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Question

An iron ball has a diameter of 6 cm and is 0.010 mm too large to pass through a hole in a brass when the ball and plate are at a temperature of $$30^{\circ}C$$. At what temperature, the same for ball and plate, will the ball just pass through the hole ?


Solution

Since the linear expansion is given by the expression $$\Delta l=l\alpha \Delta T$$

Hence, in the given situation, since both iron ball and brass plate will expand. Hence,

$$d({ \alpha  }_{ Brass }-{ \alpha  }_{ Iron })\Delta T=0.01mm$$

$$60({ \alpha  }_{ Brass }-{ \alpha  }_{ Iron })\Delta T=0.01$$

$$60(19-12) { 10 }^{ -6 }(T-30)=0.01$$

$$T={ 50.8 }^{ 0 }C$$

Answer is $${ 50.8 }^{ 0 }C$$

Physics

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