Question

The work required to be done in producing an extension of 1.0mm in a wire of length 50 cm and area of cross-section 1mm^2 will be ( Y =2×10^10Nm^-2)​

Answer 1

Answer: The work done is 0.02 J.

Explanation:

Given that,

Length l = 50 cm

Extension in wire $\Delta l= 1.0 mm$

Area of cross section $A = 1mm^2$

Young modulus $Y= 2\times10^10 Nm^2$

Using formula of young's modulus

$Y = \dfrac{F l}{A\Delta l}$

$2\times10^{10}= \dfrac{F\times50\times10^{-2}}{1\times10^{-6}\times1\times10^{-3}}$

$F = \dfrac{2\times10^{10}\times1\times10^{-6}\times1\times10^{-3}}{50\times10^{-2}}$

$F = 40 N$

The work done is

$W = \dfrac{1}{2}\times F\times \Delta l$

$W = \dfrac{1}{2}\times40\times1\times10^{-3}$

$W = 0.02\ J$

Hence, The work done is 0.02 J.