The time taken by a simple pendulum to perform 100 vibration is 8 minutes 9 sec in bombay and 8 minutes 20 sec. in pune. Calcualte the ratio of acceleration due to gravity in bombay and pune.

Calculate the period of a simple pendulum of length 0.98 m at a place where acceleration due to gravity is 9.8ms^(-2) .

Value of acceleration due to gravity is 9.8m//sec^(2) . Find its value in km//hr^(2) .

Q. Statement I Three pendulums are suspended from ceiling as shown in Fig. These three pendulums are set to oscillate as shon by arrows, and it is found that all three have same time period. Noe, all three are taken to a place where acceleration due to gravity changes to 4/9th of its value at the first place. If spring pendulum makes 60 cycles in a given time at this place, then torsion pendulum and simple pendulum will also make 60 oscillation in same (given) time interval. Statement II: Time period of torsion pendulum is independent of acceleration due to gravity.

Q. Statement I Three pendulums are suspended from ceiling as shown in Fig. These three pendulums are set to oscillate as shon by arrows, and it is found that all three have same time period. Noe, all three are taken to a place where acceleration due to gravity changes to 4/9th of its value at the first place. If spring pendulum makes 60 cycles in a given time at this place, then torsion pendulum and simple pendulum will also make 60 oscillation in same (given) time interval. Statement II: Time period of torsion pendulum is independent of acceleration due to gravity.

The acceleration due to gravity on the surface of earth is 9.8 ms^(-2) .Time period of a simple pendulum on earth and moon are 3.5 second and 8.4 second respectively. Find the acceleration due to gravity on the moon . Hint : T_(e) = 2pi sqrt((L)/(g_(e))) T_(m)= 2pi sqrt((L)/(g_(m))) (T_(e)^(2))/(T_(m)^(2))= (g_(m))/(g_(e)) g_(m) = (T_(e)^(2))/(T_(m)^(2))g_(e)

The acceleration due to gravity on the surface of earth is 9.8 ms^(-2) .Time period of a simple pendulum on earth and moon are 3.5 second and 8.4 second respectively. Find the acceleration due to gravity on the moon . Hint : T_(e) = 2pi sqrt((L)/(g_(e))) T_(m)= 2pi sqrt((L)/(g_(m))) (T_(e)^(2))/(T_(m)^(2))= (g_(m))/(g_(e)) g_(m) = (T_(e)^(2))/(T_(m)^(2))g_(e)