Answers

2015-03-14T12:50:40+05:30
time period of a simple pendulun is given by,
T=2 \pi  \sqrt{ \frac{1}{g} }
T \alpha  \sqrt{l} ....................1
l'= 2l
T' \alpha  \sqrt{l'}
T' \alpha \sqrt{2l} ...................2
DEVIDING 1 AND 2
 \frac{T}{T'} = \frac{ \sqrt{l}}{ \sqrt{2l} }
 \frac{T}{T'} = \frac{ 1}{ \sqrt{2l} }
T'= \sqrt{2} T
1 5 1
2015-03-14T14:15:06+05:30
The time period will increase

0