From a circular ring of mass ‘M’ and radius ‘R’ an arc corresponding to a 90° sector is removed. The moment of inertia of the remaining part of the ring about an axis passing through the centre of the ring and perpendicular to the plane of the ring is ‘K’ times ‘MR²’. Then the value of ‘K’ is :

For a thin circular ring of mass $M$ and radius $R$, the moment of inertia about an axis through its centre and perpendicular to its plane is

$I\_\{\backslash text\{full\}\}\; =\; M\; R^\{2\}$

Because the mass is uniformly distributed along the circumference, the mass is proportional to the length of the arc. A 90° sector represents one quarter of the full 360° circle, so the removed sector has mass

$M\_\{\backslash text\{removed\}\}\; =\; \backslash frac\{1\}\{4\}\backslash ,M$

The moment of inertia contributed by that sector is its mass times $R^\{2\}$ (the same radius applies to every element of the ring):

$I\_\{\backslash text\{removed\}\}\; =\; M\_\{\backslash text\{removed\}\}\backslash ,R^\{2\}=\; \backslash frac\{1\}\{4\}\backslash ,M\; R^\{2\}$

The remaining part of the ring therefore has moment of inertia

$I\_\{\backslash text\{remaining\}\}\; =\; I\_\{\backslash text\{full\}\}\; -\; I\_\{\backslash text\{removed\}\}\; =\; M\; R^\{2\}\; -\; \backslash frac\{1\}\{4\}\; M\; R^\{2\}\; =\; \backslash frac\{3\}\{4\}\backslash ,M\; R^\{2\}$

Thus the remaining ring’s moment of inertia can be written as $K\backslash ,M\; R^\{2\}$ with

$K\; =\; \backslash frac\{3\}\{4\}$

Hence the correct value of $K$ is 3/4.