Question

A brass disc fits into a hole in an iron plate. To remove the disc
A. the plate must be cooled
B. the plate must be heated
C. the plate may be heated (or) cooled
D. the disc must be heated

Answer 1

Hint:Go through the concept of expansion of the materials. The expansion of the materials depends on the coefficient of expansion of the materials. Check out of iron and brass which one has more coefficient of thermal expansion and hence determine what should be done in order to remove the brass disc out of the hole.

Complete answer:
The coefficient of thermal expansion for brass is greater than the coefficient of thermal expansion for iron. Therefore, if the iron plate is heated then the brass disc will expand more rapidly than the iron plate and fits in the hole more tightly.Hence, option B is incorrect.

The iron plate cannot be heated because while heating the iron plate there is a transfer the heat to the brass disc. Due to this heating, the brass plate expands and fits in the hole instead of getting out from it.Hence, the option C is incorrect.

The heating of brass disc expands the disc and it cannot be removed from the hole.Hence, the option D is incorrect.

When the iron plate is cooled, the temperature of the brass disc in contact with the iron plate also decreases. Due to this cooling, the iron plate and brass disc starts compressing. Since the coefficient of thermal expansion for brass is more than the coefficient of thermal expansion of iron. Hence, the disc gets compressed more quickly than iron plate and hence can be easily removed out of the hole.

Hence, the correct option is A.

Note: The students should not get confused between the effect of heating and cooling on the size of the materials. The students should perfectly keep in mind that on heating the materials expands and on cooling the materials contract provided that the material being heated should be a conductor of heat.