With what acceleration ′a′ shown, the elevator descends so that the block of mass M exerts a force of Mg10 on the weighing machine? [g = acceleration due to gravity]
A
0.3g
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B
0.1g
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C
0.9g
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D
0.6g
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Solution
The correct option is C0.9g as lift is moving down pseudo force acts up ⇒N=m(g−a)
Given N=mg10 mg10=m(g−a) g=10g−10a 10a=9g a=0.9g