Question

To determine Young's modulus of a wire, the formula is

Y = \frac{F}{A} \cdot \frac{L}{Δ l}

, where

L =

length,

A =

area of cross-section of the wire, and

Δ l =

change in the length of the wire when stretched with a force

F

. Find the conversion factor to change it from CGS to the MKS system.

Accepted Answer

We know that the dimension of Young's modulus is will be,

$Y = \frac{[{MLT}^{- 2}]}{[L^{2}]} \times \frac{[L]}{[L]}$

$[Y] = [{ML}^{- 1} T^{- 2}]$ , here we have, $a = 1$ , $b = - 1$ and $c = - 2$ .

CGS unit, $g {cm}^{- 1} s^{- 2}$ and MKS unit, $kg m^{- 1} s^{- 2}$

By using the conversion formula,

MKS system CGS system

$M_{1} = 1 kg$ $M_{2} = 1 g$

$L_{1} = 1 m$ $L_{2} = 1 cm$

$T_{1} = 1 s$ $T_{2} = 1 s$

$n_{2} = n_{1} {[\frac{M_{1}}{M_{2}}]}^{1} {[\frac{L_{1}}{L_{2}}]}^{- 1} {[\frac{T_{1}}{T_{2}}]}^{- 2} = {[\frac{1 g}{1 kg}]}^{1} {[\frac{1 cm}{1 m}]}^{- 1} {[\frac{1 s}{1 s}]}^{- 2}$

Hence, the conversion factor will be,

$\frac{n_{2}}{n_{1}} = {[\frac{1 g}{10^{3} g}]}^{1} {[\frac{1 cm}{10^{2} cm}]}^{- 1} {[\frac{1 s}{1 s}]}^{- 2} = \frac{1}{10} = 0.1$ .

B M Sharma Solutions for Chapter: Dimensions and Measurement, Exercise 10: ILLUSTRATION 1.9

B M Sharma Physics Solutions for Exercise - B M Sharma Solutions for Chapter: Dimensions and Measurement, Exercise 10: ILLUSTRATION 1.9

Questions from B M Sharma Solutions for Chapter: Dimensions and Measurement, Exercise 10: ILLUSTRATION 1.9 with Hints & Solutions

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