MAT Exam 2025 Paper

Concept:For a hollow cylindrical pipe, the difference between its outer and inner curved surface areas depends on the difference of radii, while the metal used equals the difference of the outer and inner cylinder volumes.Explanation:Let the outer radius be $R$ and the inner radius be $r$.The length of the pipe is $h = 42\ \mathrm{cm}$.The external curved surface area is $2\pi R h$ and the internal curved surface area is $2\pi r h$.Given, $2\pi R h - 2\pi r h = 396$.So, $2\pi h(R-r)=396$.Using $\pi = \frac{22}{7}$ and $h = 42\ \mathrm{cm}$:$$R-r=\frac{396}{2 \times \frac{22}{7} \times 42}=\frac{396\times 7}{2\times 22\times 42}=\frac{3}{2}.$$Thus, $R-r=\frac{3}{2}$. The volume of metal is the difference of the outer and inner cylinder volumes:$$\pi R^2 h-\pi r^2 h=2079.$$So, $\pi h(R^2-r^2)=2079$.Substitute $\pi=\frac{22}{7}$ and $h=42$:$$R^2-r^2=\frac{2079}{\frac{22}{7}\times 42}=\frac{63}{4}.$$Now factor the left side:$$(R+r)(R-r)=\frac{63}{4}.$$Since $R-r=\frac{3}{2}$, we get:$$(R+r)\times\frac{3}{2}=\frac{63}{4}\Rightarrow R+r=\frac{21}{2}.$$Now solve:$$R+r=\frac{21}{2}\quad \text{and}\quad R-r=\frac{3}{2}.$$Adding the two equations:$$2R=12\Rightarrow R=6\ \mathrm{cm}.$$Then,$$r=6-\frac{3}{2}=4.5\ \mathrm{cm}.$$Therefore, the outer radius is $6\ \mathrm{cm}$ and the inner radius is $4.5\ \mathrm{cm}$.Answer:Outer radius $=6\ \mathrm{cm}$, inner radius $=4.5\ \mathrm{cm}$.Correct option: C.