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10-1.Thermometry, Thermal Expansion and Calorimetry
hard
A pendulum clock (fitted with a small heavy bob that is connected with a metal rod) is $5\, seconds$ fast each day at a temperature of $15\,^oC$ and $10\,seconds$ slow at a temperature of $30\,^oC$. The temperature at which it is designed to give correct time, is ........ $^oC$
A
$18$
B
$20$
C
$24$
D
$25$
Solution
$\Delta \mathrm{t}=\frac{1}{2} \alpha \Delta \theta \mathrm{t}$
$5=\frac{1}{2} \alpha[\theta-15] \times 1 \mathrm{Day}$ $…(1)$
$10=\frac{1}{2} \alpha[30-\theta] \times 1 \mathrm{Day}$ $…(2)$
$\mathrm{Eq}-(1) \div(2)$
$\frac{1}{2}=\frac{\theta-15}{30-\theta} \Rightarrow 30-\theta=2 \theta-30 \Rightarrow \theta=20^{\circ} \mathrm{C}$
Standard 11
Physics
