A 200 kg load is raised using a 110 m long lever as shown in the diagram above. The load is 10m from the pivot P. If the efficiency of the the lever is 80%, find the effort E required to lift the load.
[Take g = 10ms-2]

250

300

450

200

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Correct answer is A

ε= 80%, L=200× 10 = 2000N, $d_L$ =10m

$d_E$ = 110-10 = 100m, E=?

ε = $\frac{work done on the load}{work done by the effort}$ × 100%

⇒ work done on load = 2000 ×10 =20,000 j

⇒ work done by effort = E × 100 = 100E

⇒ 80 = $\frac{20,000}{100E}$ × 100%

⇒ 80 = $\frac{20,000}{E}$

⇒ E = $\frac{20,000}{80}$

⇒ E = 250N

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[Take g = 10ms-2]

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[Take g = 10ms-2]