Choose the physical quantity pair with the same dimensions.
Choose the physical quantity pair with the same dimensions.
- Angular momentum and work
- Work and torque
- Potential energy and linear momentum
- Kinetic energy and velocity
Solution
Dimensional formula of work,
$\begin{aligned}
[W] & =[F] \times[s] \\
& =\left[\mathrm{MLT}^{-2}\right][\mathrm{L}]=\left[\mathrm{ML}^2 \mathrm{T}^{-2}\right]
\end{aligned}$
Similarly, dimensional formulae of
$\begin{aligned}
\text{Angular momentum }[L] & =[m][v][r]=[\mathrm{M}]\left[\mathrm{LT}^{-1}\right][\mathrm{L}] \\
& =\left[\mathrm{ML}^2 \mathrm{T}^{-1}\right] \\
\text{Torque }[\tau] & =[F] \times[r] \\
& =\left[\mathrm{MLT}^{-2}\right][\mathrm{L}]=\left[\mathrm{ML}^2 \mathrm{T}^{-2}\right]
\end{aligned}$
$\begin{aligned}
\text{Potential energy }[U] & =[m][g][h] \\
& =[\mathrm{M}]\left[\mathrm{LT}^{-2}\right][\mathrm{L}] \\
& =\left[\mathrm{ML}^2 \mathrm{T}^{-2}\right]
\end{aligned}$
$\text{Linear momentum } \begin{aligned}
[p] & =[m][v]=[\mathrm{M}]\left[\mathrm{LT}^{-1}\right] \\
& =\left[\mathrm{MLT}^{-1}\right]
\end{aligned}$
$\begin{aligned}
\text{Kinetic energy }[K] & =\frac{1}{2}[m]\left[v^2\right] \\
& =[\mathrm{M}]\left[\mathrm{LT}^{-1}\right]^2=\left[\mathrm{ML}^2 \mathrm{T}^{-2}\right] \\
\text{Velocity }[v] & =\left[\mathrm{LT}^{-1}\right]
\end{aligned}$
Clearly, dimensional formulae of work and torque are same, hence option (b) is correct.
Asked in: AP EAMCET 2020 (18 Sep Shift 1)
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