The acceleration due to gravity on the surface of a satellite is 1.5 ms-2, the time period of a simple pendulum on it, if its time period on earth is r, will be

Question

The acceleration due to gravity on the surface of a satellite is 1.5 ms-2, the time period of a simple pendulum on it, if its time period on earth is r, will be (A) ∞ (B) 3.5 t (C) 4.8 t (D) 2.56 t

Tardigrade Admin · Accepted Answer

On earth Time period, Te = 2π √(l/ge) , or t = 2π √(1/9.8) ....(i) On moon Tp = 2 π √(l/gm) or Tp = 2 π √(l/1.5) .....(ii) Dividing (ii) by (i), (Tp/t) = √(9.8/1.5) or Tp = t √(9.8/1.5) = 2.56 t

Q. The acceleration due to gravity on the surface of a satellite is $1.5 m s^{- 2}$ , the time period of a simple pendulum on it, if its time period on earth is r, will be

$\infty$

3.5 t

4.8 t

2.56 t

$O n e a r t h$ Time period, $T_{e} = 2 π \frac{l}{g e}$ , or $t = 2 π \frac{1}{9.8}$ ....(i)
$O n m oo n$ $T_{p} = 2 π \frac{l}{g _{m}}$ or $T_{p} = 2 π \frac{l}{1.5}$ .....(ii)
Dividing (ii) by (i), $\frac{T _{p}}{t} = \frac{9.8}{1.5}$ or $T_{p} = t \frac{9.8}{1.5}$ = 2.56 t

Q. The acceleration due to gravity on the surface of a satellite is 1.5ms−2, the time period of a simple pendulum on it, if its time period on earth is r, will be

∞