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Q-The length of a simple pendulum is about 100cm known to have a accuracy of 1mm.Its period of oscillation is 2s determined by measuring the time for 100oscillations using a clock of 0.1s resolution.What is the accuracy in the determined value of g? (a) 0.2% (b) 0.5% (c) 0.1% (d) 2%

Q-The length of a simple pendulum is about 100cm known to have a accuracy of 1mm.Its period of oscillation is 2s determined by measuring the time for 100oscillations using a clock of 0.1s resolution.What is the accuracy in the determined value of g?


(a) 0.2%


(b) 0.5%


(c) 0.1%


(d) 2%

Grade:11

3 Answers

SAURABH KUMAR
16 Points
11 years ago

SINCE WE KNOW THAT,

T  = 2 PI (L/G)^1/2

G = 4 (PI)^2 L/G

AS WE KNOW, THERE CAN BE ERROR IN TIME & LENGTH BUT SINCE TIME IS MEASURED BY A SCALE WHICH 0.1 S, AND MEASURED TIME IS  2 S HENCE THERE IS NO UNCERTAINITY IN TIME

WE KNOW THAT ERROR COMBINES,  2(A''/A),WHERE AIS LENGTH, = 2(0.1/100)= 0.002

PERCENTAGE ERROR= 0.002 X100 = 0.2 % ANS......

 

Atul Kumar Kuthiala
25 Points
6 years ago
Length measurement has been given with accuracy of 1mm.
As a general rule: The degree of accuracy is half a unit each side of the unit of measure.
Assuming that length has been measured with a scale whose least count is 1mm, \Rightarrow Error in measurement of length is \pm 0.5mm
Similarly the clock has a resolution of 0.1s. We know that the least count error is the error associated with the resolution of the instrument. Time of 100 oscillations has been measured.
As such error in measurement of time is \pm 0.05 /100=\pm0.0005s
Time period #T# of a pendulum is gievn by the equation
T=2pisqrt(L/g)
Squaring both sides and stating in terms of g we get
g=4pi^2L/T^2
Taking log of both sides and from theory of errors we get
(\Deltag)/g=(\DeltaL)/L+2(\DeltaT)/T
\Rightarrow(\Deltag)/g=(\pm0.5)/1000+2*(\pm0.0005)/2
\Rightarrow(\Deltag)/g=\pm0.0005+(\pm0.0005)
\Rightarrow(\Deltag)/g=\pm0.001
\Rightarrow(\Deltag)/g=\pm0.1%
In terms of given choices 0.2%, (d)
 
kartikeya
19 Points
6 years ago
L= 100 CMT= 2 SDELTA T= RESOLUTION OF MEASUREMENT/NO. OF OSCILLATION =0.1/100= 0.001DELTA L= 0.1 CMNOW T= 2PI (L/G)^1/2G= 4PI^2L/T^2DELTA G/G = DELTA L/L + 2 DELTA T /T=0.001 + 0.001=0.002DELTA G/G= 0.002= 0.2 %

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