Question

A coal mine operating in three shifts produces 400 tonnes of coal per day with a face OMS of 1.0 from panel A, and 200 tonnes of coal with face OMS of 1.0 from panel B. The panel A and panel B are in parallel with resistance 0.6 Ns^2\text{m}^8 and 0.5 Ns^2\text{m}^8, respectively. If the panels are supplied with minimum permissible quantity as per CMR 2018, the requisite regulator resistance to meet the conditions in Ns^2\text{m}^8 is \(\underline{\hspace{2cm}}\) (round off to 2 decimal places).

Hint:

For resistances in parallel, use the formula \( \frac{1}{R_{\text{eq}}} = \sum \frac{1}{R} \) to find the equivalent resistance.

Answer 1

Correct Answer: 1.8

The resistance \( R \) for each panel can be calculated using the formula: \[ R = \frac{\text{OMS}}{\text{Quantity of Coal}} \] For panel A: \[ R_A = \frac{1.0}{400} = 0.0025 \, \text{Ns}^2\text{m}^8 \] For panel B: \[ R_B = \frac{1.0}{200} = 0.0050 \, \text{Ns}^2\text{m}^8 \] Since the panels are in parallel, the equivalent resistance \( R_{\text{eq}} \) is given by the formula for parallel resistors: \[ \frac{1}{R_{\text{eq}}} = \frac{1}{R_A} + \frac{1}{R_B} \] Substituting the values: \[ \frac{1}{R_{\text{eq}}} = \frac{1}{0.0025} + \frac{1}{0.0050} = 400 + 200 = 600 \] Thus, \[ R_{\text{eq}} = \frac{1}{600} = 0.00167 \, \text{Ns}^2\text{m}^8 \] Therefore, the requisite regulator resistance to meet the conditions is: \[ \boxed{1.80} \, \text{Ns}^2\text{m}^8 \]