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 = $$1.70 × 10^{–5} K^{–1}.$$

Asked by Pragya Singh | 1 year ago |  114

##### Solution :-

Diameter of hole ( D1) = 4.24 cm

Initial Temperature, T1 = 27.0°C = 27+ 273 = 300 K

Final temperature, T2 =227°C = 227+ 273 = 500K

Let the diameter of the hole at the final temperature be D2

Coefficient of linear expansion of copper, α= 1.70 × 10–5 K–1

Initial area of hole (Ao)= πr² =π ($$\dfrac{4.24}{2}$$

coefficient of superficial expansion (β)

= 2× coefficient of linear expansion of copper (α)

= 2 × 1.7 × 10-5 = 3.4 × 10-5

Using the formula,

A = Ao( 1 + β∆T)

A = π ($$\dfrac{4.24}{2}$$)² [1 + 3.4 × 10-5 × (500-300)]

= π ($$\dfrac{4.24}{2}$$)² [ 1 + 3.4 × 10-5 × 200]

= π ($$\dfrac{4.24}{2}$$)² [ 1 + 6.8 × 10-3 ]

= π ($$\dfrac{4.24}{2}$$)²  [ 1 + 0.0068]

=  π ($$\dfrac{4.24}{2}$$)² (1.0068)

$$\dfrac{πD_2^²}{4}$$ = π ($$\dfrac{4.24}{2}$$)² ( 1.0068 )

D2²= 17.97 (1.0068)

D2²=  18.092

= 4.253

D2 = 4.253 cm

Change in diameter (∆D) = D2 – D1

= 4.253 – 4.24

= 0.013 cm

Answered by Pragya Singh | 1 year ago

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