A cylindrical rod has an outer curved surface area of 4400 cm$... | Mathematics MCQ | Examoogle

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A cylindrical rod has an outer curved surface area of 4400 cm $^{2}$ . If the length of the rod is 10 cm, then the outer radius (in cm) of the rod is: (Take $\pi = 22/7$ )

140

Correct Answer

📐 MA • Math Direct FormulaMathematicsMensuration

Option B

Quick Summary:

Observe the expression $4400 = 2 \times (22/7) \times r \times 10$ . Simplifying $4400 = (440/7) \times r$ , it is clear $r = 4400 \times 7 / 440$ , which is $10 \times 7 = 70$ .

📐MAMath SolutionDirect Formula

📋 Given

Outer curved surface area of the cylinder (CSA) = 4400 cm², length (height) of the rod h = 10 cm, value of $\pi$ = 22/7.

🔢 Formula Used

CSA=2×π×r×hCSA = 2 \times \pi \times r \times hCSA=2×π×r×h

⚡ Exam Hall Shortcut / Speed Trick

Observe the expression $4400 = 2 \times (22/7) \times r \times 10$ . Simplifying $4400 = (440/7) \times r$ , it is clear $r = 4400 \times 7 / 440$ , which is $10 \times 7 = 70$ .

⚠️ Common Student Trap / Pitfall

Students often confuse the curved surface area with the total surface area, which includes the area of the two circular bases ( $2 \times \pi \times r^2$ ).

📊 Diagram / Illustration

🔢 Step-by-Step Solution

State the formula for Curved Surface Area

The formula for the curved surface area (CSA) of a cylinder with radius $r$ and height $h$ is given by:

CSA=2πrhCSA = 2 \pi r hCSA=2πrh

Substitute the given values

Plug in the given values $CSA = 4400$ , $h = 10$ , and $\pi = 22/7$ into the formula:

4400=2×227×r×104400 = 2 \times \frac{22}{7} \times r \times 104400=2×722​×r×10

Solve for the radius r

Simplify the equation: $4400 = \frac{440}{7} \times r$ . Multiplying both sides by $7/440$ gives $r = \frac{4400 \times 7}{440}$ .

r=10×7=70r = 10 \times 7 = 70r=10×7=70

✅

B is correct because solving the equation $4400 = 2 \times (22/7) \times r \times 10$ yields $r = 70$ .

Core Concepts Used

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Mensuration Cylinder Geometry Algebraic Manipulation

💡 EXAM TIP

This concept of lateral surface area is frequently used in 3D geometry problems involving pipes, pillars, and wire dimensions in physics.

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