Effect of Variable Conductivity MCQ

Test your knowledge with important Effect of Variable Conductivity MCQ and their applications. These MCQs are beneficial for competitive exams too. Explore 30 + more Effect of Variable Conductivity MCQs at Bissoy

• “If β is less than zero, then with respect to the relation k = k0 (1 + β t), conductivity depends on surface area”.
  ক

  True

  খ

  False
• The temperatures on the two sides of a plane wall are t1 and t2 and thermal conductivity of the wall material is prescribed by the relation K = k0 e (-x/δ) Where, k0 is constant and δ is the wall thickness. Find the relation for temperature distribution in the wall?
  ক

  t 1 – t x / t 1 – t 2 = x

  খ

  t 1 – t x / t 1 – t 2 = δ

  গ

  t 1 – t x / t 1 – t 2 = δ/x

  ঘ

  t 1 – t x / t 1 – t 2 = x/δ
• The unit of thermal conductivity doesn’t contain which parameter?
  ক

  Watt

  খ

  Pascal

  গ

  Meter

  ঘ

  Kelvin
• If β is greater than zero, then choose the correct statement with respect to given relation k = k0 (1 +β t)
  ক

  k doesn’t depend on temperature

  খ

  k depends on temperature

  গ

  k is directly proportional to t

  ঘ

  Data is insufficient
• The accompanying sketch shows the schematic arrangement for measuring the thermal conductivity by the guarded hot plate method. Two similar 1 cm thick specimens receive heat from a 6.5 cm by 6.5 cm guard heater. When the power dissipation by the wattmeter was 15 W, the thermocouples inserted at the hot and cold surfaces indicated temperatures as 325 K and 300 K. What is the thermal conductivity of the test specimen material?
  ক

  0.81 W/m K

  খ

  0.71 W/m k

  গ

  0.61 W/m K

  ঘ

  0.51 W/m K
• With respect to the equation k = k0 (1 +β t) which is true if we put β = 0?
  ক

  Slope of temperature curve is constant

  খ

  Slope of temperature curve does not change

  গ

  Slope of temperature curve increases

  ঘ

  Slope of temperature curve is decreases
• The mean thermal conductivity evaluated at the arithmetic mean temperature is represented by . b) km = k0 . c) km = k0 . d) km = k0 .
  ক

  km = k0 .

  খ

  1 + β (t1 – t2)/2

  গ

  1 + β (t1 – t2)/2

  ঘ

  1 + β (t1 – t2)/2
• A plane wall of thickness δ has its surfaces maintained at temperatures T1 and T2. The wall is made of a material whose thermal conductivity varies with temperature according to the relation k = k0 T2. Find the expression to work out the steady state heat conduction through the wall?
  ক

  Q = 2A k0 (T 1 3 – T 2 3)/3 δ

  খ

  Q = A k0 (T 1 3 – T 2 3)/3 δ

  গ

  Q = A k0 (T 1 2 – T 2 2)/3 δ

  ঘ

  Q = A k0 (T 1 – T 2)/3 δ
• The inner and outer surfaces of a furnace wall, 25 cm thick, are at 300 degree Celsius and 30 degree Celsius. Here thermal conductivity is given by the relation K = (1.45 + 0.5 * 10-5 t2) KJ/m hr deg Where, t is the temperature in degree centigrade. Calculate the heat loss per square meter of the wall surface area?
  ক

  1355.3 kJ/m2 hr

  খ

  2345.8 kJ/m2 hr

  গ

  1745.8 kJ/m2 hr

  ঘ

  7895.9 kJ/m2 hr
• With variable thermal conductivity, Fourier law of heat conduction through a plane wall can be expressed as
  ক

  Q = -k0 (1 + β t) A d t/d x

  খ

  Q = k0 (1 + β t) A d t/d x

  গ

  Q = – (1 + β t) A d t/d x

  ঘ

  Q = (1 + β t) A d t/d x