Question

10. A force of 1000 newton, doubles the length of a
cord having cross-sectional area 1 mm2. The
Young's modulus of the material of the cord is
(1) 2 x 106 N/m2
(2) 5 x 105 N/m2
(3) 10° N/m2
(4) 105 N/m2​

Answer 1

GiveN :-

• A force of 1,000 Newtons doubles the length of a cord .
• The cross-sectional area of the cord is 1mm².

To FinD :-

• The Young Modulus of the material of the cord .

AnsweR :-

Given that , a force of 1,000 Newtons doubles the length of a cord . The cross-sectional area of the cord is 1mm².

We know that the Young Modulus is defined as Stress by Strain.

$\qquad\boxed{\red{\bf \gamma = \dfrac{Stress}{Strain}}}$

$\underline{\blue{\sf Stress :- }}$

• Force acting per unit Area of the material is called Stress .

$\underline{\blue{\sf Strain :- }}$

• Change in lenght by Original lenght is called strain .

$\boxed{\bf Strain = \dfrac{\Delta l }{l }}$

$\rule{200}2$

$\pink{\tt:\implies \gamma = \dfrac{Stress }{Strain}}\\\\\tt:\implies \gamma = \dfrac{\dfrac{Force}{Area}}{\dfrac{\Delta l }{l}}\\\\\tt:\implies \gamma = \dfrac{\dfrac{1000N}{1mm^2}}{\dfrac{l}{l}}\\\\\tt:\implies \gamma = \dfrac{1000 N }{1\times 10^{-6}m^2 }\\\\\tt:\implies \gamma = 10^{6}\times 1000 \\\\\underline{\boxed{\red{\tt\longmapsto Young \ Modulus = 10^9 \:\: N/m^2 }}}$

Hence the Young Modulus is 10⁹ N / m².

$\rule{200}2$

$\underline{\underline{\purple{\textsf{\textbf{ \leadsto \: Stress - Strain \:\: Graph :- }}}}}$

$\setlength{\unitlength}{1 cm}\begin{picture}(12,12)\linethickness{0.2mm}\put(0,0){\vector(1,0){5}}\put(0,0.001){\vector(0,1){5}}\put(0,0.01){\line(1,2){2}}\qbezier(2,4)(2.5,4.5)( 3, 4.5)\put(2, - 0.5){ \bf Strain }\put(-1.5,2){ \bf Stress } \end{picture}$