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. |
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