
A machine lifts a load of 2 tonnes weight with an effort of 100 kg weight. When the effort moves through 5 m the load moves through 0.2 m. What is the efficiency of the machine?
$\text{A}\text{. 20 }\!\!\%\!\!\text{ }$
$\text{B}\text{. 40 }\!\!\%\!\!\text{ }$
$\text{C}\text{. 80 }\!\!\%\!\!\text{ }$
$\text{D}\text{. 0}\text{.8 }\!\!\%\!\!\text{ }$
Answer
581.1k+ views
Hint: Efficiency of any machine is defined as the ratio of its output work done to its input energy.
Here input is the product of effort and distance it moves through while output work done is product of the load and distance it moves through.
Formula Used:
Efficiency, $\eta =\dfrac{{{W}_{out}}}{{{W}_{in}}}$; Work done, $W=\text{Force }\!\!\times\!\!\text{ displacement}$
Complete answer:
Mechanical efficiency of any machine is the ratio of the power output to the power input. It is a dimensionless quantity. Efficiency of a machine measures the effectiveness of the machine.
Mathematically, we can write efficiency of machine as
$\eta =\dfrac{\text{Power output}}{\text{Power input}}$
It can also be written as
$\eta =\dfrac{{{W}_{out}}}{{{W}_{in}}}$
Where ${{W}_{out}}$ denotes output work done by machine and ${{W}_{in}}$ is the work done on machine by some external means.
Work done on any object is defined as the product of its displacement and the force that caused the displacement. Mathematically,
$W=\text{Force }\!\!\times\!\!\text{ displacement}$
According to the question,
Machine uses weight (gravitational force) of the effort to lift the load against its weight.
Therefore
${{W}_{out}}={{F}_{out}}\times d{{x}_{load}}$ = 2000 × 0.2 kg-weight m = 400 J
${{W}_{in}}={{F}_{in}}\times d{{x}_{effort}}$ = 100 × 5 kg-weight m = 500 J
$\Rightarrow \eta =\dfrac{400}{500}=0.8=80\%$
Hence, option C is correct.
Note:
Kilogram is a unit of mass but kilogram-weight is defined as gravitational force on one kilogram of mass.
Effort is defined as the force applied to overcome the resistance or to lift the load. Magnitude of effort is smaller than magnitude of load, the resistive force to overcome with the help of effort on a machine.
One tonne equals 1000 kilograms.
Here input is the product of effort and distance it moves through while output work done is product of the load and distance it moves through.
Formula Used:
Efficiency, $\eta =\dfrac{{{W}_{out}}}{{{W}_{in}}}$; Work done, $W=\text{Force }\!\!\times\!\!\text{ displacement}$
Complete answer:
Mechanical efficiency of any machine is the ratio of the power output to the power input. It is a dimensionless quantity. Efficiency of a machine measures the effectiveness of the machine.
Mathematically, we can write efficiency of machine as
$\eta =\dfrac{\text{Power output}}{\text{Power input}}$
It can also be written as
$\eta =\dfrac{{{W}_{out}}}{{{W}_{in}}}$
Where ${{W}_{out}}$ denotes output work done by machine and ${{W}_{in}}$ is the work done on machine by some external means.
Work done on any object is defined as the product of its displacement and the force that caused the displacement. Mathematically,
$W=\text{Force }\!\!\times\!\!\text{ displacement}$
According to the question,
Machine uses weight (gravitational force) of the effort to lift the load against its weight.
Therefore
${{W}_{out}}={{F}_{out}}\times d{{x}_{load}}$ = 2000 × 0.2 kg-weight m = 400 J
${{W}_{in}}={{F}_{in}}\times d{{x}_{effort}}$ = 100 × 5 kg-weight m = 500 J
$\Rightarrow \eta =\dfrac{400}{500}=0.8=80\%$
Hence, option C is correct.
Note:
Kilogram is a unit of mass but kilogram-weight is defined as gravitational force on one kilogram of mass.
Effort is defined as the force applied to overcome the resistance or to lift the load. Magnitude of effort is smaller than magnitude of load, the resistive force to overcome with the help of effort on a machine.
One tonne equals 1000 kilograms.
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