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6. system of particles and rotational motion — Physics STD 11 Science — Question

Gujarat BoardEnglish MediumSTD 11 SciencePhysics6. system of particles and rotational motion1 Mark
MCQ
Weights of $1\,g,2\,g.....,100\,g$ are suspended from the $1 \,cm, 2 \,cm, ...... 100\, cm, \,marks$ respectively of a light metre scale. Where should it be supported for the system to be in equilibrium ...... $cm$ mark.
  • A
    $55$
  • B
    $60$
  • ✓
    $66$
  • D
    $72$
✓

Answer

Correct option: C.
$66$
c
$\cdot$ Weights are $1 \mathrm{g}, 2 \mathrm{g}, 3 \mathrm{g}, \ldots \ldots .100 \mathrm{g}$

$\cdot$ Weights are suspended at a distance $=1 \mathrm{cm}, 2 \mathrm{cm}, 3 \mathrm{cm}, \ldots \ldots .100 \mathrm{cm}$

respectively to the masses.

For the system to be in equilibrium the point must lie in the center of mass.

So center of mass of the system can be calculated as

$\mathrm{COM}=\frac {\left(1^{2}+2^{2}+3^{2} \ldots 100^{2}\right)} {(1+2+3 \ldots+100)}$

$COM = \left[ {\frac{{\frac{{\left( {100} \right)\left( {101} \right)\left( {201} \right)}}{6}}}{{\frac{{\left( {100} \right)\left( {101} \right)}}{2}}}} \right]$

$\mathrm{COM}=\frac {201}{3}=67 \mathrm{cm}$

Hence at $67 \mathrm{cm}$ from the origin or $66 \mathrm{cm}$ from the first particle, at that point system will be supported to get equilibrium.

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