A lift of mass 'm' is connected to a rope which is moving upward with maximum acceleration 'a'. For maximum safe stress, the elastic limit of the rope is 'T'. The minimum diameter of the rope is
(g = gravitational acceleration)

Options

A. `[(6M(g +a))/(piS)]^((1)/(2))`
B. `[(4m(g +a))/(piS)]^((1)/(2))`
C. `[(M(g+a))/(piS)]^((1)/(2))`
D. `[(M(g -a))/(piS)]^((1)/(2))`

Answer

C

Explanation

The maximum stress produced in a rope is given by
` sigma _ ( max ) = ( "Force")/("Area" ) = ( Mg ) /( pir ^ 2) `
As the lift is accelerating with acceleration a , then
` sigma _ ( max ) = ( M ( g pm a )) /( pi r ^ 2 ) `
` rArr r ^ 2 = ( M ( g pm a ) ) /( pi S ) " "` [ Given, ` sigma _ ( max ) = S ` ]
` ( d ^ 2 )/ ( 4 ) = ( M ( g pm a ) ) /(pi S ) " " [ because r = ( d ) / ( 2 ) ] `
` rArr d = sqrt (( 4 M ( g pm a )) /( pi S) ) `
As acceleration is maximum i.e., ` g' = g + a , `so
` d = sqrt (( 4M ( g + a ))/ ( pi S)) `