Answer :

Correct answer is : 1.265

Given, Hardness of sheet in first case = 250 HRC, Clearance provided for sheet = 1 mm, Hardness of sheet in second case = 400 HRC

$\frac{C_2}{C_1}=\frac{\sqrt{HR{C}_2}}{\sqrt{HR{C}_1}}$

$C_2=C_1\times \sqrt{\frac{HR{C}_2}{HR{C}_1}}=\left[1\times \sqrt{\frac{400}{250}}\right]$

C₂ = 1.265 mm

Hence, clearance provided in second case = 1.265 mm