(D) Given,diameter of sphere $= D$. Initial surface area,$A = 4 \pi R^2 = 4 \pi (D/2)^2 = \pi D^2$. Surface area after heating by temperature $\delta

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$\pi D^2 \cdot \alpha \delta T(\alpha \delta T+2)$

(D) Given,diameter of sphere $= D$. Initial surface area,$A = 4 \pi R^2 = 4 \pi (D/2)^2 = \pi D^2$. Surface area after heating by temperature $\delta T$ is $A' = \pi (D')^2$,where $D'$ is the new diameter. From the equation of linear expansion,$D' = D(1 + \alpha \delta T)$. Substituting $D'$ into the expression for $A'$,we get $A' = \pi [D(1 + \alpha \delta T)]^2 = \pi D^2 (1 + 2\alpha \delta T + \alpha^2 \delta T^2)$. The change in surface area $\Delta A = A' - A = \pi D^2 (1 + 2\alpha \delta T + \alpha^2 \delta T^2) - \pi D^2$. $\Delta A = \pi D^2 (2\alpha \delta T + \alpha^2 \delta T^2) = \pi D^2 \alpha \delta T (2 + \alpha \delta T)$.

Class 11 Physics 10-1.Thermometry, Thermal Expansion and Calorimetry Thermal Expansion for Solid

Easy

An iron sphere having diameter $D$ and mass $M$ is immersed in hot water so that the temperature of the sphere increases by $\delta T$. If $\alpha$ is the coefficient of linear expansion of the iron,then the change in the surface area of the sphere is:

AP EAMCET 2022 All AP EAMCET

A
$\pi D^2 \cdot \alpha \delta T(\alpha \delta T-4)$
B
$\pi D^2 \cdot \alpha \cdot \delta T(\alpha \delta T+4)$
C
$\pi D^2 \cdot \alpha \cdot \delta T(\alpha \delta T-2)$
D
$\pi D^2 \cdot \alpha \delta T(\alpha \delta T+2)$

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10-1.Thermometry, Thermal Expansion and Calorimetry

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Thermal Expansion for Solid

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$A$ pendulum clock shows the correct time at $20^{\circ}C$. During summer days,when the temperature rises to $40^{\circ}C$,what will be the time change in the clock in one day (in $.64$)? (Given: $\alpha = 10^{-5} {^{\circ}C}^{-1}$)

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Two holes of unequal diameters $d_1$ and $d_2$ $(d_1 > d_2)$ are cut in a metal sheet. If the sheet is heated,what happens to the diameters?

Easy

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$A$ student records the initial length $l$,change in temperature $\Delta T$ and change in length $\Delta l$ of a rod as follows:

$S$.No. $l (m)$ $\Delta T (^{\circ}C)$ $\Delta l (m)$

$(1)$ $2$ $10$ $4 \times 10^{-4}$

$(2)$ $1$ $10$ $4 \times 10^{-4}$

$(3)$ $2$ $20$ $2 \times 10^{-4}$

$(4)$ $3$ $10$ $6 \times 10^{-4}$

If the first observation is correct,what can you say about observations $(2)$,$(3)$ and $(4)$?

$S$.No.	$l (m)$	$\Delta T (^{\circ}C)$	$\Delta l (m)$
$(1)$	$2$	$10$	$4 \times 10^{-4}$
$(2)$	$1$	$10$	$4 \times 10^{-4}$
$(3)$	$2$	$20$	$2 \times 10^{-4}$
$(4)$	$3$	$10$	$6 \times 10^{-4}$

Medium

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$A$ solid metallic cube having a total surface area of $24 \; m^{2}$ is uniformly heated. If its temperature is increased by $10 \; ^{\circ}C$,calculate the increase in volume of the cube. (Given: $\alpha = 5.0 \times 10^{-4} \; ^{\circ}C^{-1}$)

MediumJEE MAIN 2022

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Three rods of equal lengths are joined to form an equilateral triangle $ABC$. $D$ is the mid-point of $AB$. The coefficient of linear expansion is $\alpha_1$ for the material of rod $AB$ and $\alpha_2$ for the material of rods $AC$ and $BC$. If the distance $DC$ remains constant for small changes in temperature, then:

DifficultTS EAMCET 2010

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An iron sphere having diameter $D$ and mass $M$ is immersed in hot water so that the temperature of the sphere increases by $\delta T$. If $\alpha$ is the coefficient of linear expansion of the iron,then the change in the surface area of the sphere is:

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$A$ pendulum clock shows the correct time at $20^{\circ}C$. During summer days,when the temperature rises to $40^{\circ}C$,what will be the time change in the clock in one day (in $.64$)? (Given: $\alpha = 10^{-5} {^{\circ}C}^{-1}$)

Two holes of unequal diameters $d_1$ and $d_2$ $(d_1 > d_2)$ are cut in a metal sheet. If the sheet is heated,what happens to the diameters?

$A$ solid metallic cube having a total surface area of $24 \; m^{2}$ is uniformly heated. If its temperature is increased by $10 \; ^{\circ}C$,calculate the increase in volume of the cube. (Given: $\alpha = 5.0 \times 10^{-4} \; ^{\circ}C^{-1}$)

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