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Updated on : 2022-09-05

Solution

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Correct option is A)

$⇒y1 +−x1 =−R2 $

$dtdy =(R−2x)Rdtdx +(R−2x)_{2}2Rxdtdx $

$⇒y1 =R−2 +x1 =−Rx−2x+R $

$y=R−2xRx ................(1)$

Differentiating equation (1)

$dtdy =(R−2x)Rdtdx +(R−2x)_{2}2Rxdtdx $

$dtdy =(R−2x)_{2}[R(R−2x)+2Rx]dtdx =(R−2x)_{2}R_{2}v $

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