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physics ncert solutions class 11th Physics Class 11th Thermal Properties of Matter

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.

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Payal Gupta | Contributor-Level 10

4 months ago

11.8 Initial temperature, $T_{1}$ = 27.0 °C, Initial diameter of the hole, $d_{1}$ = 4.24 cm

Final temperature, $T_{2}$ = 227.0 °C, Final diameter of the hole $= d_{2}$

Coefficient of linear expansion of copper, $α_{c o p p e r}$ = 1.70 $\times 10^{- 5}$ K–1

We know

$\frac{C h a n g e i n a r e a (? A)}{O r i g i n a l a r e a (A)}$ = $β ?$ T where $β$ is the coefficient of superficial expansion, $β = 2 α_{c o p p e r}$

$\frac{(π \frac{d_{2}^{2}}{4} - π \frac{d_{1}^{2}}{4})}{π \frac{d_{1}^{2}}{4}}$ = $2 α_{c o p p e r} ?$ T

$\frac{d_{2}^{2} - d_{1}^{2}}{d_{1}^{2}}$ = $2 α_{c o p p e r} ?$ T

$\frac{d_{2}^{2}}{d_{1}^{2}}$ - 1= $2 α_{c o p p e r} (T_{2} - T) =$ 2 $\times$ 1.70 $\times 10^{- 5}$ (227-27) = 6.8 $\times 10^{- 3}$

$d_{2}^{2} = 1.0068 \times 4.24 \times 4.24$

$d_{2}$ = 4.2544 cm

So change in diameter = 4.2544 – 4.24 = 0.01439 cm

Diameter increase by 1.44 $\times 10^{- 2}$ cm

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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.

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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.

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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.