Q.
The Young's modulus of steel is twice that of brass. Two wires of same length and of same area of cross section, one of steel and another of brass are suspended from the same roof. If we want the lower ends of the wires to be at the same level, then the weights added to the steel and brass wires must be in the ratio of
see full answer
Talk to JEE/NEET 2025 Toppers - Learn What Actually Works!
Real Strategies. Real People. Real Success Stories - Just 1 call away
An Intiative by Sri Chaitanya
a
4 : 1
b
1 : 1
c
1 : 2
d
2 : 1
answer is D.
(Unlock A.I Detailed Solution for FREE)
Ready to Test Your Skills?
Check your Performance Today with our Free Mock Test used by Toppers!
Take Free Test
Detailed Solution
Let L and A be length and area of cross section of each wire. In order to have the lower ends of the wires to be at the same level (i.e. same elongation is produced in both wires), let weights Ws and Wb are added to steel and brass wires respectively. Then By definition of Young's modulus, the elongation produced in the steel wire isΔLs=WsLYsA as Y=W/AΔL/Land that in the brass wire isΔLb=WbLYbA But ΔLs=ΔLb (given) ∴WsLYsA=WbLYbA or WsWb=YsYb As YsYb=2 (given)∴WsWb=21
Watch 3-min video & get full concept clarity