Answer :
Period of an ideal simple pendulum = 2π √(L / G)
1.87 = 2π √ (L / 9.81)
Divide each side by 2π : (1.87 / 2π) = √ (L / 9.81)
Square each side: (1.87 / 2π)² = L / 9.81
Multiply each side by 9.81 : L = (9.81) (1.87 / 2π)² = 0.869 meter
Choice 'D' is the closest one.
Answer : The correct option is, (D) 0.87 meters
Solution :
Formula used :
[tex]T=2\pi \times \sqrt{\frac{L}{g}}[/tex]
where,
T = time period of a pendulum = 1.87 seconds
L = length of the pendulum = ?
g = gravity on earth = [tex]9.8m/s^2[/tex]
Now put all the given values in the above formula, we get the length of the pendulum.
[tex]1.87s=2\times \frac{22}{7}\times \sqrt{\frac{L}{9.8m/s^2}}[/tex]
[tex]0.2975=\sqrt{\frac{L}{9.8m/s^2}}[/tex]
Now squaring on both the sides, we get
[tex]L=0.868m=0.87m[/tex]
Therefore, the length of the pendulum is, 0.87 meters.
