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Byju's Answer

Standard XII

Physics

Second Law of Motion

Particles of ...

Question

Particles of masses $m_{1}$ and $m_{2}$ are at a fixed distance apart. If the gravitational field strength at $m_{1}$ and $m_{2}$ are $l_{1}$ and $l_{2}$ respectively. Then:

m_{1} l_{1} + m_{2} l_{2} = 0

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m_{1} l_{2} + m_{2} l_{1} = 0

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m_{1} l_{1} - m_{2} l_{2} = 0

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m_{1} l_{2} - m_{2} l_{1} = 0

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Solution

The correct option is A $m_{1} l_{1} + m_{2} l_{2} = 0$
$m_{1} I_{1} = - m_{2} I_{2}$
By Newton's Third Law of Motion
Since $m_{i} I_{i}$ is force on $m_{i}$ due to other particle

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Particles of masses m1 and m2 are at a fixed distance apart. If the gravitational field strength at m1 and m2 are l1 and l2 respectively. Then:

The correct option is A m1l1+m2l2=0m1I1=−m2I2By Newton's Third Law of MotionSince miIi is force on mi due to other particle

Particles of masses $m_{1}$ and $m_{2}$ are at a fixed distance apart. If the gravitational field strength at $m_{1}$ and $m_{2}$ are $l_{1}$ and $l_{2}$ respectively. Then:

The correct option is A $m_{1} l_{1} + m_{2} l_{2} = 0$
$m_{1} I_{1} = - m_{2} I_{2}$
By Newton's Third Law of Motion
Since $m_{i} I_{i}$ is force on $m_{i}$ due to other particle