CBSE 2025, Set 3

Question

A cylindrical water container has developed a leak at the bottom. The water is leaking at the rate of $5 	ext{ cm}^{3}	ext{/s}$ from the leak. If the radius of the container is $15 	ext{ cm}$, find the rate at which the height of water is decreasing inside the container, when the height of water is $2 	ext{ metres}$.

Edvaya · Accepted Answer

The correct answer is Subjective. **Step 1:** State the given values and formula.
 Let $V$ be the volume of water, $r$ be the radius, and $h$ be the height of the water level in the cylindrical container at any time $t$.
 Given:
 * Rate of change of volume, $\frac{dV}{dt} = -5 	ext{ cm}^3	ext{/s}$ (negative because volume is decreasing).
 * Radius of the cylinder, $r = 15 	ext{ cm}$ (this is constant as it's a uniform cylinder).
 * Height of water at the instant, $h = 2 	ext{ m} = 200 	ext{ cm}$.

The volume of a cylinder is given by the formula:
 $$V = \pi r^2 h$$

**Step 2:** Substitute the constant radius into the volume equation.
 Since the radius does not change with time, we substitute $r = 15$:
 $$V = \pi (15)^2 h$$
 $$V = 225\pi h$$

**Step 3:** Differentiate with respect to time $t$.
 $$\frac{dV}{dt} = 225\pi \frac{dh}{dt}$$

**Step 4:** Substitute the known rate of volume change to find the rate of height change.
 $$-5 = 225\pi \frac{dh}{dt}$$
 $$\frac{dh}{dt} = \frac{-5}{225\pi}$$
 $$\frac{dh}{dt} = -\frac{1}{45\pi} 	ext{ cm/s}$$

The negative sign indicates that the height is decreasing over time. 
 *(Note: Because the cylinder is uniform, the rate of decrease in height is constant and independent of the current height of 2 meters.)*

Therefore, the rate at which the height of the water is decreasing is **$\frac{1}{45\pi} 	ext{ cm/s}$**.

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Mathematics > Calculus > Application Of Derivatives

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