While measuring 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 the time period. His percentage error in the measurement of the value of g will be

Question

While measuring 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 the time period. His percentage error in the measurement of the value of g will be (A) 2 % (B) 4 % (C) 7 % (D) 10 %

Tardigrade Admin · Accepted Answer

T=2π√(l/g) ⇒ T2=(4π2l/g) ⇒ g=(4π2l/T2) (Δ g/g)×100=(Δ l/l)× 100+2 (Δ T/T)×100 =1 %+2×3 %=7 %

Q. While measuring 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 the time period. His percentage error in the measurement of the value of $g$ will be

$2%$

$4%$

$7%$

$10%$

$T = 2 π \frac{l}{g} \Rightarrow T^{2} = \frac{4 π ^{2} l}{g} \Rightarrow g = \frac{4 π ^{2} l}{T ^{2}}$
$\frac{Δ g}{g} \times 100 = \frac{Δ l}{l} \times 100 + 2 \frac{Δ T}{T} \times 100$
$= 1% + 2 \times 3% = 7%$

Q. While measuring 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 the time period. His percentage error in the measurement of the value of g will be

2%

4%

7%

10%