Our ncert solutions for Exercise 13.2 Class 10 maths 13. Surface Areas and Volumes - ncert solutions - Toppers Study is the best material for English Medium students cbse board and other state boards students.

# Exercise 13.2 Class 10 maths 13. Surface Areas and Volumes - ncert solutions - Toppers Study

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## Exercise 13.2 Class 10 maths 13. Surface Areas and Volumes - ncert solutions - Toppers Study

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### You can Find Mathematics solution Class 10 Chapter 13. Surface Areas and Volumes

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### Exercise 13.2 class 10 Mathematics Chapter 13. Surface Areas and Volumes

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- Exercise 13.2 Class 10 Maths 13. Surface Areas And Volumes - Ncert Solutions - Toppers Study
- Class 10 Ncert Solutions
- Solution Chapter 13. Surface Areas And Volumes Class 10
- Solutions Class 10
- Chapter 13. Surface Areas And Volumes Exercise 13.2 Class 10

## 13. Surface Areas and Volumes

### | Exercise 13.2 |

## Exercise 13.2 Class 10 maths 13. Surface Areas and Volumes - ncert solutions - Toppers Study

**Exercise 13.2**

**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 π.**

**Solution**:

The radius of cone r = *1* cm

The height of cone = *1* cm

The radius of hemisphere r = *1* cm

The volume of solid is **π cm ^{3}**

**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: **

The height of the cone h = 2 cm

The radius of cylinder r = 1.5 cm

The height of cylinder H = 12 – 2 – 2 = 8 cm

Volume of air contained in the model = 2(volume of cone) + volume of cylinder

**Thus, the volume of air contained in the model is ****66 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. 13.15).**

**Solution: **

Diameter of hemispherical end = 2.8 cm

Then the radius of the hemispherical end r = 1.4 cm

Length of whole Gulab Jamun l = 5 cm

So the length of the cylindrical part h = 5 – (1.4 + 1.4)

= 5 – 2.8 cm

= 2.2 cm

Volume of all 45 gulab jamuns = 45(volume of hemisphere + volume of cylinder + volume of hemisphere)

Sugar syrup = 30% of 1127.280 cm3

= 338.1840 cm3

Hence, the quantity of syrup in 45 Gulab Jamuns is 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 (see Fig. 13.16).**

**Solution: **

The length of the arc l = 15 cm

The breadth of cuboid b = 10 cm

The height of the cuboid h = 3.5 cm

The radius of conical part (r) = 0.5 cm

Height (h) = 1.4 cm

Volume of whole wood of Pen holder = Volume of cuboid – 4 (Volume of conical pit)

= 525 - 1.47 cm^{3}

= 523.53 cm^{3}

The volume of the whole lotus wood is 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.**

**Solution: **

**Height of cone (h) = 8 cm**

**Radius of cone (R) = 5 cm**

**Radius of led balls (r) = 0.5 cm**

**Let the number of balls put in the vessel = n**

**Hence, the number of led balls is 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 π = 3.14)**

**Solution: **

Height (H) of the obtuse cylinder = 220 cm

Diameter (d) = 24 cm

Thus, radius (R) = 12 cm

Height of slender cylinder (h) = 60 cm

radius (r) = 8 cm

Now the volume of the iron pillar = pR^{2}H + pr^{2}h

= p(R^{2}H + r^{2}h)

= 3.14 (12×12×220 + 8×8×60)

= 3.14 (31680 + 3840)

= 3.14 (35520)

= 111532.8 cm^{3}

The mass of Iron = 111532.8 cm^{3} × 8

= 892262.4 g

Hence, the mass of the iron pillar is 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.**

**Solution: **

Height of cone of solid (h) = 120 cm

Radius of cone of solid (r) = 60 cm

Radius of hemisphere of solid (r) = 60 cm

Height of large cylinder (H) = 180 cm

Radius of the large cylinder (r) = 60 cm

Volume of remaining water = volume of large cylinder – volume of solid

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 π = 3.14.

Solution:

Diameter of circular part = 8.5 cm

Radius of circular part (R) = 8.5/2 cm

Height of cylindrical neck (h) = 8 cm

Diameter of neck (d) = 2 cm

Therefore, radius (r) = 1 cm

Volume of water that can be filled = Volume of sphere + Volume of cylinder

Hence the measurement taken by the child is not correct.

##### Other Pages of this Chapter: 13. Surface Areas and Volumes

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