Question

If the apparent weight of a cube of mass \(400\,\text{g}\) immersed in water is \(3.36\,\text{N}\), then the density of the material of the cube (in \(\text{kg m}^{-3}\)) is \[ (\text{Acceleration due to gravity}=10\,\text{m s}^{-2}) \]

• A. \(8350\)
• B. \(9250\)
• C. \(7500\)
• D. \(6250\)

Hint:

For an immersed body, \[ \boxed{\text{Apparent weight}=\text{Actual weight}-\text{Buoyant force}.} \] Also, \[ \boxed{F_B=\rho_{\text{liquid}}Vg.} \]

Answer 1

The Correct Option is D

Step 1: Calculate the actual weight of the cube. Given, \[ m=400\,\text{g}=0.4\,\text{kg}. \] Hence, \[ W=mg=0.4\times10=4\,\text{N}. \] The apparent weight is \[ W_a=3.36\,\text{N}. \] Therefore, the buoyant force is \[ F_B=W-W_a=4-3.36=0.64\,\text{N}. \]

Step 2: Apply Archimedes' principle. The buoyant force is \[ F_B=\rho_w Vg, \] where \[ \rho_w=1000\,\text{kg m}^{-3}. \] Thus, \[ 0.64=1000\times V\times10, \] \[ V=\frac{0.64}{10000}=6.4\times10^{-5}\,\text{m}^3. \]

Step 3: Calculate the density of the cube. The density is \[ \rho=\frac{m}{V} =\frac{0.4}{6.4\times10^{-5}} =6250\,\text{kg m}^{-3}. \] Hence, \[ \boxed{\rho=6250\,\text{kg m}^{-3}.} \] Therefore, the correct option is \(\boxed{(D)}\).