The diameter of a brass ring at 30 °C is 50.0 cm. To what temperature must this ring be heated to increase its diameter to 50.29 cm? [ linear expansivity of...

Assessment: WAEC SSCE - Physics - 2022 (Objective) Subject: Physics

Question 1 Report

The diameter of a brass ring at 30 °C is 50.0 cm. To what temperature must this ring be heated to increase its diameter to 50.29 cm? [ linear expansivity of brass = \(1.9 \times 10^{-5}\ \mathrm{K}^{-1}\)]

Answer Details
The problem involves the concept of linear expansivity. The diameter of the brass ring increases with an increase in temperature. We can use the formula for linear expansivity to solve the problem: ΔL = αLΔT where ΔL is the change in length, L is the original length, α is the linear expansivity and ΔT is the change in temperature. Since we are given the initial diameter, we can use it to find the original length (L) of the ring. L = πd/2 where d is the diameter. Substituting the values, we get: L = π × 50.0/2 = 78.54 cm The change in length (ΔL) is given by the difference in the final and initial lengths: ΔL = 50.29/2 - 50.0/2 = 0.145 cm Substituting the given values of α and L, we get: ΔL = αLΔT 0.145 = (1.9 × 10^-5) × 78.54 × ΔT Solving for ΔT, we get: ΔT = 0.145/(1.9 × 10^-5 × 78.54) = 116.3 °C Therefore, the temperature to which the ring must be heated is: 30 + 116.3 = 146.3 °C Rounding off to one decimal place, we get: 146.3 °C Hence, the correct option is 152.6 °C (Option A).

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