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Question

Physics Question on mechanical properties of solids

A tensile force of 2×1052\times10^5 dynes doubles the length of an elastic cord whose area of cross section is 2scm Young's modulus of the material of the cord is

A

1×105dynes/cm21\times 10^5\, dynes /cm^2

B

2×105dynes/cm22\times 10^5\, dynes /cm^2

C

0.5×105dynes/cm20.5\times 10^5\, dynes /cm^2

D

4×105dynes/cm24\times 10^5\, dynes /cm^2

Answer

1×105dynes/cm21\times 10^5\, dynes /cm^2

Explanation

Solution

γ=FlAe\gamma = \frac{Fl}{Ae}