Question

7. A conical vessel of base radius 9 cm and height
20 cm is full of water. A part of this water is now
poured into a hollow cylinder, closed at one end,
till the cylinder is completely filled with water. If
the base radius and the height of the cylinder are
6 cm and 10 cm respectively, find the volume of
water which is left in the cone. (Take a = 3.14)

please solve it correctly I request you

​

Answer 1

Answer:

Step-by-step explanation:

ANSWER:-

Given that:-

First Phase:-

• Conical vessel of radius 9 cm
• Height of the conical vessel is 20 cm

Second Phase:-

• The water is poured into a cylinder, lidless
• Base is 6 cm and height is 10 cm.

Volume of water now left in cone? What we need to do now? Suspicious!

$\setlength{\unitlength}{1.2mm}\begin{picture}(5,5)\thicklines\put(0,0){\qbezier(1,0)(12,3)(24,0)\qbezier(1,0)(-2,-1)(1,-2)\qbezier(24,0)(27,-1)(24,-2)\qbezier(1,-2)(12,-5)(24,-2)}\put(-0.5,-1){\line(1,2){13}}\put(25.5,-1){\line(-1,2){13}}\multiput(12.5,-1)(2,0){7}{\line(1,0){1}}\multiput(12.5,-1)(0,4){7}{\line(0,1){2}}\put(18,1.6){\sf{9 \ cm}}\put(9.5,10){\sf{20 \ cm}}\end{picture}$  Transferred >> $\setlength{\unitlength}{1mm}\begin{picture}(5,5)\thicklines\multiput(-0.5,-1)(26,0){2}{\line(0,1){40}}\multiput(12.5,-1)(0,3.2){13}{\line(0,1){1.6}}\multiput(12.5,-1)(0,40){2}{\multiput(0,0)(2,0){7}{\line(1,0){1}}}\multiput(0,0)(0,40){2}{\qbezier(1,0)(12,3)(24,0)\qbezier(1,0)(-2,-1)(1,-2)\qbezier(24,0)(27,-1)(24,-2)\qbezier(1,-2)(12,-5)(24,-2)}\multiput(18,2)(0,32){2}{\sf{6 \ cm}}\put(9,17.5){\sf{10 \ cm}}\end{picture}$

$\boxed{\sf{Volume \ of \ water - Volume \ of \ Cylinder = Required \ Answer}}$

Confusion gone! Let's Do!

$\boxed{\rm{Volume \ of \ Cone = \dfrac{1}{3} \pi \times r^2 \times h}}$

• Where r is 9 cm
• Where h is 20 cm

$\boxed{\bf{Volume \ of \ Cylinder = \pi \times r^2 \times h}}$

• Where r is 6 cm
• Where h is 10 cm

$\rm{Volume \ of \ Cone = \dfrac{1}{3} \times 3.14 \times (9)^2 \times 20}$

$\rm{Volume \ of \ Cone = \dfrac{1}{3} \times 3.14 \times 81 \times 20}$

$\boxed{\boxed{\rm{Volume \ of \ Cone = 1695.6 \ cm^3 }}}$

Next phase:-

$\bf{Volume \ of \ Cylinder= 3.14 \times (6)^2 \times 10}$

$\bf{Volume \ of \ Cylinder = 3.14 \times 36 \times 10}$

$\boxed{\boxed{\bf{Volume \ of \ Cylinder = 1130.4 \ cm^3}}}$

Now, let us subtract the values to get the answer.

$\boxed{\sf{Volume \ of \ water - Volume \ of \ Cylinder = Required \ Answer}}$ [Remember?]

$\implies 1695.6-1130.4$

$\boxed{\sf{\longrightarrow 565.2 \ cm^3 is \ the \ required \ answer.}}$ $| \longleftarrow$ Amount of water left.

Figure attached if not visible.

Answer 2

☣Hɪɪɪ MATᴇ☣

⇓Sᴏʟᴜᴛɪᴏɴ⇓

➣First phrase :

• Conical vessel of radius 9 Cm.
• Height of the conical vessel is 20 Cm.

➣second phrase :

• the water is poured into a cylinder lidless.
• base is 6 Cm & height is 10 Cm.

➱ Volume of water -volume of cylinder =R

➱Volume of cone =1/3 π × r²× h

• where r is 9 Cm.
• where h is 20 Cm

➱ Volume of cone =1/3 × 3.14 × (9)² × 20

➱Volume of cone =1/3 × 3.14 ×81 ×20

➱Volume of cone =1695.6 Cm³

➱Volume of cylinder =3.14 × (6)² ×10

➱volume of cylinder =3.14 × 36 ×10

➱Volume of cylinder=1130.4cm³

➱volume of water - volume of cylinder =R

➱1695.6-1130.4

➱565.2 Cm ³

☣ʜᴏᴘᴇ ɪᴛ's ʜᴇʟᴘғᴜʟ☣