A machine has a velocity ratio of 4. If It is used to raise a load of 400 N by 4 m using an effort of 20kgf, find the work done on the machine. Take g = 10 m/s $^{2}$ .

1600 J

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3200 J

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6400 J

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2000 J

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Solution

The correct option is B
3200 J

Now, $M . A . = \frac{load}{effort} = \frac{400}{200} = 2$

Efficiency of the machine = $\frac{M . A}{V . R} = \frac{2}{4} = \frac{1}{2},$

Percentage efficiency = $\frac{1}{2} \times 100$

So, work done on the machine (input) = $\frac{output}{efficiency} = \frac{1600}{(1 / 2)} = 3200 J$

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Similar questions

A machine has a velocity ratio of 4. It is used to raise a load of 400N by 4m using an effort 20kgf. Find the work done on the machine. Take g = 10 m/s $^{2}$ .

A machine has a velocity ratio of 4. It is used to raise a load of 200 N by 4 m using an effort of 20 kgf. Find the work done on the machine. Take $g = 10 m s^{- 2}$ .

In the solution to this problem "A machine has a velocity ratio of 4. It is used to raise a load of 200N by 4m using an effort 20kgf. Find the work done on the machine. Take g = 10 m/s2." effort is calculated as 200. Please explain how this was calculated & the formula for this. Also do explain the unit kgf of effort (what does f stand for).

The efficiency of a machine is the ratio of the work done on the load by the machine to the work done on the machine by the effort.

Q. An effort of

20 k g f

is applied on a machine through a distance of

80 c m

, when a load of

30 k g f

moves through a distance of

1 m

. Calculate (a) Velocity ratio (b) mechanical advantage of the machine.

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A machine has a velocity ratio of 4. If It is used to raise a load of 400 N by 4 m using an effort of 20kgf, find the work done on the machine. Take g = 10 m/s2 .

The correct option is B 3200 J Now, M.A.=loadeffort=400200=2 Efficiency of the machine = M.AV.R =24 =12, Percentage efficiency = 12×100 So, work done on the machine (input) = outputefficiency=1600(1/2)=3200 J

A machine has a velocity ratio of 4. If It is used to raise a load of 400 N by 4 m using an effort of 20kgf, find the work done on the machine. Take g = 10 m/s $^{2}$ .