NCERT Solutions Class 10 Maths Chapter 12: Surface Areas and Volumes

#### **Q1. A solid is in the shape of a cone standing on a hemisphere with both their 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 $\pi$.** #### **A1. Solution:** Given: Radius of hemisphere ($r$) = 1 cm. Radius of cone ($r$) = 1 cm. Height of cone ($h$) = Radius ($r$) = 1 cm. Volume of solid = Volume of Cone + Volume of Hemisphere $= \frac{1}{3}\pi r^2h + \frac{2}{3}\pi r^3$ $= \frac{1}{3}\pi (1)^2(1) + \frac{2}{3}\pi (1)^3$ $= \frac{1}{3}\pi + \frac{2}{3}\pi$ $= \frac{3\pi}{3} = \pi$ 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.)** #### **A2. Solution:** Given: Diameter = 3 cm $\Rightarrow$ Radius ($r$) = 1.5 cm. Total length of model = 12 cm. Height of each cone ($h_1$) = 2 cm. Height of cylinder ($h_2$) = Total length $-$ (Height of 2 cones) $h_2 = 12 - (2 + 2) = 8$ cm. Total Volume = Volume of Cylinder + $2 \times$ Volume of Cone $= \pi r^2 h_2 + 2 \times \frac{1}{3}\pi r^2 h_1$ $= \pi r^2 (h_2 + \frac{2}{3}h_1)$ $= \frac{22}{7} \times 1.5 \times 1.5 \times (8 + \frac{2}{3} \times 2)$ $= \frac{22}{7} \times 2.25 \times (8 + \frac{4}{3})$ $= \frac{22}{7} \times 2.25 \times \frac{28}{3}$ $= 22 \times 0.75 \times 4 = 66$ cm³. #### **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.** #### **A3. Solution:** Given: Diameter = 2.8 cm $\Rightarrow$ Radius ($r$) = 1.4 cm. Total length = 5 cm. Height of cylindrical part ($h$) = $5 - (1.4 + 1.4) = 2.2$ cm. Volume of 1 Gulab Jamun = Vol of Cylinder + $2 \times$ Vol of Hemisphere $= \pi r^2 h + 2 \times \frac{2}{3}\pi r^3$ $= \pi r^2 (h + \frac{4}{3}r)$ $= \frac{22}{7} \times 1.4 \times 1.4 \times (2.2 + \frac{4}{3} \times 1.4)$ $= 22 \times 0.2 \times 1.4 \times (2.2 + 1.87)$ $= 6.16 \times 4.07 \approx 25.07$ cm³. Volume of 45 Gulab Jamuns = $45 \times 25.07 = 1128.15$ cm³. Volume of syrup = $30\%$ of total volume $= \frac{30}{100} \times 1128.15 \approx 338$ cm³. #### **Q4. A pen stand made of wood is in the shape of a cuboid with four conical depressions to hold pens. The dimensions of the cuboid are 15 cm by 10 cm by 3.5 cm. The radius of each of the depressions is 0.5 cm and the depth is 1.4 cm. Find the volume of wood in the entire stand.** #### **A4. Solution:** Given: Cuboid dimensions ($l, b, h$) = $15 \times 10 \times 3.5$. Cone: Radius ($r$) = 0.5 cm, Depth ($h_c$) = 1.4 cm. Volume of wood = Volume of Cuboid $-$ $4 \times$ Volume of Cone $= (l \times b \times h) - 4 \times \frac{1}{3}\pi r^2 h_c$ $= (15 \times 10 \times 3.5) - \frac{4}{3} \times \frac{22}{7} \times 0.5 \times 0.5 \times 1.4$ $= 525 - \frac{4}{3} \times 22 \times 0.25 \times 0.2$ $= 525 - \frac{4.4}{3}$ $= 525 - 1.47 = 523.53$ cm³. #### **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.** #### **A5. Solution:** Given: Cone: Height ($H$) = 8 cm, Radius ($R$) = 5 cm. Sphere (Lead shot): Radius ($r$) = 0.5 cm. Volume of water flowed out = $\frac{1}{4} \times$ Volume of Cone. Let $n$ be the number of lead shots. $n \times$ Volume of Sphere = $\frac{1}{4} \times$ Volume of Cone $n \times \frac{4}{3}\pi r^3 = \frac{1}{4} \times \frac{1}{3}\pi R^2 H$ $n \times 4 r^3 = \frac{1}{4} R^2 H$ $n \times 4 (0.5)^3 = \frac{1}{4} (5)^2 (8)$ $n \times 4 \times 0.125 = \frac{1}{4} \times 25 \times 8$ $n \times 0.5 = 50$ $n = \frac{50}{0.5} = 100$. Number 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 $\pi = 3.14$)** #### **A6. Solution:** Given: Large Cylinder: Height ($H$) = 220 cm, Radius ($R$) = 12 cm. Small Cylinder: Height ($h$) = 60 cm, Radius ($r$) = 8 cm. Total Volume = Volume of Large Cylinder + Volume of Small Cylinder $= \pi R^2 H + \pi r^2 h$ $= 3.14 \times (12)^2 \times 220 + 3.14 \times (8)^2 \times 60$ $= 3.14 \times 144 \times 220 + 3.14 \times 64 \times 60$ $= 99475.2 + 12057.6 = 111532.8$ cm³. Mass = Volume $\times$ Density $= 111532.8 \times 8$ g $= 892262.4$ g $= 892.26$ 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.** #### **A7. Solution:** Given: Solid: Cone ($h=120, r=60$) + Hemisphere ($r=60$). Cylinder: Height ($H=180$), Radius ($R=60$). Volume of water left = Volume of Cylinder $-$ Volume of Solid $= \text{Vol of Cylinder} - (\text{Vol of Cone} + \text{Vol of Hemisphere})$ $= \pi R^2 H - (\frac{1}{3}\pi r^2 h + \frac{2}{3}\pi r^3)$ Since radii are same ($R=r=60$): $= \pi r^2 [H - (\frac{1}{3}h + \frac{2}{3}r)]$ $= \frac{22}{7} \times 60 \times 60 \times [180 - (\frac{1}{3} \times 120 + \frac{2}{3} \times 60)]$ $= \frac{22}{7} \times 3600 \times [180 - (40 + 40)]$ $= \frac{22}{7} \times 3600 \times 100$ $= \frac{7920000}{7} \approx 1131428.57$ cm³. In m³: $\approx 1.131$ m³. #### **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 $\pi = 3.14$.** #### **A8. Solution:** Given: Cylindrical neck: Height ($h$) = 8 cm, Radius ($r_1$) = 1 cm. Spherical part: Radius ($r_2$) = $\frac{8.5}{2} = 4.25$ cm. Total Volume = Volume of Sphere + Volume of Cylinder $= \frac{4}{3}\pi r_2^3 + \pi r_1^2 h$ $= 3.14 \times [\frac{4}{3} \times (4.25)^3 + (1)^2 \times 8]$ $= 3.14 \times [1.333 \times 76.765 + 8]$ $= 3.14 \times [102.35 + 8]$ $= 3.14 \times 110.35$ $= 346.51$ cm³. The child's value was 345 cm³. Since $346.51 \neq 345$, she is **incorrect**.