Question

Six objects are placed at the vertices of a regular hexagon. The geometric center of the hexagon is at the origin with objects $1$ and $4$ on the x-axis(see figure). The mass of the $k^{th}$ object is $m_k=k^{i}M|\cos {\theta }_k|$ where i is an integer, M is a constant with dimension of mass, and ${\theta }_k$ is the angular position of the $k^{th}$ vertex measured from the positive x-axis in the counter-clockwise sense. If the net gravitational force on a body at the centroid vanishes, the value of i is?

• A. $0$
• B. $1$
• C. $2$
• D. $3$

Answer 1

Answer: (A)

$0$

Ratio of masses is given by:

$m_1:m_2:m_3:m_4:m_5:m_6=1:2^i/2:3^i/2:4^i:5^i/2:6^i/2$

For $i>0$,

Due to masses $m_3$ & $m_6$, gravitational field at center will be towards $m_6$ as $m_6>m_3$

Due to masses $m_2$ & $m_5$, gravitational field at center will be towards $m_5$ as $m_5>m_2$

Due to masses $m_1$ & $m_4$, gravitational field at center will be towards $m_4$ as $m_4>m_1$

From vector addition, it can be observed that these three fields cannot add to zero. Hence, $i\notin 1,2,3,....$

For $i=0$,

Due to masses $m_3$ & $m_6$, gravitational field at center will be zero as $m_6=m_3$

Due to masses $m_2$ & $m_5$, gravitational field at center will be zero as $m_5=m_2$

Due to masses $m_1$ & $m_4$, gravitational field at center will be zero as $m_4=m_1$

Hence, net gravitational field will be zero.