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Three spherical balls of radius 2 cm, 4 cm, and 6 cm are melted to form a new spherical ball. In this process, there is a loss of 25% of the material. What is the radius (in cm) of the new ball?

Solution

✅ Correct Option: 1

Volume of a sphere =43πr3= \dfrac{4}{3}\pi r^3 where rr is the radius.


For the three original balls:

Ball 1 with radius =2= 2 cm:

V1=43π(2)3V_1 = \dfrac{4}{3}\pi(2)^3

V1=43π(8)V_1 = \dfrac{4}{3}\pi(8)

V1=32π3V_1 = \dfrac{32\pi}{3} cm³

Ball 2 with radius =4= 4 cm:

V2=43π(4)3V_2 = \dfrac{4}{3}\pi(4)^3

V2=43π(64)V_2 = \dfrac{4}{3}\pi(64)

V2=256π3V_2 = \dfrac{256\pi}{3} cm³

Ball 3 with radius =6= 6 cm:

V3=43π(6)3V_3 = \dfrac{4}{3}\pi(6)^3

V3=43π(216)V_3 = \dfrac{4}{3}\pi(216)

V3=864π3V_3 = \dfrac{864\pi}{3} cm³


Total volume of all three balls:

Total Volume =32π3+256π3+864π3= \dfrac{32\pi}{3} + \dfrac{256\pi}{3} + \dfrac{864\pi}{3}

Total Volume =(32+256+864)π3= \dfrac{(32 + 256 + 864)\pi}{3}

Total Volume =1152π3= \dfrac{1152\pi}{3}

Total Volume =384π= 384\pi cm³


Since there is a 25% loss of material, 75% of the material remains.

Volume of new ball =75%= 75\% of 384π384\pi

Volume of new ball =0.75×384π= 0.75 \times 384\pi

Volume of new ball =288π= 288\pi cm³


Let rr be the radius of the new ball.

43πr3=288π\dfrac{4}{3}\pi r^3 = 288\pi

43r3=288\dfrac{4}{3}r^3 = 288

r3=288×34r^3 = 288 \times \dfrac{3}{4}

r3=216r^3 = 216

r=2163r = \sqrt[3]{216}

r=6r = 6 cm

Therefore, the radius of the new ball is 6 cm.

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