IIT JEE , AIEEE Forum - Physics , General

pinnaka

We will look at the first question now

Given that the measurement in the diameter of the radius is with a least count of 0.01cm

Here is a proof
we know that r =2cm
d = 4+/-0.01m
now divide both the sides by 2
we
r = 2+/-0.005 cm

Now the second Question

d =1.95cm+/-0.01cm
Now divide by 2 on both the sides
consider the number 1.95cm
since the digit 1 and 9 are a certain number therefore it is significant
Where as the other .95 digits are not uncertain therefore they are not classified under the category of significant digits

So according to the rule I have written
d/2 = 0.975+/-0.005cm
but now wait for moment we have only one significant figure to start off Now only the last digit in 0.975

now we have to round off the digit 5 because on starting our significant figures were only 2
d =0.98+/-0.005

----Hope This helps

Please report any errors

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