Q.
Figure shows a graph of the extension (A) of a wire of length 1 m suspended from the top of a roof at one end and with a load W connected to the other end. If the cross-sectional area of the wire is 10-6m2 , the Young's modulus of the material of the wire is x×1011N/m2. The value of x is
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
answer is 2.
(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
Stress = force area =WA: Strain =Δll. Young's modulus is given byY= stress strain =W/AΔl/l=WΔl×lA ……..(i)Now, slope of graph is ΔlW=4×10−480m/N. Using this in i), we get (given l=1m and A=10−6m2 ) Y=804×10−4×110−6=2×1011N/m2
Watch 3-min video & get full concept clarity