149670 Two objects of mass \(10 \mathrm{~kg}\) and \(20 \mathrm{~kg}\) respectively are connected to the two ends of a rigid rod of length \(10 \mathrm{~m}\) with negligible mass. The distance of the center of the system from the \(10 \mathrm{~kg}\) mass is

149672 A uniform rod of length \(200 \mathrm{~cm}\) and mass \(500 \mathrm{~g}\) is balanced on a wedge placed at \(40 \mathrm{~cm}\) mark. A mass of \(2 \mathbf{~ k g}\) is suspended from the rod at 20 \(\mathrm{cm}\) and another unknown mass \(\mathrm{m}\) is suspended from the rod at \(160 \mathrm{~cm}\) mark as shown in the figure. Find the value of \(m\) such that the rod is in equilibrium. \(\left(g=10 \mathrm{~m} / \mathrm{s}^{2}\right)\)

149674 A homogenous semi circular plate of radius 9 \(\mathrm{cm}\) placed at the origin as shown in the figure. The coordinate of center of mass is (Assume thickness is negligible)

149675 Centre of mass (C.M.) of three particles of masses \(1 \mathrm{~kg}, 2 \mathrm{~kg}\) and \(3 \mathrm{~kg}\) lies at the point \((1,2\), 3) and C.M. of another system of particles of 3 \(\mathrm{kg}\) and \(2 \mathrm{~kg}\) lies at the point \((-1,3,-2)\). Where should we put a particle of mass \(5 \mathrm{~kg}\) so that the C.M. of entire system lies at the C.M. of the first system?

149670 Two objects of mass \(10 \mathrm{~kg}\) and \(20 \mathrm{~kg}\) respectively are connected to the two ends of a rigid rod of length \(10 \mathrm{~m}\) with negligible mass. The distance of the center of the system from the \(10 \mathrm{~kg}\) mass is

149672 A uniform rod of length \(200 \mathrm{~cm}\) and mass \(500 \mathrm{~g}\) is balanced on a wedge placed at \(40 \mathrm{~cm}\) mark. A mass of \(2 \mathbf{~ k g}\) is suspended from the rod at 20 \(\mathrm{cm}\) and another unknown mass \(\mathrm{m}\) is suspended from the rod at \(160 \mathrm{~cm}\) mark as shown in the figure. Find the value of \(m\) such that the rod is in equilibrium. \(\left(g=10 \mathrm{~m} / \mathrm{s}^{2}\right)\)

149674 A homogenous semi circular plate of radius 9 \(\mathrm{cm}\) placed at the origin as shown in the figure. The coordinate of center of mass is (Assume thickness is negligible)

149675 Centre of mass (C.M.) of three particles of masses \(1 \mathrm{~kg}, 2 \mathrm{~kg}\) and \(3 \mathrm{~kg}\) lies at the point \((1,2\), 3) and C.M. of another system of particles of 3 \(\mathrm{kg}\) and \(2 \mathrm{~kg}\) lies at the point \((-1,3,-2)\). Where should we put a particle of mass \(5 \mathrm{~kg}\) so that the C.M. of entire system lies at the C.M. of the first system?

149670 Two objects of mass \(10 \mathrm{~kg}\) and \(20 \mathrm{~kg}\) respectively are connected to the two ends of a rigid rod of length \(10 \mathrm{~m}\) with negligible mass. The distance of the center of the system from the \(10 \mathrm{~kg}\) mass is

149672 A uniform rod of length \(200 \mathrm{~cm}\) and mass \(500 \mathrm{~g}\) is balanced on a wedge placed at \(40 \mathrm{~cm}\) mark. A mass of \(2 \mathbf{~ k g}\) is suspended from the rod at 20 \(\mathrm{cm}\) and another unknown mass \(\mathrm{m}\) is suspended from the rod at \(160 \mathrm{~cm}\) mark as shown in the figure. Find the value of \(m\) such that the rod is in equilibrium. \(\left(g=10 \mathrm{~m} / \mathrm{s}^{2}\right)\)

149674 A homogenous semi circular plate of radius 9 \(\mathrm{cm}\) placed at the origin as shown in the figure. The coordinate of center of mass is (Assume thickness is negligible)

149675 Centre of mass (C.M.) of three particles of masses \(1 \mathrm{~kg}, 2 \mathrm{~kg}\) and \(3 \mathrm{~kg}\) lies at the point \((1,2\), 3) and C.M. of another system of particles of 3 \(\mathrm{kg}\) and \(2 \mathrm{~kg}\) lies at the point \((-1,3,-2)\). Where should we put a particle of mass \(5 \mathrm{~kg}\) so that the C.M. of entire system lies at the C.M. of the first system?

149670 Two objects of mass \(10 \mathrm{~kg}\) and \(20 \mathrm{~kg}\) respectively are connected to the two ends of a rigid rod of length \(10 \mathrm{~m}\) with negligible mass. The distance of the center of the system from the \(10 \mathrm{~kg}\) mass is

149672 A uniform rod of length \(200 \mathrm{~cm}\) and mass \(500 \mathrm{~g}\) is balanced on a wedge placed at \(40 \mathrm{~cm}\) mark. A mass of \(2 \mathbf{~ k g}\) is suspended from the rod at 20 \(\mathrm{cm}\) and another unknown mass \(\mathrm{m}\) is suspended from the rod at \(160 \mathrm{~cm}\) mark as shown in the figure. Find the value of \(m\) such that the rod is in equilibrium. \(\left(g=10 \mathrm{~m} / \mathrm{s}^{2}\right)\)

149674 A homogenous semi circular plate of radius 9 \(\mathrm{cm}\) placed at the origin as shown in the figure. The coordinate of center of mass is (Assume thickness is negligible)

149675 Centre of mass (C.M.) of three particles of masses \(1 \mathrm{~kg}, 2 \mathrm{~kg}\) and \(3 \mathrm{~kg}\) lies at the point \((1,2\), 3) and C.M. of another system of particles of 3 \(\mathrm{kg}\) and \(2 \mathrm{~kg}\) lies at the point \((-1,3,-2)\). Where should we put a particle of mass \(5 \mathrm{~kg}\) so that the C.M. of entire system lies at the C.M. of the first system?