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Answer:
Abhinav be seated at "2.57 cm" far. The further explanation is given below.
Explanation:
The given values are:
Deepta's weight,
= 450 N
Abhinav's weight,
= 350 N
Deepta's distance from fulcrum,
= 2cm
i.e.,
= 0.02 m
Let Abhinav's distance be "x".
As we know,
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i.e.,
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