JEE MAIN - Physics (2016 (Offline) - No. 20)
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 :
$${30^ \circ }C;\,\,\alpha = 1.85 \times {10^{ - 3}}/{}^ \circ C$$
$${55^ \circ }C;\,\,\alpha = 1.85 \times {10^{ - 2}}/{}^ \circ C$$
$${25^ \circ }C;\,\,\alpha = 1.85 \times {10^{ - 5}}/{}^ \circ C$$
$${60^ \circ }C;\,\,\alpha = 1.85 \times {10^{ - 4}}/{}^ \circ C$$
Explanation
Time lost/gained per day $$ = {1 \over 2} \propto \Delta \theta \times 86400$$ second
$$12 = {1 \over 2}\alpha \left( {40 - \theta } \right) \times 86400\,\,\,\,\,\,\,\,\,\,\,\,\,...\left( i \right)$$
$$4 = {1 \over 2}\alpha \left( {\theta - 20} \right) \times 86400\,\,\,\,\,\,\,\,\,\,\,\,\,...\left( {ii} \right)$$
On dividing we get, $$\,\,\,3 = {{40 - \theta } \over {\theta - 20}}$$
$$3\theta - 60 = 40 - \theta $$
$$4\theta = 100 \Rightarrow \theta = {25^ \circ }C$$
$$12 = {1 \over 2}\alpha \left( {40 - \theta } \right) \times 86400\,\,\,\,\,\,\,\,\,\,\,\,\,...\left( i \right)$$
$$4 = {1 \over 2}\alpha \left( {\theta - 20} \right) \times 86400\,\,\,\,\,\,\,\,\,\,\,\,\,...\left( {ii} \right)$$
On dividing we get, $$\,\,\,3 = {{40 - \theta } \over {\theta - 20}}$$
$$3\theta - 60 = 40 - \theta $$
$$4\theta = 100 \Rightarrow \theta = {25^ \circ }C$$
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