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(C) The formula for Young's modulus is $Y = \frac{F \cdot L}{A \cdot \Delta L}$.The dimensional formula for Young's modulus is $[Y] = [M L^{-1} T^{-2}

Vedclass · Accepted Answer

$0.1$

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(C) The formula for Young's modulus is $Y = \frac{F \cdot L}{A \cdot \Delta L}$.The dimensional formula for Young's modulus is $[Y] = [M L^{-1} T^{-2}]$.In the $CGS$ system,the unit is $	ext{dyne/cm}^2$,and in the $MKS$ $(SI)$ system,the unit is $	ext{N/m}^2$.$1 	ext{ N/m}^2 = 1 \frac{	ext{kg} \cdot 	ext{m/s}^2}{	ext{m}^2} = 1 	ext{ kg} \cdot 	ext{m}^{-1} \cdot 	ext{s}^{-2}$.$1 	ext{ dyne/cm}^2 = 1 \frac{	ext{g} \cdot 	ext{cm/s}^2}{	ext{cm}^2} = 1 	ext{ g} \cdot 	ext{cm}^{-1} \cdot 	ext{s}^{-2}$.Converting $1 	ext{ dyne/cm}^2$ to $MKS$:$1 	ext{ g} = 10^{-3} 	ext{ kg}$$1 	ext{ cm}^{-1} = (10^{-2} 	ext{ m})^{-1} = 10^2 	ext{ m}^{-1}$.Therefore,$1 	ext{ dyne/cm}^2 = 10^{-3} 	ext{ kg} \cdot 10^2 	ext{ m}^{-1} \cdot 	ext{s}^{-2} = 10^{-1} 	ext{ N/m}^2 = 0.1 	ext{ N/m}^2$.Thus,the conversion factor from $CGS$ to $MKS$ is $0.1$.

To determine Young's modulus of a wire,the formula is $Y = \frac{F}{A} \cdot \frac{L}{\Delta L}$,where $F/A$ is the stress and $L/\Delta L$ is the reciprocal of strain. The conversion factor to change $Y$ from $CGS$ to $MKS$ system is:

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