Consider a rectangular sheet of solid material of length $$l = 9$$ 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 sheet $$m = 0.1$$ 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 change in area of the rectangular sheet is :

First convert the given dimensions to SI units:
length $$l = 9\text{ cm} = 0.09\text{ m}$$, width $$d = 4\text{ cm} = 0.04\text{ m}$$.

The initial area of the rectangular sheet is
$$A_0 = l \times d = 0.09 \times 0.04 = 0.0036\ \text{m}^2$$ $$-(1)$$

Heat supplied to the sheet is $$Q = 8.1 \times 10^{2}\ \text{J}$$.
Using the relation $$Q = m\,C_v\,\Delta T$$, the rise in temperature is
$$\Delta T = \frac{Q}{m\,C_v} = \frac{8.1 \times 10^{2}}{0.1 \times 900} = 9\ \text{K}$$ $$-(2)$$

For an isotropic solid, the coefficient of superficial (area) expansion is
$$\beta = 2\alpha$$ $$-(3)$$
where $$\alpha = 3.1 \times 10^{-5}\ \text{K}^{-1}$$ is the linear expansion coefficient.

From $$(3)$$,
$$\beta = 2 \times 3.1 \times 10^{-5} = 6.2 \times 10^{-5}\ \text{K}^{-1}$$ $$-(4)$$

The fractional change in area due to heating through $$\Delta T$$ is $$\beta\,\Delta T$$, so the absolute change in area is
$$\Delta A = A_0 \,\beta \,\Delta T$$ $$-(5)$$

Substituting from $$(1)$$, $$(2)$$ and $$(4)$$ into $$(5)$$:
$$\Delta A = 0.0036 \times 6.2 \times 10^{-5} \times 9$$

Compute step by step:
$$6.2 \times 10^{-5} \times 9 = 5.58 \times 10^{-4}$$;
$$0.0036 \times 5.58 \times 10^{-4} = 2.01 \times 10^{-6}\ \text{m}^2$$.

Rounding to two significant figures,
$$\Delta A \approx 2.0 \times 10^{-6}\ \text{m}^2$$.

Hence, the change in area of the rectangular sheet is $$2.0 \times 10^{-6}\ \text{m}^2$$.
Option A is correct.