The mass of a beaker is found to be ( 35 . 6 ± 0 . 2 ) gram when empty, and ( 46 . 4 ± 0 . 2 ) gram when filled partially with a liquid. Find the mass of the liquid with proper uncertainty.

Mass of the liquid = Mass of filled beaker - Mass of empty beaker Mass of the liquid = ( 46 . 4 ± 0 . 2 ) - ( 35 . 6 ± 0 . 2 ) gram Here mass is subtracted from mass and error is added to error (because in case of subtraction also effective error is sum of error) Mass of the liquid = 46 . 4 - 35 . 6 ± ( 0 . 2 + 0 . 2 ) Mass of the liquid = ( 10 . 8 ± 0 . 4 ) g

Solve with due regard to significant figures. 6 . 30 × 10 - 3 + 4 . 2 × 10 - 2

6 . 30 × 10 - 3 + 4 . 2 × 10 - 2 = 0 . 630 × 10 - 2 + 4 . 2 × 10 - 2 = 4 . 830 × 10 - 2 The result of addition must be rounded off to same number of decimal places as present in the digit ( used in calculation ) having least number after decimal places. Here 4 . 2 × 10 - 2 has least decimal places, so rounding off to one decimal place, final answer will be 4 . 8 × 10 - 2 .