A clock with a metallic pendulum is $$5$$ seconds fast each day at a temperature of $$15$$∘c and $$10$$ seconds slow each day at a temperature of $$30$$∘c, find the coefficient of linear expansion of the metal?

Question

A clock with a metallic pendulum is $$5$$ seconds fast each day at a temperature of  $$15$$∘c and $$10$$ seconds slow each day at a temperature of  $$30$$∘c, find the coefficient of linear expansion of the metal?

Infinity Learn · Accepted Answer

Graduation $$temperature=$$ Tochange in time due to coefficient of linear expansion of material of pendulum is∆time $$=$$ $$12$$α(temperature$$ $$difference)x86400$$ here α is coefficient of linear expansion of metal of $$pendulum5=12$$.α.$$To-15$$×$$86400----(1)10=12$$.α$$.30-To$$×$$86400----(2)on$$ dividing eqn $$(1) with $$eqn(2)$$ $$12=T0-1530-T0$$ ⇒$$T0=200C$$ substitute $$T0$$ vallue in $$eqn(1)$$ $$5=12$$α$$(20-15)86400$$⇒α$$=2.31$$ x $$10-5$$ C $$-1$$  $$0=coefficient$$ of linear expansion of metal

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