Question #22019

1) a seconds pendulum is taken to the moon where the value of g is a sixth of the value on earth. the time period of the pendulum will be------
a) 1sec b) 2 sec c) 5sec d) none of these
2)if a force of 1 newton acts on a mass of 500 gms, acceleration will be?

Expert's answer

1)

T = 2*pi*Sqrt[l/g]

T -& the time period of the pendulum

g - acceleration due to gravity

l - the length of the pendulum

T1 = 2*pi*Sqrt[l/g] - on the Earth

T2 = 2*pi*Sqrt[l/(g/6)] - on the Moon

Therefore

T2 = T1 * Sqrt[6]

Answer: d) none of these

2)

F = m*a

F - force

m - mass

a - acceleration

a = F/m = 1H/500g = 1H /0.5 kg = 2 m/s^2

T = 2*pi*Sqrt[l/g]

T -& the time period of the pendulum

g - acceleration due to gravity

l - the length of the pendulum

T1 = 2*pi*Sqrt[l/g] - on the Earth

T2 = 2*pi*Sqrt[l/(g/6)] - on the Moon

Therefore

T2 = T1 * Sqrt[6]

Answer: d) none of these

2)

F = m*a

F - force

m - mass

a - acceleration

a = F/m = 1H/500g = 1H /0.5 kg = 2 m/s^2

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