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MODEL PAPER 2 (BASIC) — Maths STD 10 — Question

CBSE BoardEnglish MediumSTD 10MathsMODEL PAPER 2 (BASIC)5 Marks
Question
A solid consisting of a right cone standing on a hemisphere is placed upright in a right circular cylinder full of water and touches the bottom. Find the volume of water left in the cylinder, if the radius of the cylinder is $60 \ cm$ and its height is $180 \ cm,$ the radius of the hemisphere is $60 \ cm$ and height of the cone is $120 \ cm,$ assuming that the hemisphere and the cone have common base.
✓

Answer

We have radius of cylinder $=$ radius of cone $=$ radius of hemisphere $= 60 \ cm$
Height of cone $= 120 \ cm$
$\therefore$ Height of cylindrical vessel $=120+60=180 \ cm$
$\therefore V =$ Volume of water that the cylinder contains $=\pi r^2 h=\left\{\pi \times(60)^2 \times 180\right\} \ cm ^3$
Let $V_1$ be the volume of the conical part. Then,
Image
$V_1=\frac{1}{3} \pi r^2 h_1$
$\Rightarrow V_1=\frac{1}{3} \times \pi \times 60^2 \times 120 \ cm^3=\left\{\pi \times 60^2 \times 40\right\} \ cm ^3$
For hemispherical part $r =$ Radius $= 60 \ cm$
Let $V_2$ be the volume of the hemisphere.
Then, $V_2=\left\{\frac{2}{3} \pi \times 60^3\right\} \ cm ^3$
$\Rightarrow V_2=\left\{2 \pi \times 20 \times 60^2\right\} \ cm ^3=\left\{40 \pi \cdot 60^2\right\} \ cm ^3$
Let $V_3$ the the volume of the water left$-$out in the cylinder. Then,
$V_3=V-V_1-V_2$
$V_3=\left\{\pi \times 60^2 \times 180-\pi \times 60^2 \times 40-40 \pi \times 60^2\right\} \ cm ^3$
$V_3=\pi \times 60^2 \times\{180-40-40\} \ cm ^3$
$V_3=\frac{22}{7} \times 3600 \times 100 \ cm^3$
$\Rightarrow V_3=\frac{22 \times 360000}{7} \ cm^3$
$=\frac{22 \times 360000}{7 \times(100)^3} m^3$
$=\frac{22 \times 36}{700} m^3$
$=1.1314 m^3$

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