Class 1oth, Surface and Volume, Exe-13.1 (Solution 2)

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Class 10th, Surface Areas and Volumes

 Exercise: 13.1 (Solution), Q5 to Q8

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Q5. A hemispherical depression is cut out from one face of a cubical wooden block such that the diameter l of the hemisphere is equal to the edge of the cube. Determine the surface area of the remaining solid.

Answer:

The diagram :-

Now, the diameter of hemisphere = Edge of the cube = l

So, the radius of hemisphere = l/2

∴ The total surface area of solid = surface area of cube + CSA of hemisphere – Area of base of hemisphere

=> TSA of remaining solid = 6 (edge)2 + 2πr2 – πr2

= 6l2 + πr2

= 6l2 + π(l/2)2

= 6l2 + πl2/4

= l2/4 (24 + π) sq. units

Q6. A medicine capsule is in the shape of a cylinder with two hemispheres stuck to each of its ends. The length of the entire capsule is 14 mm and the diameter of the capsule is 5 mm. Find its surface area.

Answer:

Two hemisphere and one cylinder are shown in the figure given below.

Here, the diameter of the capsule = 5 mm

∴ Radius = 5/2 = 2.5 mm

Now, the length of the capsule = 14 mm

So, the length of the cylinder = 14 – (2.5 + 2.5) = 9 mm

∴ The surface area of a hemisphere = 2πr2 = 2 × 22/7 × 2.5 × 2.5

= 275/7 mm2

Now, the surface area of the cylinder = 2πrh

= 2 × 22/7 × 2.5 x 9

=> 22/7 × 45 = 990/7 mm2

Thus, the required surface area of medicine capsule will be

= 2 × surface area of hemisphere + surface area of the cylinder

= (2 × 275/7) × 990/7

=> 550/7 + 990/7 = 1540/7 = 220 mm2

Q7. A tent is in the shape of a cylinder surmounted by a conical top. If the height and diameter of the cylindrical part are 2.1 m and 4 m respectively, and the slant height of the top is 2.8 m, find the area of the canvas used for making the tent. Also, find the cost of the canvas of the tent at the rate of Rs 500 per m2. (Note that the base of the tent will not be covered with canvas.)

Answer:

It is known that a tent is a combination of cylinder and a cone.

From the question we know that

Diameter = 4 m

Slant height of the cone (l) = 2.8 m

Radius of the cone (r) = Radius of cylinder = 4/2 = 2 m

Height of the cylinder (h) = 2.1 m

So, the required surface area of tent = surface area of cone + surface area of cylinder

= πrl + 2πrh

= πr (l+2h)

= 22/7 × 2 (2.8 + 2×2.1)

= 44/7 (2.8 + 4.2)

= 44/7 × 7 = 44 m2

∴ The cost of the canvas of the tent at the rate of ₹500 per m2 will be

= Surface area × cost per m2

=> 44 × 500 = ₹22000

So, Rs. 22000 will be the total cost of the canvas.

Q8. From a solid cylinder whose height is 2.4 cm and diameter 1.4 cm, a conical cavity of the

same height and same diameter is hollowed out. Find the total surface area of the

remaining solid to the nearest cm2.

Answer:

The diagram for the question is as follows:

From the question we know the following:

The diameter of the cylinder = diameter of conical cavity = 1.4 cm

So, the radius of the cylinder = radius of the conical cavity = 1.4/2 = 0.7

Also, the height of the cylinder = height of the conical cavity = 2.4 cm

Now, the TSA of remaining solid = surface area of conical cavity + TSA of the cylinder

= πrl + (2πrh + πr2)

= πr (l + 2h + r)

= 22/7 × 0.7 (2.5 + 4.8 + 0.7)

= 2.2 × 8 = 17.6 cm2

So, the total surface area of the remaining solid is 17.6 cm2

See next blog for more Solution........ Thank you.