A) 1 : 2
B) 2 : 1
C) 4 : 1
D) 1 : 1
Correct Answer: B
Solution :
Given, \[{{Y}_{steel}}=2{{Y}_{brass}}\]and \[{{L}_{s}}={{L}_{b}}\] and \[{{A}_{a}}={{A}_{b}}\] such that \[\Delta {{L}_{s}}=\Delta {{L}_{b}}\] As we know, Young's modulus, \[Y=\frac{\text{stress}}{\text{strain}}=\frac{\frac{F}{A}}{\frac{\Delta L}{L}}=\frac{W\Delta L}{A\Delta L}\] So, \[W=\frac{FA\Delta L}{L}\propto Y\] i.e. \[\frac{{{W}_{s}}}{{{W}_{b}}}=\frac{{{Y}_{s}}}{{{Y}_{b}}}=\frac{2{{Y}_{b}}}{{{Y}_{b}}}=\frac{2}{1}\] \[\Rightarrow \,\,\,\,\,\,\,\,\,\,\,\,\,2:1\] Thus, weight added to the steel and brass wires must be in the ratio of 2 : 1.You need to login to perform this action.
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