Water flows, at 9 km per hour, through a cylindrical pipe of cross-sectional area 25 cm². If this water is collected into a rectangular cistern of dimensions 7.5 m by 5 m by 4 m; calculate the rise in level in the cistern in 1 hour 15 minutes.

Given,

Rate of water flow = 9 km/hr

= 9 × 10⁵ cm/hr. [As, 1 km = 10⁵ cm.]

1 hour 15 minutes = $1 + \dfrac{15}{60} = 1 + \dfrac{1}{4} = \dfrac{5}{4}$ hours.

Volume of water flowing in $\dfrac{5}{4}$ hours = Area of cross-section of pipe × Rate of water flow × $\dfrac{5}{4}$ hours

= 25 cm² × 9 km/hr × $\dfrac{5}{4}$ hr

= 25 cm² × (9 × 10⁵) cm/hr × $\dfrac{5}{4}$ hr

= $\dfrac{1125 \times 10^5}{4}$ cm³.

Let increase in level of water be h cm.

Given,

Length of rectangular cistern = 7.5 m = 750 cm

Breadth of rectangular cistern = 5 m = 500 cm

Volume of water increase in cistern = 750 × 500 × h cm³

We know that,

Volume of water increase in cistern = Volume of water flowing in $\dfrac{5}{4}$ hours

$\Rightarrow 750 × 500 × h = \dfrac{1125 \times 10^5}{4} \\[1em] \Rightarrow h = \dfrac{1125 \times 10^5}{4 \times 750 \times 500} \\[1em] \Rightarrow h = 0.00075 \times 10^5 \\[1em] \Rightarrow h = 75 \text{ cm}.$

Hence, there is an increase of 75 cm in level of water in cistern.