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If the volume of a sphere increases at constant rate `((dv)/(dt)=4)`. If radius of the sphere is denoted by `r` then surface area of the sphere inceases at the rate`:-`A. `(4)/(r )`B. `(8)/(r )`C. `(12)/(r )`D. `(16)/(r )` |
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Answer» Correct Answer - B `V=(4)/(3)pir^(2)rArr(dV)/(dt)=(dV)/(dt)xx(dr)/(dt)=[(4)/(3)pi(2r)](dr)/(dt)=4` `rArr(dr)/(dt)=(4)/(4pir^(2))` `5=4pir^(2)rArr(dS)/(dt)=(dS)/(dr)xx(dr)/(dt)=[4pi(2r)](dr)/(dt)` `(dS)/(dt)=(8pir)((4)/(4pir^(2)))=(8)/(r)` |
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