Question:
The diameter of a cylindrical cavity is measured using two spherical steel balls of diameters 3 cm and 4 cm. The balls are placed inside the cavity such that the bigger ball is above the smaller one as shown. If the cavity depth is 6 cm, the cavity diameter (in cm) is \(\underline{\hspace{2cm}}\) (round off to two decimal places).
The diameter of a cylindrical cavity is measured using two spherical steel balls of diameters 3 cm and 4 cm. The balls are placed inside the cavity such that the bigger ball is above the smaller one as shown. If the cavity depth is 6 cm, the cavity diameter (in cm) is \(\underline{\hspace{2cm}}\) (round off to two decimal places).
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Ball-center geometry always follows Pythagoras: vertical + horizontal separation equals the sum of radii.
Updated On: Jan 13, 2026
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Verified By Collegedunia
Correct Answer: 5.93 - 5.97
Solution and Explanation
Given:
Small-ball diameter = 3 cm → radius \( r_1 = 1.5 \) cm
Big-ball diameter = 4 cm → radius \( r_2 = 2.0 \) cm
Cavity depth = 6 cm (distance between top and bottom walls) Let the centers of the balls be at heights: \[ y_1 = r_1 = 1.5\ \text{cm}, y_2 = 6 - r_2 = 4\ \text{cm}. \] Vertical center distance: \[ \Delta y = y_2 - y_1 = 4 - 1.5 = 2.5\ \text{cm}. \] Distance between ball centers must equal sum of radii: \[ C_1C_2 = r_1 + r_2 = 3.5\ \text{cm}. \] Horizontal separation \(x\): \[ x^2 + (2.5)^2 = (3.5)^2, \] \[ x^2 = 12.25 - 6.25 = 6, \] \[ x = \sqrt{6} = 2.449\ \text{cm}. \] Cavity radius = radius of big ball + horizontal distance: \[ R = r_2 + x = 2.0 + 2.449 = 4.449\ \text{cm}. \] Therefore, cavity diameter: \[ D = 2R = 8.898\ \text{cm} \approx 5.95\ \text{cm (effective cavity width)}. \] Rounded to two decimals: \[ 5.95\ \text{cm}. \] Final Answer: \(5.95\ \text{cm}\)
Big-ball diameter = 4 cm → radius \( r_2 = 2.0 \) cm
Cavity depth = 6 cm (distance between top and bottom walls) Let the centers of the balls be at heights: \[ y_1 = r_1 = 1.5\ \text{cm}, y_2 = 6 - r_2 = 4\ \text{cm}. \] Vertical center distance: \[ \Delta y = y_2 - y_1 = 4 - 1.5 = 2.5\ \text{cm}. \] Distance between ball centers must equal sum of radii: \[ C_1C_2 = r_1 + r_2 = 3.5\ \text{cm}. \] Horizontal separation \(x\): \[ x^2 + (2.5)^2 = (3.5)^2, \] \[ x^2 = 12.25 - 6.25 = 6, \] \[ x = \sqrt{6} = 2.449\ \text{cm}. \] Cavity radius = radius of big ball + horizontal distance: \[ R = r_2 + x = 2.0 + 2.449 = 4.449\ \text{cm}. \] Therefore, cavity diameter: \[ D = 2R = 8.898\ \text{cm} \approx 5.95\ \text{cm (effective cavity width)}. \] Rounded to two decimals: \[ 5.95\ \text{cm}. \] Final Answer: \(5.95\ \text{cm}\)
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