Time period of a simple pendulum inside a lift that is accelerating up at `2ms^(-2)` is `T_(1)`. If left is retarding down at `6ms^(-2)` then time period of same pendulum is `T_(1)`. Then `T_(1)//T_(2)` is

Time period of a simple pendulum is T inside a lift when the lift is stationary. If the lift moves upwards with an acceleration g/2, the time period of pendulum will be:

The time period of simple pendulum inside a stationary lift is T. If lift starts accelerating upwards with the acceleration of g/2, then the new time period of the simple pendulum will be

The time period of simple pendulum inside a stationary lift is T. If lift starts accelerating upwards with the acceleration of g/2, then the new time period of the simple pendulum will be

The period of a simple pendulum inside a stationary lift is T. The lift accelerates upwards with an acceleration of g/3 . The time period of pendulum will be

A vibrating simple pendulum of period T is placed in a lift which is accelerating downwards. What is the effect of this on the time period of the pendulum?

The time period of a simple pendulum inside a stationery lift is T . If the lift accelerates upwards uniformly with (g)/(4) , then its time period would be

A person measured the time period of a simple pendulum inside a stationary lift and finds it to be .T.. If the lift starts accelerating down wards with an acceleration g/3, the time period of the pendulum will be

The time period of a simple pendulum inside a stationary lift is T. If it starts descending with acceleration g/5. the new period of the simple pendulum will be