Question

Water is being poured at the rate of 36 m$^3$/min into a cylindrical vessel whose circular base is of radius 3 meters. Then the water level in the cylinder increases at the rate of:

• A. \(4\pi \, \text{m/min} \)
• B. \(\frac{4}{\pi} \, \text{m/min} \)
• C. \(\frac{1}{4\pi} \, \text{m/min} \)
• D. \(\frac{\pi}{4} \, \text{m/min} \)

Hint:

For problems involving cylindrical shapes, remember to use the volume formula \(V = \pi r^2 h\) and differentiate with respect to time to find the rate of change of height.

Answer 1

The Correct Option is A

The volume of a cylinder is given by the formula: \[ V = \pi r^2 h \] Where: - \(V\) is the volume of the cylinder, - \(r\) is the radius of the base, - \(h\) is the height (or water level in this case). We are given that water is being poured at the rate of 36 m$^3$/min, which means: \[ \frac{dV}{dt} = 36 \, \text{m}^3/\text{min} \] To find the rate of change of the water level \(h\), we differentiate the volume formula with respect to time: \[ \frac{dV}{dt} = \frac{d}{dt} \left( \pi r^2 h \right) = \pi r^2 \frac{dh}{dt} \] Substitute the values of \(r = 3\) meters and \(\frac{dV}{dt} = 36\) m$^3$/min: \[ 36 = \pi (3)^2 \frac{dh}{dt} \] \[ 36 = 9\pi \frac{dh}{dt} \] Solve for \(\frac{dh}{dt}\): \[ \frac{dh}{dt} = \frac{36}{9\pi} = \frac{4}{\pi} \] Thus, the rate of change of the water level is \(\frac{4}{\pi} \, \text{m/min}\).