Radiation Shielding MCQ Quiz in मल्याळम - Objective Question with Answer for Radiation Shielding - സൗജന്യ PDF ഡൗൺലോഡ് ചെയ്യുക

Last updated on Apr 6, 2025

നേടുക Radiation Shielding ഉത്തരങ്ങളും വിശദമായ പരിഹാരങ്ങളുമുള്ള മൾട്ടിപ്പിൾ ചോയ്സ് ചോദ്യങ്ങൾ (MCQ ക്വിസ്). ഇവ സൗജന്യമായി ഡൗൺലോഡ് ചെയ്യുക Radiation Shielding MCQ ക്വിസ് പിഡിഎഫ്, ബാങ്കിംഗ്, എസ്എസ്‌സി, റെയിൽവേ, യുപിഎസ്‌സി, സ്റ്റേറ്റ് പിഎസ്‌സി തുടങ്ങിയ നിങ്ങളുടെ വരാനിരിക്കുന്ന പരീക്ഷകൾക്കായി തയ്യാറെടുക്കുക

Latest Radiation Shielding MCQ Objective Questions

Top Radiation Shielding MCQ Objective Questions

Radiation Shielding Question 1:

Two infinitely large parallel plates (I and II) are held at temperatures TI and TII (TI > TII) respectively and placed at a distance 2d apart in a vacuum. An infinitely large flat radiation shield (III) is placed in parallel in between I and II. The emissivities of all the plates are equal. The ratio of the steady-state radiative heat fluxes with and without the shield is:

22 16

  1. 0.5
  2. 0.75
  3. 0.25
  4. 0

Answer (Detailed Solution Below)

Option 1 : 0.5

Radiation Shielding Question 1 Detailed Solution

In case if there are n screens used, all of them are having the same emissivity as those of plate.

\({q_{withscreen}} = \left( {\frac{1}{{n + 1}}} \right){q_{withoutscreen}}\)

Here, n = 1

\(\therefore {q_{withscreen}} = \left( {\frac{1}{{1 + 1}}} \right){q_{withoutscreen}}\)

\(\therefore \frac{{{q_{with}}}}{{{q_{without}}}} = \frac{1}{2} = 0.5\)

Radiation Shielding Question 2:

Two long parallel surfaces each of emissivity 0.7 are maintained at different temperatures and accordingly have radiation heat exchange between them. It is desired to reduce 75% of this radiant heat transfer by inserting thin parallel shields of same emissivity on both sides. The number of shields should be

  1. One
  2. Two
  3. Three
  4. Four

Answer (Detailed Solution Below)

Option 3 : Three

Radiation Shielding Question 2 Detailed Solution

Let N be the required number of shields. When emissivities of the main radiating surface and those of parallel radiation shields are equal, then

\(\frac{{With\;shields}}{{without\;shields}} = \frac{1}{{N + 1}}\)

Given, \({Q_{shielded}} = \left( {1 - 0.75} \right){Q_{un - shieled}}\)

\(\therefore \frac{1}{{\left( {N + 1} \right)}} = 0.25\)

Or N = 3
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