Question

To determine the Young's modulus of a wire, the formula is $Y = \frac{F}{A}.\frac{L}{\Delta l};$ where L= length, A= area of cross- section of the wire, $\Delta L =$Change in length of the wire when stretched with a force F. The conversion factor to change it from CGS to MKS system is

• A. 1
• B. 10
• C. 0.1
• D. 0.01

Answer 1

Answer: (C)

0.1

Answer 2

We know that the dimension of young's modulus is $\lbrack ML^{- 1}T^{- 2}\rbrack$

C.G.S. unit : gm $cm^{- 1}\sec^{- 2}{}$ and M.K.S. unit : kg. m^–1 sec^–2

By using the conversion formula:

$n_{2} = n_{1}\left\lbrack \frac{M_{1}}{M_{2}} \right\rbrack^{1}\left\lbrack \frac{L_{1}}{L_{2}} \right\rbrack^{- 1}\left\lbrack \frac{T_{1}}{T_{2}} \right\rbrack^{- 2} = \left\lbrack \frac{gm}{kg} \right\rbrack^{1}\left\lbrack \frac{cm}{meter} \right\rbrack^{- 1}\left\lbrack \frac{\sec}{\sec}^{- 2} \right\rbrack$

$\therefore$Conversion factor $\frac{n_{2}}{n_{1}} = \left\lbrack \frac{gm}{10^{3}gm} \right\rbrack^{1}\left\lbrack \frac{cm}{10^{2}cm} \right\rbrack^{- 1}\left\lbrack \frac{\sec}{\sec}^{- 2} \right\rbrack$

$= \frac{1}{10} = 0.1$