{"id":42936,"date":"2020-09-28T15:02:19","date_gmt":"2020-09-28T09:32:19","guid":{"rendered":"https:\/\/cracku.in\/blog\/?p=42936"},"modified":"2020-09-28T15:02:19","modified_gmt":"2020-09-28T09:32:19","slug":"volume-questions-for-railways-pdf","status":"publish","type":"post","link":"https:\/\/cracku.in\/blog\/volume-questions-for-railways-pdf\/","title":{"rendered":"Volume questions for Railways PDF"},"content":{"rendered":"

Volume questions for Railways PDF<\/strong><\/span><\/h2>\n

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.<\/p>\n

Download Volume questions for Railways PDF <\/a><\/p>\n

Download Time and Work question and answers for Railway Exams PDF<\/a><\/p>\n\n

Download RRB Group-D Previous Papers PDF<\/a><\/p>\n

Take a RRB Group-D free mock test<\/a><\/p>\n

Question 1:\u00a0<\/b>The total surface area of a sphere is $8\\pi$ square unit. The volume of the sphere is<\/p>\n

a)\u00a0$\\frac{8}{3}\\pi$<\/p>\n

b)\u00a0$8\\sqrt{3}\\pi$<\/p>\n

c)\u00a0$\\frac{8\\sqrt{3}}{5}\\pi$<\/p>\n

d)\u00a0$\\frac{8\\sqrt{2}}{3}\\pi$<\/p>\n

Question 2:\u00a0<\/b>The perimeter of one face of a cube is 20 cm. Its volume will be<\/p>\n

a)\u00a0625 cm<\/p>\n

b)\u00a0100 cm<\/p>\n

c)\u00a0125 cm<\/p>\n

d)\u00a0400 cm<\/p>\n

Question 3:\u00a0<\/b>If the volume of a sphere is numerically equal to its surface area then its diameter is<\/p>\n

a)\u00a06cm<\/p>\n

b)\u00a04 cm<\/p>\n

c)\u00a02 cm<\/p>\n

d)\u00a03 cm<\/p>\n

Question 4:\u00a0<\/b>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 \u0394DAB = 90\u00b0. If the volume of the prism be 2070 cm3, then the area of the lateral surface is<\/p>\n

a)\u00a0720 cm2<\/p>\n

b)\u00a0810 cm2<\/p>\n

c)\u00a01260 cm2<\/p>\n

d)\u00a02070 cm2<\/p>\n

Question 5:\u00a0<\/b>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<\/p>\n

a)\u00a03 : 1<\/p>\n

b)\u00a01 : 3<\/p>\n

c)\u00a06 : 1<\/p>\n

d)\u00a01 : 6<\/p>\n

Question 6:\u00a0<\/b>The perimeter of the base of a right circular cylinder is \u2018a\u2019 unit. If the volume of the cylinder is V cubic unit, then the height of the cylinder is<\/p>\n

a)\u00a0$\\frac{4a^2v}{\\pi}$<\/p>\n

b)\u00a0$\\frac{4\\pi a^2}{v}$<\/p>\n

c)\u00a0$\\frac{\\pi a^2v}{v}$<\/p>\n

d)\u00a0$\\frac{4\\pi v}{a^2}$<\/p>\n

RRB Group D previous year papers<\/a><\/p>\n

Daily Free RRB Online Test<\/a><\/p>\n

Question 7:\u00a0<\/b>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 :<\/p>\n

a)\u00a0$1232 Cm^{3}$<\/p>\n

b)\u00a0$1848 Cm^{3}$<\/p>\n

c)\u00a0$1632 Cm^{3}$<\/p>\n

d)\u00a0$1078 Cm^{3}$<\/p>\n

Question 8:\u00a0<\/b>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:<\/p>\n

a)\u00a0$108 \\pi cm^{3}$<\/p>\n

b)\u00a0$\\frac{112}{\\pi}cm^3$<\/p>\n

c)\u00a0$112 \\pi cm^{3}$<\/p>\n

d)\u00a0$\\frac{108}{\\pi} cm^{3}$<\/p>\n

Question 9:\u00a0<\/b>The radius of the base and the height of a right circular cone are doubled. The volume of the cone will be<\/p>\n

a)\u00a08 times of the previous volume<\/p>\n

b)\u00a0three times of the previous volume<\/p>\n

c)\u00a0 3\u221a2 times of the previous volume<\/p>\n

d)\u00a06 times of the previous volume<\/p>\n

Question 10:\u00a0<\/b>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<\/p>\n

a)\u00a08 : 3<\/p>\n

b)\u00a03 : 8<\/p>\n

c)\u00a04 : 3<\/p>\n

d)\u00a03 : 4<\/p>\n

RRB Group D previous year papers<\/a><\/p>\n

Daily Free RRB Online Test<\/a><\/p>\n

Question 11:\u00a0<\/b>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 :<\/p>\n

a)\u00a08.8 sq.m<\/p>\n

b)\u00a04.4 sq.m<\/p>\n

c)\u00a044 sq.m<\/p>\n

d)\u00a016.8 sq.m<\/p>\n

Question 12:\u00a0<\/b>A bicycle wheel makes 5000 revolutions in moving 11km.Then the radius of the wheel (in cm) is<\/p>\n

a)\u00a070<\/p>\n

b)\u00a035<\/p>\n

c)\u00a017.5<\/p>\n

d)\u00a0140<\/p>\n

Question 13:\u00a0<\/b>If the radius of the cylinder and of a sphere having the same volume as that cylinder is \u2018r\u2019 then what is the height of that cylinder?<\/p>\n

a)\u00a03\/2 r<\/p>\n

b)\u00a02\/3 r<\/p>\n

c)\u00a04\/3 r<\/p>\n

d)\u00a03\/4 r<\/p>\n

Question 14:\u00a0<\/b>Volume of a cylinder is 2310 cubic cm. If circumference of its base is 44 cm, find the curved surface area of the cylinder?<\/p>\n

a)\u00a0660 sq cm<\/p>\n

b)\u00a01320 sq cm<\/p>\n

c)\u00a01980 sq cm<\/p>\n

d)\u00a0330 sq cm<\/p>\n

Question 15:\u00a0<\/b>If the curved surface area of a right circular cone is 10010 sq cm and its radius is 35 cm, find its volume?<\/p>\n

a)\u00a06930 cubic cm<\/p>\n

b)\u00a0107800 cubic cm<\/p>\n

c)\u00a03465 cubic cm<\/p>\n

d)\u00a027720 cubic cm<\/p>\n

RRB Group D previous year papers<\/a><\/p>\n

Daily Free RRB Online Test<\/a><\/p>\n

Answers & Solutions:<\/strong><\/span><\/p>\n

1)\u00a0Answer\u00a0(D)<\/strong><\/p>\n

Let the radius of the sphere be ‘r’
\nHence $4\\pi r^2$ = $8\\pi$
\n=> r = $\\sqrt{2}$
\nVolume of the cube = $\\frac{4}{3}\\pi r^3$ = $\\frac{4}{3}\\pi \\sqrt{2}^3$ = $\\frac{8\\sqrt{2}}{3}\\pi$<\/p>\n

2)\u00a0Answer\u00a0(C)<\/strong><\/p>\n

Let each side of the cube be $a$<\/p>\n

One face of a cube is a square with perimeter 20 cm<\/p>\n

=> 4 * $a$ = 20<\/p>\n

=> $a$ = 5<\/p>\n

Volume of cube = $a^3$<\/p>\n

= $5^3$ = 125 $cm^3$<\/p>\n

3)\u00a0Answer\u00a0(A)<\/strong><\/p>\n

Let the radius of sphere = $r$<\/p>\n

Since, volume of sphere = surface area<\/p>\n

=> $\\frac{4}{3} \\pi r^3 = 4 \\pi r^2$<\/p>\n

=> $r = 3$<\/p>\n

=> Diameter = 2 * $r$ = 2*3 = 6 cm<\/p>\n

4)\u00a0Answer\u00a0(A)<\/strong><\/p>\n

<\/p>\n

In right $\\triangle$DAB, using Pythagoras theorem,<\/p>\n

=> $BD = \\sqrt{(AD)^2 + (AB)^2}$<\/p>\n

=> $BD = \\sqrt{12^2 + 9^2} = \\sqrt{225}$<\/p>\n

=> $BD = 15 cm$<\/p>\n

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

Area of $\\triangle$BCD = $\\sqrt{s(s-a)(s-b)(s-c)}$<\/span><\/p>\n

where, $s = \\frac{a+b+c}{2}$<\/span><\/p>\n

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

=> Area of quad ABCD = area of $\\triangle$DAB + area of $\\triangle$BCD<\/span><\/span><\/span><\/span><\/p>\n

= 54+84 = $138 cm^2$<\/span><\/span><\/span><\/span><\/p>\n

Volume of prism = base area * height<\/span><\/span><\/span><\/span><\/p>\n

=> 2070 = 138 * height<\/span><\/span><\/span><\/span><\/p>\n

=> height = 15 cm<\/span><\/span><\/span><\/span><\/p>\n

Lateral surface area of prism = perimeter of base * height<\/span><\/span><\/span><\/span><\/p>\n

= (12 + 9 + 14 + 13) * 15 = 48 * 15<\/span><\/span><\/span><\/span><\/p>\n

= 720 $cm^2$<\/span><\/span><\/span><\/span><\/p>\n

5)\u00a0Answer\u00a0(D)<\/strong><\/p>\n

Let the radius and height of right circular cylinder be r and h respectively.<\/p>\n

Let radius of sphere is R.<\/p>\n

The radius of the cylinder and the diameter of the sphere are equal.<\/p>\n

Therefore, r = 2R<\/p>\n

The volumes of a right circular cylinder and a sphere are equal.<\/p>\n

=> $\\pi r^2h = \\frac{4}{3} \\pi R^3$<\/p>\n

=> $3 r^2h = 4(\\frac{r}{2})^3$<\/p>\n

=> $6 r^2h = r^3$<\/p>\n

=> $6h = r$<\/p>\n

=> $\\frac{h}{r} = \\frac{1}{6}$<\/p>\n

RRB Group-D Important Questions (download PDF)<\/a><\/p>\n

General Science Notes for RRB Exams (PDF)<\/a><\/p>\n\n

6)\u00a0Answer\u00a0(D)<\/strong><\/p>\n

Perimeter of base = $2 \\pi r = a$<\/p>\n

=> $r = \\frac{a}{2 \\pi}$<\/p>\n

Volume of cylinder = $\\pi r^2 h = V$<\/p>\n

=> $\\pi (\\frac{a}{2 \\pi})^2 h = V$<\/p>\n

=> $a^2 h = 4 \\pi V$<\/p>\n

=> $h = \\frac{4 \\pi V}{a^2}$<\/p>\n

7)\u00a0Answer\u00a0(D)<\/strong><\/p>\n

Let radius of cylinder be $r$ and height be $h$<\/p>\n

=> $\\frac{2 \\pi rh}{2 \\pi rh + 2 \\pi r^2} = \\frac{1}{2}$<\/p>\n

=> $\\frac{h}{h + r} = \\frac{1}{2}$<\/p>\n

=> $h = r$<\/p>\n

Total surface area = $2 \\pi rh + 2 \\pi r^2 = 616$<\/p>\n

=> $2 \\pi r^2 + 2 \\pi r^2 = 616$<\/p>\n

=> $r^2 = \\frac{154 * 7}{22}$<\/p>\n

=> $r = 7 = h$<\/p>\n

Now, volume of cylinder = $\\pi r^2 h$<\/p>\n

= $\\frac{22}{7} * 7^2 * 7$<\/p>\n

= $1078 cm^3$<\/p>\n

8)\u00a0Answer\u00a0(B)<\/strong><\/p>\n

Let the radius of the cone be $r$ and height = 21<\/p>\n

=> Perimeter of base = $2 \\pi r = 8$<\/p>\n

=> $r = \\frac{4}{\\pi}$<\/p>\n

Volume of cone = $\\frac{1}{3} \\pi r^2 h$<\/p>\n

= $\\frac{1}{3} \\pi (\\frac{4}{\\pi})^2 21$<\/p>\n

= $\\frac{112}{\\pi} cm^3$<\/p>\n

9)\u00a0Answer\u00a0(A)<\/strong><\/p>\n

Let original radius be $r$ and height be $h$<\/p>\n

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

New radius = $2r$ and new height = $2h$<\/p>\n

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

= $8 * \\frac{1}{3} \\pi r^2h$<\/p>\n

=> New volume of cone = 8 times the previous one.<\/p>\n

10)\u00a0Answer\u00a0(A)<\/strong><\/p>\n

Let the radii of two cones be $r_1 = x$ and $r_2 = 2x$<\/p>\n

Let the heights of the two cones be $h_1$ and $h_2$<\/p>\n

=> Ratio of volumes :<\/p>\n

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

=> $\\frac{x^2 h_1}{4x^2 h_2} = \\frac{2}{3}$<\/p>\n

=> $\\frac{h_1}{h_2} = \\frac{8}{3}$<\/p>\n

=> Required ratio = 8 : 3<\/p>\n

11)\u00a0Answer\u00a0(C)<\/strong><\/p>\n

We know that, Surface area of cylinder = \u03c0 \u00d7 d \u00d7 L Where, d and L are diameter and length of the cylinder
\nGiven, Diameter of the roller = 1.4 m & Length of the roller = 2 m<\/p>\n

\u2234 Surface area of the roller = \u03c0 \u00d7 1.4 \u00d7 2 = (22\/7) \u00d7 1.4 \u00d7 2 = 8.8 m2
\n<\/sup>\u2234 Area covered by 5 revolution = 5 \u00d7 8.8 m2 <\/sup>= 44 m2<\/sup><\/p>\n

12)\u00a0Answer\u00a0(B)<\/strong><\/p>\n

The distance of 11km is covered by the perimeter of tyre by 5000 revolutions.
\nHence, $11\\times1000\\times100 = 5000\\times$ \u03c0<\/em><\/span>D
\nD = 70cm
\nr = 35cm<\/p>\n

13)\u00a0Answer\u00a0(C)<\/strong><\/p>\n

volume of sphere = $\\frac{4}{3}$ $\\pi r^{3}$<\/p>\n

volume of cylinder =\u00a0$\\pi r^{2}h$<\/p>\n

given that radius and volume of both are same<\/p>\n

so\u00a0\u00a0$\\frac{4}{3}$ $\\pi r^{3}$ =\u00a0$\\pi r^{2}h$<\/p>\n

$\\Rightarrow\\frac{4}{3}r$\u00a0 =\u00a0 $h$<\/p>\n

so the answer is option C.<\/p>\n

14)\u00a0Answer\u00a0(A)<\/strong><\/p>\n

Let radius of base of cylinder = $r$ cm and height = $h$ cm<\/p>\n

Circumference of base = $2 \\pi r = 44$<\/p>\n

=> $2 \\times \\frac{22}{7} \\times r = 44$<\/p>\n

=> $r = 44 \\times \\frac{7}{44} = 7$ cm<\/p>\n

Now, volume of cylinder = $\\pi r^2 h = 2310$<\/p>\n

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

=> $22 \\times 7 \\times h = 2310$<\/p>\n

=> $h = \\frac{2310}{154} = 15$ cm<\/p>\n

$\\therefore$ Curved Surface area of cylinder = $2 \\pi r h$<\/p>\n

= $44 \\times 15 = 660 cm^2$<\/p>\n

15)\u00a0Answer\u00a0(B)<\/strong><\/p>\n

Let slant height of cone = $l$ cm and radius = 35 cm<\/p>\n

Curved surface area of cone = $\\pi r l = 10010$<\/p>\n

=> $\\frac{22}{7} \\times 35 \\times l = 10010$<\/p>\n

=> $22 \\times 5 \\times l = 10010$<\/p>\n

=> $l = \\frac{10010}{110} = 91$<\/p>\n

Let height of cone = $h$ cm<\/p>\n

=> $(h)^2 = (l)^2 – (r)^2$<\/p>\n

=> $(h)^2 = (91)^2 – (35)^2$<\/p>\n

=> $(h)^2 = 8281 – 1225 = 7056$<\/p>\n

=> $h = \\sqrt{7056} = 84$<\/p>\n

$\\therefore$ Volume of cone = $\\frac{1}{3} \\times \\pi r^2 h$<\/p>\n

= $\\frac{1}{3} \\times \\frac{22}{7} \\times (35)^2 \\times 84$<\/p>\n

= $22 \\times (35)^2 \\times 4 = 107800 cm^3$<\/p>\n\n

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We hope this Volume questions\u00a0 and answers PDF for RRB Group-D Exam will be highly useful for your preparation.<\/p>\n","protected":false},"excerpt":{"rendered":"

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. Download RRB Group-D Previous Papers PDF Take a RRB Group-D free mock test Question 1:\u00a0The total surface area […]<\/p>\n","protected":false},"author":49,"featured_media":42944,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"om_disable_all_campaigns":false,"_mi_skip_tracking":false,"footnotes":""},"categories":[3167,169,125,3171,1679,1897,1605,1603],"tags":[166,492,1635,1593,4211],"class_list":{"0":"post-42936","1":"post","2":"type-post","3":"status-publish","4":"format-standard","5":"has-post-thumbnail","7":"category-downloads-en","8":"category-downloads","9":"category-featured","10":"category-railways-en","11":"category-rrb-group-d","12":"category-rrb-group-d-2","13":"category-rrb-je","14":"category-rrb-ntpc","15":"tag-railways","16":"tag-rrb-group-d","17":"tag-rrb-je","18":"tag-rrb-ntpc","19":"tag-volume"},"better_featured_image":{"id":42944,"alt_text":"Volume questions for Railways PDF","caption":"Volume questions for 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