Question

Question 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 x 10-5K-1.​

Answer 1

Initial temperature, T1 = 27.0°C

Diameter of the hole at T1, d1 = 4.24 cm

Final temperature, T2 = 227°C

Diameter of the hole at T2 = d2

Co-efficient of linear expansion of copper, αCu= 1.70 × 10–5 K–1

For co-efficient of superficial expansion β,and change in temperature ΔT, we have the relation:

Change in area (∆) / Original area (A) = β∆T

[ (πd22/ 4) – (πd12 / 4) ] / (πd11 / 4) = ∆A / A

∴ ∆A / A = (d22 – d12) / d12

But β = 2α

∴ (d22 – d12) / d12 = 2α∆T

(d22 / d12) – 1 = 2α(T2 – T1)

d22 / 4.242 = 2 × 1.7 × 10-5 (227 – 27) +1

d22 = 17.98 × 1.0068 = 18.1

∴ d2 = 4.2544 cm

Change in diameter = d2 – d1 = 4.2544 – 4.24 = 0.0144 cm

Hence, the diameter increases by 1.44 × 10–2 cm.

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Answer 2

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ɪɴɪᴛɪᴀʟ ᴛᴇᴍᴘᴇʀᴀᴛᴜʀᴇ, ᴛ₁ = 27.0°ᴄ

ᴅɪᴀᴍᴇᴛᴇʀ ᴏꜰ ᴛʜᴇ ʜᴏʟᴇ ᴀᴛ ᴛ₁, ᴅ₁ = 4.24 ᴄᴍ

ꜰɪɴᴀʟ ᴛᴇᴍᴘᴇʀᴀᴛᴜʀᴇ, ᴛ₂ = 227°ᴄ

ᴅɪᴀᴍᴇᴛᴇʀ ᴏꜰ ᴛʜᴇ ʜᴏʟᴇ ᴀᴛ ᴛ₂ = d₂

ᴄᴏ-ᴇꜰꜰɪᴄɪᴇɴᴛ ᴏꜰ ʟɪɴᴇᴀʀ ᴇxᴘᴀɴꜱɪᴏɴ ᴏꜰ ᴄᴏᴘᴘᴇʀ, Αᴄᴜ= 1.70 × 10⁻⁵ ᴋ⁻¹

ꜰᴏʀ ᴄᴏ-ᴇꜰꜰɪᴄɪᴇɴᴛ ᴏꜰ ꜱᴜᴘᴇʀꜰɪᴄɪᴀʟ ᴇxᴘᴀɴꜱɪᴏɴ Β,ᴀɴᴅ ᴄʜᴀɴɢᴇ ɪɴ ᴛᴇᴍᴘᴇʀᴀᴛᴜʀᴇ Δᴛ, ᴡᴇ ʜᴀᴠᴇ ᴛʜᴇ ʀᴇʟᴀᴛɪᴏɴ:

ᴄʜᴀɴɢᴇ ɪɴ ᴀʀᴇᴀ (∆)  /  ᴏʀɪɢɪɴᴀʟ ᴀʀᴇᴀ (ᴀ)  =β∆T

[ (πd₂²/ 4) – (πd₁² / 4) ] / (πd₁¹ / 4)  =  ∆A / A

∴ ∆A / A = (d₂² – d₁²) / d₁²

But β = 2α

∴ (d₂² – d₁²) / d₁² = 2α∆T

(d₂² / d₁²) – 1  =  2α(T₂ – T₁)

d₂² / 4.242 = 2 × 1.7 × 10-5 (227 – 27) +1

d₂² = 17.98 × 1.0068  =  18.1

∴ d₂ = 4.2544 cm

Change in diameter = d₂ – d₂= 4.2544 – 4.24 = 0.0144 cm

Hence, the diameter increases by 1.44 × 10⁻²cm.

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