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The surface area of a balloon being inflated changes at constant rate . If initially, its radius was 5 unit and after 3 seconds it is 7 units, find the radius after t seconds, 
1. r = \(\rm\sqrt{8t +81}\)
2. r = \(\rm\sqrt{8t +9}\)
3. r = \(\rm\sqrt{8t +25}\)
4. r = \(\rm\sqrt{8t + 5}\)

1 Answer

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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. 

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