Question

A tensile force of 2 x 10${}^5$ dynes doubles the length of an elastic cord whose area of cross-section is 2 sq. cm. Young's modulus of the material of the cord is

• A. 1 x 10${}^5$ dynes /cm${}^2$
• B. 2 x 10${}^5$ dynes /cm${}^2$
• C. 0.5 x 10${}^5$ dynes /cm${}^2$
• D. 4 x 10${}^5$ dynes /cm${}^2$

Answer 1

The correct option is A 1 x 10${}^5$ dynes /cm${}^2$

Length is doubled, So, elongation is equal to the length, So, $\frac{△l}{l}=1$
$Y=\frac{Fl}{A△l}$
$=\frac{2\times {10}^5}{2}(\because \frac{△l}{l}=1)$
$={10}^{5}dynes/c{m}^2$