A) 0.01 mm
B) 0.02 mm
C) 0.04 mm
D) 0.03 mm
Correct Answer: D
Solution :
(d) |
\[l=60\text{ }cm,\text{ }m=6\text{ }g,\text{ }A=1\text{ }m{{m}^{2}},\text{ }v=90\text{ }m/s\] |
\[v=\sqrt{\frac{T}{m}\times l}\Rightarrow T=\frac{m{{v}^{2}}}{l}\] |
\[\therefore \Delta L=\frac{Tl}{YA}=\frac{m{{v}^{2}}\times l}{l(YA)}\] |
\[=\frac{6\times {{10}^{-3}}\times {{90}^{2}}}{16\times {{10}^{11}}\times {{10}^{-6}}}=3\times {{10}^{-4}}m\] |
\[=0.03\,\,mm\] |
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