Consider a rectangular sheet of solid material of length $ \ell = 9 $ cm and width $ d = 4 $ cm. The coefficient of linear expansion is $ \alpha = 3.1 \times 10^{-5} $ K$^{-1}$ at room temperature and one atmospheric pressure. The mass of the sheet is $ m = 0.1 $ kg and the specific heat capacity $ C_v = 900 $ J kg$^{-1}$K$^{-1}$. If the amount of heat supplied to the material is $ 8.1 \times 10^2 $ J, then the change in area of the rectangular sheet is:
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- \( 2.0 \times 10^{-6} \, \text{m}^2 \)
- \( 3.0 \times 10^{-7} \, \text{m}^2 \)
- \( 6.0 \times 10^{-7} \, \text{m}^2 \)
- \( 4.0 \times 10^{-7} \, \text{m}^2 \)
The Correct Option is A
Approach Solution - 1
The problem involves calculating the change in area of a rectangular sheet upon heating. First, we need to determine the change in temperature of the material.
- Calculate the change in temperature, \(\Delta T\):
Given:
- Heat supplied, \(Q = 8.1 \times 10^2 \, \text{J}\)
- Mass of sheet, \(m = 0.1 \, \text{kg}\)
- Specific heat capacity, \(C_v = 900 \, \text{J kg}^{-1}\text{K}^{-1}\)
The relationship for heat transfer is:
\(Q = m \cdot C_v \cdot \Delta T\)
Rearranging gives:
\(\Delta T = \frac{Q}{m \cdot C_v}\)
Substitute the given values:
\(\Delta T = \frac{8.1 \times 10^2}{0.1 \cdot 900} = \frac{8.1 \times 10^2}{90} = 9 \, \text{K}\)
- Calculate the linear expansion and resulting change in area:
The change in area due to thermal expansion is related to the coefficients of linear expansion:
The coefficient of linear expansion is given by \(\alpha = 3.1 \times 10^{-5} \, \text{K}^{-1}\).
For a rectangular sheet, the change in area \(\Delta A\) can be approximated by:
\(\Delta A = A_0 \cdot 2 \alpha \cdot \Delta T\)
where \(A_0\) is the original area.
Calculate \(A_0\) (original area):
\(A_0 = \ell \cdot d = 9 \, \text{cm} \times 4 \, \text{cm} = 36 \, \text{cm}^2 = 36 \times 10^{-4} \, \text{m}^2\)
Substitute the values to find \(\Delta A\):
\(\Delta A = 36 \times 10^{-4} \cdot 2 \cdot 3.1 \times 10^{-5} \cdot 9\)
Calculate:
\(\Delta A = 36 \times 10^{-4} \cdot 6.2 \times 10^{-5} \cdot 9= 36 \cdot 6.2 \cdot 9 \times 10^{-9}\)
\(\Delta A = 2008.8 \times 10^{-9} \, \text{m}^2 = 2.0088 \times 10^{-6} \, \text{m}^2\)
Therefore, the change in area is approximately \(2.0 \times 10^{-6} \, \text{m}^2\).
The correct answer is:
\(2.0 \times 10^{-6} \, \text{m}^2\)
Approach Solution -2
We are given a rectangular sheet of solid material with specific properties, and we need to calculate the change in its area when heat is supplied.
1. Initial Area of the Sheet:
The initial area \( A \) of the sheet is the product of its length \( \ell \) and width \( d \): \[ A = \ell \times d = 9 \, \text{cm} \times 4 \, \text{cm} = 36 \, \text{cm}^2 = 36 \times 10^{-4} \, \text{m}^2 \]
2. Heat Supplied and Temperature Change:
The temperature change \( \Delta T \) can be found using the heat equation: \[ Q = m C_v \Delta T \] where: - \( Q = 8.1 \times 10^2 \, \text{J} \) is the heat supplied, - \( m = 0.1 \, \text{kg} \) is the mass of the sheet, - \( C_v = 900 \, \text{J/kg} \cdot \text{K} \) is the specific heat capacity. Rearranging the equation to solve for \( \Delta T \): \[ \Delta T = \frac{Q}{m C_v} = \frac{8.1 \times 10^2}{0.1 \times 900} = 9 \, \text{K} \]
3. Change in Area:
The change in area \( \Delta A \) is related to the temperature change \( \Delta T \) by the following formula: \[ \Delta A = A \alpha \Delta T \] Substituting the values: \[ \Delta A = 36 \times 10^{-4} \times 3.1 \times 10^{-5} \times 9 \] \[ \Delta A = 2.0 \times 10^{-6} \, \text{m}^2 \]
Thus, the change in area is \( 2.0 \times 10^{-6} \, \text{m}^2 \), and the correct answer is (1).
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