The temperature gradient across a copper rod of thickness 0.02m, maintained at two temperature junctions of 20°C and 80°C respectively is
3.0 x 102Km-1
3.0 x 103Km-1
5.0 x 103Km-1
3.0 x 104Km-1
Explanation
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Discussions (42)
Quick tip for figuring out how to solve a problem is by looking at the units in the options. Like this one now, I saw K/m and figured out that it was Temperature in K divided by length or thickness in M and so I looked at the options and the power range and thought to subtract the temperature and then divide it by the 0.02m given in the question.
So next time you don't know what to do with a question just look at the units and you might be able to figure out what is expected of you.
Thanks for reading.👋👋
60/0.02 is same 6000/2 cos d numerator n denominator are multiplied by 100 2 get whole number for easy division.
It lyk sayin 1/0.5=10/5=2
or beta stil convert the 600 and 0.2 to standard form n carryout ur working using d law of indicies
I.e (6x10^1)/(2x10^-2)
=(6/2) x 10^(1-(-2))
=3x 10^3.
now how would you justify the units used here? the width is in meters but the options are given in /km???? isn't there supposed to be a conversion from meter to kilometer?
This ansa is wrong, pls myschool should do sometin about it bf it tarnishes their image
what concerns kilometre in the above question.
Again how degree Celsius changes to Kelvin is still confusing.
