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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


one year ago

## Answers : (2)

Arun
25353 Points
							 SINCE WE KNOW THAT,T  = 2 PI (L/G)^1/2G = 4 (PI)^2 L/GAS 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 TIMEWE KNOW THAT ERROR COMBINES,  2(A''/A),WHERE AIS LENGTH, = 2(0.1/100)= 0.002PERCENTAGE ERROR= 0.002 X100 = 0.2 %

one year ago
venkatesh allam
21 Points
							ANSWER: 0.2% or 0.002 length of a simple pendulum “L”=100cm have an accuracy of  $\Delta$L=1mm=0.1cmTime Period of oscillations “t”=2sNumber of oscillations “n”=100Clock resolution $\Delta$t=0.1s$T=2\pi \sqrt{L/g}$g=4*pi^2*L/T^2Apply log on both sideslog g =log4*pi^2  + log L – 2 log TApply derivative on both sides$\frac{\Delta g}{g} = \frac{\Delta L}{L} +2 \frac{\Delta T}{T}$     …..............................(1)T=t/nlog 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; $\frac{\Delta g}{g} =0.002$

one year ago
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