A mass falls from a height 'h' and its time of fall 'l' is recorded in terms of time period T of a simple pendulum. On the surface of earth it is found that I = 2T. The entire set up is taken on the surface of another planet whose mass is half of that of earth and radius the same. Same experiment is repeated and corresponding times noted as t' and T'.

Question

A mass falls from a height 'h' and its time of fall 'l' is recorded in terms of time period T of a simple pendulum. On the surface of earth it is found that I = 2T. The entire set up is taken on the surface of another planet whose mass is half of that of earth and radius the same. Same experiment is repeated and corresponding times noted as t' and T'. (A) t ' = √ 2 T ' (B) t ' > 2 T' (C) t ' < 2 T' (D) t ' = 2 T'

Tardigrade Admin · Accepted Answer

Time of flight = √(2h/g) propto (1/√g) Time period of pendulum = 2π √(1/g) propto (1/√g) Ratio of time of flight time period of pendulum is independent of g. Hence t' = 2T

t ' = $2$ T '

t ' > 2 T'

t ' < 2 T'

t ' = 2 T'

Time of flight $= \frac{2 h}{g} \propto \frac{1}{g}$
Time period of pendulum $= 2 π \frac{1}{g} \propto \frac{1}{g}$
Ratio of time of flight & time period of pendulum is independent of g. Hence $t^{'} = 2 T$

t ' = 2​ T '

t ' > 2 T'

t ' < 2 T'

t ' = 2 T'

Time of flight =g2h​​∝g​1​ Time period of pendulum =2πg1​​∝g​1​ Ratio of time of flight & time period of pendulum is independent of g. Hence t′=2T

t ' = $2$ T '

Time of flight $= \frac{2 h}{g} \propto \frac{1}{g}$
Time period of pendulum $= 2 π \frac{1}{g} \propto \frac{1}{g}$
Ratio of time of flight & time period of pendulum is independent of g. Hence $t^{'} = 2 T$