Question

Young's modulus of steel is $19\times {10}^{10}\frac{N}{m^2}$. Express it in $\frac{dyne}{c{m}^2}$. Here dyne is the CGS unit of force.

• A. $19\times {10}^5\frac{dyne}{c{m}^2}$
• B. $19\times {10}^{10}\frac{dyne}{c{m}^2}$
• C. $19\times {10}^{12}\frac{dyne}{c{m}^2}$
• D. None of these

Answer 1

The correct option is D

None of these

The unit of young's modulus is $\frac{N}{m^2}$.

This suggests that it has dimensions of $\left(\frac{Force}{(distance{)}^2}\right)$

Thus, $\left[Y\right]=\frac{\left[F\right]}{L^2}=\frac{ML{T}^{-2}}{L^2}=M{L}^{-1}{T}^{-2}$

$1\frac{N}{m^2}$is in SI units.

So, $1\frac{N}{m^2}=\left(1kg\right)\left(1m{)}^{-1}\right(1s{)}^{-2}$

And $1\frac{dyne}{c{m}^2}=\left(1gm\right)\left(1cm{)}^{-1}\right(1s{)}^{-2}$

So, $\frac{1\frac{N}{m^2}}{1\frac{dyne}{c{m}^2}}=\left(\frac{1kg}{1g}\right){\left(\frac{1m}{1cm}\right)}^{-1}{\left(\frac{1s}{1s}\right)}^{-2}$

= $1000\times \frac{1}{100}\times 1$ = 10

or, $1\frac{N}{m^2}$ = 10$\frac{dyne}{c{m}^2}$

or, $19\times {10}^{10}\frac{N}{m^2}=19\times {10}^{11}\frac{dyne}{c{m}^2}$