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Choose the physical quantity pair with the same dimensions.
Options:
Solution:
1172 Upvotes
Verified Answer
The correct answer is:
Work and torque
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.
\(\begin{aligned}
{[W] } & =[F] \times[s] \\
& =\left[\mathrm{MLT}^{-2}\right][\mathrm{L}]=\left[\mathrm{ML}^2 \mathrm{~T}^{-2}\right]
\end{aligned}\)
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.
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