Two objects of masses `m_(1) and m_(2)` fall from the heights `h_(1) and h_(2)` respectively. The ratio of the magntidue of their momenta when they hit the ground is

Answer

A, B, D

Explanation

`(Q/t)=(KA(T_1-T_2))/L`
`T_2=(KA(T_1-T_2))/(Lms)`
`Fall in temperature in T_1=(KA(T_1-T_2))/(Lm_1s_1)`
`Final tempareture in T_1=T_1-=(KA(T_1-T_2))/(Lm_1 s_1)`
`Final temprature in `
` T_2=T_2+(KA(T_1-T_2))/(Lm_2 s_2)`
`Change in temparature `
` T_1-(KA(T_1-T_2))/(Lm_1 s_1)`
`=(T_2+(KA(T_1-T_2))/(Lm_2 s_2)`
`=(T_1-T_2)`
`-[(KA(T_1-T_2))/(Lm_1 s_1)+(KA(T_1-T_2))/(Lm_2 s_2)]`
`IndT=(KA)/(L)((M_2)(S_2) +(M_1)(S_1)/(M_1)(S_1)(M_2)(S_2))`
So difference in tempareture `=(T_2-T_1)e^(-lambda t)`
`where (lambda)=(KA)/(l)((m_1)(s_1)+(m_2)(s_2))/((m_1)(s_1)(m_2)(s_2))`.