Saturday, 23 January 2016

APPROXIMATIONS AND SMALL ERRORS

APPROXIMATIONS AND SMALL ERRORS

PREVIOUS EAMCET BITS

1.             There is an error of ± 0.04 cm in the measurement of the diameter of a sphere. When the radius is

10 cm, the percentage error in the volume of the sphere is
[EAMCET 2009]
1) ± 1.2
2) ± 1.0
3) ± 0.8
4) ± 0.6
Ans: 4




Sol.  r = 10cm; δr = 0.02

δrr × 1000.2

δVV × 100 = 3 × ( ± 0.2)0.6

2.             The circumference of a circle is measured as 56 cm with an error 0.02 cm. The percentage error

in its area is












[EAMCET 2007]


1)

1







2)
1


3)

1


4)

1





7














28



14


56



Ans: 3






















Sol.  radius = r, circumference = x; Area = A
















x = 2πr r =
x
; δx = 0.02







































2π















A = πr
2
=

x2


















4π





















x



















δA =



.δx
















2π



δA









































Percentage error in A =

×100



























x















A









.δx






1

















=

2π


×100 =











































2



14




















x




























































4π

























3.
The radius of a circular plate is increasing at the rate of 0.01 cm/sec when the radius is 12 cm.

















www








[EAMCET 2005]


Then the rate at which the area increases is









1) 0.24 π sq.cm /sec 2) 60 π sq.cm /sec
3) 24 π sq.cm /sec
4) 1.2 π sq.cm /sec


Ans: 1

























Sol.
r =12, dr
= 0.01/ sec






















dt























A r2

























dA = 2πr dr = 24π×0.01












dr






dt















= 0.24π sq.cm / sec














4.    The approximate value of (1 .0002)3000
is







[EAMCET 2002]


1) 1.2






2) 1.4

3) 1.6

4) 1.8


































                Ans: 3

             Sol.  Let y = f ( x )= x3000

   here x = 1, δ= 0.0002

    δy = f( x )δx = 3000x2999δx

    = ( 3000)( 0.0002)

    = 0.6

   ∴ f ( x x )= y y =1+0.6 =1.6