Question 41: Units , Dimensions and Error Analysis

Question

Keeping the significant figures in view, the sum of the physical quantities 52.01 m, 153.2 m and 0.123 m is :

Accepted Answer

In addition, or subtraction, the final result should be reported to the same number of decimal places as the measurement with the least number of decimal places. This ensures the result is no more precise than the least precise measurement.
The given measured values are,
$52.01 \mathrm{~m}: 2$ decimal places.
153.2 m : 1 decimal place.
0.123 m : 3 decimal places.
The measurement with the least number of decimal places is 153.2 m ( 1 decimal place). Therefore, our final answer must be rounded to one decimal place.
$$ 52.01+153.2+0.123=205.333 \mathrm{~m} $$
Now, we round 205.333 to one decimal place :
The digit at the first decimal place is $3 \rightarrow 205.33$
The digit following it is 3 (which is less than 5 ) Ã¢ÂÂ 205.3
Therefore, the rounded result is 205.3 m . Hence, the correct option is (A)

Answer

205.3 m

Answer

205 m

Answer

205.333 m

Answer

205.33 m

Units , Dimensions and Error Analysis

Keeping the significant figures in view, the sum of the physical quantities 52.01 m, 153.2 m and 0.123 m is :

Choose the correct answer: