The edge of a cube is measured using a vernier callipers. [`9` divisions of the main scale is equal to `10` divisions of the vernier scale and `1` main scale dividsion is `1 mm`]. The main scale division reading is `10` and `1st` division of vernier scale was found to be coinciding with the main scale. The mass of the cube is `2.736 g`. Calculate the density in `g//cm^3` upto correct significant figures.

Explanation

To solve the problem of calculating the density of the cube, we will follow these steps:

### Step 1: Determine the Least Count of the Vernier Calipers
Given that 9 divisions of the main scale are equal to 10 divisions of the vernier scale, we can find the least count (LC) using the formula:
\[ \text{Least Count} = \text{1 Main Scale Division} - \text{1 Vernier Scale Division} \]
Since 1 main scale division (MSD) is equal to 1 mm, we first need to find the value of 1 Vernier Scale Division (VSD):
\[ \text{1 VSD} = \frac{9}{10} \text{MSD} = \frac{9}{10} \text{mm} = 0.9 \text{mm} \]
Now, we can calculate the least count:
\[ \text{LC} = 1 \text{mm} - 0.9 \text{mm} = 0.1 \text{mm} = 0.01 \text{cm} \]

### Step 2: Calculate the Length of the Edge of the Cube
The main scale reading (MSR) is given as 10 mm, and the first division of the vernier scale (VSR) coinciding with the main scale is 1. Therefore, the total length (L) of the edge of the cube is calculated as:
\[ L = \text{MSR} + \text{VSR} \times \text{LC} \]
Substituting the values:
\[ L = 10 \text{mm} + 1 \times 0.1 \text{mm} = 10 \text{mm} + 0.1 \text{mm} = 10.1 \text{mm} = 1.01 \text{cm} \]

### Step 3: Calculate the Volume of the Cube
The volume (V) of the cube is given by the formula:
\[ V = L^3 \]
Substituting the length of the edge:
\[ V = (1.01 \text{cm})^3 = 1.01 \times 1.01 \times 1.01 \approx 1.030301 \text{cm}^3 \]

### Step 4: Calculate the Density of the Cube
The density (ρ) is calculated using the formula:
\[ \rho = \frac{\text{Mass}}{\text{Volume}} \]
Given that the mass of the cube is 2.736 g, we can substitute the values:
\[ \rho = \frac{2.736 \text{g}}{1.030301 \text{cm}^3} \approx 2.655534 \text{g/cm}^3 \]

### Step 5: Round to Correct Significant Figures
The mass (2.736 g) has four significant figures, and the volume (1.01 cm) has three significant figures. Therefore, the density should be reported to three significant figures:
\[ \rho \approx 2.66 \text{g/cm}^3 \]

### Final Answer
The density of the cube is approximately **2.66 g/cm³**.