Question
A pendulum clock loses $$12\,s$$ a day if the temperature is $${40^ \circ }C$$ and gains $$4\,s$$ a day if the temperature is $${20^ \circ }C.$$ The temperature at which the clock will show correct time, and the co-efficient of linear expansion $$\left( \alpha \right)$$ of the metal of the
pendulum shaft are respectively:
A.
$${30^ \circ }C;\,\alpha = \frac{{1.85 \times {{10}^{ - 3}}}}{{^ \circ C}}$$
B.
$${55^ \circ }C;\,\alpha = \frac{{1.85 \times {{10}^{ - 2}}}}{{^ \circ C}}$$
C.
$${25^ \circ }C;\,\alpha = \frac{{1.85 \times {{10}^{ - 5}}}}{{^ \circ C}}$$
D.
$${60^ \circ }C;\,\alpha = \frac{{1.85 \times {{10}^{ - 4}}}}{{^ \circ C}}$$
Answer :
$${25^ \circ }C;\,\alpha = \frac{{1.85 \times {{10}^{ - 5}}}}{{^ \circ C}}$$
Solution :
$$\frac{{{\text{Time lost}}}}{{{\text{gained per day}}}} = \frac{1}{2} \propto \,\Delta \theta \times 86400\,{\text{second}}$$
$$\eqalign{
& 12 = \frac{1}{2}\alpha \left( {40 - \theta } \right) \times 86400\,\,\,.....\left( {\text{i}} \right) \cr
& 4 = \frac{1}{2}\alpha \left( {\theta - 20} \right) \times 86400\,\,\,\,\,.....\left( {{\text{ii}}} \right) \cr} $$
On dividing we get, $$3 = \frac{{40 - \theta }}{{\theta - 20}}$$
$$\eqalign{
& 3\theta - 60 = 40 - \theta \cr
& 4\theta = 100 \cr
& \Rightarrow \,\,\theta = {25^ \circ }C \cr} $$