The time period of oscillations of a second's pendulum on the surface of a planet having mass and radius double. those of earth is

Question

The time period of oscillations of a second's pendulum on the surface of a planet having mass and radius double. those of earth is (A) 4 s (B) 1 s (C) √2 s (D) 2 √2 s

Tardigrade Admin · Accepted Answer

g1=(G m/R2) g2=(G × 2 m/4 R2)=(g1/2) T1=2 π √(l/g1) T2=2 π √(l/g2) T2=√2 T1 Since T1 is time period of seconds pendulum T1=2. Hence T2=2 √2

Q. The time period of oscillations of a second's pendulum on the surface of a planet having mass and radius double. those of earth is

$4 s$

$1 s$

$2 s$

$22 s$

$g_{1} = \frac{G m}{R ^{2}}$
$g_{2} = \frac{G \times 2 m}{4 R ^{2}} = \frac{g _{1}}{2}$
$T_{1} = 2 π \frac{l}{g _{1}}$
$T_{2} = 2 π \frac{l}{g _{2}}$
$T_{2} = 2 T_{1}$
Since $T_{1}$ is time period of seconds pendulum $T_{1} = 2$ .
Hence $T_{2} = 22$