The work done in lifting a body of mass 20 kg and specific gravity 3.2 to a height of 8m in water is `(g=10ms^(-2))`

Options

A. 110J
B. 1600 J
C. 500 j
D. 1100 J

Answer

To solve the problem of calculating the work done in lifting a body of mass 20 kg and specific gravity 3.2 to a height of 8 m in water, we will follow these steps:

### Step 1: Understand Specific Gravity
Specific gravity (SG) is defined as the ratio of the density of a substance to the density of water. Given that the specific gravity of the body is 3.2, we can express the density of the body (ρ_b) in terms of the density of water (ρ_w).

\[ \rho_b = SG \times \rho_w \]

Where:
- ρ_w (density of water) = 1000 kg/m³
- SG = 3.2

### Step 2: Calculate the Density of the Body
Using the formula from Step 1, we can calculate the density of the body:

\[ \rho_b = 3.2 \times 1000 \, \text{kg/m}^3 = 3200 \, \text{kg/m}^3 \]

### Step 3: Determine the Weight of the Body
The weight (W) of the body can be calculated using the formula:

\[ W = m \cdot g \]

Where:
- m = mass of the body = 20 kg
- g = acceleration due to gravity = 10 m/s²

Calculating the weight:

\[ W = 20 \, \text{kg} \times 10 \, \text{m/s}^2 = 200 \, \text{N} \]

### Step 4: Calculate the Work Done in Lifting the Body
The work done (W_d) in lifting the body in a fluid can be calculated using the formula:

\[ W_d = mgh \left(1 - \frac{\rho_w}{\rho_b}\right) \]

Where:
- h = height = 8 m
- ρ_w = density of water = 1000 kg/m³
- ρ_b = density of the body = 3200 kg/m³

Substituting the values into the formula:

\[ W_d = 20 \, \text{kg} \times 10 \, \text{m/s}^2 \times 8 \, \text{m} \left(1 - \frac{1000}{3200}\right) \]

### Step 5: Simplify the Calculation
First, calculate the fraction:

\[ \frac{1000}{3200} = 0.3125 \]

Now substituting this back into the work done formula:

\[ W_d = 20 \times 10 \times 8 \left(1 - 0.3125\right) \]

Calculating \(1 - 0.3125\):

\[ 1 - 0.3125 = 0.6875 \]

Now substituting this value:

\[ W_d = 20 \times 10 \times 8 \times 0.6875 \]

Calculating the product:

\[ W_d = 1600 \times 0.6875 = 1100 \, \text{J} \]

### Final Answer
The work done in lifting the body is **1100 Joules**.

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