Q.50. A tube is filled with water and a wooden cube 10 cm 10cm 10cm is placed in the water. The wo

Q: What is the correct answer for this exam Physics Q50?

The correct answer is 387. Sol. The wooden cube has dimensions $10 \text{ cm} \times 10 \text{ cm} \times 10 \text{ cm}$. Its base area is $A = 10 \text{ cm} \times 10 \text{ cm} = 100 \text{ cm}^2$. When the metal coin is placed on the wooden cube, the total weight supported by buoyancy increases. The additional buoyant force must be equal to the weight of the coin. The increase in submerged part is $h = 3.87 \text{ cm}$.

Q: Q.50. A tube is filled with water and a wooden cube 10 cm 10cm 10cm is placed in the water. The wooden cube is found to float on the water with a part of it submerged in water. When a metal coin is…

The correct answer is 387. Sol. The wooden cube has dimensions $10 \text{ cm} \times 10 \text{ cm} \times 10 \text{ cm}$. Its base area is $A = 10 \text{ cm} \times 10 \text{ cm} = 100 \text{ cm}^2$. When the metal coin is placed on the wooden cube, the total weight supported by buoyancy increases. The additional buoyant force must be equal to the weight of the coin. The increase in submerged part is $h = 3.87 \text{ cm}$.

Question

Q.50. A tube is filled with water and a wooden cube 10 cm  10cm  10cm is placed in the water. The wooden cube is found to float on the water with a part of it submerged in water. When a metal coin is placed on the wooden cube, the submerged part is increased by 3.87 cm. The mass of the metal coin is _______ gram. (Take water density as 1g/cm and density of wood as 0.4 g/cm)

FirstInTest · Accepted Answer

Sol. The wooden cube has dimensions $10 	ext{ cm} 	imes 10 	ext{ cm} 	imes 10 	ext{ cm}$. Its base area is $A = 10 	ext{ cm} 	imes 10 	ext{ cm} = 100 	ext{ cm}^2$. When the metal coin is placed on the wooden cube, the total weight supported by buoyancy increases. The additional buoyant force must be equal to the weight of the coin. The increase in submerged part is $h = 3.87 	ext{ cm}$. The additional volume submerged is $\Delta V = A 	imes h = 100 	ext{ cm}^2 	imes 3.87 	ext{ cm} = 387 	ext{ cm}^3$. The buoyant force is $F_B = ho_{water} g \Delta V$. The weight of the coin is $m_{coin} g$. Equating these, $m_{coin} g = ho_{water} g \Delta V$. So $m_{coin} = ho_{water} \Delta V$. Given $ho_{water} = 1 	ext{ g/cm}^3$. $m_{coin} = 1 	ext{ g/cm}^3 	imes 387 	ext{ cm}^3 = 387 	ext{ g}$.

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