansion of copper, $α_{Cu} = 1.70 \times 10^{- 5} K^{- 1}$

Co-efficient of areal expansion, $β = 2 α$

For co-efficient of areal expansion β, and change in temperature ΔT,

$\frac{Change in area (∆ A)}{Oringnal area (A)} = β ∆ T \frac{(π \frac{d_{2}^{2}}{4} - π \frac{d_{1}^{2}}{4})}{(π \frac{d_{1}^{2}}{4})} = 2 α ∆ T ∴ \frac{d_{2}^{2} - d_{1}^{2}}{d_{1}^{2}} = 2 α ∆ T \frac{d_{2}^{2}}{d_{1}^{2}} - 1 = 2 α (T_{2} - T_{1}) \frac{d_{2}^{2}}{{(4.24)}^{2}} = 2 α (T_{2} - T_{1}) d_{2}^{2} = 17.98 \times 1.0068 = 18.1 ∴ d_{2} = 5.2544 cm$

Change in diameter = d $_{2}$ – d $_{1}$ = 4.2544 – 4.24 = 0.0144 cm

The diameter increases by 1.44 × 10 $^{- 2}$ cm.

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