(D) The time period of a simple pendulum is given by $T = 2\pi \sqrt{\frac{L}{g}}$.Taking the derivative,the fractional change in time period is $\fra

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$\frac{1}{2}\alpha t \times 86400$

(D) The time period of a simple pendulum is given by $T = 2\pi \sqrt{\frac{L}{g}}$.Taking the derivative,the fractional change in time period is $\frac{\Delta T}{T} = \frac{1}{2} \frac{\Delta L}{L}$.Since $\Delta L = L \alpha \Delta t$,we have $\frac{\Delta L}{L} = \alpha \Delta t$.Therefore,the fractional change in time is $\frac{\Delta T}{T} = \frac{1}{2} \alpha t$.This represents the time lost per unit time.To find the time lost per day,we multiply by the total number of seconds in a day,which is $24 \times 60 \times 60 = 86400 \, s$.Thus,the time loss per day is $\Delta T_{day} = \frac{1}{2} \alpha t \times 86400$ seconds.

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

Difficult

$A$ pendulum clock keeps correct time at $0\,^{\circ}C$. The thermal coefficient of linear expansion of the material of the pendulum is $\alpha$. If the temperature rises to $t\,^{\circ}C$,then the clock loses per day by (in seconds):

A
$\alpha t$
B
$\frac{1}{2}\alpha t$
C
$\alpha t \times 86400$
D
$\frac{1}{2}\alpha t \times 86400$

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1 Thermal Properties of Matter: Thermometry, Expansion and Calorimetry

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Thermal Expansion for Solid

Two rods $A$ and $B$ of identical dimensions are at temperature $30\,^{\circ}C$. If $A$ is heated up to $180\,^{\circ}C$ and $B$ up to $T\,^{\circ}C$,then the new lengths are the same. If the ratio of the coefficients of linear expansion of $A$ and $B$ is $4:3$,then the value of $T$ is........$^{\circ}C$.

MediumJEE MAIN 2019

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The coefficient of apparent expansion of mercury in a glass vessel is $153 \times 10^{-6} /{ }^{\circ} C$ and in a steel vessel is $144 \times 10^{-6} /{ }^{\circ} C$. If $\alpha$ for steel is $12 \times 10^{-6} /{ }^{\circ} C$, then the coefficient of linear expansion $\alpha$ for glass is:

MediumAIIMS 2019

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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:

EasyAP EAMCET 2022

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$A$ copper rod of $88\; cm$ and an aluminum rod of unknown length have their increase in length independent of increase in temperature. The length of the aluminum rod is....$cm$.
$(\alpha_{Cu} = 1.7 \times 10^{-5}\; K^{-1}$ and $\alpha_{Al} = 2.2 \times 10^{-5}\; K^{-1})$

MediumNEET 2019

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An equilateral triangle $ABC$ is formed by two copper rods $AB$ and $BC$ and one aluminium rod $AC$. The system is heated such that the temperature of each rod increases by $\Delta T$. Find the change in the angle $\angle ABC$. (Coefficient of linear expansion for copper is $\alpha_1$ and for aluminium is $\alpha_2$).

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$A$ pendulum clock keeps correct time at $0\,^{\circ}C$. The thermal coefficient of linear expansion of the material of the pendulum is $\alpha$. If the temperature rises to $t\,^{\circ}C$,then the clock loses per day by (in seconds):

Similar Questions

The coefficient of apparent expansion of mercury in a glass vessel is $153 \times 10^{-6} /{ }^{\circ} C$ and in a steel vessel is $144 \times 10^{-6} /{ }^{\circ} C$. If $\alpha$ for steel is $12 \times 10^{-6} /{ }^{\circ} C$, then the coefficient of linear expansion $\alpha$ for glass is:

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:

$A$ copper rod of $88\; cm$ and an aluminum rod of unknown length have their increase in length independent of increase in temperature. The length of the aluminum rod is....$cm$.$(\alpha_{Cu} = 1.7 \times 10^{-5}\; K^{-1}$ and $\alpha_{Al} = 2.2 \times 10^{-5}\; K^{-1})$

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$A$ copper rod of $88\; cm$ and an aluminum rod of unknown length have their increase in length independent of increase in temperature. The length of the aluminum rod is....$cm$.
$(\alpha_{Cu} = 1.7 \times 10^{-5}\; K^{-1}$ and $\alpha_{Al} = 2.2 \times 10^{-5}\; K^{-1})$