Class 11 physics, NCERT chapter 1. Attempt each question, check your answer, then open the step-by-step solution.
The dimensional formula of Planck's constant $h$ is
Answer: (B) $[ML^2T^{-1}]$
Use $E = h\nu$, so $h = E/\nu$.
$$[h] = \frac{[ML^2T^{-2}]}{[T^{-1}]} = [ML^2T^{-1}]$$
This is also the dimension of angular momentum.
Solution by Sreeraj P, M.Sc Physics
Consider a modified Bernoulli equation.
$$\left(P + \frac{A}{Bt^2}\right) + \rho g(h + Bt) + \frac{1}{2}\rho V^2 = \text{constant}$$
If $t$ has the dimension of time then the dimensions of $A$ and $B$ are _____, _____ respectively.
Answer: (D) $[\mathrm{ML^0T^{-1}}]$ and $[\mathrm{M^0LT^{-1}}]$
Only quantities with the same dimensions can be added, so every term inside a bracket must match the other term in that bracket.
**Finding $B$:** $Bt$ is added to the height $h$, so $[Bt] = [\mathrm{L}]$:
$$[B] = \frac{[\mathrm{L}]}{[\mathrm{T}]} = [\mathrm{M^0LT^{-1}}]$$
**Finding $A$:** $\dfrac{A}{Bt^2}$ is added to the pressure $P$, so
$$\left[\frac{A}{Bt^2}\right] = [P] = [\mathrm{ML^{-1}T^{-2}}]$$
$$[A] = [\mathrm{ML^{-1}T^{-2}}]\,[B]\,[t^2] = [\mathrm{ML^{-1}T^{-2}}][\mathrm{LT^{-1}}][\mathrm{T^2}] = [\mathrm{ML^0T^{-1}}]$$
So $A = [\mathrm{ML^0T^{-1}}]$ and $B = [\mathrm{M^0LT^{-1}}]$, which is option (D).
Solution by Sreeraj P, M.Sc Physics
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