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    Question

    A solid metal cylinder of 12 cm height and 7 cm radius

    is melted and recast into two cones in the proportion of 1:2 (volume), keeping the height 12 cm. What would be the percentage change in the flat surface area before and after [use ╧А =22/7]?
    A 60% Correct Answer Incorrect Answer
    B 25% Correct Answer Incorrect Answer
    C 50% Correct Answer Incorrect Answer
    D 70% Correct Answer Incorrect Answer

    Solution

    Volume of Cylinder = ╧Аr┬▓h = ╧А ├Ч 72 ├Ч 12 = 588╧А Now, the solid metal cylinder is re-cast into two cones in the proportion 1: 2 i.e. the volumes of cone 1 and cone 2 is 196 ╧А and 392 ╧А respectively. So, flat Surface area of cylinder before melting = 2╧Аr┬▓ = 2 ├Ч 72 = 98╧А ┬а Volume of cone 1 =1/3 ╧Аr1┬▓h= 196╧А ┬а r12 = (196├Ч3/ 12) = 49 Volume of cone 2 = 1/3╧Аr2┬▓h = 392╧А ┬а r22 = 392├Ч3 /12 = 98 Flat surface area of cones = ╧А (r12 + r22) = ╧А (49 + 98) = 147╧А ┬а Percentage change in surface area = (147-98╧А) /98╧А ├Ч 100 = 1/2 ├Ч 100 = 50%

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