Let the junction temperature be $T$. In steady state, the heat current entering the junction equals the heat current leaving it:

$H_{cu} = H_{br} + H_{st}$

$\frac{K_c A(100 - T)}{L_c} = \frac{K_b A(T - 0)}{L_b} + \frac{K_s A(T - 0)}{L_s}$

Substituting the values:

$\frac{0.92 \times 4 \times (100 - T)}{46} = \frac{0.26 \times 4 \times T}{13} + \frac{0.12 \times 4 \times T}{12}$

$0.02(100 - T) = 0.02T + 0.01T \implies 2 - 0.02T = 0.03T \implies T = 40^\circ\text{C}$

Rate of heat flow in copper rod $H_{cu} = 0.02 \times 4 \times (100 - 40) = 0.08 \times 60 = 4.8 \text{ cal/s}$.

The standard relation between $\alpha$ and $\beta$ is $\beta = \frac{\alpha}{1-\alpha}$, which can be rearranged to $\alpha = \frac{\beta}{1+\beta}$.

• Option B is correct.
• Option D is simply a rearranged form: $\frac{1}{\alpha} = \frac{1+\beta}{\beta} = \frac{1}{\beta} + 1$.
• Option C is the square of Option B: $\alpha = \left(\frac{\beta}{1+\beta}\right)^2$. While not the standard linear relation, Option A is explicitly wrong because $\alpha$ must always be less than 1, and $\frac{\beta}{1-\beta}$ would result in a value greater than 1.

Under accrual accounting, revenue is recognized when it is earned, not when cash is received. A credit sale of Rs. 350 means the business has earned the revenue, so it is recorded in the income statement as sales revenue. The receivable (Rs. 350) is recorded on the balance sheet as an asset, but the sale itself appears on the income statement. Cash-basis accounting would delay recognition until cash is received, but accrual accounting does not.

Interest on drawings = $40,000 $\times$ 5% = $2,000.
Let Opening Balance = $x$.
$x$ (Opening) + $47,500 (Profit) - $40,000 (Drawings) - $2,000 (Interest) = $4,500 (Closing).
$x$ + $5,500 = $4,500 $\Rightarrow$ $x = -1,000$.
A negative result for a credit-balance calculation indicates a $1,000 debit balance.