The acceleration due to gravity on the surface of moon is 1.7 ms-2. The time period of a simple pendulum on the surface of moon if its time period on the surface of earth being 3.5 s is (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 ms-2. The time period of a simple pendulum on the surface of moon if its time period on the surface of earth being 3.5 s is (g on the surface of earth is 9.8 m s-2). (A) 8.4 s (B) 4.8 s (C) 48 s (D) 84 s

Tardigrade Admin · Accepted Answer

Earth, Time period, T = 2 π √(l/g) i.e. 3.5 = 2 π √(l/9.8) Moon, Here, g = 1.7 m s-2 T' = √(l/1.7) Dividing, (T'/3.5) = 2 π √(l/1.7)/ 2π √(l/9.8) T' = √(9.8/1.7) × 3.5 = 8.4 s

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

8.4 s

4.8 s

48 s

84 s

$E a r t h$ , Time period, T = $2 π \frac{l}{g}$ i.e. 3.5 = $2 π \frac{l}{9.8}$
$M oo n$ , Here, $g = 1.7 m s^{- 2}$
$T^{'} = \frac{l}{1.7}$
Dividing, $\frac{T ^{'}}{3.5} = 2 π \frac{l}{1.7} /2 π \frac{l}{9.8}$
$T^{'} = \frac{9.8}{1.7} \times 3.5 = 8.4 s$

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