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If surface area of a cube is changing at...

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`.

A

`5 m//s`

B

`5sqrt(3) m//s`

C

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

D

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

Text Solution

Verified by Experts

The correct Answer is:
C, D
Surface area of cube `S=6a^(2)` (where a=side of cube)
Body diagonal `1=sqrt(3)a`. Therefore `S=2l^(2)`
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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