Physics Q87 – AP-EAMCET 2026

Home Physics Work, Energy & PowerQuestion #639

A motor of efficiency

80 %

lifts water from a tank and delivers it from the end of a pipe which is

3.2 m

vertically above the level from which water is drawn. If the cross-sectional area of the pipe is

12 {cm}^{2}

and water leaves the end of the pipe at a speed of

6 {m s}^{- 1}

, then the power of the motor is (acceleration due to gravity =

10 {m s}^{- 2}

)

Clue 1 of 3: Energy Conservation and Power

The power supplied by the motor is used to increase both the potential and kinetic energy of the water, considering the motor's efficiency.

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Step 1: Identify Given Parameters and Required Conversions.✦ Active

List all given parameters and ensure they are in consistent SI units. The efficiency $η = 80 % = 0.8$ . The height $h = 3.2 m$ . The cross-sectional area of the pipe $A = 12 {cm}^{2}$ . Convert this to square meters: $A = 12 \times (10^{- 2} m)^{2} = 12 \times 10^{- 4} m^{2}$ . The speed of water $v = 6 {m s}^{- 1}$ . The acceleration due to gravity $g = 10 {m s}^{- 2}$ . The density of water $ρ$ is typically $1000 {kg m}^{- 3}$ if not specified.

💡 Teacher's Secret Hint

Always check units carefully, especially when area is given in ${cm}^{2}$ . A common mistake is to forget to square the conversion factor.

Step 2: Calculate Mass Flow Rate of Water.○ Expand

First, calculate the volume flow rate ( $\dot{V}$ ) using the cross-sectional area ( $A$ ) and the speed of water ( $v$ ). Then, calculate the mass flow rate ( $\dot{m}$ ) using the density of water ( $ρ$ ). The formulas are:

\dot{V} = A \times v

\dot{m} = ρ \times \dot{V}

Substituting the values:

\dot{V} = (12 \times 10^{- 4} m^{2}) \times (6 m/s) = 72 \times 10^{- 4} m^{3} / s

\dot{m} = (1000 {kg/m}^{3}) \times (72 \times 10^{- 4} m^{3} / s) = 7.2 kg/s

Step 3: Determine the Power Delivered to the Water (Output Power).○ Expand

The power delivered to the water ( $P_{o u t}$ ) is the rate at which its potential and kinetic energy increase. This can be expressed in terms of the mass flow rate ( $\dot{m}$ ), height ( $h$ ), and velocity ( $v$ ). The formula is:

P_{o u t} = \dot{m} g h + \frac{1}{2} \dot{m} v^{2}

Substitute the calculated mass flow rate and other given values:

P_{o u t} = (7.2 kg/s) (10 {m/s}^{2}) (3.2 m) + \frac{1}{2} (7.2 kg/s) (6 m/s)^{2}

P_{o u t} = 230.4 W + 129.6 W

P_{o u t} = 360 W

💡 Teacher's Secret Hint

Remember to account for both potential and kinetic energy changes. If the water was delivered at rest, only the potential energy term would be needed.

Step 4: Calculate the Power of the Motor (Input Power).○ Expand

The efficiency ( $η$ ) of the motor relates the output power ( $P_{o u t}$ ) to the input power (power of the motor, $P_{m o t o r}$ ) by the formula:

η = \frac{P_{o u t}}{P_{m o t o r}}

Rearrange the formula to solve for $P_{m o t o r}$ and substitute the values:

P_{m o t o r} = \frac{P_{o u t}}{η} = \frac{360 W}{0.8}

P_{m o t o r} = 450 W

💡 Teacher's Secret Hint

Efficiency is always a fraction less than 1 (or a percentage less than 100%). Ensure you divide by efficiency to find the total input power, as $P_{m o t o r}$ will always be greater than $P_{o u t}$ .

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Physics Question 87 – AP-EAMCET 2026

Clue 1 of 3: Energy Conservation and Power

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