JEE Mains · Physics · STD 11 - 6. system of particles and rotational motion
Two blocks of masses \(10 \; kg\) and \(30 \; kg\) are placed on the same straight line with coordinates \((0,0) \; cm\) and \(( x , 0) \; cm\) respectively. The block of \(10 \; kg\) is moved on the same line through a distance of \(6 \; cm\) towards the other block. The distance through which the block of \(30 \; kg\) must be moved to keep the position of centre of mass of the system unchanged is
- A \(4 \; cm\) towards the \(10 \; kg\) block
- B \(2 \; cm\) away from the \(10 \; kg\) block
- C \(2 \; cm\) towards the \(10 \; kg\) block
- D \(4 \; cm\) away from the \(10 \; kg\) block
Answer & Solution
Correct Answer
(C) \(2 \; cm\) towards the \(10 \; kg\) block
Step-by-step Solution
Detailed explanation
\(\Delta x_{G}=\frac{m_{1} \Delta x_{1}+m_{2} \Delta x_{2}}{m_{1}+m_{2}}\) \(0=\frac{10 \times 6+30\left(\Delta x_{2}\right)}{40}\) \(\Delta x_{2}=-2 \; c m\) Block of mass \(30 \; kg\) will to move towards \(10 \; kg\).
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