In what time would a cistern be filled by three pipes whose diameters are 1cm, \(\frac{4}{3}\) cm, 2 cm, running together, when the largest alone will fill it in 61 minutes, the amount of water flowing in by each pipe being proportional to square of its diameter ?

Explanation: Let the diameters be \(d_1 = 1\) cm, \(d_2 = \frac{4}{3}\) cm, \(d_3 = 2\) cm. The flow rate (\(R\)) is proportional to the square of the diameter (\(d^2\)). \(R \propto d^2\). Let \(k\) be the constant of proportionality. \(R = k d^2\). The largest pipe is \(d_3 = 2\) cm. Its flow rate is \(R_3 = k (2^2) = 4k\). This pipe fills the cistern in 61 minutes. Volume of cistern = \(R_3 \times 61 = 4k \times 61 = 244k\). The flow rates for the other pipes are: \(R_1 = k (1^2) = k\). \(R_2 = k (\frac{4}{3})^2 = k \frac{16}{9}\). Combined flow rate (\(R_{total}\)) when all three pipes run together: \(R_{total} = R_1 + R_2 + R_3 = k + k \frac{16}{9} + 4k = k (1 + \frac{16}{9} + 4) = k (\frac{9 + 16 + 36}{9}) = k \frac{61}{9}\). Time taken (\(T\)) to fill the cistern = \(\frac{\text{Volume}}{\text{Combined Flow Rate}}\). \(T = \frac{244k}{k \frac{61}{9}} = \frac{244}{1} \times \frac{9}{61}\). \(244 \div 61 = 4\). \(T = 4 \times 9 = 36\) minutes. (The provided solution does not include \(k\), but the calculation is mathematically sound and reaches the correct answer).