Sushant has a vessel, of the form of an inverted cone, open at the top, of height 11 cm and
radius of top as 2.5 cm and is full of water. Metallic spherical balls each of diameter 0.5 cm are
put in the vessel due to which 25th of the water in the vessel flows out. Find how many balls
were put in the vessel. Sushant made the arrangement so that the water that flows out irrigates
the flower beds. What value has been shown by Sushant?
Answers
Answered by
10
dimensions of vessel
radius r = 2.5 Cm
Height H = 11 Cm
Thus
L = root [ r^2 + H^2]
= root [ 121 + 6.25 ]
= root 127.25
= 11.28 Cm
Thus
volume of cone = pi r L
= 22/7 * 2.5 * 11.28
= 88.54 Cm^2
According to question 2/5 th of water flows out
Thus water flowed out = 2/5 * 88.54
= 35.41
Thus
4/3 pi (0.5)^3 * n = 35.41
n= 35.41/0.5
n = 70
radius r = 2.5 Cm
Height H = 11 Cm
Thus
L = root [ r^2 + H^2]
= root [ 121 + 6.25 ]
= root 127.25
= 11.28 Cm
Thus
volume of cone = pi r L
= 22/7 * 2.5 * 11.28
= 88.54 Cm^2
According to question 2/5 th of water flows out
Thus water flowed out = 2/5 * 88.54
= 35.41
Thus
4/3 pi (0.5)^3 * n = 35.41
n= 35.41/0.5
n = 70
Anandt2a0m2i0l:
Vol. Of cone is 1/3pir^2h jack$#%*
ha ha
The answer given is 440
answer is 440
Answered by
9
Answer:
H=11
R=5/2
r=1/4
2/5*1/3*πR^2H=4/3πr^3*n
2/5*1/3*5/2*5/2*11=4*1/4*1/4*1/4*n
55*8=n
n=440
Step-by-step explanation:
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