Question

The inner diameter of a brass ring at 273 K
is 5 cm. To what temperature should it be
heated for it to accommodate a ball 5.01 cm
in diameter. (a=2 x 10-5 /°C)
1) 273 K 2)372 K 3) 437 K 4) 173K​

Answer 1

Answer:

See the explanation below (its really easy, you can find out the answer yourself)

Explanation:

1. d1=5 cm at T= 273k
2. Accomodating a ball 5.01 cm in diameter means we have get the diameter of ring as 5.01cm
3. by the formula of thermal expansion, a2=a1(1+at)
4. Here a2=required area, a1=initial area, a= coefficient of expansion of the ring and t= change in temperature
5. so putting a = πd, the formula can be written as πd2=πd1(1+at)
6. hence putting all the values we can easily find out t
7. here t = 273+ T because we are raising the temperature after 273k
8. So the value of T will be the final answer