General Physics Ques 10
- Match the physical quantities given in Column I with dimensions expressed in terms of mass $(M)$, length $(L)$, time $(T)$, and charge $(Q)$ given in Column II and write the correct answer against the matched quantity in a tabular form in your answer book.
(1993, $6 \mathrm{M}$ )
$ \begin{array}{ll} \hline \text { Column I } & \text { Column II } \ \hline \text { Angular momentum } & {\left[\mathrm{ML}^2 \mathrm{~T}^{-1}\right]} \\ \hline \text { Latent heat } & {\left[\mathrm{ML}^2 \mathrm{T}^{-2}\right]} \ \hline \text { Torque } & {\left[\mathrm{ML}^2 \mathrm{~T}^{-2}\right]} \ \hline \text { Capacitance } & {\left[\mathrm{ML}^2 \mathrm{~T}^{-1} \mathrm{Q}^{-2}\right]} \\ \hline \text { Inductance } & {\left[\mathrm{M} \mathrm{~L}^2 \mathrm{~T}^{-2} \mathrm{Q}^{-2}\right]} \\ \hline \text { Resistivity } & {\left[\mathrm{L} \mathrm{\Omega}\right]} \\ \hline \end{array} $
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Answer:
- (See the solution)
Solution:
- Angular momentum, $L=I \omega$ $ \therefore[\mathrm{L}]=[\mathrm{I} \omega]=\left[\mathrm{ML}^2\right]\left[\mathrm{T}^{-1}\right]=\left[\mathrm{ML}^2 \mathrm{~T}^{-1}\right] $
$ \begin{alignedat} & \text { Latent heat, } L=\frac{Q}{m} \quad(\text { as } Q=m L \text { ) } \\ & \Rightarrow[\mathrm{L}]=\left[\frac{\mathrm{Q}}{\mathrm{m}}\right]=\left[\frac{\mathrm{ML}^2 \mathrm{~T}^{-2}}{\mathrm{M}}\right]=\left[\mathrm{L}^2 \mathrm{~T}^{-3}\right] \\ & \end{aligned} $
Torque, $\tau=F \times r_{\perp}$
$ \begin{aligned} \therefore[\tau] & =\left[\mathrm{F} \times \mathrm{r}_{\perp}\right]=\left[\mathrm{MLT}^{-2}\right][\mathrm{L}] \\ & =\left[\mathrm{ML}^2 \mathrm{~T}^{-2}\right] \end{aligned} $
Capacitance, $C=\frac{q}{U}$
$ \left(\text { as } U=\frac{1}{2} \frac{q^2}{C}\right) $
$ \begin{alignedat} \therefore[C] & =\left[\frac{q^2}{U}\right]=\left[\frac{Q^2}{\mathrm{ML}^2 \mathrm{~T}^{-2}}\right] \\ & =\left[\mathrm{M}^{-1} \mathrm{~L}^{-2} \mathrm{~T}^2 \mathrm{Q}^2\right] \end{aligned} $
BETA
