logo

Ask Questions, Get Answers

 
X
 Search
Want to ask us a question? Click here
Browse Questions
Ad
Home  >>  JEEMAIN and AIPMT  >>  Physics  >>  Class11  >>  Units and Measurement
0 votes

The distance of a point p from a point source of light $S_1$ is $100. 1243\;cm$ and from another point source $S_2$ is $100.1664\;cm$ Find $(S_2P-S_1P)$ and $(S_2P +S_1P)$ correct upto the same number of significant figures.

$\begin{array}{1 1}(A)\;200\;cm \\(B)\;23 \;cm \\(C)\;98 \;cm \\(D)\;25\;cm \end{array} $

Can you answer this question?
 
 

1 Answer

0 votes
$S_2P- S_1P= 100.1664\;cm -100.1243\;cm$
$\qquad= 0.0421\;cm$
The difference of two length have 3 significant figure. Now
$S_2P +S_1P =100.1664\;cm+100.1243\;cm$
$\qquad= 200.2907\;cm$
Since the number of significant figures in difference is 3, the number of significant figures in sum should also be 3 Thus,
$S_2P+S_1P=200\;cm$
Hence A is the correct answer.
answered Jul 11, 2014 by meena.p
 

Related questions

Ask Question
student study plans
x
JEE MAIN, CBSE, NEET Mobile and Tablet App
The ultimate mobile app to help you crack your examinations
Get the Android App
...