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    A metallic globe has a radius 12.7 cm.

    Question
    SLPaper 2

    A metallic globe has a radius 12.7 cm.

    1.

    Determine the volume of the globe expressing your answer in the form a×10ka \times {10^k}a×10k, where 1⩽a<101 \leqslant a < 101⩽a<10 and k∈Zk \in \mathbb{Z}k∈Z.

    [3]
    Verified
    Solution
    • Using formula for volume of sphere: V=43πr3V = \frac{4}{3}\pi r^3V=34​πr3 with r=12.7r = 12.7r=12.7 A1

    • Substituting and calculating:

    43π(12.7)3=8580.24\frac{4}{3}\pi(12.7)^3 = 8580.2434​π(12.7)3=8580.24 A1

    • Converting to required form: V=8.58×103V = 8.58 \times 10^3V=8.58×103 A1

    Remember to express final answer in scientific notation with 1⩽a<101 \leqslant a < 101⩽a<10 and k∈Zk \in \mathbb{Z}k∈Z

    2.

    The globe is to be melted down and reshaped into the form of a cone with a height of 14.814.814.8 cm.

    Determine the radius of the base of the cone, correct to 2 significant figures.

    [3]
    Verified
    Solution
    • Recognize volume of cone equals volume of sphere (conservation of volume) M1

    • Use cone volume formula: 13πr2(14.8)=8580.24\frac{1}{3}\pi r^2(14.8) = 8580.2431​πr2(14.8)=8580.24 A1

    • Solve for rrr:

      • r2=8580.24×3π×14.8r^2 = \frac{8580.24 \times 3}{\pi \times 14.8}r2=π×14.88580.24×3​
      • r=23.529...r = 23.529...r=23.529...
      • r=24r = 24r=24 cm (2 sig fig) A1

    Accept equivalent methods that show volume conservation and correct working

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