BITSAT 2019 Slot 2 Solved Paper

Given, Y=2.2×108m∕s2
Area of cross-section of steel wire,
A=4cm2=4×10−4m2
Mass of elevator, m=1000kg
The maximum tension in the steel wire that can support elevator is given by
T=

1

4

× stress × area of cross - section
=

1

4

×2.2 ×108×4×10−4
=2.2×104N∕m2
If f be the maximum upward acceleration of the elevator, then
⇒T=m(g+f)
2.2×104=1000(10+f)
⇒f=12m∕s2