Solve the followings Questions.

Unless stated otherwise, use π =22/7

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:

Total Volume = Volume of cone + Volume of Hemisphere

We know that

Volume of cone = 1/3 πr²h

Volume of hemisphere = 2/3πr³

= 1/3πr²h + 2/3πr³

= 1/3 πr²(h+2r)

= 1/3 × π × 1 × 1 × (1+2)

= π cm³

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:

Height of cylinder = 12 – 4 = 8 cm, radius = 1.5 cm, height of cone = 2 cm

Volume of cylinder = πr²h

= π × 1.5² × 8

= 18π cm³

Volume of a cone = 1/3 π × 1.5² × 2

= 1.5π cm³

Total volume = Volume of cylinder + 2 Volume of a cone

= 1.5π + 1.5π + 18π

= 21π = 66 cm³

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 figure).

Answer:

It is known that the gulab jamuns are similar to a cylinder with two hemispherical ends.

So, the total height of a gulab jamun = 5 cm.

Diameter = 2.8 cm

So, radius = 1.4 cm

∴ The height of the cylindrical part = 5 cm–(1.4+1.4) cm

=2.2 cm

Now, total volume of One Gulab Jamun = Volume of Cylinder + Volume of two hemispheres

= πr²h+(4/3)πr³

= 4.312π+(10.976/3) π

= 25.05 cm³

We know that the volume of sugar syrup = 30% of total volume

So, volume of sugar syrup in 45 gulab jamuns = 45×30%(25.05 cm³)

= 45×7.515 = 338.184 cm³

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 id 1.4 cm. Find the volume of wood in the entire stand (see figure).

Answer.

Volume of cuboid = length x width x height

We know the cuboid’s dimensions as 15 cmx10 cmx3.5 cm

So, the volume of the cuboid = 15x10x3.5 = 525 cm³

Here, depressions are like cones and we know,

Volume of cone = (⅓)πr²h

Given, radius (r) = 0.5 cm and depth (h) = 1.4 cm

∴ Volume of 4 cones = 4x(⅓)πr²h

= 1.46 cm²

Now, volume of wood = Volume of cuboid – 4 x volume of cone

= 525-1.46 = 523.54 cm²

5. A vessel is in the form of inverted cone. Its height is 8 cm and the radius of the 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:

It is known that,

Volume of cone = volume of water in the cone

= ⅓πr2h = (200/3)π cm3

Now,

Total volume of water overflown= (¼)×(200/3) π =(50/3)π

Volume of lead shot

= (4/3)πr³

= (1/6) π

Now,

The number of lead shots = Total Volume of Water over flown/ Volume of Lead shot

= (50/3)π/(⅙)π

= (50/3)×6 = 100

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 8 g mass. ( π = 3.14 ).

Answer:

Given that, the height of the big cylinder (H) = 220 cm

Radius of the base (R) = 24/2 = 12 cm

So, the volume of the big cylinder = πR2H

= π(12)2 × 220 cm³

= 99565.8 cm³

Now, the height of smaller cylinder (h) = 60 cm

Radius of the base (r) = 8 cm

So, the volume of the smaller cylinder = πr2h

= π(8)2×60 cm³

= 12068.5 cm³

∴ Volume of iron = Volume of the big cylinder+ Volume of the small cylinder

= 99565.8 + 12068.5

=111634.5 cm³

We know,

Mass = Density x volume

So, mass of the pole = 8×111634.5

= 893 Kg (approx.)

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:

Here, the volume of water left will be = Volume of cylinder – Volume of solid

Given,

Radius of cone = 60 cm,

Height of cone = 120 cm

Radius of cylinder = 60 cm

Height of cylinder = 180 cm

Radius of hemisphere = 60 cm

Now,

Total volume of solid = Volume of Cone + Volume of hemisphere

Volume of cone = 1/3πr2h = 1/3 × π×602×120cm3 = 144×103π cm³

Volume of hemisphere = (⅔)×π×603 cm3 = 144×103π cm³

So, total volume of solid =  144×103π cm3 + 144×103π cm3 = 288 ×103π cm³

Volume of cylinder = π×602×180 = 648000 = 648×103 π cm³

Now, volume of water left will be = Volume of cylinder – Volume of solid

= (648-288) × 103×π = 1.131 m³

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 cm² . Check whether she is correct, taking the above as the inside measurements and π = 3.14.

Answer:

=

Given,

For the cylinder part, Height (h) = 8 cm and Radius (R) = (2/2) cm = 1 cm

For the spherical part, Radius (r) = (8.5/2) = 4.25 cm

Now, volume of this vessel = Volume of cylinder + Volume of sphere

= π×(1)²×8+(4/3)π(4.25)³

= 346.51 cm³

CHAPTER NAME	OLD NCERT	NEW NCERT
Real Numbers	EXERCISE 1.1
	EXERCISE 1.2	1.1	CLICK HERE
	EXERCISE 1.3	1.2	CLICK HERE
	EXERCISE 1.4
Polynomials	EXERCISE 2.1	2.1	CLICK HERE
	EXERCISE 2.2	2.2	CLICK HERE
	EXERCISE 2.3
	EXERCISE 2.4
Pair of Linear Equations in Two Variables	EXERCISE 3.1
	EXERCISE 3.2	3.1	CLICK HERE
	EXERCISE 3.3	3.2	CLICK HERE
	EXERCISE 3.4	3.3	CLICK HERE
	EXERCISE 3.5
	EXERCISE 3.6
	EXERCISE 3.7
Quadratic Equations	EXERCISE 4.1	4.1	CLICK HERE
	EXERCISE 4.2	4.2	CLICK HERE
	EXERCISE 4.3
	EXERCISE 4.4	4.3	CLICK HERE
Arithmetic Progressions	EXERCISE 5.1	5.1	CLICK HERE
	EXERCISE 5.2	5.2	CLICK HERE
	EXERCISE 5.3	5.3	CLICK HERE
	EXERCISE 5.4	5.4 (Optional)	CLICK HERE
Triangles	EXERCISE 6.1	6.1	CLICK HERE
	EXERCISE 6.2	6.2	CLICK HERE
	EXERCISE 6.3	6.3	CLICK HERE
	EXERCISE 6.4
	EXERCISE 6.5
	EXERCISE 6.6
Coordinate Geometry	EXERCISE 7.1	7.1	CLICK HERE
	EXERCISE 7.2	7.2	CLICK HERE
	EXERCISE 7.3
	EXERCISE 7.4
Introduction to Trigonometry	EXERCISE 8.1	8.1	CLICK HERE
	EXERCISE 8.2	8.2	CLICK HERE
	EXERCISE 8.3
	EXERCISE 8.4	8.3	CLICK HERE
Some Applications of Trigonometry	EXERCISE 9.1	9.1	CLICK HERE
Circles	EXERCISE 10.1	10.1	CLICK HERE
	EXERCISE 10.2	10.2	CLICK HERE
Construction
Areas Related to Circles	EXERCISE 12.1
	EXERCISE 12.2	11.1	CLICK HERE
	EXERCISE 12.3
Surface Areas and Volumes	EXERCISE 13.1	12.1	CLICK HERE
	EXERCISE 13.2	12.2	CLICK HERE
	EXERCISE 13.3
	EXERCISE 13.4
	EXERCISE 13.5
Statistics	EXERCISE 14.1	13.1	CLICK HERE
	EXERCISE 14.2	13.2	CLICK HERE
	EXERCISE 14.3	13.3	CLICK HERE
	EXERCISE 14.4
Probability	EXERCISE 15.1	14.1	CLICK HERE
	EXERCISE 15.2