Courses
Courses for Kids
Free study material
Offline Centres
More
Store

# A $5m$ long aluminium wire ($Y = 7 \times {10^{10}}\;N/{m^2}$ ) of diameter $3\;mm$ supports a $40\;kg$ mass. In order to have the same elongation in the copper wire ($Y = 12 \times {10^{10}}\;N/{m^2}$) of the same length under the same weight, the diameter should now be (in mm):A) 1.75B) 1.5C) 2.5D) 5.0

Last updated date: 17th Apr 2024
Total views: 35.4k
Views today: 0.35k
Verified
35.4k+ views
Hint: Stress and Strain are the two terms in Physics that describe the forces causing the deformation of the objects. Deformation is known as the change of the shape of an object by applications of force. Very small forces can also cause deformation. The object experiences it due to external forces; for example, the forces might be like squeezing, squashing, twisting, shearing, ripping, or pulling the objects apart.

We will now write the relation between $d$ and $E$.
${d^2} \propto \dfrac{1}{E}$
$d_1^2{E_1} = d_2^2{E_2}$
$\begin{array}{l} {3^2} \times 7 \times {10^{10}} = d_2^2 \times 12 \times {10^{10}} \end{array}$
${d_2} = 2.29\;mm$