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Please explain pressures of which two points are equated....
Dear student The lower point pressure and the pressure at distance r will be equated, Good Luck ….............................................
The air bubble is in the shape of a sphere.Now, the volume of an air bubble (V) with radius (r) is given by,\begin{align} V = \frac{4}{3}\pi r^3 \end{align}The rate of change of volume (V) with respect to time (t) is given by,\begin{align} \frac{dV}{dt} = \frac{4}{3}\pi \frac{d}{dr}(r^3).\frac{dr}{dt} \;\;\;[By\; Chain\; Rule] \end{align}\begin{align} = \frac{4}{3}\pi (3r^2).\frac{dr}{dt} \end{align}\begin{align} = \frac{4}{3}\pi r^2.\frac{dr}{dt} \end{align}It is given that\begin{align} \frac{dr}{dt}=\frac{1}{2} cm/s .\end{align}Therefore, when r = 1 cm,\begin{align} \frac{dV}{dt}=4\pi(1)^2.(\frac{1}{2})=2\pi\; cm^3/s \end{align}Hence, the rate at which the volume of the bubble increases is 2π cm3/s.
The air bubble is in the shape of a sphere.
Now, the volume of an air bubble (V) with radius (r) is given by,
\begin{align} V = \frac{4}{3}\pi r^3 \end{align}
The rate of change of volume (V) with respect to time (t) is given by,
\begin{align} \frac{dV}{dt} = \frac{4}{3}\pi \frac{d}{dr}(r^3).\frac{dr}{dt} \;\;\;[By\; Chain\; Rule] \end{align}
\begin{align} = \frac{4}{3}\pi (3r^2).\frac{dr}{dt} \end{align}
\begin{align} = \frac{4}{3}\pi r^2.\frac{dr}{dt} \end{align}
It is given that
\begin{align} \frac{dr}{dt}=\frac{1}{2} cm/s .\end{align}
Therefore, when r = 1 cm,
\begin{align} \frac{dV}{dt}=4\pi(1)^2.(\frac{1}{2})=2\pi\; cm^3/s \end{align}
Hence, the rate at which the volume of the bubble increases is 2π cm3/s.
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