Question

16 The length of second's pendulum where g = 9.8 m/s²
is 1 metre. The length of second's pendulum, on a
planet, where g = 4.9 m/s2 will be :​

Answer 1

Answer:

0.5 sec is the answer

Explanation:

in above pic

Answer 2

Answer:

The length of seconds' pendulum where $g=9.8m/s^{2}$

is $1m$ The length of seconds' pendulum, on a

planet, where $g=4.9m/s^{2}$ will be $l=0.5m$

Explanation:

The seconds' pendulum is a type of pendulum whose time period is two seconds $T=2s$

The time period for a normal pendulum is given by the formula

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

From the question, it is given that length of the seconds' pendulum where $g=9.8m/s^{2}$ is $l=1m$

substituting these values in the time period formula we get

$4=4\pi ^{2}(\frac{1}{9.8} )$

If the pendulum is placed on a planet where $g=4.9m/s^{2}$ the time period is $2$ only that is,$4=4\pi ^{2} (\frac{l}{4.9})$

By equating these two equations we get

$1/9.8=l/4.9\\l=4.9/9.8=0.5m$