While measuring the acceleration due to gravity by a simple pendulum, a student makes a positive error of $2 %$ in the length of the pendulum and a positive error of $1 %$ in the value of time period. His actual percentage error in the measurement of the value of g will be

3 %

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0 %

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4 %

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5 %

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Solution

The correct option is C $4 %$
For simple pendulum
$T = 2 π \sqrt{\frac{ϑ}{g}}$
$g = \frac{4 π^{2} ϑ}{T^{2}}$
$\frac{Δ g}{g} = \frac{Δ I}{L} + \frac{2 Δ T}{T}$
$\frac{Δ g}{g} \times 100 = 2 + 2 (1)$
$\frac{Δ g}{g} \times 100 = 4$
Percentage error $= 4$ %

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Q. While measuring the acceleration due to gravity by a simple pendulum, a student makes a positive error of 1% in the length of the pendulum and a negative error of 3% in the value of time period. His percentage error in the measurement of g by the relation

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While measuring the acceleration due to gravity by a simple pendulum, a student makes a positive error of 2% in the length of the pendulum and a positive error of 1% in the value of time period. His actual percentage error in the measurement of the value of g will be

The correct option is C 4%For simple pendulumT=2π√ϑgg=4π2ϑT2Δgg=ΔIL+2ΔTTΔgg×100=2+2(1)Δgg×100=4Percentage error =4%

While measuring the acceleration due to gravity by a simple pendulum, a student makes a positive error of $2 %$ in the length of the pendulum and a positive error of $1 %$ in the value of time period. His actual percentage error in the measurement of the value of g will be

The correct option is C $4 %$
For simple pendulum
$T = 2 π \sqrt{\frac{ϑ}{g}}$
$g = \frac{4 π^{2} ϑ}{T^{2}}$
$\frac{Δ g}{g} = \frac{Δ I}{L} + \frac{2 Δ T}{T}$
$\frac{Δ g}{g} \times 100 = 2 + 2 (1)$
$\frac{Δ g}{g} \times 100 = 4$
Percentage error $= 4$ %