?marbles of diameter 1.4 cm are dropped into a cylindrical vessel of d...
Solution :
Diameter of the Spherical ball ( d ) = 1.4cm
radius of the ball ( r ) = d/2
r = 0.7 cm
Volume of the Sphere = ( 4/3 )πr³ cm³
v = ( 4/3 ) π ( 0.7 )³ ----( 1 )
ii ) Diameter of the cylinder ( D ) = 7cm
radius of the cylinder ( R ) = D/2
R = 7/2 cm
Suppose water level rises by h= 5.6 cm
in the cylindrical beacker .
Then ,
Volume of the cylinder of height
h = 5.6 cm and radius 7/2 cm
V = πR²h
V = [ π ×( 7/2 )² × 5.6 ]cm³ ----( 2 )
Let the number of balls dropped in the
beaker = n
n = V /v
n = [π ×(7/2)²×5.6cm³]/ [4/3×π×(0.7)³cm³]
after cancellation ,we get
n = 150
?marbles of diameter 1.4 cm are dropped into a cylindrical vessel of d...
The answer will be approx 150 marbles.
Here, we see in this question,, there is a cylindrical vessel and spherical marbles...
so, we have,,
• volume of cylindrical vessel =πr²h. (π=22/7)
radius(r)=7/2=3.5cm, height(h)=5.6cm.
•volume of spherical marble =4/3πr³(r=14/2=0.7cm)
therefore, the no. of marbles =the volume of cylindrical vessel /the volume of spherical marble
=(22/7* 3.5 * 3.5 * 5.6 )÷ (4/3 * 22/7 * 14/2 * 14/2 * 14.2) =approx 150 marble.
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