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If surface area of a cube is changing at a rate of 5 m^(2)//s, find the rate of change of body diagonal at the moment when side length is 1 m.

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Answer Text

`5 m//s``5sqrt(3) m//s``5/2sqrt(3) m//s``5/(4sqrt(3)) m//s`

Answer :
C::D
Solution :
Surface area of cube `S=6a^(2)` (where a=side of cube) <br> Body diagonal `1=sqrt(3)a`. Therefore `S=2l^(2)` <br> Differentiating it w.r.t. time `(dS)/(dt)=2(2l)(dl)/(dt)rArr (dl)/(dt)=(1)/(4(sqrt(3)a))(dS)/(dt)=5/(4sqrt(3)) m//s`

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