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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, directly from above. The pendulum is performing small oscillations in a liquid of refractive index $μ$ . The equation of path of bob as seen by observer is best given as

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$μ^{2} x^{2} + y^{2} = l^{2}$

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

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

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

answer is D.

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Detailed Solution

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For small oscillation eye is almost vertically above the bob.
$y= \frac{lcosθ}{μ} and x=lsinθ$
Hence equation of trajectory is

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

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