Explanation:
When a ray of light enters a denser medium from a rarer medium, it bends towards the normal, and when it emerges from a denser medium to a rarer medium, it bends away from the normal. This bending of light is known as refraction. When a light ray enters a glass sphere, it bends towards the normal, and when it emerges out of the sphere, it bends away from the normal.

Formula:
Let R be the radius of the sphere, and d be the actual distance of the mark from the surface of the sphere. Let x be the distance of the mark from the diametrically opposite side of the sphere. Then, using the formula for refraction, we have:

1.5 x = 2R - d
x + 10 = 2R

Solving these two equations, we get:

R = (3x + 30 - 1.5d)/4


Find x:
As the mark is viewed from a diametrically opposite side, the line joining the mark and the observer passes through the center of the sphere. Let O be the center of the sphere, and A be the mark. Let B be the point where the line joining A and the observer intersects the surface of the sphere. Then, we have:

OA = OB = R
AB = x
OB + AB = 2R - 10 (as the mark appears to be at a distance of 10 cm from its actual position)

Substituting the values, we get:

x = R - 5

Find d:
Let C be the point where the ray of light entering the sphere from the observer intersects the surface of the sphere. Then, we have:

OC = R
BC = d
OB = 2R - x - 10

Using Pythagoras theorem in triangle OBC, we get:

d^2 + (2R - x - 10)^2 = R^2

Substituting the value of x, we get:

d^2 + (3R - 15)^2 = R^2

Simplifying this equation, we get:

d^2 = 6R^2 - 90R + 225

Find R:
Substituting the value of x and d in the formula for R, we get:

R = (3x + 30 - 1.5d)/4

Substituting the values, we get:

R = (3(R - 5) + 30 - 1.5(√(6R^2 - 90R + 225)))/4

Simplifying this equation, we get:

3R^2 - 51R + 75 = 0

Solving this quadratic equation, we get:

R = 5 or R = 10/3

As the radius of the sphere cannot be negative, the radius of the sphere is 5 cm.

Radius = 15 cm ... am I right or wrong??