$p=P(\text{ace})=\dfrac{4}{52}=\dfrac{1}{13}$, $q=\dfrac{12}{13}$. Binomial with $n=2$.

$P(X=1)=\binom{2}{1}\cdot\dfrac{1}{13}\cdot\dfrac{12}{13}=\dfrac{24}{169}$

$P(X=2)=\binom{2}{2}\cdot\left(\dfrac{1}{13}\right)^2=\dfrac{1}{169}$

$P(X=1)+P(X=2)=\dfrac{24}{169}+\dfrac{1}{169}=\dfrac{25}{169}$

Let the three consecutive integers be $n-1,\ n,\ n+1$.

$(n-1) + n + (n+1) = 3n = -84$

$n = -28$

The three integers are $-29,\ -28,\ -27$.

The smallest is $-29$.

Quantity A $= -29$, Quantity B $= -28$.

Since $-29 < -28$, Quantity B is greater.

Thermal strain produced in the rod if it were free to expand is $\frac{\Delta L}{L} = \alpha \Delta T$.

Young's modulus is defined as $Y = \frac{\text{Stress}}{\text{Strain}} = \frac{F/A}{\Delta L/L}$.

Substituting the thermal strain:

$Y = \frac{F/A}{\alpha \Delta T} = \frac{F}{A\alpha \Delta T}$

Assuming the gas is monoatomic (as is standard in such problems unless specified otherwise), the internal energy $U$ is given by:

$U = \frac{3}{2} nRT = \frac{3}{2} PV$

$U = \frac{3}{2} \times (3 \times 10^6 \text{ Pa}) \times (2 \text{ m}^3)$

$U = 3 \times 3 \times 10^6 = 9 \times 10^6 \text{ J}$.

The three slabs are placed side-by-side (parallel combination), each occupying one-third of the area $A/3$ and the full separation $d$.

Total capacitance:

$C = \frac{\varepsilon_0 (A/3)}{d}(K_1 + K_2 + K_3) = \frac{\varepsilon_0 A}{3d}(10+12+14) = \frac{36\varepsilon_0 A}{3d} = \frac{12\varepsilon_0 A}{d}$

For a single dielectric $K$:

$C = \frac{K\varepsilon_0 A}{d}$

$\Rightarrow K = 12$

The equivalent dielectric constant is simply the arithmetic mean: $(10+12+14)/3 = 12$.

The internal energy $U$ of a monoatomic gas is $U = \frac{3}{2} nRT = \frac{3}{2} PV$.

First, find the volume $V$ using mass and density: $V = \frac{\text{mass}}{\text{density}} = \frac{2}{8} = 0.25 \text{ m}^3$.

$U = \frac{3}{2} \times (4 \times 10^4) \times 0.25 = 1.5 \times 10^4 = 15000 \text{ J}$.

The order of magnitude is $10^4 \text{ J}$.