Verified Solution • 2025–26 Session moderate

QJEEM19Ph9ed8: JEE Main 2019 Physics - Class 11 JEEM19Ph9ed8 Solution: In a simple pendulum experiment for determination of…

In a simple pendulum experiment for determination of acceleration due to gravity (g), time taken for 20 oscillations is measured by using a watch of 1 second least count. The mean value of time taken comes out to be 30 s. The length of pendulum is measured by using a meter scale of least count 1 mm and the value obtained is 55.0 cm. The percentage error in the determination of g is close to :-

Multiple Choice Options

  • A 0.20%
  • B 3.50%
  • C 0.70%
  • D 6.80%

Step-by-Step Verified Solution

Time period of a pendulum (T) = 2πlg2\pi \sqrt {{l \over g}}

⇒\Rightarrow T2 = 4π2lg4{\pi ^2}{l \over g}

⇒\Rightarrow g=4π2lT2g = {{4{\pi ^2}l} \over {{T^2}}}

Fractional change

(dgg)×100=(dll)×100−(2dTT)×100\left( {{{dg} \over g}} \right) \times 100 = \left( {{{dl} \over l}} \right) \times 100 - \left( {2{{dT} \over T}} \right) \times 100

∴\therefore Maximum possible percentage error,

(dgg)×100=(dll)×100+(2dTT)×100\left( {{{dg} \over g}} \right) \times 100 = \left( {{{dl} \over l}} \right) \times 100 + \left( {2{{dT} \over T}} \right) \times 100

Error in time period(dT) = least count of time = 1 second

and T = 30 second

Error in length(dl) = least count of length = 1 mm

and ll = 55.0 cm

∴\therefore (dgg)×100=\left( {{{dg} \over g}} \right) \times 100 = (0.155)×100+2(130)×100\left( {{{0.1} \over {55}}} \right) \times 100 + 2\left( {{1 \over {30}}} \right) \times 100 = 6.8%

Related Learning Resources & Exam Preparation

JEE Main PYQ Question Bank
Browse year-wise authentic previous years' questions with solutions
Authentic CBT Mock Tests
Simulate official exam environment with live timer and percentile scoring