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 $H_{cu} = 0.02 \times 4 \times (100 - 40) = 0.08 \times 60 = 4.8 \text{ cal/s}$.

Pakistan's Expanded Programme on Immunization (EPI) currently protects children against 12 vaccine-preventable diseases:

1. Tuberculosis (BCG)
2. Poliomyelitis (OPV/IPV)
3. Diphtheria
4. Pertussis (Whooping cough)
5. Tetanus
6. Hepatitis B
7. Haemophilus influenzae type b (Hib)
8. Pneumococcal disease (PCV)
9. Rotavirus diarrhea
10. Measles
11. Rubella
12. Typhoid (recently added)

As a Head Nurse, knowing the current EPI schedule is mandatory for supervising vaccination services at community level.

Centripetal force $F = \frac{mv^2}{R} \propto \frac{1}{R^n}$.

• $v^2 \propto \frac{1}{R^{n-1}} \implies v \propto R^{(1-n)/2}$
• $T = \frac{2\pi R}{v} \propto \frac{R}{R^{(1-n)/2}} = R^{1 - (\frac{1-n}{2})} = R^{(n+1)/2}$.