Volume questions for Railways PDF

0
141

Volume questions for Railways PDF

Download Top-15 RRB Group-D Volume questions and answers PDF. RRB Group – D Volume questions and answers based on asked questions in previous exam papers very important for the Railway Group-D exam.

Question 1: The total surface area of a sphere is $8\pi$ square unit. The volume of the sphere is

a) $\frac{8}{3}\pi$

b) $8\sqrt{3}\pi$

c) $\frac{8\sqrt{3}}{5}\pi$

d) $\frac{8\sqrt{2}}{3}\pi$

Question 2: The perimeter of one face of a cube is 20 cm. Its volume will be

a) 625 cm

b) 100 cm

c) 125 cm

d) 400 cm

Question 3: If the volume of a sphere is numerically equal to its surface area then its diameter is

a) 6cm

b) 4 cm

c) 2 cm

d) 3 cm

Question 4: The base of a right prism is a quadrilateral ABCD. Given that AB = 9 cm, BC = 14 cm, CD = 13 cm, DA = 12 cm and ΔDAB = 90°. If the volume of the prism be 2070 cm3, then the area of the lateral surface is

a) 720 cm2

b) 810 cm2

c) 1260 cm2

d) 2070 cm2

Question 5: The volumes of a right circular cylinder and a sphere are equal. The radius of the cylinder and the diameter of the sphere are equal. The ratio of height and radius of the cylinder is

a) 3 : 1

b) 1 : 3

c) 6 : 1

d) 1 : 6

Question 6: The perimeter of the base of a right circular cylinder is ‘a’ unit. If the volume of the cylinder is V cubic unit, then the height of the cylinder is

a) $\frac{4a^2v}{\pi}$

b) $\frac{4\pi a^2}{v}$

c) $\frac{\pi a^2v}{v}$

d) $\frac{4\pi v}{a^2}$

Question 7: The curved surface area and the total surface area of a cylinder are in the ratio 1:2 If the total surface area of the right cylinder is 616 cm , then its volume is :

a) $1232 Cm^{3}$

b) $1848 Cm^{3}$

c) $1632 Cm^{3}$

d) $1078 Cm^{3}$

Question 8: The perimeter of the base of a right circular cone is 8 cm. If the height of the cone is 21 cm, then its volume is:

a) $108 \pi cm^{3}$

b) $\frac{112}{\pi}cm^3$

c) $112 \pi cm^{3}$

d) $\frac{108}{\pi} cm^{3}$

Question 9: The radius of the base and the height of a right circular cone are doubled. The volume of the cone will be

a) 8 times of the previous volume

b) three times of the previous volume

c)  3√2 times of the previous volume

d) 6 times of the previous volume

Question 10: If the ratio of volumes of two cones is 2 : 3 and the ratio of the radii of their bases is 1 : 2, then the ratio of their heights will be

a) 8 : 3

b) 3 : 8

c) 4 : 3

d) 3 : 4

Question 11: The diameter of a garden roller is 1.4 metre and it is 2 metre long. The area covered by the roller in 5 revolutions is :

a) 8.8 sq.m

b) 4.4 sq.m

c) 44 sq.m

d) 16.8 sq.m

Question 12: A bicycle wheel makes 5000 revolutions in moving 11km.Then the radius of the wheel (in cm) is

a) 70

b) 35

c) 17.5

d) 140

Question 13: If the radius of the cylinder and of a sphere having the same volume as that cylinder is ‘r’ then what is the height of that cylinder?

a) 3/2 r

b) 2/3 r

c) 4/3 r

d) 3/4 r

Question 14: Volume of a cylinder is 2310 cubic cm. If circumference of its base is 44 cm, find the curved surface area of the cylinder?

a) 660 sq cm

b) 1320 sq cm

c) 1980 sq cm

d) 330 sq cm

Question 15: If the curved surface area of a right circular cone is 10010 sq cm and its radius is 35 cm, find its volume?

a) 6930 cubic cm

b) 107800 cubic cm

c) 3465 cubic cm

d) 27720 cubic cm

Let the radius of the sphere be ‘r’
Hence $4\pi r^2$ = $8\pi$
=> r = $\sqrt{2}$
Volume of the cube = $\frac{4}{3}\pi r^3$ = $\frac{4}{3}\pi \sqrt{2}^3$ = $\frac{8\sqrt{2}}{3}\pi$

Let each side of the cube be $a$

One face of a cube is a square with perimeter 20 cm

=> 4 * $a$ = 20

=> $a$ = 5

Volume of cube = $a^3$

= $5^3$ = 125 $cm^3$

Let the radius of sphere = $r$

Since, volume of sphere = surface area

=> $\frac{4}{3} \pi r^3 = 4 \pi r^2$

=> $r = 3$

=> Diameter = 2 * $r$ = 2*3 = 6 cm

In right $\triangle$DAB, using Pythagoras theorem,

=> $BD = \sqrt{(AD)^2 + (AB)^2}$

=> $BD = \sqrt{12^2 + 9^2} = \sqrt{225}$

=> $BD = 15 cm$

Now, area of $\triangle$DAB = $\frac{1}{2} * 9 * 12 = 54 cm^2$

Area of $\triangle$BCD = $\sqrt{s(s-a)(s-b)(s-c)}$

where, $s = \frac{a+b+c}{2}$

=> Area of $\triangle$BCD = $\sqrt{21 * 6 * 7 * 8} = 84 cm^2$

=> Area of quad ABCD = area of $\triangle$DAB + area of $\triangle$BCD

= 54+84 = $138 cm^2$

Volume of prism = base area * height

=> 2070 = 138 * height

=> height = 15 cm

Lateral surface area of prism = perimeter of base * height

= (12 + 9 + 14 + 13) * 15 = 48 * 15

= 720 $cm^2$

Let the radius and height of right circular cylinder be r and h respectively.

Let radius of sphere is R.

The radius of the cylinder and the diameter of the sphere are equal.

Therefore, r = 2R

The volumes of a right circular cylinder and a sphere are equal.

=> $\pi r^2h = \frac{4}{3} \pi R^3$

=> $3 r^2h = 4(\frac{r}{2})^3$

=> $6 r^2h = r^3$

=> $6h = r$

=> $\frac{h}{r} = \frac{1}{6}$

Perimeter of base = $2 \pi r = a$

=> $r = \frac{a}{2 \pi}$

Volume of cylinder = $\pi r^2 h = V$

=> $\pi (\frac{a}{2 \pi})^2 h = V$

=> $a^2 h = 4 \pi V$

=> $h = \frac{4 \pi V}{a^2}$

Let radius of cylinder be $r$ and height be $h$

=> $\frac{2 \pi rh}{2 \pi rh + 2 \pi r^2} = \frac{1}{2}$

=> $\frac{h}{h + r} = \frac{1}{2}$

=> $h = r$

Total surface area = $2 \pi rh + 2 \pi r^2 = 616$

=> $2 \pi r^2 + 2 \pi r^2 = 616$

=> $r^2 = \frac{154 * 7}{22}$

=> $r = 7 = h$

Now, volume of cylinder = $\pi r^2 h$

= $\frac{22}{7} * 7^2 * 7$

= $1078 cm^3$

Let the radius of the cone be $r$ and height = 21

=> Perimeter of base = $2 \pi r = 8$

=> $r = \frac{4}{\pi}$

Volume of cone = $\frac{1}{3} \pi r^2 h$

= $\frac{1}{3} \pi (\frac{4}{\pi})^2 21$

= $\frac{112}{\pi} cm^3$

Let original radius be $r$ and height be $h$

=> Original volume of cone = $\frac{1}{3} \pi r^2h$

New radius = $2r$ and new height = $2h$

=> New volume = $\frac{1}{3} \pi (4r^2) (2h)$

= $8 * \frac{1}{3} \pi r^2h$

=> New volume of cone = 8 times the previous one.

Let the radii of two cones be $r_1 = x$ and $r_2 = 2x$

Let the heights of the two cones be $h_1$ and $h_2$

=> Ratio of volumes :

=> $\frac{\frac{1}{3} \pi r_1^2 h_1}{\frac{1}{3} \pi r_2^2 h_2} = \frac{2}{3}$

=> $\frac{x^2 h_1}{4x^2 h_2} = \frac{2}{3}$

=> $\frac{h_1}{h_2} = \frac{8}{3}$

=> Required ratio = 8 : 3

We know that, Surface area of cylinder = π × d × L Where, d and L are diameter and length of the cylinder
Given, Diameter of the roller = 1.4 m & Length of the roller = 2 m

∴ Surface area of the roller = π × 1.4 × 2 = (22/7) × 1.4 × 2 = 8.8 m2
∴ Area covered by 5 revolution = 5 × 8.8 m2 = 44 m2

The distance of 11km is covered by the perimeter of tyre by 5000 revolutions.
Hence, $11\times1000\times100 = 5000\times$ πD
D = 70cm
r = 35cm

volume of sphere = $\frac{4}{3}$ $\pi r^{3}$

volume of cylinder = $\pi r^{2}h$

given that radius and volume of both are same

so  $\frac{4}{3}$ $\pi r^{3}$ = $\pi r^{2}h$

$\Rightarrow\frac{4}{3}r$  =  $h$

so the answer is option C.

Let radius of base of cylinder = $r$ cm and height = $h$ cm

Circumference of base = $2 \pi r = 44$

=> $2 \times \frac{22}{7} \times r = 44$

=> $r = 44 \times \frac{7}{44} = 7$ cm

Now, volume of cylinder = $\pi r^2 h = 2310$

=> $\frac{22}{7} \times (7)^2 \times h = 2310$

=> $22 \times 7 \times h = 2310$

=> $h = \frac{2310}{154} = 15$ cm

$\therefore$ Curved Surface area of cylinder = $2 \pi r h$

= $44 \times 15 = 660 cm^2$

Let slant height of cone = $l$ cm and radius = 35 cm

Curved surface area of cone = $\pi r l = 10010$

=> $\frac{22}{7} \times 35 \times l = 10010$

=> $22 \times 5 \times l = 10010$

=> $l = \frac{10010}{110} = 91$

Let height of cone = $h$ cm

=> $(h)^2 = (l)^2 – (r)^2$

=> $(h)^2 = (91)^2 – (35)^2$

=> $(h)^2 = 8281 – 1225 = 7056$

=> $h = \sqrt{7056} = 84$

$\therefore$ Volume of cone = $\frac{1}{3} \times \pi r^2 h$

= $\frac{1}{3} \times \frac{22}{7} \times (35)^2 \times 84$

= $22 \times (35)^2 \times 4 = 107800 cm^3$

We hope this Volume questions  and answers PDF for RRB Group-D Exam will be highly useful for your preparation.