For the four sets of three measured physical quantities as given below. Which of the following options is correct? (i) A _ 1 = 24.36, B _ 1 = 0.0724, C _ 1 = 256.2 (ii) A _ 2 = 24.44, B _ 2 = 16.082, C _ 2 = 240.2 (iii)

Question

Quantrex Academy · Accepted Answer

Correct answer: C. A _ 1 + B _ 1 + C _ 1 < A _ 2 + B _ 2 + C _ 2 = A _ 3 + B _ 3 + C _ 3 < A _ 4 + B _ 4 + C _ 4

In addition or subtractions, final answer must contain the equal numbers of digits after decimal as the adding or subtracting quantities among which is having the lowest numbers of digits after decimal.

${A}_{1} + {B}_{1} + {C}_{1} = 24.36 + 0.0724 + 256.2 = 280.6324 = 280 . 6$

The Least number of significant digit after decimal in the above expression is 1 So, Sum should conatian 1 significant digit after decimal.
${A}_{2} + {B}_{2} + {C}_{2} = 24.44 + 16.082 + 240.2 = 280.722 = 280.7$

The Least number of significant digit after decimal in the above expression is 1 So, Sum should conatian 1 significant digit after decimal
${A}_{3} + {B}_{3} + {C}_{3} = 25.2 + 19.2812 + 236.183 = 280.6642 = 280 . 7$

The Least number of significant digit after decimal in the above expression is 1 So, Sum should conatian 1 significant digit after decimal ${A}_{4} + {B}_{4} + {C}_{4} = 25 + 236.191 + 19.5 = 280.691 = 281$

The Least number of significant digit after decimal in the above expression is 0 So, Sum should conatian 0 significant digit after decimal
Answer should be ${A}_{1} + {B}_{1} + {C}_{1} < {A}_{2} + {B}_{2} + {C}_{2} = {A}_{3} + {B}_{3} + {C}_{3} < {A}_{4} + {B}_{4} + {C}_{4}$

For the four sets of three measured physical quantities as given...

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