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BITSAT 2021
Exam
BITSAT
Year
2021
Subject
Physics
Chapter
Center of Mass, Momentum and Collision

BITSAT 2021 · Physics · Center of Mass, Momentum and Collision

Three bricks each of length L and mass M are arranged as shown from...

Three bricks each of length LL and mass MM are arranged as shown from the wall. The distance of the centre of mass of the system from the wall is Three bricks each of length L and mass M are arranged as shown from the wall. The distance of the centre of mass of the

Options

  1. AL4\frac{\mathrm{L}}{4}
  2. BL2\frac{\mathrm{L}}{2}
  3. C(32)L\left(\frac{3}{2}\right) \mathrm{L}
  4. D(1112)L\left(\frac{11}{12}\right) \mathrm{L}

Correct answer

D. (1112)L\left(\frac{11}{12}\right) \mathrm{L}

Step-by-step solution

Three bricks each of length L and mass M are arranged as shown from the wall. The distance of the centre of mass of the
x1=L2,x2=L,x3=5L4∴xCM=m1x1+m2x2+m3x3m1+m2+m3=M×L2+M×L+M×5L4M+M+M=11L12x_{1}=\frac{\mathrm{L}}{2}, x_{2}=\mathrm{L}, x_{3}=\frac{5 \mathrm{ L}}{4}\therefore x_{\mathrm{CM}}=\frac{m_{1} x_{1}+m_{2} x_{2}+m_{3} x_{3}}{m_{1}+m_{2}+m_{3}}=\frac{\mathrm{M} \times \frac{\mathrm{L}}{2}+\mathrm{M} \times \mathrm{L}+\mathrm{M} \times \frac{5 \mathrm{ L}}{4}}{\mathrm{M}+\mathrm{M}+\mathrm{M}}=\frac{11 \mathrm{ L}}{12}
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