मैसेज राइटिंग कन्वरसेशन बिटवीन वेदर एंड विनीता​

Explanation:Answer:Rate at which volume of the bubble is increasing = 72 π cm³/sStep-by-step explanation:Given:Radius of an air bubble is increasing at the rate of 2 cm/sTo Find:The rate at which the volume of the bubble is increasing when the radius of the air bubble is 3 cm.Solution:Let the radius of the air bubble be r cm and volume be V.Here the air bubble is in the shape of a sphere.Volume of a sphere = 4/3 × π × r³Now the rate of volume change with respect to time is given by,\sf \dfrac{dV}{dt} = \dfrac{d}{dt} (\dfrac{4}{3}\: \pi r^{3}) dtdV = dtd ( 34 πr 3 )Using chain rule,\sf \dfrac{dV}{dt} =\dfrac{d}{dr} (\dfrac{4}{3}\: \pi r^{3} ).\dfrac{dr}{dt} dtdV = drd ( 34 πr 3 ). dtdr Differentiating,\sf \dfrac{dV}{dt} =\dfrac{4}{3}\: \pi \times 3r^{2} .\dfrac{dr}{dt} dtdV = 34 π×3r 2 . dtdr Now the rate of increase of radius of the air bubble is given as,\sf \dfrac{dr}{dt} =2\:cm/s dtdr =2cm/sAlso by given, the radius of the air bubble is 3 cm.Substitute the data,\sf \dfrac{dV}{dt} =\dfrac{4}{3}\: \pi \times 3\times 3^{2} \times 2 dtdV = 34 π×3×3 2 ×2Simplifying we get,\sf \dfrac{dV}{dt} =4\times \pi \times 9\times 2 dtdV =4×π×9×2⇒ 72 π cm³/sTherefore the rate at which the volume of the air bubble is increasing is 72 π cm³/s.