KCET · Physics · Units and Dimensions
If \(C\) be the capacitance and \(V\) be the electric potential, then the dimensional formula of \(C V^{2}\) is
- A \(\left[\mathrm{ML}^{2} \mathrm{~T}^{-2} \mathrm{~A}^{0}\right]\)
- B \(\left[\mathrm{MLT}^{-2} \mathrm{~A}^{-1}\right]\)
- C \(\left[\mathrm{M}^{0} \mathrm{LT}{ }^{2} \mathrm{~A}^{0}\right]\)
- D \(\left[\mathrm{ML}{ }^{3} \mathrm{TA}\right]\)
Answer & Solution
Correct Answer
(A) \(\left[\mathrm{ML}^{2} \mathrm{~T}^{-2} \mathrm{~A}^{0}\right]\)
Step-by-step Solution
Detailed explanation
We know,
Energy, \(E=C V^{2}\)
Dimensions of \(C V^{2}=\) Dimensions of energy
\(=\left[\mathrm{ML}^{2} \mathrm{~T}^{-2}\right]\)
Energy, \(E=C V^{2}\)
Dimensions of \(C V^{2}=\) Dimensions of energy
\(=\left[\mathrm{ML}^{2} \mathrm{~T}^{-2}\right]\)
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