Verified answer

Given : An electric motor of 1hp is used to run water pump. The water pump takes 10 minutes to fill an overhead tank. The tank has a capacity of 800 liters and height of 15 m.

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To Find : Find the work done by motor and Efficiency of the motor .

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SolutioN :

[tex] \maltese [/tex] Formula Used :

• [tex] {\underline{\boxed{\pmb{\sf{ Power = \dfrac{Work \; Done}{Time} }}}}} [/tex]

• [tex] {\underline{\boxed{\pmb{\sf{ Efficiency = \dfrac{W}{ W' } \times 100 }}}}} [/tex]

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Where :

• W = Work Done
• W' = Required form of Output

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[tex] \maltese [/tex] Calculating the Work Done :

• Converting the Values :

[tex] \mapsto \; \; \sf{ 1 \; hp = \green{746 \; Watt} } [/tex]

[tex] \mapsto \; \; \sf{ 10 \; min = \green{600 \; s} } [/tex]

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• Let's Go :

[tex] {\dashrightarrow{\qquad{\sf{ Work \; Done = Power \times Time }}}} \\ \\ \\ \\ \ {\dashrightarrow{\qquad{\sf{ Work \; Done = 746 \times 600 }}}} \\ \\ \\ \\ \ {\qquad \; \; {\dashrightarrow \; {\underline{\boxed{\pmb{\sf{ Work \; Done = 447600 \; Watt }}}}}}} \; {\red{\bigstar}} [/tex]

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[tex] \maltese [/tex] Calculating the Efficiency :

• Converting the Values :

[tex] \begin{gathered} \qquad \; \mapsto \; \sf { W' = mgh } \\ \\ \\ \end{gathered} [/tex]

[tex] \begin{gathered} \qquad \; \mapsto \; \sf { W' = 800 \times 10 \times 15 } \\ \\ \\ \end{gathered} [/tex]

[tex] \begin{gathered} \qquad \; \mapsto \; \sf { W' = 800 \times 150 } \\ \\ \\ \end{gathered} [/tex]

[tex] \begin{gathered} \qquad \; \mapsto \; {\underline{\boxed{\pmb{\sf{ W' = 120000 \; Joules }}}}} \; {\orange{\bigstar}} \\ \\ \\ \end{gathered} [/tex]

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• Let's Go :

[tex] {\dashrightarrow{\qquad{\sf{ Efficiency = \bigg\lgroup \dfrac{120000}{447600} \bigg\rgroup \times 100 }}}} \\ \\ \\ \\ \ {\dashrightarrow{\qquad{\sf{ Efficiency = \bigg\lgroup \dfrac{120000}{4476\cancel{00}} \bigg\rgroup \times \cancel{100} }}}} \\ \\ \\ \\ \ {\dashrightarrow{\qquad{\sf{ Efficiency = \bigg\lgroup \dfrac{120000}{4476} \bigg\rgroup }}}} \\ \\ \\ \\ \ {\dashrightarrow{\qquad{\sf{ Efficiency = \bigg\lgroup \cancel\dfrac{120000}{4476} \bigg\rgroup }}}} \\ \\ \\ \\ \ {\qquad \; \; {\dashrightarrow \; {\underline{\boxed{\pmb{\sf{ Efficiency = 26.8 \; \% }}}}}}} \; {\purple{\bigstar}} [/tex]

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[tex] \therefore \; [/tex] Work Done by the motor is 447600 Watt and Efficiency of the motor is 26.8 % .

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Power of pump, P = 1 hp = 746 W

time, t = 10 min = (10*60) s = 600 s

height, h = 15 m

mass, m = 800 kg (800 L; 1 L = 1 kg)

∴ Input Energy = P x t = 746 x 600 = 447600 J

Now, output energy = work done to lift water

∴ output energy or actual work done = mgh = 800 x 10 x 15 = 120000 J

So, now,
Efficiency = {Output Energy/Input Energy}*100

                = {120000/447600}*100

∴ Efficiency = 26.8 %