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Grade: 12th pass
        
The length of a simple pendulum is about 100 cm known to have an accuracy of 1mm.its time period of oscillation is 2 s determined by measuring the time for 100 oscillations using a click of 0.1s resolution. What is the accuracy in the determined value of g
 
5 months ago

Answers : (2)

Arun
22958 Points
							
 

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 %

5 months ago
venkatesh allam
21 Points
							
ANSWER: 0.2% or 0.002
 length of a simple pendulum “L”=100cm
 have an accuracy of  \DeltaL=1mm=0.1cm
Time Period of oscillations “t”=2s
Number of oscillations “n”=100
Clock resolution \Deltat=0.1s
T=2\pi \sqrt{L/g}
g=4*pi^2*L/T^2
Apply log on both sides
log g =log4*pi^2  + log L – 2 log T
Apply derivative on both sides
\frac{\Delta g}{g} = \frac{\Delta L}{L} +2 \frac{\Delta T}{T}     …..............................(1)
T=t/n
log T =(1/n)* logt
\frac{\Delta T}{T} = \frac{\Delta t}{nt}
\frac{\Delta T}{T}*100 = \frac{\Delta t}{nt} *100
n=100
\frac{\Delta T}{T}*100 = \frac{\Delta t}{t} …..................................(2)
from (1)
\frac{\Delta g}{g} *100= \frac{\Delta L}{L} *100+2 \frac{\Delta T}{T}*100
\frac{\Delta g}{g} *100= \frac{\Delta L}{L} *100+2 *\frac{\Delta t}{t}
\frac{\Delta g}{g} *100= \frac{0.1}{100}*100 +2 *\frac{0.1}{2}
\frac{\Delta g}{g} *100=0.1+0.1=0.2
\frac{\Delta g}{g} *100=0.2
\frac{\Delta g}{g} =0.002
the accuracy in the determined value of g; \frac{\Delta g}{g} =0.002
 
 
5 months ago
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