Question

An iron ball has a diameter of 5 cm and 0.01mm too large to pass through a hole in a brass plate when the ball and the plate are at a temperature of 30 degree at what temperature will the ball just pass through the hole

Answer 1

The problem is that the ball can’t fit through the hole. If we’re allowed to change the temperature of the brass plate and the iron ball independently, we just have to heat the brass plate so it expands by 1/5000 (0.0002) or cool the iron ball so it contracts by 1/5000 (0.0002).

Coefficients of linear thermal expansion:

brass: 1.8 x 10−5−5 per K

iron: 1.2 x 10−5−5 per K

Since brass is more sensitive to temperature changes, I’ll choose to heat the brass.

0.00021.8×10−5=11K0.00021.8×10−5=11K

That’s our delta-T, so we have to heat the brass up to 41 °C.

However, if we assume that the plate and the ball have to be at the same temperature, things get a bit dodgier.

Again, the dimensions of the brass will change faster than the iron, and we want the clearance to increase, so we still have to heat.

The dimension upon a change of temperature will be given by

brass hole=4.999+1.8×10−5ΔTbrass hole=4.999+1.8×10−5ΔT

iron ball=5.000+1.2×10−5ΔTiron ball=5.000+1.2×10−5ΔT

They’ll both be getting bigger as we heat them up, but the brass will get bigger faster, and eventually catch up to the iron.

Set the two expressions equal, and solve the above for ΔT and add it to the 30 °C we’re starting at.

Answer 2

Answer:

The temperature at which the ball just pass through the hole is 63.3°c

Explanation:

Given ::

• Diameter of iron ball= 5 cm

As other terms are in millimetres we need to convert this also in millimetres we get,

5×10 = 50 mm

• This is orginal length
• Change in length =0.01mm
• Both iron and brass are at temperature 30°c

As both iron and brass get expanded we need to calculate the change in coefficient of linear expansion of both iron and brass.

Change in Coefficients of linear expansion of Brass and iron is given by

= Alpha( Brass) - Alpha(Iron) i.e

• Here Alpha is coefficient of linear expansion = (18×10^-6-12×10^-6)

=6×10^-6.

• And Change in length is given by

L" = L ×Change in Coefficents ×(T-30)

where,

L'' is change in length

L is orginal length

T is temperature

0.01 = 50×6×10^-6(T-30).

0.01=3×10^-4(T-30)

1/3×10^2=T-30

33.3=T-30

T=63.3°C.

Hence,

The temperature at which the ball just pass through the hole is 63.3°c