Multiple choice

The volume of a sphere is increasing at the rate of 1200c.cm/sec.The rate of increase in its surface area when the radius is 10 cm is.

  1. 120 sq.cm/sec

  2. 240 sq.cm.sec

  3. 200 sq.cm.sec

  4. 100 sq.cm.sec
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

For a sphere, dV/dt = 4pi r^2 dr/dt and dS/dt = 8pi r dr/dt. Thus dS/dt = (2/r)(dV/dt) = (2/10)(1200) = 240 square cm/sec.

AI explanation

The volume of a sphere is V = (4/3) * pi * r^3 and its rate of change dV/dt = 4 * pi * r^2 * (dr/dt). Setting this equal to 1200 with r = 10 gives 1200 = 4 * pi * 100 * (dr/dt), so dr/dt = 3 / pi. The surface area is S = 4 * pi * r^2 and its rate of change dS/dt = 8 * pi * r * (dr/dt). Substituting the known values gives dS/dt = 8 * pi * 10 * (3 / pi), which results in 240 sq.cm/sec.

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