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 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).
The volume \( V \) of the cubical block is given by the cube of its side length:
$$ V = L^3 = (0.1 \, \text{m})^3 = 0.001 \, \text{m}^3 $$
The density of the wooden block \( \rho_s \) is calculated using its specific gravity:
$$ \rho_s = SG_s \times \rho_w = 0.75 \times 1000 \, \text{kg/m}^3 = 750 \, \text{kg/m}^3 $$
The weight \( W \) of the block acting downwards is:
$$ W = \rho_s \cdot V \cdot g $$ $$ W = 750 \times 0.001 \times 9.8 = 7.35 \, \text{N} $$
The buoyant force \( F_B \) is the weight of the fluid displaced. Since the block is completely submerged in water, we use the density of water:
$$ F_B = \rho_w \cdot V \cdot g $$ $$ F_B = 1000 \times 0.001 \times 9.8 = 9.8 \, \text{N} $$
For the block to be in equilibrium, the sum of downward forces must equal the sum of upward forces. The upward force is the buoyant force, and the downward forces are the weight and the tension \( T \) in the wire:
$$ \sum F_{up} = \sum F_{down} $$ $$ F_B = W + T $$
Solving for tension \( T \):
$$ T = F_B - W $$ $$ T = 9.8 \, \text{N} - 7.35 \, \text{N} $$ $$ T = 2.45 \, \text{N} $$
The tension in the wire is \( 2.45 \, \text{N} \).
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