An open frustum-shaped water tank has end radii 1.2 m and 0.5 m and a depth of 2.4 m. Its curved outer surface is to be coated at ₹ 90 per m². Find the cost. (Take \π = 3.14 .)

Question

Accepted Answer

₹ 1201.05. l = √2.4^2 + 0.7^2 = √6.25 = 2.5 m. CSA = 3.14 × 2.5 × 1.7 = 13.345 m², so cost = 13.345 × 90 = ₹ 1201.05. ₹ 1153.01 uses the depth 2.4 m as the slant height; ₹ 494.55 uses (R - r) instead of (R + r); ₹ 1607.99 also coats the base, but only the curved surface was asked for.

An open frustum-shaped water tank has end radii 1.2 $m$ and 0.5 $m$ and a depth of 2.4 $m$ . Its curved outer surface is to be coated at ₹ 90 per $m^{2}$ . Find the cost. (Take $π = 3.14.$ )

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More Class 10 Frustum of a Cone practice

An open frustum-shaped water tank has end radii 1.2 m and 0.5 m and a depth of 2.4 m. Its curved outer surface is to be coated at ₹ 90 per m2. Find the cost. (Take π=3.14.)

Step-by-Step Solution

Master this chapter in 30 minutes a day.

More Class 10 Frustum of a Cone practice

An open frustum-shaped water tank has end radii 1.2 $m$ and 0.5 $m$ and a depth of 2.4 $m$ . Its curved outer surface is to be coated at ₹ 90 per $m^{2}$ . Find the cost. (Take $π = 3.14.$ )