Stokes' law states that the viscous drag forc | vedclass

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(A) The given equation is $\frac{V}{t} = k \left( \frac{p}{l} \right)^a \eta^b r^c$.The dimensions of the quantities are:$[V/t] = [L^3 T^{-1}]$$[p/l]

Vedclass · Accepted Answer

$a=1, b=-1, c=4$

Vedclass · Answer

(A) The given equation is $\frac{V}{t} = k \left( \frac{p}{l} \right)^a \eta^b r^c$.The dimensions of the quantities are:$[V/t] = [L^3 T^{-1}]$$[p/l] = [M L^{-1} T^{-2} / L] = [M L^{-2} T^{-2}]$$[\eta] = [M L^{-1} T^{-1}]$$[r] = [L]$Substituting these into the equation:$[L^3 T^{-1}] = [M L^{-2} T^{-2}]^a [M L^{-1} T^{-1}]^b [L]^c$$[L^3 T^{-1}] = M^{a+b} L^{-2a-b+c} T^{-2a-b}$Equating the powers of $M, L,$ and $T$ on both sides:For $M$: $a + b = 0 \Rightarrow b = -a$For $T$: $-2a - b = -1$Substituting $b = -a$ into the $T$ equation: $-2a - (-a) = -1 \Rightarrow -a = -1 \Rightarrow a = 1$.Since $b = -a$,we get $b = -1$.For $L$: $-2a - b + c = 3$Substituting $a = 1$ and $b = -1$: $-2(1) - (-1) + c = 3 \Rightarrow -2 + 1 + c = 3 \Rightarrow -1 + c = 3 \Rightarrow c = 4$.Thus,the values are $a=1, b=-1, c=4$.

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