A hole is drilled in a metal sheet. At 27^ C, the diameter of the hole is 5 cm. When the sheet is heated to 177^ C, the change in the diameter of the hole is d. The value of d will be, if the…

Given: Initial diameter of the hole is d0 = 5 cm, coefficient of linear expansion is = 1.6 10^-5/^ C, and temperature change is T = 177^ C - 27^ C = 150^ C. Find: The change in diameter d. For a hole in a metal sheet, the diameter expands as if it were made of the same material. Hence, d = d0 T Substituting the given values, d = 5 1.6 10^-5 150 = 12 10^-3 cm Therefore, d = 0.012 cm the solution states the final answer as 12, but the working gives 12 10^-3 cm = 0.012 cm. Hence the numerical value from the extracted solution working is 0.012.

A hole is drilled in a metal sheet. At 27^ C, the diameter of the… — Solution | JEE Physics

NVAEasyJEE 2023Heat, Temperature & Thermal Expansion

A hole is drilled in a metal sheet. At $27^\circ \text{C}$ , the diameter of the hole is $5 \, \text{cm}$ . When the sheet is heated to $177^\circ \text{C}$ , the change in the diameter of the hole is $\Delta d$ . The value of $\Delta d$ will be, if the coefficient of linear expansion of the metal is $1.6 \times 10^{-5}/^\circ \text{C}$ :

Answer

Correct answer:12

Step-by-step solution

Standard Method

Given: Initial diameter of the hole is $d_0 = 5 \, \text{cm}$ , coefficient of linear expansion is $\alpha = 1.6 \times 10^{-5}/^\circ \text{C}$ , and temperature change is $\Delta T = 177^\circ \text{C} - 27^\circ \text{C} = 150^\circ \text{C}$ .

Find: The change in diameter $\Delta d$ .

For a hole in a metal sheet, the diameter expands as if it were made of the same material. Hence,

\Delta d = d_0 \alpha \Delta T

Substituting the given values,

\Delta d = 5 \times 1.6 \times 10^{-5} \times 150

= 12 \times 10^{-3} \, \text{cm}

Therefore,

\Delta d = 0.012 \, \text{cm}

the solution states the final answer as $12$ , but the working gives $12 \times 10^{-3} \, \text{cm} = 0.012 \, \text{cm}$ . Hence the numerical value from the extracted solution working is $0.012$ .

Common mistakes

Using the idea that the hole shrinks on heating is incorrect. A hole in an expanding sheet also expands as if the missing material were present. Use the same linear expansion relation for the hole diameter.
Forgetting to compute the temperature rise correctly leads to error. The relevant quantity is $\Delta T = 177 - 27 = 150^\circ \text{C}$ , not the final temperature alone.
Misplacing the decimal in $1.6 \times 10^{-5}$ gives a wrong order of magnitude. Multiply carefully and convert $12 \times 10^{-3}$ to $0.012$ , not $12$ or $0.12$ .

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