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%

Question

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%

SAURABH KUMAR · Answer

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 · Answer

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 $\pm 0.5$ mm

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 $\pm 0.05 /100=\pm0.0005$ s

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
( $\Delta$ g)/g=( $\Delta$ L)/L+2( $\Delta$ T)/T
$\Rightarrow$ ( $\Delta$ g)/g=( $\pm$ 0.5)/1000+2*( $\pm$ 0.0005)/2
$\Rightarrow$ ( $\Delta$ g)/g= $\pm$ 0.0005+( $\pm$ 0.0005)
$\Rightarrow$ ( $\Delta$ g)/g= $\pm$ 0.001
$\Rightarrow$ ( $\Delta$ g)/g= $\pm$ 0.1%

In terms of given choices 0.2%, (d)

kartikeya · Answer

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 %

Thank you for registering.

Register Now and Win Upto 25% Scholorship for a Full Academic Year !

Enter Your Details

Registration done!

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%

3 Answers

Think You Can Provide A Better Answer ?

Provide a better Answer & Earn Cool Goodies See our forum point policy

Other Related Questions on Mechanics

ASK QUESTION

Answer and earn gift vouchers

View courses by askIITians

Design classes One-on-One in your own way with Top IITians/Medical Professionals

Complete Self Study Package designed by Industry Leading Experts

Live 1-1 coding classes to unleash the Creator in your Child

a Complete All-in-One Study package Fully Loaded inside a Tablet!

Register Now & Win Upto 25% Scholorship