Two spherical steel balls of diameters 3 and 4 cm are placed in a cylindrical cavity as shown. The bigger ball is above the smaller one. Cavity depth is 6 cm. Find cavity diameter in cm to two decimals.
The small-ball center is 1.5 cm above the bottom; the large-ball center is 6−2 = 4 cm above it. Vertical separation = 2.5 cm and center distance = 1.5+2 = 3.5 cm. Horizontal separation is $\sqrt{3.5^2-2.5^2}=\sqrt6$ cm. Opposite wall contacts give cavity diameter $D=1.5+2+\sqrt6=\mathbf{5.95}$ cm.
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