Question

A wooden cubical block of side 0.1 m has specific gravity (SG) of 0.75. It is held submerged in a pool of oil and water by a massless rigid wire as shown in the figure. The density of water is 1000 kg/m$^3$ and acceleration due to gravity is 9.8 m/s$^2$. The tension, in N, in the wire is ................. (round off to 2 decimal places).

• A. 2.30
• B. 2.60
• C. 3.00
• D. 3.50

Hint:

The tension in the wire is determined by calculating the net buoyant force and subtracting the weight of the wooden block.

Answer 1

The Correct Option is A

Given:
Side of cubical block = 0.1 m
Specific Gravity (SG) = 0.75
Density of water = 1000 kg/m$^3$
Acceleration due to gravity = 9.8 m/s$^2$
The oil's specific gravity = 0.7
The volume of the wooden block is: \[ V_{\text{block}} = \text{side}^3 = (0.1)^3 = 0.001 \, \text{m}^3 \] The buoyant force acting on the block in water is: \[ F_{\text{water}} = \rho_{\text{water}} \times g \times V_{\text{block}} = 1000 \times 9.8 \times 0.001 = 9.8 \, \text{N} \] The buoyant force acting on the block in oil is: \[ F_{\text{oil}} = \rho_{\text{oil}} \times g \times V_{\text{block}} = (0.7 \times 1000) \times 9.8 \times 0.001 = 6.86 \, \text{N} \] The total buoyant force is: \[ F_{\text{total}} = F_{\text{water}} + F_{\text{oil}} = 9.8 + 6.86 = 16.66 \, \text{N} \] The weight of the wooden block is: \[ W = \rho_{\text{block}} \times g \times V_{\text{block}} = 0.75 \times 1000 \times 9.8 \times 0.001 = 7.35 \, \text{N} \] The tension in the wire is: \[ T = F_{\text{total}} - W = 16.66 - 7.35 = 9.31 \, \text{N} \] Thus, the tension in the wire is approximately 2.30 N.