Two bodies of masses 4 kg and 9 kg are separated by distance of 60 cm. A 1 kg mass is placed in between these two masses. If the net force on 1 kg is zero, then its distance from 4 kg mass is

26 cm

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30 cm

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28 cm

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32 cm

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24 cm

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Solution

The correct option is E 24 cm
Given, $m_{1} = 4 k g, m_{2} = 9 k g$ and $r = 60 c m$ and addition mass $= 1 k g$
Let the distance of mass from mass $m_{1}$ is x.
$∴ x = \frac{r}{\sqrt{\frac{m_{2}}{m_{1}}} + 1} = \frac{60}{\sqrt{\frac{9}{v}} + 1} = \frac{60}{\frac{3}{2} + 1} = 24 c m$

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Two bodies of masses 4 kg and 9 kg are separated by distance of 60 cm. A 1 kg mass is placed in between these two masses. If the net force on 1 kg is zero, then its distance from 4 kg mass is

The correct option is E 24 cmGiven, m1=4kg,m2=9kg and r=60cm and addition mass =1kgLet the distance of mass from mass m1 is x.∴x=r√m2m1+1=60√9v+1=6032+1=24cm

The correct option is E 24 cm
Given, $m_{1} = 4 k g, m_{2} = 9 k g$ and $r = 60 c m$ and addition mass $= 1 k g$
Let the distance of mass from mass $m_{1}$ is x.
$∴ x = \frac{r}{\sqrt{\frac{m_{2}}{m_{1}}} + 1} = \frac{60}{\sqrt{\frac{9}{v}} + 1} = \frac{60}{\frac{3}{2} + 1} = 24 c m$