JEE Mains · Physics · STD 11 - Units,dimensions and measurement
- JEE Mains 2026
- Hard
- MCQ
Consider a modified Bernoulli equation.
\(\left( P +\frac{ A }{ Bt ^2}\right)+\rho g ( h + Bt )+\frac{1}{2} \rho V^2=\) constant
If \(t\) has the dimension of time then the dimensions of A and B are _________ , _________ respectively.
Pick an option to check your answer.
Answer & Solution
Correct Answer
(B) \( [ML^{0}T^{-1}] \) and \( [M^{0}LT^{-1}] \)
Step-by-step Solution
Detailed explanation
\(\Rightarrow\lfloor P \rfloor=\left[\frac{ A }{ Bt ^2}\right]\)....(1)
\(\Rightarrow[ h ]=[ Bt ]\) ....(2)
\(\Rightarrow\lfloor B \rfloor=\left[\frac{ h }{ t }\right]=\left[\frac{ L }{ T }\right]=\left[ LT ^{-1}\right]\)
Putting B is equation (1)
\(\left[ ML ^{-1} T^{-2}\right]=\left[\frac{ A }{ LT ^{-1} \times T ^2}\right]\)
\([ A ]=\left[ ML ^0 T^{-1}\right]\)
\(\Rightarrow[ h ]=[ Bt ]\) ....(2)
\(\Rightarrow\lfloor B \rfloor=\left[\frac{ h }{ t }\right]=\left[\frac{ L }{ T }\right]=\left[ LT ^{-1}\right]\)
Putting B is equation (1)
\(\left[ ML ^{-1} T^{-2}\right]=\left[\frac{ A }{ LT ^{-1} \times T ^2}\right]\)
\([ A ]=\left[ ML ^0 T^{-1}\right]\)
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