SURFACE AREAS AND VOLUMES-Exercise 12.2

Q1. A solid is in the shape of a cone standing on a hemisphere with both its radii being equal to 1 cm and the height of the cone is equal to its radius. Find the volume of the solid in terms of \(\pi\).

Solution:

Radius and height of cone = 1 cm
Radius of hemisphere = 1 cm
Volume \(V_{t}\) = Volume of cone \(V_{c}\) + Volume of hemisphere \(V_{h}\)

$$\require{cancel}\begin{aligned}V_{t}&=V_{c}+V_{h}\\\\ &=\dfrac{1}{3}\pi r^{2}h+\dfrac{2}{3}\pi r^{3}\\\\ &=\dfrac{1}{3}\pi r^{2}\left( h+2r\right) \\\\ (r&=1)\\\\ &\Rightarrow \dfrac{1}{3}\pi \cdot 1^{2}\cdot \left( 1+2\cdot 1\right) \\\\ &=\dfrac{\pi }{3}\left( 1+2\right) \\\\ &=\dfrac{\pi }{\cancel{3}}\times \cancel{3}\\\\ &=\pi\ \mathrm{cm^{3}}\end{aligned}$$

Volume of the solid = pi cm³

Q2. Rachel, an engineering student, was asked to make a model shaped like a cylinder with two cones attached at its two ends by using a thin aluminium sheet. The diameter of the model is 3 cm and its length is 12 cm. If each cone has a height of 2 cm, find the volume of air contained in the model that Rachel made. (Assume the outer and inner dimensions of the model to be nearly the same.)

Solution:

Length of Cylinder (h1) = 12 cm
height of cone (h2) = 2 cm
Diameter of model = 3 cm
Thus Radius of the model = 3/2=1.5 cm
Air Containd in the model \(V_{t}\) = Volume of Cone \(V_{cn}\) + Volume of Cylinder \(V_{cl}\)

$$\require{cancel}\begin{aligned}V_{t}&=V_{cn}+V_{cl}+V_{cn}\\\\ &=2V_{cn}+V_{cl}\\\\ &=2\cdot \dfrac{1}{3}\pi r ^{2}h_{2}+\pi r^{2}h_{1}\\\\ &=\dfrac{2}{3}\pi r^{2}\left[ h_{2}+h_{1}\right] \\\\ &=\dfrac{2}{3}\pi 1.5\times 1.5\left[ 12+2\right] \\\\ &\scriptsize=\dfrac{2}{\cancel{3}}\times \dfrac{22}{\cancel{7}}\times \cancelto{0.5}{1.5}\times 1.5\times \cancelto{2}{14}\\\\ &=2\times 22\times 0.5\times 1.5\times 2\\\\ &=44\times 1.5\\\\ &=66\ \mathrm{cm^{3}}\end{aligned}$$

Volume contained in model is \(= 66\ \mathrm{cm^{3}}\)

Q3. A gulab jamun, contains sugar syrup up to about 30% of its volume. Find approximately how much syrup would be found in 45 gulab jamuns, each shaped like a cylinder with two hemispherical ends with length 5 cm and diameter 2.8 cm (see Fig. 12.15).

Solution:

No of Gulab Jamun = 45
Amount of sugar Syrum in Gulab Jamun = 30% of its Volume Gulab Jaman shaped as in figure
Diameter of the cylinder and hemisphere = 2.8 cm
Thus Radius = 2.8/2 = 1.4 cm
length of cylinder = total length-2 (radius)

$$\begin{aligned}h&=5-2\left( 1.4\right) \\ &=5-2.8\\ &=2.2cm\end{aligned}$$

Volume of 1 Gulab Jamun = 2 \(\times\) Volume of hemisphere + Volume of cylinder

$$\require{cancel}\begin{aligned}V&=\dfrac{4}{3}\pi r^{3}+\pi r^{2}h\\\\ &=\pi r^{2}\left[ \dfrac{4r}{3}+h\right] \\\\ &=\scriptsize\dfrac{22}{\cancel{7}}\times \cancelto{0.2}{1.4}\times 1.4\left( \dfrac{4}{3}\times 1.4+2.2\right) \\\\ &=22\times 0.2\times 1.4\left( \dfrac{5.6}{3}+2.2\right) \\\\ &=4.4\times 1.4\times 4.06\\\\ &=25.009\\\\ &\approx 25\end{aligned}$$

Volume of 45 Gulab jamun =

$$25\times 45=1125cm^{3}$$

Amount of sugar syrup =

$$\begin{aligned}&1125\times \dfrac{30}{100}\\\\ &=337.50\\\\ &\approx 338\ \mathrm{cm^{3}}\end{aligned}$$

45 Gulab Jamun have \(338\ \mathrm{cm^{3}}\) of Sugar Syrup

Q5. A vessel is in the form of an inverted cone. Its height is 8 cm and the radius of its top, which is open, is 5 cm. It is filled with water up to the brim. When lead shots, each of which is a sphere of radius 0.5 cm are dropped into the vessel, one-fourth of the water flows out. Find the number of lead shots dropped in the vessel.

Solution:

Height of the conical vessel \((h)\) = 8 cm
Radius \((r)\) = 5 cm
Radius of spherical lead shots = 0.5 cm
Volume of conical vessel

$$\begin{aligned}V_{cm}&=\dfrac{1}{3}\pi r^{2}h\\ &=\dfrac{1}{3}\cdot \pi \times 5\times 5\times 8\\ &=\dfrac{200}{3}\pi \end{aligned}$$

Volume of one spherical leadshot \(V_{s}\)

$$\begin{aligned}V_{s}&=\dfrac{4}{3}\pi r^{3}\\ &=\dfrac{4}{3}\times \dfrac{22}{7}\times \left( 0.5\right) ^{3}\\ &=0.125\times \dfrac{4\pi }{3}\\ &=\dfrac{0.5\pi }{3}\\ &=\dfrac{\pi }{6}\end{aligned}$$

Let number of lead shots dropped into the vessel = \(n\)
Total volume of lead shot = \(n\times V_s\) \(n\times V_s = \frac{1}{4}\) of the volume of conical Vessel
Since water flows out will be equal to the total volume of the lead shots, therefore,

$$\require{cancel}\begin{aligned}n\cdot \dfrac{\pi }{6}&=\dfrac{1}{4}\left( \dfrac{200}{3}\pi \right) \\\\ n&=\dfrac{1}{\cancel{4}}\times \dfrac{\cancelto{50}{200}\times \cancelto{2}{6}}{\cancel{3}}\\\\ &=100\end{aligned}$$

No of lead shots = 100

Q6. A solid iron pole consists of a cylinder of height 220 cm and base diameter 24 cm, which is surmounted by another cylinder of height 60 cm and radius 8 cm. Find the mass of the pole, given that 1 cm3 of iron has approximately 8g mass. (Use \(\pi\)= 3.14)

Solution:

Height of the cylinder \((h_1)\) = 220 cm
Base diameter = 24cm
Base Radius \((r_1)\) = 24/2 = 12 cm
Surmounted cylinder Height \((h_2)\) = 60cm
and radius \((r_2)\) = 8cm
Total volume of cylinders

$$\begin{aligned}V&=\pi r_{1}^{2}h_{1}+\pi r_{2}^{2}h_{2}\\\\ &\scriptsize=\pi \left[ \left( 12\times 12\times 220\right) +\left( 8\times 8\times 60\right) \right] \\\\ &=\pi \left[ \left( 144\times 220\right) +\left( 64\times 60\right) \right] \\\\ &=\pi \left( 31680+3840\right) \\\\ &=3.14\times 35520\\\\ &=111532.80\end{aligned}$$

Weight of 1cm³ = 8 gm
therefore weight of 111532.8cm³

$$\begin{aligned}&=111532\cdot 8\times 8\ \mathrm{gm}\\\\ &=\dfrac{111532.8\times 8}{1000}\ \mathrm{kg}\\\\ &=892.26\ \mathrm{kg}\end{aligned}$$

Mass of the pole is \(892.26\ \mathrm{kg}\)

Q7. A solid consisting of a right circular cone of height 120 cm and radius 60 cm standing on a hemisphere of radius 60 cm is placed upright in a right circular cylinder full of water such that it 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.

Solution:

Height of right circular cone = 120 cm
Radius = 60 cm
Radius of hemisphere = 60 cm
when solid is placed in the cylinder, it will displace water, which will be equal to the solid's volume.
Hence, removing water will be equal to Volume of Cylinder-(Volume of cone + Volume of hemisphere) Volume of Cylinder (r = 60cm, h = 180cm)

$$\require{cancel}\begin{aligned}V&=\pi r^{2}h\\ &=\pi \left( 60\times 60\times 180\right) \\ &=\pi \left( 3600\times 180\right) \\ &=648000\pi \end{aligned}$$

Volume of cone \(V_{cn}\)

$$\begin{aligned}V_{cn}&=\dfrac{1}{3}\pi r^{2}h\\ &=\dfrac{1}{3}\pi\times 60^{2}\times 120\\ &=\dfrac{1}{\cancel{3}}\pi\times \cancelto{20}{60} \times 60\times 120\\ &=\pi\times 1200\times 120\\ &=144000\pi \end{aligned}$$

Volume of hemisphere \(V_{h}\)

$$\require{cancel}\begin{aligned}V_{h}&=\dfrac{2}{3}\pi r^{3}\\ &=\dfrac{2}{\cancel{3}}\times \pi \times \cancelto{20}{60}\times 60\times 60\\ &=2\pi \times 20\times 60\times 60\\ &=\pi \times 40\times 60\times 60\\ &=144000\pi \end{aligned}$$

Sum of volumes of Cone and hemisphere

$$\begin{aligned}V&=V_{cn}+V_{h}\\ &=144000\pi +144000\pi \\ &=288000\pi \end{aligned}$$ Volume of water left $$\begin{aligned}&=648000\pi -288000\pi \\ &=\pi \left( 360000\right) \\ &=1130973.355cm^{3}\\ &\approx 1.131 \ \mathrm{m^{3}}\end{aligned}$$

Volume of water left in the cylinder is \(\approx 1.131 \ \mathrm{m^{3}}\)

Q8. A spherical glass vessel has a cylindrical neck 8 cm long, 2 cm in diameter; the diameter of the spherical part is 8.5 cm. By measuring the amount of water it holds, a child finds its volume to be 345 cm3. Check whether she is correct, taking the above as the inside measurements, and \(\pi\) = 3.14.

Solution:

Length of cylindrical neck = 8 cm
and diameter = 2 cm
Hence Radius = 2/2 = 1 cm
Diameter of spherical part = 8.5 cm
Hence Radius = 8.5/2 = 4.25 cm
Spherical vessel can hold water equal to the volume of the spherical part + the volume of cylindrical neck

$$\begin{aligned}V&=\dfrac{4}{3}\pi r^{3}+\pi r^{2}h\\\\ &=3.14\left[ \dfrac{4}{3}\left( 4.25\right) ^{3}+1^{2}\times 8\right] \\\\ &=3.14\left[ \dfrac{4\times \left( 4.25\right) ^{3}}{3}+8\right] \\\\ &=346.51\ \mathrm{cm^{2}}\end{aligned}$$

She is not correct as Volume is \(346.51\ \mathrm{cm^{2}}\)