A) 1.75
B) 1.5
C) 2.5
D) 5.0
Correct Answer: C
Solution :
\[l=\frac{FL}{\pi {{r}^{2}}Y}\Rightarrow {{r}^{2}}\propto \frac{1}{Y}\] (F,L and l are constant) \[\frac{{{r}_{2}}}{{{r}_{1}}}={{\left( \frac{{{Y}_{1}}}{{{Y}_{2}}} \right)}^{1/2}}={{\left( \frac{7\times {{10}^{10}}}{12\times {{10}^{10}}} \right)}^{1/2}}\] Þ \[{{r}_{2}}=1.5\times {{\left( \frac{7}{12} \right)}^{1/2}}\]= 1.145 mm \ dia = 2.29 mmYou need to login to perform this action.
You will be redirected in
3 sec