Correct Answer - Option 3 : r =
\(\rm\sqrt{8t +25}\)
Concept:
The surface area of a sphere having radius r, S = 4πr2
Rate of change of the surface area of a sphere, \(\rm \frac{\mathrm{d} S}{\mathrm{d} t} = 8π r \ \frac{\mathrm{d} r}{\mathrm{d} t}\)
Calculation:
The surface area of a sphere, having radius r, S = 4πr2
Differentiate both sides w.r.t time, we get
\(\rm \frac{\mathrm{d} S}{\mathrm{d} t} = 8π r \ \frac{\mathrm{d} r}{\mathrm{d} t}\) .... (i)
It is given that , rate of change of area is constant ,i.e. \(\rm \frac{\mathrm{d} S}{\mathrm{d} t} = constant \) = K ( assume ) .
Putting \(\rm \frac{\mathrm{d} S}{\mathrm{d} t} = K\) in (i) , we get
K = 8πr \(\rm\frac{\mathrm{d} r}{\mathrm{d} t}\)
⇒ 8πr dr = K dt
Integrating both sides, we get
\(\rm \int 8π r\ dr = \int K \ dt\)
⇒ 4πr2 = Kt +C .... (ii)
We have given that at t = 0 , r = 5 , Putting this in (ii) , we get
100π = K × 0 +C .... (iii)
Also , given that at t = 3 , r = 7 , Putting this in (ii) , we get
196π = 3K + C .... (iv)
On solving (iii) and (iv), we get , C = 100π and K = 32π
Putting value of C and K in (ii), we get
4πr2 = 32π t + 100π
⇒ r2 = 8t + 25
⇒ r = \(\rm\sqrt{8t + 25}\) .
The correct option is 3.