Question 29 of 82intermediate🔧 ApplyMCQ1 mark

A cylindrical tank of radius 10 m is being filled with wheat at the rate of 314 cubic metre per hour. Then the depth of the wheat is increasing at the rate of

Exercise: Miscellaneous Exercise on Chapter 6 | Q: 16 | (Chapter: 39)
For More Understanding

Explanation

This is a related rates problem from Application of Derivatives. We need to find the rate of change of depth (height) with respect to time, given the rate of change of volume. Since the tank is cylindrical with fixed radius, we use the volume formula V = πr²h and differentiate with respect to time using the Chain Rule as shown in Example 31 of the context.

Solution Steps

  1. Step 1: Let V be the volume of wheat in the cylindrical tank. For a cylinder, V = πr²h, where r = 10 m (constant) and h is the depth of wheat.

  2. Step 2: Since r is constant, V = π(10)²h = 100πh.

  3. Step 3: Differentiating both sides with respect to time t using Chain Rule: dV/dt = 100π × dh/dt.

  4. Step 4: Given: dV/dt = 314 m³/h (rate at which wheat is being filled).

  5. Step 5: Substituting: 314 = 100π × dh/dt, therefore dh/dt = 314/(100π).

  6. Step 6: Using π = 3.14, we get dh/dt = 314/(100×3.14100 \times 3.14) = 314/314 = 1 m/h.

  7. Step 7: Thus, the depth of wheat is increasing at the rate of 1 m/h. The correct option is (A).

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