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NTA Abhyas 2020
Exam
NTA Abhyas
Year
2020
Subject
Physics
Chapter
Center of Mass, Momentum and Collision

NTA Abhyas 2020 · Physics · Center of Mass, Momentum and Collision

Particles of masses m, 2m, 3m, . , nm are placed on the same line...

Particles of masses m,2m,3m,….,nmm, 2m, 3m, \ldots. , nm are placed on the same line at distance L,2L,3L,..nLL, 2L, 3L,.. nL from O.O. The distance of centre of mass from OO isParticles of masses m, 2m, 3m, . , nm are placed on the same line at distance L, 2L, 3L,.. nL from O. The distance of

Options

  1. A\dfrac{\left(2n + 1\right)L}{4}
  2. B\dfrac{L}{\left(2n + 1\right)}
  3. C\dfrac{n\left(n^{2} + 1\right)L}{2}
  4. D\dfrac{\left(2n + 1\right)L}{3}

Correct answer

D. \dfrac{\left(2n + 1\right)L}{3}

Step-by-step solution

Particles of masses m, 2m, 3m, . , nm are placed on the same line at distance L, 2L, 3L,.. nL from O. The distance of L

∵xCM=mL+2m×2L+…+nm×nLm+2m+…+nm∵ {x}_{CM} = \dfrac{mL + 2m \times 2L + \ldots + nm \times nL}{m + 2m + \ldots + nm}

=L(12+22+…+n2)(1+2+…+n)= \dfrac{L\left({1}^{2} + {2}^{2} + \ldots + n^{2}\right)}{\left(1 + 2 + \ldots + n\right)}

=L∑n2∑n=Ln(n+1)(2n+1)6n(n+1)2= L\dfrac{∑ n^{2}}{∑ n} = L\dfrac{\dfrac{n\left(n + 1\right)\left(2n + 1\right)}{6}}{\dfrac{n\left(n + 1\right)}{2}}

=(2n+1)L3= \dfrac{\left(2n + 1\right)L}{3}
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