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NEET Class neet Answered

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Asked by jhajuhi19 | 18 Jul, 2021, 12:06: AM
answered-by-expert Expert Answer
Speed u of water flow at the exit of small hole is given as
 
begin mathsize 14px style u space equals space square root of 2 space g space h end root end style 
where g is acceleration due to gravity and h is height of water column above small hole.
 
Water leaked from small hole in a time duration T is given as
 
begin mathsize 14px style integral subscript 0 superscript T left parenthesis space u space cross times space a space right parenthesis space d t space equals space a integral subscript 0 superscript T square root of 2 space g space h end root space d t space equals space A left parenthesis increment h right parenthesis space equals space A left parenthesis 81 minus 49 right parenthesis space equals space 32 A end style  .........................(1)
where a is area of small hole and A is area of crosssection of container.
 
It is given that A = ( 10 a ) , hence by substituting g = 9.8 m/s2 , above eqn.(1) is simplified as
 
begin mathsize 14px style integral subscript 0 superscript T square root of h space d t space equals space 72.281 end style ................................(2)
Let us consider height h of water column decrases linearly with respect to time.
 
Hence , we have ,  h = a - b t  .....................(3)
 
if we use the conditions , at t = 0 , h = 81  and at t = T , h = 49 , we rewrite the eqn.(3) as
 
h = 81 -  b t  .......................... (4)
 
where b = (32/T) 
 
From eqn.(4) , we get ,  dh = (-b) dt   ......................(5)
 
Using eqn.(5) , let us rewrite eqn.(2) after changing the limits of integration as
 
begin mathsize 14px style space integral subscript 49 superscript 81 square root of h space d h space equals space 72.281 space b space equals space fraction numerator 72.281 cross times 32 over denominator T end fraction space equals space 2313 over T end style
begin mathsize 14px style 2 over 3 cross times open square brackets open parentheses 81 close parentheses to the power of 3 divided by 2 end exponent space minus space open parentheses 49 close parentheses to the power of 3 divided by 2 end exponent close square brackets space equals space 257.33 space equals space 2313 over T end style
Hence T ≈ 9 s
Answered by Thiyagarajan K | 18 Jul, 2021, 12:25: PM
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