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NEET PHYSICSEasy

Three masses are placed on the x-axis: 300 g300 \text{ g} at the origin, 500 g500 \text{ g} at x=40 cmx = 40 \text{ cm}, and 400 g400 \text{ g} at x=70 cmx = 70 \text{ cm}. The distance of the center of mass from the origin is:

A

40 cm40 \text{ cm}

B

45 cm45 \text{ cm}

C

50 cm50 \text{ cm}

D

30 cm30 \text{ cm}

Step-by-Step Solution

The position of the center of mass XcmX_{cm} for a system of particles on the x-axis is given by: Xcm=m1x1+m2x2+m3x3m1+m2+m3X_{cm} = \frac{m_1 x_1 + m_2 x_2 + m_3 x_3}{m_1 + m_2 + m_3} Given: m1=300 gm_1 = 300 \text{ g}, x1=0 cmx_1 = 0 \text{ cm} (at the origin) m2=500 gm_2 = 500 \text{ g}, x2=40 cmx_2 = 40 \text{ cm} m3=400 gm_3 = 400 \text{ g}, x3=70 cmx_3 = 70 \text{ cm} Substituting the values: Xcm=300(0)+500(40)+400(70)300+500+400X_{cm} = \frac{300(0) + 500(40) + 400(70)}{300 + 500 + 400} Xcm=0+20000+280001200X_{cm} = \frac{0 + 20000 + 28000}{1200} Xcm=480001200=40 cmX_{cm} = \frac{48000}{1200} = 40 \text{ cm}

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