The weight of an object at the center of the earth of radius R is
(A) Zero
(B) Infinite
(C) R times the weight at the surface of the earth
(D) $\dfrac{1}{{{R^2}}}$ times the weight at surface of the earth

Hint The weight of the object depends on acceleration due to gravity which in turn depends on depth. Acceleration due to gravity is given by, $g' = g\left[ {1 - \dfrac{d}{R}} \right]$ . At the center of the earth R is equal to d so the acceleration due to gravity is zero. Since weight of the object is the product of mass and acceleration due to gravity, weight is also zero.

Complete step-by-step solution
The acceleration in the motion of a body under the effect of gravity is called acceleration due to gravity, it is denoted by g.
g depends on various factors and depth is one of them.

Acceleration due to gravity at a depth d from the earth’s surface is given by
 $g' = g\left[ {1 - \dfrac{d}{R}} \right]$
Here,
g’ is the acceleration due to gravity at depth d of earth of radius R.
So, at the center of the earth R = d
 $
  g' = g\left[ {1 - 1} \right] \\
  g' = 0 \\
 $
Therefore, g’ becomes zero.
The weight is given by
 \[W = mg\]
The acceleration due to gravity is zero hence the weight of the object is zero as well.

The correct option is A.

Note The acceleration due to gravity is also influenced by the shape of earth; the value of acceleration due to gravity is more at the poles than the equator. It also varies with height and is given by
 $g' = g{\left( {\dfrac{R}{{R + h}}} \right)^2}$
The acceleration due to gravity rotation of the earth, as the earth rotates a body placed on its surface moves along the circular path and experiences a centrifugal force, due to which the apparent weight of the body decreases.