Question

A man inside a lift holds a jar filled with a liquid of density ρ . The lower end of jar is at a height H above floor of lift and has a small hole. With lift at rest the liquid emerging out hits floor of lift at a horizontal distance D from a point below the hole on lift. The lift 

(a) accelerates up with an acceleration g/10 

(b) accelerates downwards with an acceleration g/5 

The corresponding distance now is D'. Then 

(1) (a) D´ = D (b) D´ = D 

(2) (a) D´ > D (b) D´ < D 

(3) (a) D´ = D (b) D´ > D 

(4) (a) D´ = D (b) D´ < D

Answer 1

(1) (a) D´ = D (b) D´ = D 

When lift is at rest

v = The speed of efflux of liquid from O = √2gH₁

D = v x √2H/g

= √2gH₁ x √2H/g

= 2√H₁H ...(1)

When lift accelerates up or down effective value of g changes. 

In case (a) g_eff > g and case (b) g_eff < g. 

From Eqn. (1); D does not depend on value of g. 

Therefore in both case (a) and (b); D´ = D.