Question

refer to the attachment for the question​

Answer 1

Answer:

c=12750

i think my answer is correct

Answer 2

$\\\;\underbrace{\underline{\sf{Understanding\;the\;Concept\;:-}}}$

Here the Concept of Areas of Trapezium has been used. We know that from graph, number of revolutions is equal to the area under the revolution - time graph. So firstly we will find the the parallel sides of the trapezium and then its height. Since the graph is in the shape of Trapezium, we can apply value and find the answer.

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★ Formula Used :-

$\\\;\boxed{\sf{Number\;of\;Revolutions\;=\;\bf{Area\;under\;Revolution\;-\;Time\:graph}}}$

$\\\;\boxed{\sf{Area\;of\;Trapezium\;=\;\bf{\dfrac{1}{2}\;\times\;(Sum\;of\;||^{el}\;sides)\;\times\;Height}}}$

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★ Solution :-

Given,

» 1 square division on X - axis = 0.5 seconds

» 1 square division on Y - axis = 500 revolutions

» Shape of the graph = Trapezium

» Parallel side of graph on X - axis = 10 divisions = 10 × 0.5 = 5 seconds

» Parallel side of graph on Y - axis = 5 divisions = 5 × 0.5 = 2.5 seconds

» Height of the graph = 6 divisions = 6 × 500 = 3000 revolutions per second

*Note :- The values of Y - axis will be should be multiplied by 10 for their correct value after first square division.

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~ For the Number of Revolutions made by Compressor ::

$\\\;\;\sf{:\mapsto\;\;Number\;of\;Revolutions\;=\;\bf{Area\;under\;Revolution\;-\;Time\:graph}}$

Also,

$\\\;\;\sf{:\mapsto\;\;Area\;of\;Trapezium_{(Graph)}\;=\;\bf{\dfrac{1}{2}\;\times\;(Sum\;of\;||^{el}\;sides)\;\times\;Height}}$

Combining these both equations, we get,

$\\\;\;\sf{:\mapsto\;\;Number\;of\;Revolutions\;=\;\bf{\dfrac{1}{2}\;\times\;(Sum\;of\;||^{el}\;sides)\;\times\;Height}}$

$\\\;\;\sf{:\mapsto\;\;Number\;of\;Revolutions\;=\;\bf{\dfrac{1}{2}\;\times\;(2.5\;+\;5)\;\times\;3000}}$

$\\\;\;\sf{:\mapsto\;\;Number\;of\;Revolutions\;=\;\bf{\dfrac{1}{2}\;\times\;7.5\;\times\;3000}}$

$\\\;\;\sf{:\mapsto\;\;Number\;of\;Revolutions\;=\;\bf{7.5\;\times\;1500}}$

$\\\;\;\underline{\underline{\bf{:\mapsto\;\;Number\;of\;Revolutions\;=\;\bf{11250\;\;revolutions}}}}$

So, option d.) 11250 is correct option.

$\\\;\underline{\boxed{\tt{Number\;\;of\;\;revolutions\;\;made\;\;during\;\;test\;=\;\bf{11250}}}}$

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★ More Formulas to know :-

$\\\;\sf{\leadsto\;\;Rotational\;K.E.\;=\;\dfrac{1}{2}\:I\:\omega^{2}}$

$\\\;\sf{\leadsto\;\;L\;=\;I\omega}$

$\\\;\sf{\leadsto\;\;\tau\;=\;I \alpha}$

$\\\;\sf{\leadsto\;\;\tau\;=\;Fd}$

$\\\;\sf{\leadsto\;\;\Delta W\;=\;\tau \Delta \theta}$