JEE Main Physics Class 11 Properties of Solids and Liquids Questions Part 2 Questions

Given, breaking stress of wires 1 and 2 are σ1=σ2=1.25×109N∕m2
Cross-sectional area of wire 1 ,
A1=8×10−7m2
Cross-sectional area of wire 2 ,
A2=4×10−7m2
Mass hanging from first wire, m1=20kg Mass hanging from second wire,
m2=10kg
Acceleration due to gravity, g=10ms−2
Let m be the maximum mass placed in pan without breaking the wire.
According to free body diagram of given figure,
Since, stress(σ)=‌

‌ Tension ‌(T)

Area(A)

∴‌‌T1=σA1
=1.25×109×8×10−7
=10.00×102=1000N
and T2=σA2
=1.25×109×4×10−7
=5.00×102=500N
By using concept of tension in string
T2=(10+m)g
⇒‌‌500=(10+m)10
⇒‌‌m=50−10=40kg