PSEB Solutions for Class 10 Maths Chapter 13 Surface Areas and Volumes Ex 13.2 Guide

PSEB Solutions for Class 10 Maths Chapter 13 Surface Areas and Volumes Ex 13.2 Guide

Question1. 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 it.

Solution:
Radius of cone = Radius of hemisphere = 1 cm

Question 2. 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 the Rachel made. (Assume the outer and inner dimensions of the model to be nearly the same.)

Solution:

Hence, Volume of air in cylinder =66 cm³.

Question 3. 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.

Solution:
Gulab Jamun is in the shape of cylinder
Diameter of cylinder = Diameter of hemisphere = 2.8 cm

Question 4. 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.

Question 5. A vessel is in the form of an inverted cone. Its height ¡s 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:

Radius of cone (R) = 5 cm
Height of cone (H) = 8 cm
Radius of each spherical lead shot (r) = 0.5 cm
Let number of shot put into the cone = N
According to Question,
N [Volume of one lead shot] = 1/4 Volume of water in cone

= 10 × 10 = 100
Hence, Number of lead shots = 100.

Question 6. A solid iron pole consists of a cylinder of height 220 cfi 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 8 g mass. (Use n = 3.14)

Solution:

Diameter of lower cylinder = 24 cm
Radius of lower cylinder (R) = 12 cm
Height of lower cylinder (H) = 220 cm
Radius of upper cylinder (r) = 8 cm
Height of upper cylinder (h) = 60 cm
Volume of pole = Volume of Lower cylinder + volume of upper cylinder
= πR²H + πr²h
= 3.14 × 12 × 12 × 220 + 3.14 × 8 × 8 × 60
= 99475.2 + 12057.6

Volume of pole = 111532.8 cm³
Mass of 1 cm³ = 8 gm
Mass of 111532.8 cm³ = 8 × 111532.8 = 892262.4 gm
= 892262.4/1000 kg = 892.2624 kg
Hence, Mass of Pole = 892.2624 kg.

Question 7. 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:
Radius of cone = Radius of hemisphere = Radius of cylinder

Question 8. 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 π = 3.14.

Solution:

Diameter of neck (cylindrical Portion) = 2 cm

Conclusion

In conclusion, the PSEB Solutions for Class 10 Maths Chapter 13, Surface Areas and Volumes, Exercise 13.2, provide students with a comprehensive guide to understanding and solving problems related to surface areas and volumes of different geometric shapes. This exercise focuses on important concepts such as the surface area and volume of cones, spheres, and hemispheres. By solving the questions in this exercise, students can develop a strong foundation in these topics and enhance their problem-solving skills.

The solutions provided in this guide are detailed and well-explained, ensuring that students can easily understand the steps involved in finding the surface area and volume of various shapes. Additionally, the guide offers alternative methods and shortcuts, making it easier for students to approach complex problems. The step-by-step explanations and solved examples enable students to grasp the underlying concepts and apply them to different scenarios.

Overall, the PSEB Solutions for Class 10 Maths Chapter 13, Exercise 13.2, serve as a valuable resource for students, helping them to build confidence in solving surface area and volume problems. By utilizing these solutions, students can strengthen their mathematical skills and excel in their examinations.