A rod has length 3 m and its mass acting per unit length is directly proportional to distance x from one of its end, then its centre of gravity from that end will be at

A rod of length 3 meters has a mass that increases along its length, with the mass per unit length directly proportional to the distance from one end. This means that as you move further along the rod from the starting point, the density of the rod increases, making the far end heavier compared to the starting end.

To find the center of gravity of this rod, we will consider the balance point where the total mass on either side is equal. The rod’s density increases with distance, so its weight is concentrated more toward the far end. Thus, the center of gravity will not be at the midpoint, which is at 1.5 meters, but will shift closer to the heavier end.

By considering the mass distribution and the balance point, it is thus determined that the center of gravity is 2.5 meters from the starting end of the rod. This position then balances the rod just right in account of its mass variability along its length.

Therefore, the center of gravity shifts more towards the denser side, thereby showing how the mass distribution impacts an object’s balancing point.

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