A) 0.00141 m
B) 0.00282 m
C) 0.005m
D) 0.01m
Correct Answer: D
Solution :
[d]: When the lift is accelerated upwards with acceleration a, then tension in the wire is \[T=m\left( g+a \right)=1000\left( 9.8+1.2 \right)=11000\text{ }N\] Now, stress\[=\frac{F}{A}=\frac{T}{\pi {{r}^{2}}}\]or\[r=\frac{1}{200}\] \[\therefore \]Minimum diameter of the wire \[D=2r=\frac{1}{100}=0.01m\]You need to login to perform this action.
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