Asilver sphere with radius 1.3611 cm at 23.0°c must slip through a brass ring that has an internal radius of 1.3590 cm at the same temperature. to what temperature must the ring be heated so that the sphere, still at 23.0°c, can just slip through?

The temperature must the ring be heated so that the sphere can just slip through is 106.165 °C.

Explanation:

For brass:

Radius = 1.3590 cm

Initial temperature = 23.0 °C

The sphere of radius 1.3611 cm must have to slip through the brass. Thus, on heating the brass must have to attain radius of 1.3611 cm

So,

Δ r = 1.3611 cm - 1.3590 cm = 0.0021 cm

The linear thermal expansion coefficient of a metal is the ratio of the change in the length per 1 degree temperature to its length.

Thermal expansion for brass = 19×10⁻⁶ °C⁻¹

Thus,

Also,

So,

Solving for final temperature as:

Final temperature = 106.165 °C

the ima (higher than 1m) will be greater than the ima with the height of 1m.