To solve the problem of finding the tension in the string holding the block of wood submerged in water, we can follow these steps:
### Step 1: Calculate the Volume of the Block
The specific gravity (SG) of the block is given as 0.75. Specific gravity is the ratio of the density of the substance to the density of water. Therefore, we can express the density of the block as:
\[
\text{Density of block} = \text{SG} \times \text{Density of water} = 0.75 \times 1000 \, \text{kg/m}^3 = 750 \, \text{kg/m}^3
\]
Next, we can find the volume of the block using its weight (W) and the formula:
\[
W = \text{Volume} \times \text{Density} \times g
\]
Rearranging the formula to find the volume (V):
\[
V = \frac{W}{\text{Density} \times g}
\]
Given that the weight of the block \( W = 71.2 \, \text{N} \) and \( g = 9.81 \, \text{m/s}^2 \):
\[
V = \frac{71.2 \, \text{N}}{750 \, \text{kg/m}^3 \times 9.81 \, \text{m/s}^2} = \frac{71.2}{7357.5} \approx 0.00967 \, \text{m}^3
\]
### Step 2: Calculate the Upthrust (Buoyant Force)
The upthrust (or buoyant force) acting on the block when it is fully submerged can be calculated using Archimedes' principle:
\[
\text{Upthrust} = \text{Volume} \times \text{Density of water} \times g
\]
Substituting the values:
\[
\text{Upthrust} = 0.00967 \, \text{m}^3 \times 1000 \, \text{kg/m}^3 \times 9.81 \, \text{m/s}^2 \approx 94.9 \, \text{N}
\]
### Step 3: Apply the Equilibrium Condition
In equilibrium, the sum of the forces acting on the block must equal zero. The forces acting on the block are the weight of the block (downward) and the upthrust (upward), along with the tension in the string (also upward). Therefore, we can write:
\[
\text{Upthrust} = \text{Weight} + \text{Tension}
\]
Rearranging for tension (T):
\[
T = \text{Upthrust} - \text{Weight}
\]
Substituting the values we have:
\[
T = 94.9 \, \text{N} - 71.2 \, \text{N} = 23.7 \, \text{N}
\]
### Final Answer
The tension in the string is approximately \( 23.7 \, \text{N} \).
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