On applying a stress of 20 × 108 Nm -2, the length of a perfectly elastic wire is doubled. Its Young's modulus will be

Question

On applying a stress of 20 × 108 Nm -2, the length of a perfectly elastic wire is doubled. Its Young's modulus will be (A) 40 × 108 Nm -2 (B) 20 × 108 Nm -2 (C) 10 × 108 Nm -2 (D) 5 × 108 Nm -2

Tardigrade Admin · Accepted Answer

Given, stress F=20 × 108 Nm -2 Young's modulus =( text Stress / text Strain ) As the length of wire gets doubled, therefore strain =1 [∵ text strain =( text change in length / text original length )=(2 L-L/L)=1] ∴ Y= text stress =20 × 108 Nm -2

Q. On applying a stress of $20 \times 1 0^{8} N m^{- 2}$ , the length of a perfectly elastic wire is doubled. Its Young's modulus will be

$40 \times 1 0^{8} N m^{- 2}$

$20 \times 1 0^{8} N m^{- 2}$

$10 \times 1 0^{8} N m^{- 2}$

$5 \times 1 0^{8} N m^{- 2}$

Given, stress $F = 20 \times 1 0^{8} N m^{- 2}$
Young's modulus $= \frac{Stress}{Strain}$
As the length of wire gets doubled, therefore strain $= 1$
$[∵ strain = \frac{change in length}{original length} = \frac{2 L - L}{L} = 1]$
$∴ Y = stress = 20 \times 1 0^{8} N m^{- 2}$

Q. On applying a stress of 20×108Nm−2, the length of a perfectly elastic wire is doubled. Its Young's modulus will be

40×108Nm−2

20×108Nm−2

10×108Nm−2

5×108Nm−2