JEE MAIN - Physics (2023 - 15th April Morning Shift - No. 12)
The speed of a wave produced in water is given by $v=\lambda^{a} g^{b} \rho^{c}$. Where $\lambda, g$ and $\rho$ are wavelength of wave, acceleration due to gravity and density of water respectively. The values of $a, b$ and $c$ respectively, are :
$\frac{1}{2}, 0, \frac{1}{2}$
$1,1,0$
$1,-1,0$
$\frac{1}{2}, \frac{1}{2}, 0$
Explanation
$$
\begin{aligned}
& v=\lambda^{\mathrm{a}} \mathrm{g}^{\mathrm{b}} \rho^{\mathrm{c}} \\\\
& \text {using dimension formula} \\\\
& \Rightarrow\left[\mathrm{M}^0 \mathrm{~L}^1 \mathrm{~T}^{-1}\right]=\left[\mathrm{L}^1\right]^{\mathrm{a}}\left[\mathrm{L}^1 \mathrm{~T}^{-2}\right]^{\mathrm{b}}\left[\mathrm{M}^1 \mathrm{~L}^{-3}\right]^{\mathrm{c}} \\\\
& \Rightarrow\left[\mathrm{M}^0 \mathrm{~L}^1 \mathrm{~T}^{-1}\right]=\left[\mathrm{M}^{\mathrm{c}} \mathrm{L}^{\mathrm{a}+\mathrm{b}-\mathrm{c}} \mathrm{T}^{-2 \mathrm{~b}}\right] \\\\
& \therefore \mathrm{c}=0, \mathrm{a}+\mathrm{b}-3 \mathrm{c}=1,-2 \mathrm{~b}=-1 \Rightarrow \mathrm{b}=\frac{1}{2} \\\\
& \text { Now } \mathrm{a}+\mathrm{b}-3 \mathrm{c}=1 \\\\
& \Rightarrow \mathrm{a}+\frac{1}{2}-0=1 \\\\
& \Rightarrow \mathrm{a}=\frac{1}{2} \\\\
& \therefore \mathrm{a}=\frac{1}{2}, \mathrm{~b}=\frac{1}{2}, \mathrm{c}=0
\end{aligned}
$$
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