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Air is being pumped into a spherical balloon so that its volume increases at a rate of

100 cm^3 /s. How fast is the radius of the balloon increasing
when the diameter is 50 cm.



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Volume of a Sphere (V) = \frac{4}{3}  \pi  r^{3}
Differentiating both sides wrt 't'
 \frac{dV}{dt} = \frac{4}{3}  \pi 3 r^{2}  \frac{dr}{dt}

 \frac{dV}{dt} =100cm{3}/s, diameter=50cm, radius(r)=25cm

Putting the values accordingly,
100= \frac{4}{3} × \pi ×3×25×25× \frac{dr}{dt}
 \frac{dr}{dt}= \frac{100×3}{4 \pi ×3×25×25}
Cancelling the numerator and denominator in a appropriate manner,
we get,
 \frac{dr}{dt} =  \frac{1}{25 \pi }

Hence, the answer is  \frac{1}{25 \pi } (Rate of increase of radius)

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