A block of aluminium of mass `1 kg` and volume `3.6 xx 10^-4 m^3` is suspended from a string and then completely immersed in a container of water. The decrease in tension in the string after immersion is.

Options

A. `9.8 N`
B. `6.2 N`
C. `3.6 N`
D. `1.0 N`

Explanation

To solve the problem step by step, we can follow these instructions:

### Step 1: Understand the Given Information
We have an aluminum block with:
- Mass \( m = 1 \, \text{kg} \)
- Volume \( V = 3.6 \times 10^{-4} \, \text{m}^3 \)

### Step 2: Calculate the Weight of the Block
The weight \( W \) of the block can be calculated using the formula:
\[ W = m \cdot g \]
where \( g \) (acceleration due to gravity) is approximately \( 10 \, \text{m/s}^2 \).
\[ W = 1 \, \text{kg} \cdot 10 \, \text{m/s}^2 = 10 \, \text{N} \]

### Step 3: Calculate the Buoyant Force
The buoyant force \( F_B \) acting on the block when it is completely immersed in water can be calculated using Archimedes' principle:
\[ F_B = \text{density of water} \cdot g \cdot V \]
The density of water is approximately \( 1000 \, \text{kg/m}^3 \).
\[ F_B = 1000 \, \text{kg/m}^3 \cdot 10 \, \text{m/s}^2 \cdot 3.6 \times 10^{-4} \, \text{m}^3 \]
\[ F_B = 1000 \cdot 10 \cdot 3.6 \times 10^{-4} = 3.6 \, \text{N} \]

### Step 4: Determine the Tension in the String After Immersion
Before immersion, the tension in the string \( T_1 \) is equal to the weight of the block:
\[ T_1 = W = 10 \, \text{N} \]
After immersion, the new tension \( T_2 \) can be found using the relationship:
\[ T_2 + F_B = W \]
Rearranging gives:
\[ T_2 = W - F_B \]
Substituting the values:
\[ T_2 = 10 \, \text{N} - 3.6 \, \text{N} = 6.4 \, \text{N} \]

### Step 5: Calculate the Decrease in Tension
The decrease in tension \( \Delta T \) can be calculated as:
\[ \Delta T = T_1 - T_2 \]
Substituting the values:
\[ \Delta T = 10 \, \text{N} - 6.4 \, \text{N} = 3.6 \, \text{N} \]

### Final Answer
The decrease in tension in the string after immersion is \( 3.6 \, \text{N} \).

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