The result after adding 3.8 times 10^{-6} and | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(C) To add these numbers,we first express them with the same power of $10$: $3.8 	imes 10^{-6} = 0.38 	imes 10^{-5}$ Now,add the numbers: $0.38 	im

Vedclass · Accepted Answer

$4.6 	imes 10^{-5}$

Vedclass · Answer

(C) To add these numbers,we first express them with the same power of $10$: $3.8 	imes 10^{-6} = 0.38 	imes 10^{-5}$ Now,add the numbers: $0.38 	imes 10^{-5} + 4.2 	imes 10^{-5} = (0.38 + 4.2) 	imes 10^{-5} = 4.58 	imes 10^{-5}$ According to the rules of significant figures for addition,the result should be reported to the same number of decimal places as the measurement with the fewest decimal places. $0.38$ has two decimal places and $4.2$ has one decimal place. Therefore,the result must be rounded to one decimal place. Rounding $4.58$ to one decimal place gives $4.6$. Thus,the final result is $4.6 	imes 10^{-5}$.

The result after adding $3.8 \times 10^{-6}$ and $4.2 \times 10^{-5}$ with due regard to significant figures is:

Similar Questions

What is the number of significant figures in $11.118 \times 10^{-6} \ V$?

The length,breadth,and thickness of a block are given by $l = 12 \, cm$,$b = 6 \, cm$,and $t = 2.45 \, cm$. The volume of the block according to the rules of significant figures should be:

If the side length of a cube is $7.203 \ m$,then its volume is ............ $m^3$.

An object of mass $4.237 \, g$ occupies a volume of $1.72 \, cm^3$. The density of the object to appropriate significant figures is ......... $g \, cm^{-3}$.

$A$ piece of length $3.532 \ m$ is cut from a rod of length $43.4 \ m$. The length of the remaining rod in metre is (up to correct significant figures)

Vedclass Test Series

Exam Paper Generator

Online Exam Module