The difference between the outside and the inside surface area of a cylindrical pipe 14\text{ cm} long is 44\text{ cm}^2. The pipe is made of 99\text{ cm}^3 of metal. If R is the outer radius and r is the inner radius of the pipe, then what is (R+r) equal to? (Take \pi=\frac{22}{7})
- A. 9 cm
- B. 7.5 cm
- C. 6 cm
- D. 4.5 cm ✓
Correct Answer: D. 4.5 cm
Explanation
Surface area difference: 2\pi h(R-r) = 44 \implies 2\pi(14)(R-r) = 44 \implies R-r = \frac{1}{2}. Volume of metal: \pi(R^2-r^2)h = 99 \implies \pi(R-r)(R+r)h = 99. Substituting R-r = 0.5 gives \frac{22}{7}(0.5)(R+r)(14) = 99 \implies 22(R+r) = 99 \implies R+r = 4.5\text{ cm}.
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