If speed $(V)$, acceleration $(A)$ and force $(F)$ are considered as fundamental units, the dimension of Young’s modulus will be
${V^{ - 2}}{A^2}{F^{ - 2}}$
${V^{ - 2}}{A^2}{F^2}$
${V^{ - 4}}{A^{ - 2}}F$
${V^{ - 4}}{A^{2}}F$
A function $f(\theta )$ is defined as $f(\theta )\, = \,1\, - \theta + \frac{{{\theta ^2}}}{{2!}} - \frac{{{\theta ^3}}}{{3!}} + \frac{{{\theta ^4}}}{{4!}} + ...$ Why is it necessary for $f(\theta )$ to be a dimensionless quantity ?
If velocity$(V)$, force$(F)$ and time$(T)$ are chosen as fundamental quantities then dimensions of energy are
Let us consider an equation
$\frac{1}{2} m v^{2}=m g h$
where $m$ is the mass of the body. velocity, $g$ is the acceleration do gravity and $h$ is the height. whether this equation is dimensionally correct.