Mechanical Engineering - Engineering Mechanics - Discussion

Discussion Forum : Engineering Mechanics - Section 2 (Q.No. 13)
13.
A weight of 1000 N can be lifted by an effort of 80 N. If the velocity ratio is 20, the machine is
reversible
non-reversible
ideal
Answer: Option
Explanation:
No answer description is available. Let's discuss.
Discussion:
32 comments Page 1 of 4.

Shashidhar hiremath said:   2 years ago
Here, 1000N/80N = 12.5N
Then, Effcny = (12.5/V. R) * 100 = 62.5,
So, the eff. of a machine is 62.5% and 62.5% > 50%.
If the efficiency is greater than 50%, it is known as a "reversible machine".
(4)

Akhilesha said:   4 years ago
I disagree with the answer. I think the answer is non-reversible because the maximum efficiency is 62.5 which greater than 50.
(2)

Omprakash Yadav said:   8 years ago
Mechanical Efficiency MA=load Lifted/Effort.

1000/80 =12.5,
Velocity ratio =20,

MA=VR 12.5 /20 *100 =62.5 %.
So it's Reversible, because >50 % is Reversible, <50% Irreversible i.e Self-locking, used in screw jack.
100% ideal machine.
(2)

Subhash said:   6 years ago
It is Less than 50%.
(2)

Chander kant said:   1 decade ago
MA = W/P = 1000/80 = 12.5.

Efficiency = 12.5/20*100 = 62.5%.

So its efficiency is above 50% so it is a reversible machine.
(1)

AKSHAY KALE said:   8 years ago
Less than 50% is irreversible.

Greater than 50% is reversible.
(1)

Entekarunakar said:   7 years ago
Thanks to all for proving the correct explanation.
(1)

Jithin.k said:   7 years ago
What about the universal machine?
(1)

Aishu said:   7 years ago
Why should we multiply 100 with 20 in calculating velocity ratio? Please explain.
(1)

Bhuvnesh Rana said:   7 years ago
No machine is an ideal machine in practice.

And the efficiency of a machine can never be 100% in practice.
(1)


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