JEE MAIN - Physics (2023 - 24th January Evening Shift - No. 17)
The frequency ($$\nu$$) of an oscillating liquid drop may depend upon radius ($$r$$) of the drop, density ($$\rho$$) of liquid and the surface tension (s) of the liquid as $$\nu=r^a\rho^b s^c$$. The values of a, b and c respectively are
$$\left( {{3 \over 2},{1 \over 2}, - {1 \over 2}} \right)$$
$$\left( { - {3 \over 2}, - {1 \over 2},{1 \over 2}} \right)$$
$$\left( {{3 \over 2}, - {1 \over 2},{1 \over 2}} \right)$$
$$\left( { - {3 \over 2},{1 \over 2},{1 \over 2}} \right)$$
Explanation
$[v]=\left[\mathrm{T}^{-1}\right]$
$$ \begin{aligned} & {[r]=\mathrm{L} \quad[s]=\left[\frac{\mathrm{MLT}^{-2}}{\mathrm{~L}}\right]} \\\\ & {[\rho]=\left[\frac{\mathrm{M}}{\mathrm{L}^{3}}\right]=\left[\mathrm{ML}^{-3}\right]} \\\\ & \Rightarrow v=r^{a} \rho^{b} \mathrm{~s}^{c} \\\\ & \Rightarrow \mathrm{T}^{-1}=\mathrm{L}^{a} \mathrm{M}^{b} \mathrm{~L}^{-3 b} \mathrm{M}^{c} \mathrm{~T}^{-2 c} \\\\ & \Rightarrow \mathrm{T}^{-1}=\mathrm{M}^{(b+c)} \mathrm{L}^{(a-3 b)} \mathrm{T}^{-2 c} \\\\ & -2 c=-1 \Rightarrow c=\frac{1}{2} \\\\ & b+c=0 \\\\ & \Rightarrow b=-\frac{1}{2} \\\\ & a-3 b=0 \Rightarrow 3 b=a \Rightarrow a=-\frac{3}{2} \\\\ & (a, b, c)=\left(-\frac{3}{2},-\frac{1}{2}, \frac{1}{2}\right) \end{aligned} $$
$$ \begin{aligned} & {[r]=\mathrm{L} \quad[s]=\left[\frac{\mathrm{MLT}^{-2}}{\mathrm{~L}}\right]} \\\\ & {[\rho]=\left[\frac{\mathrm{M}}{\mathrm{L}^{3}}\right]=\left[\mathrm{ML}^{-3}\right]} \\\\ & \Rightarrow v=r^{a} \rho^{b} \mathrm{~s}^{c} \\\\ & \Rightarrow \mathrm{T}^{-1}=\mathrm{L}^{a} \mathrm{M}^{b} \mathrm{~L}^{-3 b} \mathrm{M}^{c} \mathrm{~T}^{-2 c} \\\\ & \Rightarrow \mathrm{T}^{-1}=\mathrm{M}^{(b+c)} \mathrm{L}^{(a-3 b)} \mathrm{T}^{-2 c} \\\\ & -2 c=-1 \Rightarrow c=\frac{1}{2} \\\\ & b+c=0 \\\\ & \Rightarrow b=-\frac{1}{2} \\\\ & a-3 b=0 \Rightarrow 3 b=a \Rightarrow a=-\frac{3}{2} \\\\ & (a, b, c)=\left(-\frac{3}{2},-\frac{1}{2}, \frac{1}{2}\right) \end{aligned} $$
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