# PSEB Solutions for Class 10 Maths Chapter 13 Surface Areas and Volumes Ex 13.5

PSEB Solutions for Class 10 Maths Chapter 13 Surface Areas and Volumes Ex 13.5

## PSEB 10th Class Maths Solutions Chapter 13 Surface Areas and Volumes Ex 13.5

Question 1.

A copper wire 3 mm in diameter is wound about a cylinder whose length is 12 cm, and diameter 10 cm, so as to cover the curved surface of the cylinder. Find the length and mass of the wire, assuming the density of copper to 8.88 g per cm^{3}.

Solution:

Question 2.

A right triangle, whose sides are 3 cm and 4 cm (other than hypotenuse) is made to revolve about its hypotenuse. Find the volume and surface area the double cone so formed. (Choose value of t as found

appropriate.)

Solution:

Question 3.

A cistern, Internally measuring 150 cm × 120 cm × 110 cm, has 129600 cm3 of water in it. Porous bricks are placed in the water until the cistern is full to the brim. Each brick absorbs one-seventeenth of its own volume of water. How many bricks can be put in without the water overflowing, each brick being 22.5 cm × 7.5 cm × 6.5 cm?

Solution:

Volume of one brick = 22.5 × 7.5 × 6.5 cm^{3} = 1096.87 cm^{3}

Volume of cistern = 150 × 120 × 110 cm^{3} = 1980000 cm^{3}

Let number of bricks used = n

Total volume of n bricks = n (volume of one brick) = n [1096.871] cm^{3}

Volume of water = 129600 cm^{3}

Volume of water available for bricks = 1980000 – 129600 = 1850400 cm^{3}

Question 4.

In one fortnight of a given month, there was a rainfall of 10 cm in a river valley. If the area of the valley is 97280 km^{2}, show that the total rainfall was approximately equivalent to the addition to the normal water of three rivers each 1072 km long, 75 m wide and 3 m deep.

Solution:

Question 5.

An oil funnel made of tin sheet consists of a cylindrical portion 10 cm long attached to a frust.un of a cone. if the total height is 22 cm, diameter of the cylindrical portion is 8 cm and the diameter of the top of the funnel is 18 cm, find the area of the tin sheet required to make the funnel.

Solution:

Diameter of top of funnel = 18 cm

∴ Radius of top of funnel (R) = 182 cm = 9 cm

Diameter of bottom of funnel = 8 cm

Radius of bottom of funnel (r) = 4 cm

Height of cylindrical portion (h) = 10 cm

Height of frustum (H) = (22 – 10) = 12 cm

Slant height of frustum (l)

= \sqrt{\mathrm{H}^{2}+(\mathrm{R}-r)^{2}}

Question 6.

Derive the formula for the curved surface area and total surface area of a frustum of a cone, given to you in Section 13.5, using the symbols as explamed.

Solution:

A frustum of a right circular cone has two unequal flat circular bases and a curved surface. Let ACDB be the frustum of the cone obtained by removing the portion VCÐ. The line-segment OP joining the centres of two bases is called the height of the frustum. Each of the line-segments AC and BD of frustum ACDB is called its slant height.

Let R and r be the radii of the circular end faces (R > r) of the frustum ACDB of the cone (V, AB). We complete the conical part VCD. Let h and l be the vertical height and slant height respectively. Then OP = h and AC = BD = l.

The frustum of the right circular cone can be viewed as the difference of the two right circular VAB and VCD.

Let the height of the cone VAB be h_{1} and its slant height l_{1} i.e. VP = h_{1} and VA = VB = l_{1}.

Now from right ∆DEB, we have

DB^{2} = DE^{2} + BE^{2}

Curved surface area of the frustum of cone = πRl_{1} – πr(l_{1} – l)

[Curved surface area of a cone = π × r × 1]

Question 7.

Derive the formula for the volume of the frustum of a cone, given to you in section 13.5, using the symbols as explained.

Solution:

A frustum of a right circular cone has two unequal flat circular bases and a curved surface. Let ACDB be the frustum of the cone obtained by removing the portion VCD. The line-segment OP joining the centres of two bases is called the height of the frustum. Each of the line-segments AC and BD of frustum ACDB is called its slant height.

Ler R and r be the radii of the circular end faces (R > r) of the fnistum ACDB of the cone (V, AB). We complete the conical part VCD. Let h and l be the vertical height and slant height respectively. Then OP = h and AC = BD = l. The frustum of the right circular cone can be viewed as the difference of the two right circular VAB and VCD.

Let theheight of the cone VAB be h_{1} and its slant height l_{1} i.e.VP = h_{1} and VA = VB = l_{1}.

∴ The height of the cone VCD = VP – OP = h_{1} – h

Since right ∆s VOD and VPB are similar

Volume of the frustrum ACDB of the cone (V,AB) = Volume of the cone (V, AB) – Volume of the cone (V, CD)

Again if A_{1} and A_{2} (A_{1} > A_{2}) are the surface areas of the two circular bases, then A_{1} = πR^{2} andA_{2} = πr^{2}

Now volume of the frustum of cone,

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