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The electrical resistance in ohms of a certain thermometer varies with temperature ac cording to the approximate law: `R =R_(0)[1+alpha(T-T_(0))]` The resistances is `101.6 Omega` at the triple-point of water `273.16K`, and `165.5 Omega` at the normal melting point of lead `(600.5K)`. What is the temperature when the resistance is `123.4 Omega`? |
Answer» Here, `R_(0)=101.6 Omega , T_(0)=273.16 K` Case (i) `R_(1)=165.5 Omega , T_(1)=600.5K`, Case(ii) `R_(2)=123.4 Omega , T_(2)=?` Using the relation `R =R_(0)[1+alpha(T-T_(0))]` case (i) `165.5 = 101.6[1+alpha(600.5-273.16)]` `alpha = (165.5-101.6)/(101.6xx(600.5-273.16)) = (63.9)/(101.6xx327.34)` case(ii) `123.4 = 101.6[1+(63.9)/(101.6 xx 327.34)(T_(2)-273.16)] = 101.6+(63.9)/(327.34) (T_(2)-273.16)` or, `T_(2)=((123.4-101.6)xx327.34)/(63.9)+273.16 = 111.67+273.16 = 384.83 K`. |
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