Question

There is 4% error in measuring the period of a simple pendulum. The approximate percentage error in length is.......(Hint : T = 2π√1/g) Select correct option from the given options.
(a) 4%
(b) 8%
(c) 2%
(d) 6%

Answer 1

time period is given by , $T=2\pi\sqrt{\frac{l}{g}}$
squaring both sides,
$T^2=4\pi^2\frac{l}{g}$

$T^2=\{\frac{4\pi^2}{g}\}l$

$T^2=Kl$..........(1)

where K = 4π²/g , it is constant term.

now, differentiate both sides ,

$2T.dT=K.dl$

dividing T² by both sides,

so, $2\frac{dT}{T}=\frac{K.dl}{T^2}$

from equation (1),

$2\frac{dT}{T}=\frac{dl}{l}$

multiplying 100 with both sides,

$2\frac{dT}{T}times100=\frac{dl}{l}\times100$

hence, 2 × % error in T = % error in l

given, % error in time period = 4%
so, % error in length = 2 × % error in time period
= 2 × 4 % = 8 %

hence, option (b) is correct

Answer 2

Dear Student:

Time period is given by , T = 2π√l/g

squaring both sides,

$T^{2} =4\pi ^{2} l/g$

differentiate both sides ,

option (b) is correct.

See the attachment.