Solution:
Since length, load, and extension are the same: \(Y = \frac{FL}{\pi R^2 \Delta L} ⇒ R^2 \propto \frac{1}{Y} ⇒ \frac{R_B}{R_S} = \sqrt{\frac{Y_S}{Y_B}} = \sqrt{\frac{2 \times 10^{11}}{1 \times 10^{11}}} = \sqrt{2} : 1\).
The Young’s modulus of brass and steel are \(1 \times 10^{11}\text{ N/m}^2\) and \(2 \times 10^{11}\text{ N/m}^2\) respectively. If wires of both materials, having same length, are loaded with same weight, then they both extend by 4 mm. Ratio of the radii of two wires \(R_B : R_S\) is
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