Step 1: The Rs. 18,000 was wrongly expensed. Reversing this increases profit: $83{,}600 + 18{,}000 = 1{,}01{,}600$.

Step 2: Now charge the correct depreciation on the van: $18{,}000 \times 25\% = Rs.\ 4{,}500$.

Step 3: Subtract depreciation: $1{,}01{,}600 - 4{,}500 = \mathbf{Rs.\ 97{,}100}$

The net effect is: profit increases by Rs. 18,000 (expense reversed) but decreases by Rs. 4,500 (depreciation charged) — a net increase of Rs. 13,500 over the originally reported profit.

Rewrite: $\frac{dy}{dx} = \frac{y^2-x^2}{2xy}$. This is a homogeneous ODE. Let $y=vx$:

$\frac{1-v^2}{2v}dx + x\,dv/... $

After solving, the general solution is $x^2 + y^2 = Cy$ (circles with centres on the $y$-axis).

Substituting $(1,1)$: $1+1=C\cdot1 \Rightarrow C=2$, giving $x^2+y^2=2y$, i.e. $x^2+(y-1)^2=1$.

This is a circle centred on the $y$-axis at $(0,1)$.

Areal velocity is defined as the area swept by the position vector per unit time: $\frac{dA}{dt}$.

In polar coordinates, $dA = \frac{1}{2} r^2 d\theta$.
So, $\frac{dA}{dt} = \frac{1}{2} r^2 \frac{d\theta}{dt} = \frac{1}{2} r^2 \omega$.

Since angular momentum $L = m r^2 \omega$, we can substitute $r^2 \omega = \frac{L}{m}$.

Therefore, $\text{Areal Velocity} = \frac{1}{2} \left( \frac{L}{m} \right) = \frac{L}{2m}$.

• Moles of $O_{2} = 3.12 / 32 = 0.0975 \text{ mol}$
• $V = \frac{0.0975 \times 0.0821 \times 300}{1} \approx 2.4 \text{ L}$
• Volume per gram of adsorbent = Total Volume / Mass of Pt
• $2.4 \text{ L} / 1.2 \text{ g} = 2.0 \text{ L/g}$