. 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 1° C)
Answers
Step-by-step explanation:
Hint: The question involves the thermal properties of matter. We have to find the temperature required to expand the ring of 5 cm 5 cm diameter to accommodate a ring of the radius 5.01 cm 5.01 cm . When solids are heated, the length of the solid, the surface area and volume will increase. The increase in length of the solid for the rise in temperature is called linear expansion.
Formula used:
α=ΔllΔTα=ΔllΔT
Where αα stands for the coefficient of linear expansion.
l l stands for the original length of the material.
Δl Δl is the change in length for the temperature rise.
ΔT ΔT is the temperature rise.
Complete step by step solution:
Let us take the inner radius of the ring as d1=5cm d1=5cm
The diameter of the ball to be accommodated inside the ring d2=5.01cm d2=5.01cm
The difference between the diameters d2−d1=Δl=5.01−5=0.01cm d2−d1=Δl=5.01−5=0.01cm
The coefficient of linear expansion is defined as the ratio of change in length to the original length for a one-degree rise of temperature.
The coefficient of linear expansion, α=ΔllΔT