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 $g = 4 π^{2} (l / T^{2})$ will be

2 %

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4 %

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7 %

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10 %

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Solution

The correct option is D $7 %$
Error always gets added so,
$g = 4 π^{2} (\frac{l}{T^{2}})$
$\frac{Δ g}{g} = \frac{Δ 1}{L} + \frac{2 Δ T}{T}$
$\frac{Δ g}{g} \times 100 = 1 + 2 (3)$
$\frac{Δ g}{g} \times 100 = 7$
Percentage error $= 7$ %

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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 g=4π2(l/T2) will be

The correct option is D 7%Error always gets added so,g=4π2(lT2)Δgg=Δ1L+2ΔTTΔgg×100=1+2(3)Δgg×100=7Percentage error =7%

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 $g = 4 π^{2} (l / T^{2})$ will be

The correct option is D $7 %$
Error always gets added so,
$g = 4 π^{2} (\frac{l}{T^{2}})$
$\frac{Δ g}{g} = \frac{Δ 1}{L} + \frac{2 Δ T}{T}$
$\frac{Δ g}{g} \times 100 = 1 + 2 (3)$
$\frac{Δ g}{g} \times 100 = 7$
Percentage error $= 7$ %