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If $C, R, L$ and $I$ denote capacity, resistance, inductance and electric current respectively, the quantities having the same dimensions of time are :
(1) $C R$
(2) $\frac{L}{R}$
(3) $\sqrt{L C}$
(4) $L I^2$
Options:
(1) $C R$
(2) $\frac{L}{R}$
(3) $\sqrt{L C}$
(4) $L I^2$
Solution:
1791 Upvotes
Verified Answer
The correct answer is:
(1), (2) and (3) only
$[C]=\left[\mathrm{M}^{-1} \mathrm{~L}^{-2} \mathrm{~T}^4 \mathrm{~A}^2\right],[R]=\left[\mathrm{M} \mathrm{L}^2 \mathrm{~T}^{-3} \mathrm{~A}^{-2}\right]$
$[L]=\left[\mathrm{ML}^2 \mathrm{~T}^{-2} \mathrm{~A}^{-2}\right]$ and $[I]=\left[\mathrm{M}^0 \mathrm{~L}^0 \mathrm{~T}^0 \mathrm{~A}\right]$
(1) $[C R]=\left[\mathrm{M}^{-1} \mathrm{~L}^{-2} \mathrm{~T}^4 \mathrm{~A}^2\right]\left[\mathrm{ML}^2 \mathrm{~T}^{-3} \mathrm{~A}^{-2}\right]$
$=\left[\mathrm{M}^0 \mathrm{~L}^0 \mathrm{TA}^0\right]$
(2) $\left[\frac{L}{R}\right]=\frac{\left[\mathrm{ML}^2 \mathrm{~T}^{-2} \mathrm{~A}^{-2}\right]}{\left[\mathrm{ML}^2 \mathrm{~T}^{-3} \mathrm{~A}^{-2}\right]}=\left[\mathrm{M}^0 \mathrm{~L}^0 \mathrm{TA}^0\right]$
(3) $[\sqrt{L C}]=\left(\left[\mathrm{ML}^2 \mathrm{~T}^{-2} \mathrm{~A}^{-2}\right] \times\left[\mathrm{M}^{-1} \mathrm{~L}^{-2} \mathrm{~T}^4 \mathrm{~A}^2\right]\right)^{1 / 2}$
$=\left[\mathrm{M}^0 \mathrm{~L}^0 \mathrm{TA}^0\right]$
(4) $\left[L I^2\right]=\left[\mathrm{ML}^2 \mathrm{~T}^{-2} \mathrm{~A}^{-2}\right]\left[\mathrm{M}^0 \mathrm{~L}^0 \mathrm{~T}^0 \mathrm{~A}^0\right]^2$
$=\left[\mathrm{ML}^2 \mathrm{~T}^{-2} \mathrm{~A}^0\right]$
Hence, option (d) is correct.
$[L]=\left[\mathrm{ML}^2 \mathrm{~T}^{-2} \mathrm{~A}^{-2}\right]$ and $[I]=\left[\mathrm{M}^0 \mathrm{~L}^0 \mathrm{~T}^0 \mathrm{~A}\right]$
(1) $[C R]=\left[\mathrm{M}^{-1} \mathrm{~L}^{-2} \mathrm{~T}^4 \mathrm{~A}^2\right]\left[\mathrm{ML}^2 \mathrm{~T}^{-3} \mathrm{~A}^{-2}\right]$
$=\left[\mathrm{M}^0 \mathrm{~L}^0 \mathrm{TA}^0\right]$
(2) $\left[\frac{L}{R}\right]=\frac{\left[\mathrm{ML}^2 \mathrm{~T}^{-2} \mathrm{~A}^{-2}\right]}{\left[\mathrm{ML}^2 \mathrm{~T}^{-3} \mathrm{~A}^{-2}\right]}=\left[\mathrm{M}^0 \mathrm{~L}^0 \mathrm{TA}^0\right]$
(3) $[\sqrt{L C}]=\left(\left[\mathrm{ML}^2 \mathrm{~T}^{-2} \mathrm{~A}^{-2}\right] \times\left[\mathrm{M}^{-1} \mathrm{~L}^{-2} \mathrm{~T}^4 \mathrm{~A}^2\right]\right)^{1 / 2}$
$=\left[\mathrm{M}^0 \mathrm{~L}^0 \mathrm{TA}^0\right]$
(4) $\left[L I^2\right]=\left[\mathrm{ML}^2 \mathrm{~T}^{-2} \mathrm{~A}^{-2}\right]\left[\mathrm{M}^0 \mathrm{~L}^0 \mathrm{~T}^0 \mathrm{~A}^0\right]^2$
$=\left[\mathrm{ML}^2 \mathrm{~T}^{-2} \mathrm{~A}^0\right]$
Hence, option (d) is correct.
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