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A metal rod of 2 cm diameter has a conductivity of 40W/mK, which is tobe insulated with an insulating material of conductivity of 0.1 W/m K. Ifthe convective heat transfer coefficient with the ambient atmosphere is5 W/m2K, the critical thickness of insulation will be:  
  • a)
    1 cm
  • b)
    2 cm
  • c)
    7 cm
  • d)
    8 cm
Correct answer is option 'A'. Can you explain this answer?
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A metal rod of 2 cm diameter has a conductivity of 40W/mK, which is to...
Given parameters:
- Diameter of metal rod = 2 cm
- Conductivity of metal rod = 40 W/mK
- Conductivity of insulating material = 0.1 W/mK
- Convective heat transfer coefficient with ambient atmosphere = 5 W/m2K

To find: Critical thickness of insulation

Formula:
- Heat transfer rate = (Temperature difference between rod and ambient) / (Thermal resistance)
- Thermal resistance = (Length of rod) / (Area of rod * Conductivity of rod) + (Length of insulation) / (Area of rod * Conductivity of insulation) + (1 / Convective heat transfer coefficient)

Step-by-step solution:
1. Cross-sectional area of rod = π*(diameter/2)^2 = π*(2/2)^2 = 3.14 cm^2
2. Length of rod is not given, so we can assume it to be 1 meter for simplicity.
3. Thermal resistance of the system without insulation = (1 m) / (3.14 cm^2 * 40 W/mK) + (1 / 5 W/m2K) = 0.00025 m^2K/W
4. To find the critical thickness of insulation, we need to solve for the length of insulation that would result in the same thermal resistance as the system without insulation.

Let x be the length of insulation in meters. Then:

Thermal resistance of system with insulation = (1 m) / (3.14 cm^2 * 40 W/mK) + (x m) / (3.14 cm^2 * 0.1 W/mK) + (1 / 5 W/m2K)

0.00025 m^2K/W = (1 m) / (3.14 cm^2 * 40 W/mK) + (x m) / (3.14 cm^2 * 0.1 W/mK) + (1 / 5 W/m2K)

Solving for x:

x = 0.01 m = 1 cm

Therefore, the critical thickness of insulation is 1 cm.
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A metal rod of 2 cm diameter has a conductivity of 40W/mK, which is tobe insulated with an insulating material of conductivity of 0.1 W/m K. Ifthe convective heat transfer coefficient with the ambient atmosphere is5 W/m2K, the critical thickness of insulation will be:a)1 cmb)2 cmc)7 cmd)8 cmCorrect answer is option 'A'. Can you explain this answer?
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