The coordinates of the positions of particles of mass $7,\,4{\rm{  and 10}}\,gm$ are ${\rm{(1,}}\,{\rm{5,}}\, - {\rm{3),}}\,\,{\rm{(2,}}\,5,7{\rm{) }}$ and ${\rm{(3, 3, }} - {\rm{1)}}\,cm$ respectively. The position of the centre of mass of the system would be

  • A

    $\left( { - \frac{{15}}{7},\frac{{85}}{{17}},\frac{1}{7}} \right){\rm{  }}cm$

  • B

    $\left( {\frac{{15}}{7}, - \frac{{85}}{{17}},\frac{1}{7}} \right){\rm{  }}cm$

  • C

    $\left( {\frac{{15}}{7},\frac{{85}}{{21}}, - \frac{1}{7}} \right){\rm{  }}cm$

  • D

    $\left( {\frac{{15}}{7},\frac{{85}}{{21}},\frac{7}{3}} \right){\rm{  }}cm$

Similar Questions

The centre of mass of two masses $m$ and $m'$ moves by distance $\frac{x}{5}$ when mass $m$ is moved by distance $x$ and $m'$ is kept fixed. The ratio $\frac{m'}{m}$ is

Figure shows a composite system of two uniform rods of lengths as indicated. Then the coordinates of the centre of mass of the system of rods are ...........

Find the centre of mass of a uniform L-shaped lamina (a thin flat plate) with dimensions as shown. The mass of the lamina is $3 \;kg$.

A spherical cavity of radius $r$ is curved out of a uniform solid sphere of radius $R$ as shown in the figure below. The distance of the centre of mass of the resulting body from that of the solid sphere is given by

  • [KVPY 2009]

$A$ slender uniform rod of length $\lambda$ is balanced vertically at a point $P$ on a horizontal surface having some friction. If the top of the rod is displaced slightly to the right, the position of its centre of mass at the time when the rod becomes horizontal :