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Miscellaneous Exercise 6(II) · Q145

Q.The volume of a spherical balloon being inflated changes at a constant rate. If initially its radius is 3 units and after 3 seconds it is 6 units, find the radius of the balloon after tt seconds.

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Constant rate of volume change means dVdt=k\dfrac{dV}{dt}=k (constant), so V(t)=V0+ktV(t)=V_0+kt where V0=43π(3)3=36πV_0=\dfrac43\pi(3)^3=36\pi. At t=3t=3, r=6r=6: V(3)=43π(6)3=288πV(3)=\dfrac43\pi(6)^3=288\pi, so 288π=36π+3k⇒k=84π288\pi=36\pi+3k\Rightarrow k=84\pi. So V(t)=36π+84πtV(t)=36\pi+84\pi t. Since V=43πr3V=\dfrac43\pi r^3, $r^3=\dfrac{3V}{4\pi}=\dfrac{3(36\pi+84\pi t)}{4\pi}=27+ …

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