Verified Solution • 2025–26 Session moderate

QJEEM21Phead6: JEE Main 2021 Physics - Class 11 JEEM21Phead6 Solution: The time period of a simple pendulum is given by . The measured …

The time period of a simple pendulum is given by T=2πlgT = 2\pi \sqrt {{l \over g}} . The measured value of the length of pendulum is 10 cm known to a 1mm accuracy. The time for 200 oscillations of the pendulum is found to be 100 second using a clock of 1s resolution. The percentage accuracy in the determination of 'g' using this pendulum is 'x'. The value of 'x' to be nearest integer is :-

Multiple Choice Options

  • A 2%
  • B 3%
  • C 5%
  • D 4%

Step-by-Step Verified Solution

T=2πlgT = 2\pi \sqrt {{l \over g}}

T2=2π(lg){T^2} = 2\pi \left( {{l \over g}} \right)

g=2πlT2g = 2\pi {l \over {{T^2}}}

Δgg=Δll+2ΔTT{{\Delta g} \over g} = {{\Delta l} \over l} + {{2\Delta T} \over T}

Δgg=1×10−31×10−2+2×1100{{\Delta g} \over g} = {{1 \times {{10}^{ - 3}}} \over {1 \times {{10}^{ - 2}}}} + {{2 \times 1} \over {100}}

Δgg=0.02+0.01=0.03{{\Delta g} \over g} = 0.02 + 0.01 = 0.03

100×Δgg=0.03×100=3%100 \times {{\Delta g} \over g} = 0.03 \times 100 = 3\%

Δgg×100=3%{{\Delta g} \over g} \times 100 = 3\%

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