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Solution

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Correct option is A)

$βy1β+βx1β=βR2β$

$dtdyβ=(Rβ2x)Rdtdxββ+(Rβ2x)_{2}2Rxdtdxββ$

$βy1β=Rβ2β+x1β=βRxβ2x+Rβ$

$y=Rβ2xRxβ................(1)$

Differentiating equation (1)

$dtdyβ=(Rβ2x)Rdtdxββ+(Rβ2x)_{2}2Rxdtdxββ$

$dtdyβ=(Rβ2x)_{2}[R(Rβ2x)+2Rx]dtdxββ=(Rβ2x)_{2}R_{2}vβ$

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