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If mass and specific heat capacity of calormeter is negligible what would be maximum permissible error is `S` Use the data mentioned below `m_(1) rarr 0, S_(c) rarr 0, m_(2) = 1.00kg V = 10.0 V I = 10.0A` `t = 1.00 xx 10^(2) sec theta_(0) = 15^[email protected])C` Corrected `theta_(t) = 65^(@)C` (A) `4%` (B) `5%` (C) `8%` (D) `12%` . |
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Answer» If `m_(1) rarr 0, S_(c) rarr0` `S_(l) = (Vit)/(m_(2)(theta_(f)-theta_(0))` `(DeltaS_(l))/(S_(l))=(DeltaV)/(V)+(DeltaI)/(I)+(Deltat)/(t)+(Deltam_(2))/(m_(2))+(Deltatheta_(f)+Deltatheta_(0))/(theta_(f)-theta_(0))=(0.1)/(10.0)+(0.01xx10^(2))/(1.00xx10^(2))+(0.01)/(1.00)+(1+1)/(50)=8%` . |
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