A hole is drilled in a copper sheet. The diameter of hole is 4.24 cm at ${27.0}^{o}$ C. Diameter of the hole when it is heated to $35^{o}$ C is

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Solution

Coefficient of thermal expansion of copper $α = 17 \times 10^{- 6} /^{o} C$
Given : $T_{1} = 27^{o} C$ $T_{2} = 35^{o} C$
Diameter of hole $D_{2} = D_{1} (1 + α (T_{2} - T_{1}))$
Or $D_{2} = 4.24 \times (1 + 17 \times 10^{- 6} \times (35 - 27))$
$⟹ D_{2} = 4.24057 c m$

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^{\circ}

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^{\circ}

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^{- 5}

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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A hole is drilled in a copper sheet. The diameter of hole is 4.24 cm at 27.0oC. Diameter of the hole when it is heated to 35oC is

Coefficient of thermal expansion of copper α=17×10−6/oCGiven : T1=27oC T2=35oCDiameter of hole D2=D1(1+α(T2−T1))Or D2=4.24×(1+17×10−6×(35−27))⟹ D2=4.24057 cm

A hole is drilled in a copper sheet. The diameter of hole is 4.24 cm at ${27.0}^{o}$ C. Diameter of the hole when it is heated to $35^{o}$ C is

Coefficient of thermal expansion of copper $α = 17 \times 10^{- 6} /^{o} C$
Given : $T_{1} = 27^{o} C$ $T_{2} = 35^{o} C$
Diameter of hole $D_{2} = D_{1} (1 + α (T_{2} - T_{1}))$
Or $D_{2} = 4.24 \times (1 + 17 \times 10^{- 6} \times (35 - 27))$
$⟹ D_{2} = 4.24057 c m$