
The Velocity Ratio of a machine is more than 1. It
A. Acts at a force multiplier
B. Gives loss of speed
C. A and B
D. Either A and B
Answer
589.5k+ views
Hint: Velocity ratio is the ratio of velocities or displacement of effort to velocities or displacement of load. If the velocity ratio of a machine is greater than one, it implies that the machine is a force multiplier and causes loss of speed.
Complete Step By Step Solution:
Anything that simplifies our work is a simple machine. Effort is the amount of force put in to achieve the required task. Load is the force exerted by a machine for completing the task. So both effort and load are a type of force.
Velocity Ratio is defined as the ratio of displacement of point of application of effort to displacement of point of application of load. If we denote displacement of point of application of effort by ${{D}_{E}}$ and displacement of point of application of load by ${{D}_{L}}$ then
Velocity Ratio = \[\dfrac{{{D}_{E}}}{{{D}_{L}}}\]
WE are given that velocity ratio is greater than 1, so
\[\dfrac{{{D}_{E}}}{{{D}_{L}}}>1\]
Efficiency (e) is the ratio of amount of work by load to work done by effort. Work is the product of force and displacement of point of application of force. So work done by load (${{W}_{L}}$) is equal to
${{W}_{L}}=L\times {{D}_{L}}$
Similarly work done by load (${{W}_{E}}$) is equal to
${{W}_{E}}=E\times {{D}_{E}}$
So,
$e=\dfrac{{{W}_{L}}}{{{W}_{E}}}$
$e=\dfrac{L\times {{D}_{L}}}{E\times {{D}_{E}}}$
$e=\dfrac{\dfrac{L}{E}}{\dfrac{{{D}_{E}}}{{{D}_{L}}}}$
For ideal machine we consider efficiency of that machine to be 1
So,
E=1
$\dfrac{\dfrac{L}{E}}{\dfrac{{{D}_{E}}}{{{D}_{L}}}}=1$
$\dfrac{L}{E}=\dfrac{{{D}_{E}}}{{{D}_{L}}}$
Now as\[\dfrac{{{D}_{E}}}{{{D}_{L}}}>1\], we can write
$\dfrac{L}{E}>1$
$L>E$
This implies that force at output (load) is greater than force at input (effort), so the machine acts as a force multiplier.
Also the distance moved at load is smaller than the distance moved at effort at the same time, this implies that the machine has caused loss of speed.
Hence option C is correct.
Note: Velocity Ratio can have any value. None of them is bound to have its value less than 1. This is often a misconception among students, but actually it is efficiency which is bound to have its value less than one in practical life and equal to one in ideal condition.
Complete Step By Step Solution:
Anything that simplifies our work is a simple machine. Effort is the amount of force put in to achieve the required task. Load is the force exerted by a machine for completing the task. So both effort and load are a type of force.
Velocity Ratio is defined as the ratio of displacement of point of application of effort to displacement of point of application of load. If we denote displacement of point of application of effort by ${{D}_{E}}$ and displacement of point of application of load by ${{D}_{L}}$ then
Velocity Ratio = \[\dfrac{{{D}_{E}}}{{{D}_{L}}}\]
WE are given that velocity ratio is greater than 1, so
\[\dfrac{{{D}_{E}}}{{{D}_{L}}}>1\]
Efficiency (e) is the ratio of amount of work by load to work done by effort. Work is the product of force and displacement of point of application of force. So work done by load (${{W}_{L}}$) is equal to
${{W}_{L}}=L\times {{D}_{L}}$
Similarly work done by load (${{W}_{E}}$) is equal to
${{W}_{E}}=E\times {{D}_{E}}$
So,
$e=\dfrac{{{W}_{L}}}{{{W}_{E}}}$
$e=\dfrac{L\times {{D}_{L}}}{E\times {{D}_{E}}}$
$e=\dfrac{\dfrac{L}{E}}{\dfrac{{{D}_{E}}}{{{D}_{L}}}}$
For ideal machine we consider efficiency of that machine to be 1
So,
E=1
$\dfrac{\dfrac{L}{E}}{\dfrac{{{D}_{E}}}{{{D}_{L}}}}=1$
$\dfrac{L}{E}=\dfrac{{{D}_{E}}}{{{D}_{L}}}$
Now as\[\dfrac{{{D}_{E}}}{{{D}_{L}}}>1\], we can write
$\dfrac{L}{E}>1$
$L>E$
This implies that force at output (load) is greater than force at input (effort), so the machine acts as a force multiplier.
Also the distance moved at load is smaller than the distance moved at effort at the same time, this implies that the machine has caused loss of speed.
Hence option C is correct.
Note: Velocity Ratio can have any value. None of them is bound to have its value less than 1. This is often a misconception among students, but actually it is efficiency which is bound to have its value less than one in practical life and equal to one in ideal condition.
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