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10-1.Thermometry, Thermal Expansion and Calorimetry
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A seconds pendulum clock has a steel wire. The clock shows correct time at $25^{\circ} C$. .......... $s$ time does the clock lose or gain, in one week, when the temperature is increased to $35^{\circ} C$ ? $\left(\alpha_{\text {toel }}=1.2 \times 10^{-5} /{ }^{\circ} C \right)$
A
$321.5$
B
$3.828$
C
$82.35$
D
$36.28$
Solution
(d)
$\frac{\Delta T}{T}=\frac{1}{2} \alpha \theta$
$=\frac{1}{2} \times 1.2 \times 10^{-5} \times 10$
$\frac{\Delta T}{T}=6.0 \times 10^{-5}$
$\text { Hence time lost in } 1 \text { week }=6.0 \times 10^{-5} \times T$
$=6.0 \times 10^{-5} \times 7 \times 24 \times 3600$
$=36.28 \,s$
Standard 11
Physics
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