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(B) Given, the time period of oscillation $T$ is $T \propto p^\alpha S^\beta E^\gamma$ or $T = k p^\alpha S^\beta E^\gamma$.Substituting the dimension

Vedclass · Accepted Answer

$\alpha=-\frac{5}{6}, \beta=\frac{1}{2}, \gamma=\frac{1}{3}$

Vedclass · Answer

(B) Given, the time period of oscillation $T$ is $T \propto p^\alpha S^\beta E^\gamma$ or $T = k p^\alpha S^\beta E^\gamma$.Substituting the dimensions of $T, p, S$, and $E$:$[M^0 L^0 T^1] = [ML^{-1} T^{-2}]^\alpha [ML^{-3}]^\beta [ML^2 T^{-2}]^\gamma$$[M^0 L^0 T^1] = [M^{\alpha+\beta+\gamma} L^{-\alpha-3\beta+2\gamma} T^{-2\alpha-2\gamma}]$Equating the powers of $M, L$, and $T$ on both sides, we get:$1) \alpha + \beta + \gamma = 0$$2) -\alpha - 3\beta + 2\gamma = 0$$3) -2\alpha - 2\gamma = 1$From equation $(3)$, $\alpha + \gamma = -\frac{1}{2}$.Substituting this into equation $(1)$, $-\frac{1}{2} + \beta = 0 \implies \beta = \frac{1}{2}$.Now, substitute $\beta = \frac{1}{2}$ into equation $(2)$:$-\alpha - 3(\frac{1}{2}) + 2\gamma = 0 \implies -\alpha + 2\gamma = \frac{3}{2}$.We have the system:$i) \alpha + \gamma = -\frac{1}{2}$$ii) -\alpha + 2\gamma = \frac{3}{2}$Adding $(i)$ and $(ii)$, $3\gamma = 1 \implies \gamma = \frac{1}{3}$.Substituting $\gamma = \frac{1}{3}$ into $(i)$, $\alpha + \frac{1}{3} = -\frac{1}{2} \implies \alpha = -\frac{1}{2} - \frac{1}{3} = -\frac{5}{6}$.Thus, $\alpha = -\frac{5}{6}, \beta = \frac{1}{2}, \gamma = \frac{1}{3}$.

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