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Question

Two blocks masses '3m' and '2m' are in contact on a smooth table. A force P is first applied horizontally on block of mass '3m' and then on mass '2m'. The contact force between the two block sin the two cases are in the ratio.
1129375_c061e72cbd2f4c1084ebfecfa14b348b.PNG


A
1 : 2
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B
2 : 3
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C
3 : 2
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D
5 : 3
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Solution

The correct option is B 2 : 3

$$f_1 = (2m)(acc.)$$

                      $$ = 2m \times \dfrac{P}{5m} = \dfrac{2}{5} P$$

                $$ f_2 = (3m) (acc.)$$
                     $$ = 3m \times  \dfrac{P}{5m} = \dfrac{3}{5} P$$
$$ \dfrac{f_1}{f_2} = \dfrac{ 2 P/5}{3P/5} = \dfrac{2}{3}$$ 


Physics

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