Civil Engineering - Applied Mechanics - Discussion
Discussion Forum : Applied Mechanics - Section 2 (Q.No. 34)
34.
For lifting a load of 50 kg through a distance of 2.5 cm, an effort of 12.5 kg is moved through a distance of 40 cm. The efficiency of the lifting machine, is
Discussion:
7 comments Page 1 of 1.
Hitman said:
7 years ago
I agree with 25%.
(2)
Anurag said:
7 years ago
Efficiency = Output/Input.
Output = load * distance moved by load (wx),
Input = effort * distance moved by effort(py),
Efficiency = (50 * 2.5/ 12.5 * 40)*100 = 25%.
Output = load * distance moved by load (wx),
Input = effort * distance moved by effort(py),
Efficiency = (50 * 2.5/ 12.5 * 40)*100 = 25%.
(3)
Kumar said:
8 years ago
25% is the right answer.
(2)
Sagar said:
8 years ago
25% is correct. I agree.
(2)
Kiran said:
8 years ago
Yes, I agree with you @Sanjeev Kumar.
Sanjeev kumar said:
8 years ago
Load = 50kg,
Effort = 12.5kg,
M.A = 50kg/12.5kg,
M.A = 500/125,
M.A = 4.
Now, V.R = distance moved by effort/distance moved by load.
V.R = 40/2.5,
V.R = 400/25 = 16.
Now, efficiency = M.A /V.RX 100 = 4/16X100 = 25%.
So option are not correct.
Effort = 12.5kg,
M.A = 50kg/12.5kg,
M.A = 500/125,
M.A = 4.
Now, V.R = distance moved by effort/distance moved by load.
V.R = 40/2.5,
V.R = 400/25 = 16.
Now, efficiency = M.A /V.RX 100 = 4/16X100 = 25%.
So option are not correct.
Juhi rao said:
8 years ago
The efficiency of machine =ma /vr , ma =2 and velocity ratio is 8.
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