A piece of paper (shown in figure-1) is in form of a square. Two corners of this square are folded t

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A piece of paper (shown in figure-1) is in form of a square. Two corners of this square are folded to make it appear like figure-2. Both corners are put together at centre of square 'O'. If O is taken to be (0, 0), the centre of mass of new system will be at

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Sol.$\Delta x = a - \left( \frac{a}{3} + \frac{a}{3} ight) = \frac{a}{3}$During the process shift in of triangular (heads)(∆x1) = ∆x2 = a/3∴ (∆x)cm = $\frac{m_{1}.\Delta x_{1} + m_{2}\Delta x_{2} + m_{3} 	imes 0}{m_{1} + m_{2} + m_{3}}$⇒ $(\Delta x)_{cm} = \frac{\frac{M}{8}.\frac{a}{3} + \frac{M}{8}.\frac{a}{3}}{M} = \frac{a}{12}$towards left∴ New COM = $\left( - \frac{a}{12},0 ight)$

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