Advertisements
Advertisements
प्रश्न
Calculate the length of the second’s pendulum on the surface of the moon when acceleration due to gravity on the moon is 1.63 ms−2.
उत्तर
Length of second’s pendulum = l =?
Acceleration due to gravity on the surface of the moon
= gm = 1.63 ms−2
Time period = T = 2 s
T = `2πsqrt("l"/"g")`
2 = `2xx22/7xxsqrt("l"/1.63)`
`sqrt("l"/1.63)=7/22`
Squaring both sides
`"l"/1.63=49/484`
l = `49/484xx1.63`
l = 0.165 m
APPEARS IN
संबंधित प्रश्न
Find the length of a second's pendulum at a place where g = 10 m s-2 (Take π = 3.14)
Define simple pendulum.
Write an expression for the time period of a simple pendulum.
State two factors which determine the time period of a simple pendulum.
Define the following in connection with a simple pendulum.
Effective length
What do you understand by the term graph?
State two important precautions for drawing a graph line.
What do you understand by the term constant of proportionality?
If a pendulum takes 0.5 s to travel from A to B as shown in the following figure, find its time period and frequency.