IIT JEE , AIEEE Forum - Physics , General - pls ans this question on measurement........rates for sure........

krithika.r

The period of oscillation of a pendulum has to be increased by 10%. The length of pendulum should increase by ________ %.

please give me a answer with detalied solution. rates for sure!!!

gauravtargetsiit

krithika.r

i need answers with steps

spideyunlimited

T = 2pi.root(L/g)
1.1 T = 1.1 x 2pi.root(L/g)
take 1.1 inside the root and u get 1.21 L which means L should be increased by 21%

Protyush_Sahu

t= 2*pi*rootl/g----------------1

l=t^2*g/4*pi^2

when time period is increased

t+10%of t = t*11/10

putting the new value of t in eq. 1

we will get that (new) =

so (new) = 121/100 *l

Difference = = l(21/100)

% increase = (diff / l)*100=21%

rudra.panda

$t=2\pi\sqrt{\frac{l}{g}}$

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

$l=\frac{t^2\cdot g}{4\pi^2}$ ..................................................................(1)

New time:

$t+\frac{t}{10}=\frac{11t}{10}$

Now substituing the value of "t" in equation (1)

WE have,

$l_2=\frac{t^2\cdot g}{4\pi^2}\times\frac{121}{100}$

$l_2=\frac{121}{100}l$

$l-l_2=\frac{121}{100}l-l=\frac{21l}{100}$

Incerease in %=

$\frac{l-l_2}{l}\times 100=\frac{21l}{100l}\times 100=21\%$