A) \[F\]
B) \[4F\]
C) \[6F\]
D) \[9F\]
Correct Answer: D
Solution :
[d] \[{{A}_{1}}{{l}_{1}}={{A}_{2}}{{l}_{2}}\] \[\Rightarrow {{l}_{2}}=\frac{{{A}_{1}}{{l}_{1}}}{{{A}_{2}}}=\frac{A\times {{l}_{1}}}{3A}\] \[\Rightarrow \frac{{{l}_{1}}}{{{l}_{2}}}=3\] \[\Delta {{x}_{1}}=\frac{{{F}_{1}}}{A\gamma }\times {{l}_{1}}\] \[\Rightarrow {{x}_{2}}=\frac{{{F}_{2}}}{3A\gamma }\times {{l}_{1}}\] Here \[\Delta {{x}_{1}}=\Delta {{x}_{2}}\] \[\frac{{{F}_{2}}}{3A\gamma }{{l}_{2}}=\frac{{{F}_{1}}}{A\gamma }{{l}_{1}}\Rightarrow {{F}_{2}}=3{{F}_{1}}\times \frac{{{l}_{1}}}{{{l}_{2}}}=3{{F}_{1}}\times 3=9F\]
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