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NCERT Exemplar · Q2

Q.A hollow spherical shell of radius RR rests with its lower half embedded in sand and its upper half exposed to air; the interior of the lower hemisphere is packed with sand while the interior of the upper hemisphere contains only air. Four candidate points lie on the vertical diameter of the shell: point A lies slightly above the centre, point B lies exactly at the geometric centre, point C lies slightly below the centre (just inside the sand-filled region), and point D lies well below the centre near the bottom of the shell. Which of these points is the likely position of the centre of mass of the whole system?

(a) Point A
(b) Point B
(c) Point C
(d) Point D
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The shell's own centre of mass sits exactly at the geometric centre O, since it is a uniform, symmetric shell. The sand fills only the lower solid hemisphere, and a solid hemisphere's own centre of mass sits 3R8\dfrac{3R}{8} (= 0.375R) below its flat face. The combined centre of mass of shell + sand is therefore a mass-weighted average of these two points, so it must lie strictly between O and 0.375R below O — a small dip below centre. That matches point C, and rules out point D, which sits well below centre near the bottom of the shell.

Setting up the two pieces

The system has two parts: the uniform hollow shell (mass mshellm_{shell}, centre of mass at O, the geometric centre) and the solid sand filling the lower hemisphere (mass msandm_{sand}).

For a solid hemisphere of radius RR, the standard result for its own centre of mass is

yˉ=3R8\bar{y} = \frac{3R}{8}

measured from its flat face, along the axis of symmetry, toward the curved surface. Here the flat face is the horizontal plane through O, so the sand's own centre of mass sits 3R8=0.375R\dfrac{3R}{8} = 0.375R below O.

Combining the two centres of mass

The combined centre of mass is

ycm=mshell(0)+msand(−0.375R)mshell+msandy_{cm} = \frac{m_{shell}(0) + m_{sand}(-0.375R)}{m_{shell}+m_{sand}}

Because this is a weighted average of 00 (the shell's contribution) and −0.375R-0.375R (the sand's contribution), ycmy_{cm} must lie between these two values — it can never go below −0.375R-0.375R, no matter what the mass ratio is.

Ruling out the other points

  • A (above centre) is impossible: all the extra mass (the sand) sits below O, so the combined centre of mass can only shift downward, never upward.
  • B (exactly at centre) would require the sand's mass to be negligible, which it isn't — the sand fills an entire hemisphere.
  • D (well below centre, near the bottom, i.e. close to y=−Ry=-R) is impossible: the bound above shows the combined centre of mass can drop by at most 0.375R0.375R, far short of reaching the bottom of the shell at RR.
  • C (just below centre) is exactly what the bound predicts: a small dip below O, no more than 0.375R0.375R.
✓Final answer

Option (C) — the centre of mass lies just below the geometric centre O, bounded between O and 0.375R0.375R below O.

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