Students can Download Class 10 Maths Chapter 15 Surface Areas and Volumes Ex 15.2 Questions and Answers, Notes Pdf, KSEEB Solutions for Class 10 Maths helps you to revise the complete Karnataka State Board Syllabus and to clear all their doubts, score well in final exams.

## Karnataka State Syllabus Class 10 Maths Chapter 15 Surface Areas and Volumes Ex 15.2

Question 1.

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 π.

Answer:

Height of cone is equal to its radius.

h = r = 1 cm.

radius of hemisphere = r = 1 cm

Volume of solid = Volume of cone + Volume of hemisphere

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

Answer:

diameter = 3 cm and radius = \(\frac{3}{2}\)cm

h_{2} = height of cone = 2 cm

h_{1} = height of cylinder = 12 – 2 – 2 = 8 cm.

Volume of air contained in the model = Volume of cylinder portion + volume of two conical ends.

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 (see Fig.15.15).

Answer:

Diameter of cylindrical portion and hemispherical portion of a gulab jamun = 2.8 m

∴ r = \(\frac{2.8}{2}\) = 1.4cm

Length of cylindrical portion = 5 – 1.4 – 1.4 = 2.2 cm

Volume of the gulab jamun = Volume of cylindrical portion + volume of hemispherical ends.

Quantity of syrup in 45 gulab jamuns = 30% of their volume

= \(\frac{30}{100} \times 45 \times \frac{22}{7} \times 1.96 \times \frac{12.2}{3}\)

= 338.18 cm^{3}

= 338 cm^{3} (approx)

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 (see Fig. 15.16).

Answer:

Cuboid:

length = l = 15 cm, breadth = b = 10 cm and height = h = 3.5 cm, radius of conical depression = 0.5 cm and depth = h = 1.4 cm

Volume of wood in pen stand = Volume of cuboid – 4 × Volume of a conical depression.

= lbh – 4 × \(\frac{1}{3}\)πr^{2}

= (15 × 10 × 3.5) – \(\frac{4}{3} \times \frac{22}{7}\) × (0.5)^{2} × 1.4

= 525 – 1.47

= 523.53 cm^{3}

Question 5.

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.

Answer:

Height of conical vessel = h = 8 cm and radius = r_{1} = 5 cm.

Volume of cone = Volume of water in cone.

Volume of water flows out = Volume of lead shots = 1/4 × volume of water in cone.

radius of the lead shots = r_{2} = 0.5 cm = 1/2 cm

Number of lead shots dropped in the vessel

∴ Number of lead shots are 100

Question 6.

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 π = 3.14)

Answer:

long cylinder:

d_{1} = 24 cm, r_{1} = \(\frac{d_{1}}{2}=\frac{24}{2}\) = 12cm

h_{1} = 220 cm.

small cylinder:

radius = r_{2} = 8 cm and height = h_{2} = 60 cm

Volume of solid iron pipe = Volume of long cylinder + Volume of small cylinder

= π [(12)^{2} × 220 + (8)^{2} 60]

= π [144 × 220 + 64 × 60]

= π [31680 + 3840]

= π [35520]

= 3.14 × 35520

= 111532.8 cm^{3}

mass of the pole

= 111532.8 × 8 [ 1 cm^{3} = 8 g]

= 892262.4 grams

= 892.2624 kg [ 1 kg = 1000 grams]

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.

Answer:

radius of hemisphere = r = 60 cm

radius of cone = 60 cm

radius of cylinder = 60 cm

height of cone = h_{1} = 120 cm

height of cylinder = h_{2} = 180 cm

Volume of water left = volume of cylinder – volume of solid

Hence volume of water left in the cylinder is 1.131 m^{3}.

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 cm^{3}. Check whether she is correct, taking the above as the inside measurements, and π = 3.14.

Answer:

diameter of cylindrical neck = 2 cm

radius = r_{1} = 2/2 = 1 cm and height = h = 8 cm

diameter of spherical part = 8.5 cm

∴ radius = r_{2} = \(\frac{d}{2}=\frac{8.5}{2}\) = 4.25 cm.

Volume of spherical vessel = volume of cylinder + volume of sphere.

∴ answer find by the child is incorrect.

∴ correct answer is 346.51 cm^{3}

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