A spherical iron ball 10cm in radius is coated with a layer of ice of uniform thickness that melts at a rate of `50c m^3//m in` . When the thickness of ice is 5cm, then find the rate at which the thickness of ice decreases.

Options

A. `(1)/(9pi)`
B. `(1)/(18pi)`
C. `(1)/(38pi)`
D. `(5)/(6pi)`

Answer

B

Explanation

Let the thickness of layer of ice is x cm, the volume of spherical ball (only ice layer ) is
` V= ( 4) /(3) pi [ (10 + x ) ^(3) - 10^(3)]" " `… (i)
On diferentiating Eq. (i) w.r.t. `'t'`, we get
`" " (d V) /(dt) = ( 4) /(3) pi ( 3 (10 + x )^(2)) (dx ) /(dt) = - 50` [ given ]
[- ve sign indicate that volume is decreasing as time passes ].
`rArr " " 4pi ( 10 + x ) ^(2) (dx) /(dt) = -50 `
At `x = 5` cm
` (dx) /(dt) [ 4pi ( 10 + 5)^(2) ] = - 50 `
`rArr (dx) /(dt) = - (50 ) /( 225( 4pi )) =- (1) /(9( 2pi)) = - (1) /( 18 pi )` cm /min
So, the thickness of the ice decreases at the rate of `(1) /( 18 pi )` cm / min.