A) 200 N
B) 400 N
C) 600 N
D) 800 N
Correct Answer: D
Solution :
\[Y=\frac{FL}{Al}\] or \[F=\frac{YAl}{L}\] or \[F\propto A\,or\,F\propto {{r}^{2}}\,or\,F\propto {{d}^{2}}\] \[\therefore \] \[\frac{{{F}_{1}}}{{{F}_{2}}}=\frac{d_{1}^{2}}{d_{2}^{2}}\] Given, di \[{{d}_{1}}=d,\,{{d}_{2}}=2d,\,{{F}_{1}}=200\,N\] \[\therefore \] \[\frac{200}{{{F}_{2}}}=\frac{{{(d)}^{2}}}{{{(2d)}^{2}}}=\frac{1}{4}\] or \[{{F}_{2}}=4\times 200=800\,N\]You need to login to perform this action.
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