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Q.

The relation between the displacement X of an object produced by the application of the variable force F is represented by a graph shown in the figure. If the object undergoes a displacement from X = 0.5 m to X = 2.5 m the work done will be approximately equal to

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a

16 J

b

32 J

c

1.6 J

d

8 J

answer is A.

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Detailed Solution

Work done = Area under curve and displacement axis
= Area of trapezium
= $\frac{1}{2}$ ×(sum of two parallel lines) × distance between them
= $\frac{1}{2} (10 + 4) \times (2 .5 - 0 .5)$
= $\frac{1}{2} 14 \times 2$ = 14 J
As the area actually is not trapezium so work done will be more than 14 J i.e. approximately 16 J

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