(A) Let the initial diameter be $d_0 = 5\,cm$ at temperature $T_1 = 27^{\circ}C$.When the sheet is heated to $T_2 = 177^{\circ}C$,the change in temper

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(A) Let the initial diameter be $d_0 = 5\,cm$ at temperature $T_1 = 27^{\circ}C$.When the sheet is heated to $T_2 = 177^{\circ}C$,the change in temperature is $\Delta T = T_2 - T_1 = 177^{\circ}C - 27^{\circ}C = 150^{\circ}C$.The change in diameter $\Delta d$ is given by the formula for linear expansion: $\Delta d = d_0 \alpha \Delta T$.Substituting the given values: $\Delta d = 5\,cm \times (1.6 \times 10^{-5} /^{\circ}C) \times 150^{\circ}C$.$\Delta d = 5 \times 1.6 \times 150 \times 10^{-5}\,cm$.$\Delta d = 1200 \times 10^{-5}\,cm = 12 \times 10^{-3}\,cm$.Comparing this with $d \times 10^{-3}\,cm$,we get $d = 12$.

Class 11 Physics 1 Thermal Properties of Matter: Thermometry, Expansion and Calorimetry Thermal Expansion for Solid

Medium

$A$ hole is drilled in a metal sheet. At $27^{\circ}C$,the diameter of the hole is $5\,cm$. When the sheet is heated to $177^{\circ}C$,the change in the diameter of the hole is $d \times 10^{-3}\,cm$. The value of $d$ will be $...........$ if the coefficient of linear expansion of the metal is $1.6 \times 10^{-5} /^{\circ}C$.

JEE MAIN 2023 All JEE MAIN

A
$12$
B
$11$
C
$10$
D
$9$

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1 Thermal Properties of Matter: Thermometry, Expansion and Calorimetry

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Thermal Expansion for Solid

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