Mechanical Properties of Solids

JEE Main level, single correct, medium

50 of 118

Question 50 of 118

The area of cross section of the rope used to lift a load by a crane is 2.5×10−4m22.5\times {10}^{-4}{m}^{2}. The maximum lifting capacity of the crane is 1010 metric tons. To increase the lifting capacity of the crane to 25 metric tons, the required area of cross section of the rope should be
(take g=10ms−2g=10{ms}^{-2})

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Answer: (A) 6.25×10−4m26.25\times {10}^{-4}{m}^{2}

Since breaking stress (Maximum lifting capacity) is the property of material, so it will remain same.
Now, breaking stress=MaximumliftingcapacityAreaofcrosssectionofrope=\dfrac{Maximumliftingcapacity}{Areaofcrosssectionofrope}
⇒102.5×10−4=25A\Rightarrow \dfrac{10}{2.5\times {10}^{-4}}=\dfrac{25}{A}
A=625×10−6A=625\times {10}^{-6}
=6.25×10−4m2=6.25\times {10}^{-4}{m}^{2}

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