The same horizontal force acts on two blocks of masses $M$ and $m$ resting on a smooth horizontal surface. The ratio of the displacement of the blocks in equal intervals of time, is :

M : m

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m : M

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\sqrt{m} : \sqrt{M}

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\sqrt{M} : \sqrt{m}

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Solution

The correct option is B $m : M$
$S_{1} = \frac{1}{2} a_{1} t^{2}$ [ $∴$ initially at rest, u $=$ 0]
$S_{2} = \frac{1}{2} a_{2} t^{2}$ [a $_{1}$ & a $_{2}$ are two acceleration on two respective bodies M & m]
So, $\frac{S_{1}}{S_{2}} = \frac{a_{1}}{a_{2}}$
Now, force applied on the bodies is same $F$ .
$F = M a_{1} = m a_{2}$
$\frac{S_{1}}{S_{2}} = \frac{a_{1}}{a_{2}} = \frac{F / M}{F / m} = \frac{m}{M}$
$S_{1} : S_{2} = m : M$

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The same horizontal force acts on two blocks of masses M and m resting on a smooth horizontal surface. The ratio of the displacement of the blocks in equal intervals of time, is :

The correct option is B m:MS1=12a1t2 [∴ initially at rest, u = 0]S2=12a2t2 [a1 & a2 are two acceleration on two respective bodies M & m]So, S1S2=a1a2Now, force applied on the bodies is same F.F=Ma1=ma2S1S2=a1a2=F/MF/m=mMS1:S2=m:M

The same horizontal force acts on two blocks of masses $M$ and $m$ resting on a smooth horizontal surface. The ratio of the displacement of the blocks in equal intervals of time, is :