The acceleration due to gravity on the surface of moon is 1.7 m s -2 What is the time period of a simple pendulum. on the surface of moon if its time period on the surface of earth is 3.5 s ? (g on the surface of earth is 9.8 m s -2)

Question

The acceleration due to gravity on the surface of moon is 1.7 m s -2 What is the time period of a simple pendulum. on the surface of moon if its time period on the surface of earth is 3.5 s ? (g on the surface of earth is 9.8 m s -2) (A) 6.4 s (B) 7.4 s (C) 8.4 s (D) 9.4 s

Tardigrade Admin · Accepted Answer

Here, gm=1.7 m s -2, ge=9.8 m s -2, Te=3.5 s As Te=2 π √(l/ge) and Tm=2 π √(l/gm) ∴ (Tm/Te)=√(ge/gm) or Tm=Te √(ge/gm)=3.5 √(9.8/1.7)=8.4 s

Q. The acceleration due to gravity on the surface of moon is $1.7 m s^{- 2}$ What is the time period of a simple pendulum. on the surface of moon if its time period on the surface of earth is $3.5 s$ ? (g on the surface of earth is $9.8 m s^{- 2})$

6.4 s

7.4 s

8.4 s

9.4 s

Here, $g_{m} = 1.7 m s^{- 2}, g_{e} = 9.8 m s^{- 2}, T_{e} = 3.5 s$
As $T_{e} = 2 π \frac{l}{g _{e}}$
and $T_{m} = 2 π \frac{l}{g _{m}}$
$∴ \frac{T _{m}}{T _{e}} = \frac{g _{e}}{g _{m}}$
or $T_{m} = T_{e} \frac{g _{e}}{g _{m}} = 3.5$
$\frac{9.8}{1.7} = 8.4 s$

Q. The acceleration due to gravity on the surface of moon is 1.7ms−2 What is the time period of a simple pendulum. on the surface of moon if its time period on the surface of earth is 3.5s ? (g on the surface of earth is 9.8ms−2)