Step 1: Understanding the Concept
The solid sphere has mass \(m\) and radius \(r\). The shell has mass \(m\) and radius \(2r\).
Step 2: At 2.5 r from the centre
This point is outside the shell, so the total mass \(2m\) acts at the centre.\n\(g=\dfrac{G(2m)}{(2.5r)^2}=\dfrac{2Gm}{6.25r^2}=\dfrac{8}{25}\dfrac{Gm}{r^2}\).
Step 3: Other statements
At \(1.5r\) the shell contributes zero but the solid sphere gives \(\dfrac{Gm}{(1.5r)^2}\neq0\).\nBetween the sphere and the shell the field falls as \(1/x^2\), so it is not uniform.\nAt \(15r\) the field would be \(\dfrac{2Gm}{225r^2}\), not \(\dfrac{2}{9}\dfrac{Gm}{r^2}\).
Step 4: Result
Only the statement about \(2.5r\) is correct.
Final Answer:
At 2.5 r from the centre the field is \(\dfrac{8}{25}\dfrac{Gm}{r^2}\).
\[ \boxed{\text{(B) }\text{at a distance of }(2.5)r\text{ from the centre is }\dfrac{8}{25}\dfrac{Gm}{r^{2}}} \]