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A pendulum of length $l$ is free to oscillate in vertical plane about point O. An observer is viewing the bob of the pendulum directly from above. The pendulum is performing small oscillation in water (refractive index is $μ$ ) about its equilibrium position. The equation of trajectory of bob as seen by observer is (x and y coordinates are with respect to observer)

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$μ^{2} x^{2} + y^{2} = l^{2}$

$\frac{x^{2}}{l^{2}} + \frac{y^{2}}{μ^{2} l^{2}} = 1$

$x^{2} + y^{2} = {(\frac{l}{μ})}^{2}$

$\frac{x^{2}}{l^{2}} + \frac{y^{2}}{{(l / μ)}^{2}} = 1$

answer is D.

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$x^{2} + {(μ y)}^{2} = l^{2}$

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