A) \[2:1\]
B) \[4:1\]
C) \[1:1\]
D) \[1:2\]
Correct Answer: A
Solution :
Young's modulus \[Y=\frac{W}{A}.\frac{l}{\Delta l}\] |
\[\frac{{{W}_{1}}}{{{Y}_{1}}}=\frac{{{W}_{2}}}{{{Y}_{2}}}\] [\[\because \,\,\,A,l,\Delta l\] same for both brass and steel] |
\[\frac{{{W}_{1}}}{{{W}_{2}}}=\frac{{{Y}_{1}}}{{{Y}_{2}}}=2\] \[[{{Y}_{steel}}/{{Y}_{brass}}=2\,given]\] |
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