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एक ट्यूबवेल का पम्प 2400 किग्रा. पानी प्रति मिनट ऊपर फेंकता है। यदि पानी के बाहर आने की चाल 3 मी./से. हो तो पम्प की शक्ति ज्ञात कीजिये। यदि पम्प 5 घंटे तक चले तो पम्प द्वारा किये गये कार्य का मान ज्ञात करिये।

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दिया गया है:

m = 2400 किग्रा.

t = 1 मिनट = 60 सेकण्ड

v = 3 मीटर / सेकण्ड

\(\therefore \ \)पम्प की शक्ति  \(=\frac{1}{2}\frac{mv^2}{t}\)

\(=\frac{2400\times 3 \times 3}{2\times 60}\)

यदि पम्प 5 घंटे चलता है तब किया गया कार्य

W = P × t

= 180 × 5 × 60 x 60

= 18 × 5 × 6 × 6 × 103 जूल

= 3240 × 103

या W = 3.24 x 106 जूल

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