The ratio of the rate of flow of water in pipes varies inversely as the square of the radius of the pipes. What is the ratio of the rates of flow in two pipes of diameters 2 cm and 4 cm?

### Concept & Inverse Square Variation When a quantity varies inversely as the square of another, their relationship is given by $y = \frac{k}{x^2}$. Consequently, the ratio of two states is $\frac{y_1}{y_2} = \left(\frac{x_2}{x_1}\right)^2$. Keep in mind that the ratio of diameters is identical to the ratio of radii. ### Step-by-Step Solution 1. Let the rate of flow be $R$ and the radius of the pipe be $r$. 2. The problem states $R \propto \frac{1}{r^2}$, which gives: $$ R = \frac{k}{r^2} $$ 3. We are given the diameters of two pipes: $D_1 = 2$ cm and $D_2 = 4$ cm. 4. Convert diameters to radii ($r = D/2$): $$ r_1 = \frac{2}{2} = 1 \text{ cm} $$ $$ r_2 = \frac{4}{2} = 2 \text{ cm} $$ 5. The rates of flow for the two pipes are: $$ R_1 = \frac{k}{r_1^2} = \frac{k}{1^2} = k $$ $$ R_2 = \frac{k}{r_2^2} = \frac{k}{2^2} = \frac{k}{4} $$ 6. The required ratio of the rates of flow is: $$ \frac{R_1}{R_2} = \frac{k}{k / 4} = \frac{4}{1} $$ 7. Thus, the ratio is $4 : 1$. ### Exam Strategy & Shortcut **Direct Ratio Calculation:** For inverse square variation, $\frac{R_1}{R_2} = \left(\frac{r_2}{r_1}\right)^2$. Since the ratio of radii is the same as the ratio of diameters, we can just use the diameters directly: $$ \frac{R_1}{R_2} = \left(\frac{D_2}{D_1}\right)^2 = \left(\frac{4}{2}\right)^2 = (2)^2 = 4 $$ Which translates to a ratio of $4 : 1$. ### Common Pitfall Noticing that the second diameter is twice the first and hastily choosing $1 : 2$ or $2 : 1$, completely forgetting to square the values as dictated by the word "square" in the prompt. Also, taking the ratio as $R_1/R_2 = (r_1/r_2)^2$ (direct square variation) giving $1:4$ instead of $4:1$. ### Final Answer Therefore, the correct answer is **4 : 1**.