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# while measuring acceleration due to gravity with simple pendulum if percentage errors in the measurement of length and time period are 2% and 1% respectively, then maximum percentage error in the measurement of acceleration due to gravity is

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## The percentage errors in the measurement of length and time period of a simple pendulum are 1% and 2% respectively. Then, the maximum error in the measurement of the acceleration due to gravity is

The percentage errors in the measurement of length and time period of a simple pendulum are 1% and 2% respectively. Then, the maximum error in the measurement of the acceleration due to gravity is Finding Significant Figures

The percentag... Question

The percentage errors in the measurement of length and time period of a simple pendulum are 1%and2% respectively. Then, the maximum error in the measurement of the acceleration due to gravity is

A 8% B 3% C 4% D 6% E 5% Open in App Solution

The correct option is E

5%

Step 1: Given Data:

The percentage errors in the measurement of length=1%

The percentage errors in the measurement of the time period=2%

Step 2: Formula Used:

Timeperiod(T)=2πlg, where g is the acceleration due to gravity

Step 3: Calculating the maximum error in the measurement of the acceleration due to gravity:

The time period of a simple pendulum is given by-

Timeperiod(T)=2πlg ΔTT=12(Δll−Δgg) Δgg=Δll−2ΔTT

Maximum percentage error in the measurement of the acceleration due to gravity is-

Δgg×100=Δll×100+2ΔTT×100

=1×100+2×2×100 =5×100=5%

Hence option E is the correct answer.

Suggest Corrections 3 SIMILAR QUESTIONS

Q.

The percentage errors in the measurements of the length of a simple pendulum and its time period are 3% and 5% respectively. The maximum error in the value of the acceleration due to gravity obtained from these measurements is

Q. Length

l

and Acceleration due to gravity

g are measured with ± 6 % and ± 4 %

errors respectively. The percentage error in determination of the Time period of a simple pendulum will be,

Q. The acceleration due to gravity is measured by using simple pendulum . If 'a ' and 'b' are relative errors in me?surement of length and time period respectively then percentage error in the measurement of acceleration due to gravity will be??Q. The acceleration due to gravity is measured on the surface of earth by using a simple pendulum. If Alfa and beta are relative errors in the measurement of length and time period respectively, then percentage error in the measurement of acceleration due to gravity is?Q. In an experiment to determine acceleration due to gravity by simple pendulum, a student commits 1 % positive error in the measurement of length and 3% negative error in the measurement of time period. The percentage error in the value of will be :

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स्रोत : byjus.com

## The percentage errors in the measurement of length and time period of a simple pendulum are 1

Click here👆to get an answer to your question ✍️ The percentage errors in the measurement of length and time period of a simple pendulum are 1 Question

A

B

C

D

## 5%

Medium Open in App Solution Verified by Toppr

Correct option is D) 93 23

स्रोत : www.toppr.com

## The percentage errors in the measurement of length and time period of a simple pendulum are 1% and 2% respectively. Then, the maximum error in the measurement of acceleration due to gravity is

Time period of simple pendulum is T=2pisqrt((l)/(g)) or (DeltaT)/(T)==(1)/(2)((Deltal)/(l)-(Deltag)/(g)) implies(Deltag)/(g)=(Deltal)/(l)-(2DeltaT)/(l) therefore Maximum percentage error in g, (Deltag)/(g)xx100=(Deltal)/(l)xx100+(2DeltaT)/(T)xx100 =1xx100+2xx2xx100=5xx100=5% Home > English > Class 12 > Physics > Chapter > Practice Set 10 >

The percentage errors in the m...

The percentage errors in the measurement of length and time period of a simple pendulum are 1% and 2% respectively. Then, the maximum error in the measurement of acceleration due to gravity is

Updated On: 27-06-2022

Solution

Time period of simple pendulum is

`T=2pisqrt((l)/(g))` or `(DeltaT)/(T)==(1)/(2)((Deltal)/(l)-(Deltag)/(g))`

`implies(Deltag)/(g)=(Deltal)/(l)-(2DeltaT)/(l)`

`therefore` Maximum percentage error in g,

`(Deltag)/(g)xx100=(Deltal)/(l)xx100+(2DeltaT)/(T)xx100`

`=1xx100+2xx2xx100=5xx100=5%` Step by step solution by experts to help you in doubt clearance & scoring excellent marks in exams.

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