In a Kater's pendulum the time periods about the knife edges at distance `l_(1) and l_(2)` from the centre of gravity are t and t+pi is very small. Show that `(4 pi^(2)(l_(1)+l_(2)))/(g)=t(t+(2l_(2)tau)/(l_(2)-l_(1)))`

The chord of contact of tangents from a point P to a circle passes through Q. If l_(1) and l_(2) are the length of the tangents from P and Q to the circle,then PQ is equal to (l_(1)+l_(2))/(2) (b) (l_(1)-l_(2))/(2)sqrt(l_(1)^(2)+l_(2)^(2))(d)2sqrt(l_(1)^(2)+l_(2)^(2))

If the time period of a simple pendulum is T = 2pi sqrt(l//g) , then the fractional error in acceleration due to gravity is

If l_(1) " and " l_(2) are the lengths of a seconds pendulum at the poles and the equator of the earth, then

If the time period of a pendulum is 1sec , then what is the length of the pendulum at point of intersection of l-T and l-T^(2) graph.

The length of a metal wire is l_(1) when the tension in it is T_(1) and is l_(2) when the tension is T_(2) . Then natural length of the wire is

A simple pendulum of length l_(1) has a time period of 4 s and another simple pendulum of length l_(2) has a time period 3 s. Then the time period of another pendulum of length (l_(1)-l_(2)) is

If c is small in comparision with l then ((l)/(l+c))^((1)/(2))+((l)/(l-c))^((1)/(2))=

If l_(1),l_(2) and l_(3) are the orders of the currents through our nerves, domestic appliances and average lightening, then the correct order of currents is (1) l_(1) gt l_(2) gt l_(3) (2) l_(1) gt l_(3) gt l_(2) (3) l_(1) lt l_(2) lt l_(3) (4) l_(1) = l_(2) = l_(3)