A small hole is made at a height of h ′ ′ = (1/ 
) m from the bottom of a cylindrical water tank and at a depth of h = 
m form the upper level of water in the tank. The distance where the water emerging from the hole strikes the ground is:
m
Text Solution
Verified by ExpertsC
Applying Bernoulli’s Principle between the point 1 and 2
ρ ρ gh 1 + P 1 + 1/2 P 
= P 2 + 1/2 ρ ρ 
+ ρ ρ gh 2
Substituting h 1 = h 2 = h ′ ′ and since v 1 = a/A v 2 and a << A
v 1 ~ 0 , P 1 = P 0 + ρ ρ gh, P 2 = P 0 ,
we obtain,
ρ ρ gh = 1/2 ρ ρ 
v 2 = 
= v (say), P 0 is the atmospheric
pressure. The range R = v 2 × t, where t = time of fall can be given by
h 1 = 1/2 gt 2 ⇒ ⇒ t = 
⇒ ⇒ R = v 2 
; putting v 2 = 
,
we obtain R = 2 
putting h ′ ′ = 
m and h = 
m, we obtain R = 2 m
Prepare Smarter with CGP Edu
Get practice questions, solutions, and test series in one place.
Write a Review
Share your experience with this question and solution.
Commentary
Send your comment, doubt, correction, or feedback to admin.
Similar Questions
Explore conceptually related problems