{"id":26238,"date":"2019-03-18T18:39:39","date_gmt":"2019-03-18T13:09:39","guid":{"rendered":"https:\/\/cracku.in\/blog\/?p=26238"},"modified":"2019-03-18T18:56:15","modified_gmt":"2019-03-18T13:26:15","slug":"mensuration-questions-for-ssc-chsl-pdf","status":"publish","type":"post","link":"https:\/\/cracku.in\/blog\/mensuration-questions-for-ssc-chsl-pdf\/","title":{"rendered":"Mensuration Questions for SSC CHSL PDF"},"content":{"rendered":"

Mensuration Questions for SSC CHSL PDF:<\/strong><\/span><\/h2>\n

SSC CHSL Mensuration Questions download PDF based on previous year question paper of SSC CHSL exam. 25 Very important Mensuration questions for SSC CHSL Exam.<\/p>\n

Download Title: Mensuration Questions for SSC CHSL PDF<\/a><\/p>\n

25 SSC CHSL MOCKS FOR JUST RS. 149<\/a><\/p>\n

FREE SSC EXAM YOUTUBE VIDEOS<\/a><\/p>\n

Download Topic Wise SSC CHSL Important Questions PDF<\/a><\/p>\n

SSC CHSL Study Material<\/a> (FREE Tests)<\/p>\n

\"\"<\/a><\/p>\n

Question 1:\u00a0<\/b>The edge of an ice cube is 14 cm. The volume of the largest cylindrical ice cube that can be formed out of it is<\/p>\n

a)\u00a02200 cu. cm<\/p>\n

b)\u00a02000 cu. cm<\/p>\n

c)\u00a02156 cu. cm<\/p>\n

d)\u00a02400 cu. cm<\/p>\n

e)\u00a0None of these<\/p>\n

Question 2:\u00a0<\/b>A rope makes 125 rounds of a cylinder with base radius 15 cm. How many times can it go round a cylinder with base radius 25 cm?<\/p>\n

a)\u00a0100<\/p>\n

b)\u00a075<\/p>\n

c)\u00a080<\/p>\n

d)\u00a065<\/p>\n

e)\u00a0None of these<\/p>\n

Question 3:\u00a0<\/b>What is the area of the triangle whose vertices are (0,0), (5,0) and (0,6)?<\/p>\n

a)\u00a030 sq units<\/p>\n

b)\u00a015 sq units<\/p>\n

c)\u00a020 sq units<\/p>\n

d)\u00a010 sq units<\/p>\n

e)\u00a0None of the above<\/p>\n

Question 4:\u00a0<\/b>A rectangular field of dimensions 30 m x 20 m is surrounded by a footpath of uniform width. If the area of the footpath is 216 $m^2$, what is the width of the footpath?<\/p>\n

a)\u00a02 m<\/p>\n

b)\u00a03 m<\/p>\n

c)\u00a01 m<\/p>\n

d)\u00a02.5 m<\/p>\n

e)\u00a0None of the above<\/p>\n

Question 5:\u00a0<\/b>Which of the following cannot be the ratio of angles of a right-angled triangle?<\/p>\n

a)\u00a01:2:3<\/p>\n

b)\u00a01:1:2<\/p>\n

c)\u00a05:6:7<\/p>\n

d)\u00a01:5:6<\/p>\n

e)\u00a0None of the above<\/p>\n

SSC CHSL PREVIOUS PAPERS<\/a><\/p>\n

Question 6:\u00a0<\/b>In a triangle ABC, a line is drawn parallel to BC such that it cuts AB at D and AC at E. If DE\/BC = \u00bd, what is the ratio of the area of triangle ABC to the area of the triangle ADE?<\/p>\n

a)\u00a04:1<\/p>\n

b)\u00a01:4<\/p>\n

c)\u00a02:1<\/p>\n

d)\u00a01:2<\/p>\n

e)\u00a0None of the above<\/p>\n

Question 7:\u00a0<\/b>The sides of a right-angled triangle are in Arithmetic Progression. If the length of the largest side is 10 cm, what is the area of the triangle?<\/p>\n

a)\u00a048 $cm^2$<\/p>\n

b)\u00a060 $cm^2$<\/p>\n

c)\u00a072 $cm^2$<\/p>\n

d)\u00a024 $cm^2$<\/p>\n

e)\u00a0None of the above<\/p>\n

Question 8:\u00a0<\/b>The length of a rectangle is increased by 25% and the breadth of the rectangle is decreased by 25%. What is the percentage change in the area of the rectangle?<\/p>\n

a)\u00a05% decrease<\/p>\n

b)\u00a05% increase<\/p>\n

c)\u00a06.25% increase<\/p>\n

d)\u00a06.25% decrease<\/p>\n

e)\u00a0None of the above<\/p>\n

Question 9:\u00a0<\/b>The length of a rectangle is increased by 25% and the breadth of the rectangle is decreased by 25%. What is the percentage change in its perimeter?<\/p>\n

a)\u00a05% increase<\/p>\n

b)\u00a05% decrease<\/p>\n

c)\u00a010% increase<\/p>\n

d)\u00a010% decrease<\/p>\n

e)\u00a0Can\u2019t be determined<\/p>\n

Question 10:\u00a0<\/b>A triangle ABC has lengths AB = 10 cm, BC = 9 cm and CA = 8 cm. What is the length of the median BD?<\/p>\n

a)\u00a0$\\sqrt{74.5}$<\/p>\n

b)\u00a0$\\sqrt{84.5}$<\/p>\n

c)\u00a0$\\sqrt{64.5}$<\/p>\n

d)\u00a0$\\sqrt{54.5}$<\/p>\n

e)\u00a0$\\sqrt{94.5}$<\/p>\n

\"\"<\/a><\/p>\n

SSC CHSL FREE MOCK TEST<\/a><\/p>\n

Question 11:\u00a0<\/b>The area of a square is 100 sq. cms. What is the approximate length of the diagonal?<\/p>\n

a)\u00a010<\/p>\n

b)\u00a014.1<\/p>\n

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

d)\u00a020<\/p>\n

e)\u00a025<\/p>\n

Question 12:\u00a0<\/b>The length of the side of a square is equal to the radius of a circle. If the area of the square is 100 sq metres, what is the area of the circle?<\/p>\n

a)\u00a0314 sq metres<\/p>\n

b)\u00a0414 sq metres<\/p>\n

c)\u00a0514 sq metres<\/p>\n

d)\u00a0614 sq metres<\/p>\n

e)\u00a0None of the above.<\/p>\n

Question 13:\u00a0<\/b>The area of a square is the same as that of a circle. What is the ratio of the side of the square to the radius of the circle?<\/p>\n

a)\u00a01:pi<\/p>\n

b)\u00a01: sq.root pi<\/p>\n

c)\u00a0sq.root pi : 1<\/p>\n

d)\u00a0pi square : 1<\/p>\n

e)\u00a0None of the above<\/p>\n

Question 14:\u00a0<\/b>In a right angled triangle, two sides are of the same length. Which of the options is one of the angles of that triangle?<\/p>\n

a)\u00a030<\/p>\n

b)\u00a045<\/p>\n

c)\u00a060<\/p>\n

d)\u00a075<\/p>\n

e)\u00a0None of the above<\/p>\n

Question 15:\u00a0<\/b>The length of a rectangle is twice its breadth. If the area of the rectangle is 50 metre square, what is the length of the rectangle?<\/p>\n

a)\u00a05 metres<\/p>\n

b)\u00a010 metres<\/p>\n

c)\u00a025 metres<\/p>\n

d)\u00a050 metres<\/p>\n

e)\u00a0None of the above<\/p>\n

FREE SSC MATERIAL – 18000 FREE QUESTIONS<\/a><\/p>\n

Question 16:\u00a0<\/b>The length of a rectangle is 24 metres and its perimeter is 84 metres. Find the area of a triangle whose base is equal to the breadth of the rectangle and height equal to the rectangle\u2019s diagonal?<\/p>\n

a)\u00a0200<\/p>\n

b)\u00a0210<\/p>\n

c)\u00a0240<\/p>\n

d)\u00a0270<\/p>\n

e)\u00a0None of the above<\/p>\n

Question 17:\u00a0<\/b>The circumference of a circle is 100 $\\pi$ metres. Find its area?<\/p>\n

a)\u00a02500 $\\pi$<\/p>\n

b)\u00a03000 $\\pi$<\/p>\n

c)\u00a03500 $\\pi$<\/p>\n

d)\u00a03600 $\\pi$<\/p>\n

e)\u00a0None of the above<\/p>\n

Question 18:\u00a0<\/b>If the radius of a circle is equal to the side of a square, what is the ratio of the perimeters of the circle to the square?<\/p>\n

a)\u00a0$\\pi$:1<\/p>\n

b)\u00a0$\\pi$:2<\/p>\n

c)\u00a0$\\pi$:3<\/p>\n

d)\u00a02:$\\pi$<\/p>\n

e)\u00a0None of the above<\/p>\n

Question 19:\u00a0<\/b>In a right angled traingle, two of the sides are of equal length. Which of the following is certainly one of the angles in the triangle?<\/p>\n

a)\u00a030<\/p>\n

b)\u00a045<\/p>\n

c)\u00a060<\/p>\n

d)\u00a075<\/p>\n

e)\u00a0Can’t be determined<\/p>\n

Question 20:\u00a0<\/b>The length of a rectangle is thrice it’s breadth. The perimeter of the rectangle is 544 metres.
\nWhat is it’s area?<\/p>\n

a)\u00a018372<\/p>\n

b)\u00a023872<\/p>\n

c)\u00a013872<\/p>\n

d)\u00a0Can’t be determined<\/p>\n

e)\u00a0None of the above<\/p>\n

\"\"<\/a><\/p>\n

DOWNLOAD APP TO ACESSES DIRECTLY ON MOBILE<\/a><\/p>\n

Question 21:\u00a0<\/b>If the diameter of a circle is equal to the diagonal of a square of area 100 sq. cms, what is the area of the circle?<\/p>\n

a)\u00a025 $\\pi$<\/p>\n

b)\u00a050 $\\pi$<\/p>\n

c)\u00a075 $\\pi$<\/p>\n

d)\u00a0100 $\\pi$<\/p>\n

e)\u00a0None of the above<\/p>\n

Question 22:\u00a0<\/b>In a regular polygon, the number of diagonals is 3 more than the number of sides. What is the number of sides of the polygon?<\/p>\n

a)\u00a03<\/p>\n

b)\u00a04<\/p>\n

c)\u00a05<\/p>\n

d)\u00a06<\/p>\n

e)\u00a07<\/p>\n

Question 23:\u00a0<\/b>If the height of a hollow cylinder is increased by 25%, what will be the ratio its new area to its area before increment in height?<\/p>\n

a)\u00a05:4<\/p>\n

b)\u00a05:3<\/p>\n

c)\u00a06:5<\/p>\n

d)\u00a07:3<\/p>\n

e)\u00a04:7<\/p>\n

Question 24:\u00a0<\/b>If radius of a circle is increased by 66.66%, then what will be percentage increment in the area of circle?<\/p>\n

a)\u00a0200<\/p>\n

b)\u00a0188.88<\/p>\n

c)\u00a0199.99<\/p>\n

d)\u00a0177.76<\/p>\n

e)\u00a0162.22<\/p>\n

Question 25:\u00a0<\/b>If length of a rectangle is increased by 10% and breadth of a rectangle is increased by 20%, then what will be the increment in area of rectangle?<\/p>\n

a)\u00a033.33%<\/p>\n

b)\u00a055%<\/p>\n

c)\u00a032%<\/p>\n

d)\u00a064%<\/p>\n

e)\u00a066.66%<\/p>\n

\"\"<\/a><\/p>\n

FREE MOCK TEST FOR SSC STENOGRAPHER<\/a><\/p>\n

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

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

Radius of the cylinder = r = $\\frac{14}{2}$ = 7<\/p>\n

Height of the cylinder = h =14<\/p>\n

Volume = Pi x r2 <\/sup>x h<\/p>\n

= $\\frac{22}{7}$ x 7 x 7 x 14 = 2156<\/p>\n

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

Let the required number of rounds be x.<\/p>\n

More radius, less rounds (Inverse Proportion)<\/p>\n

Radius in cm == \u00a0Round<\/p>\n

15 == 125<\/p>\n

25 == x<\/p>\n

\u27f9 x=(15 * 125)\/25 = 75 rounds<\/p>\n

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

This is a right-angled triangle with the perpendicular sides being 5 and 6. So, area = \u00bd * 5 * 6 = 15 sq units<\/p>\n

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

Let the width of the footpath be w.
\nSo, area of the footpath = (30+2w)*(20+2w) – 20*30 = 216
\nFrom the options, w = 2 satisfies the equation. So, option a) is the answer.<\/p>\n

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

One of the angles of a right-angled triangle should be 90 degrees. If the angles are in the ratio 5:6:7, the angles will be 50 degrees, 60 degrees and 70 degrees. So, 5:6:7 cannot be the ratio.<\/p>\n

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

Triangle ADE is similar to triangle ABC. If the sides are in the ratio a:b, the areas will be in the ratio $a^2 : b^2$.
\nHere, the ratio of sides BC and DE is 2:1, so the ratio of areas of triangle ABC and triangle ADE will be 4:1.<\/p>\n

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

If the sides of a right-angled triangle are in AP, then the sides are in the ratio 3:4:5. Since the largest side is 10 cm, the lengths of the other sides are 6 cm and 8 cm. So, the area of the triangle is \u00bd * 6 * 8 = 24 $cm^2$<\/p>\n

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

Let the sides be 10l and 10b. So, area = 100lb.
\nSides after the change = 12.5l and 7.5b
\nNew area = 12.5l*7.5b = 93.75lb
\nSo, percentage change in area = 6.25%<\/p>\n

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

Let the sides be 10l and 10b.
\nPerimeter = 20l + 20b
\nNew sides = 12.5l and 7.5b
\nNew perimeter = 25l+15b
\nChange in perimeter = 5l – 5b
\nPercentage change in perimeter = (5l-5b)\/(20l+20b)
\nSo, the absolute change in percentage cannot be determined.<\/p>\n

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

By Appolonius\u2019 theorem, $AB^2$ + $BC^2$ = 2*($AD^2$ + $BD^2$)
\n100 + 81 = 2*($BD^2$ + 16) => BD = $\\sqrt{90.5 – 16}$ = $\\sqrt{74.5}$<\/p>\n

GK Questions And Answers PDF<\/a><\/p>\n

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

Let the side be s cms long. Diagonal = $ s \\sqrt{2}$ and area = $s^{2}$
\nbut area = 100. So, s = 10. So, diagonal = $ 10\\sqrt {2}$<\/p>\n

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

The area of a circle is pi*r*r. Area of the square = r*r. So, the area of the circle = pi*area of square = 3.14*100 = 314 sq.m<\/p>\n

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

$S^{2} = pi * r^{2} $. So, s\/r = $\\sqrt \\pi$<\/p>\n

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

If two sides of the right angled triangle are equal, it is a right angled isosceles triangle. So, the angles of the triangle are 45, 45 and 90.<\/p>\n

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

If the breadth is s, length is 2s and area = 2s * s = 50. So, s is 5 and 2s is 10.<\/p>\n

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

If the perimeter of the rectangle is 84, 2 * (l + b) = 84. So, breadth is 18 metres since length is 24. $ diagonal^{2} = length^{2} + breadth^{2}$. So, diagonal = 30. Area of the triangle = \u00bd*base*height = \u00bd* 18*30 = 270<\/p>\n

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

Circumference = 2*$\\pi$*r = 100 $\\pi$. So, r = 50 metres. Area = $\\pi$* r * r = 2500 $\\pi$<\/p>\n

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

The required ratio is 2*$\\pi$*r : 4*r = $\\pi$:2<\/p>\n

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

This is a symmetric right angled traingle. So, the angles will be 45, 45 and 90.<\/p>\n

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

If the breadth is x, length = 3x. Perimeter = 2*(x+3x) = 544. so, breadth = 68 and length = 204. Area = 68*204<\/p>\n

FREE SSC CHSL MOCK TEST SERIES<\/a><\/p>\n

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

length of diagonal = $\\sqrt{100}$*$\\sqrt{2}$ = diameter. Radius = diameter\/2. Area = $\\pi$*$radius^{2}$<\/p>\n

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

The number of diagonals of an n-sided polygon is $^nC_2 – n$. So, $^nC_2 – n = n + 3$ => $n^2 – n = 4n + 6$. Solving this, we get n = 6.<\/p>\n

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

Ratio of area will be equal to ratio of height i.e. equal to = 1.25h\/h = 5\/4<\/p>\n

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

Area is directly proportional to the square of its radius; Let\u2019s say radius is r and area is $\\pi r^2$
\nHence new radius is 5r\/3 , and new area = $25\\pi r^2\/9$
\nIncrement in area = $16\\pi r^2\/9$
\nPercentage increment will be = 177.76<\/p>\n

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

Let\u2019s say length is l and breadth is b so area will be lb
\nnew length will be 1.25l and breadth will be 1.10l so new area will be 1.32lb
\npercentage increase in area will be = 32%<\/p>\n

\"\"<\/a><\/p>\n

FREE SSC STUDY MATERIAL<\/a><\/p>\n","protected":false},"excerpt":{"rendered":"

Mensuration Questions for SSC CHSL PDF: SSC CHSL Mensuration Questions download PDF based on previous year question paper of SSC CHSL exam. 25 Very important Mensuration questions for SSC CHSL Exam. Download Topic Wise SSC CHSL Important Questions PDF SSC CHSL Study Material (FREE Tests) Question 1:\u00a0The edge of an ice cube is 14 cm. […]<\/p>\n","protected":false},"author":21,"featured_media":26241,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"om_disable_all_campaigns":false,"_mi_skip_tracking":false,"footnotes":""},"categories":[169,125,378],"tags":[1525],"class_list":{"0":"post-26238","1":"post","2":"type-post","3":"status-publish","4":"format-standard","5":"has-post-thumbnail","7":"category-downloads","8":"category-featured","9":"category-ssc-chsl","10":"tag-ssc-chsl-2019"},"better_featured_image":{"id":26241,"alt_text":"Mensuration Questions for SSC CHSL PDF","caption":"Mensuration Questions for SSC CHSL PDF","description":"Mensuration Questions for SSC CHSL PDF","media_type":"image","media_details":{"width":900,"height":472,"file":"2019\/03\/fig-18-03-2019_13-08-06.jpg","sizes":{"thumbnail":{"file":"fig-18-03-2019_13-08-06-150x150.jpg","width":150,"height":150,"mime-type":"image\/jpeg","source_url":"https:\/\/cracku.in\/blog\/wp-content\/uploads\/2019\/03\/fig-18-03-2019_13-08-06-150x150.jpg"},"medium":{"file":"fig-18-03-2019_13-08-06-300x157.jpg","width":300,"height":157,"mime-type":"image\/jpeg","source_url":"https:\/\/cracku.in\/blog\/wp-content\/uploads\/2019\/03\/fig-18-03-2019_13-08-06-300x157.jpg"},"medium_large":{"file":"fig-18-03-2019_13-08-06-768x403.jpg","width":768,"height":403,"mime-type":"image\/jpeg","source_url":"https:\/\/cracku.in\/blog\/wp-content\/uploads\/2019\/03\/fig-18-03-2019_13-08-06-768x403.jpg"},"tiny-lazy":{"file":"fig-18-03-2019_13-08-06-30x16.jpg","width":30,"height":16,"mime-type":"image\/jpeg","source_url":"https:\/\/cracku.in\/blog\/wp-content\/uploads\/2019\/03\/fig-18-03-2019_13-08-06-30x16.jpg"},"td_80x60":{"file":"fig-18-03-2019_13-08-06-80x60.jpg","width":80,"height":60,"mime-type":"image\/jpeg","source_url":"https:\/\/cracku.in\/blog\/wp-content\/uploads\/2019\/03\/fig-18-03-2019_13-08-06-80x60.jpg"},"td_100x70":{"file":"fig-18-03-2019_13-08-06-100x70.jpg","width":100,"height":70,"mime-type":"image\/jpeg","source_url":"https:\/\/cracku.in\/blog\/wp-content\/uploads\/2019\/03\/fig-18-03-2019_13-08-06-100x70.jpg"},"td_218x150":{"file":"fig-18-03-2019_13-08-06-218x150.jpg","width":218,"height":150,"mime-type":"image\/jpeg","source_url":"https:\/\/cracku.in\/blog\/wp-content\/uploads\/2019\/03\/fig-18-03-2019_13-08-06-218x150.jpg"},"td_265x198":{"file":"fig-18-03-2019_13-08-06-265x198.jpg","width":265,"height":198,"mime-type":"image\/jpeg","source_url":"https:\/\/cracku.in\/blog\/wp-content\/uploads\/2019\/03\/fig-18-03-2019_13-08-06-265x198.jpg"},"td_324x160":{"file":"fig-18-03-2019_13-08-06-324x160.jpg","width":324,"height":160,"mime-type":"image\/jpeg","source_url":"https:\/\/cracku.in\/blog\/wp-content\/uploads\/2019\/03\/fig-18-03-2019_13-08-06-324x160.jpg"},"td_324x235":{"file":"fig-18-03-2019_13-08-06-324x235.jpg","width":324,"height":235,"mime-type":"image\/jpeg","source_url":"https:\/\/cracku.in\/blog\/wp-content\/uploads\/2019\/03\/fig-18-03-2019_13-08-06-324x235.jpg"},"td_324x400":{"file":"fig-18-03-2019_13-08-06-324x400.jpg","width":324,"height":400,"mime-type":"image\/jpeg","source_url":"https:\/\/cracku.in\/blog\/wp-content\/uploads\/2019\/03\/fig-18-03-2019_13-08-06-324x400.jpg"},"td_356x220":{"file":"fig-18-03-2019_13-08-06-356x220.jpg","width":356,"height":220,"mime-type":"image\/jpeg","source_url":"https:\/\/cracku.in\/blog\/wp-content\/uploads\/2019\/03\/fig-18-03-2019_13-08-06-356x220.jpg"},"td_356x364":{"file":"fig-18-03-2019_13-08-06-356x364.jpg","width":356,"height":364,"mime-type":"image\/jpeg","source_url":"https:\/\/cracku.in\/blog\/wp-content\/uploads\/2019\/03\/fig-18-03-2019_13-08-06-356x364.jpg"},"td_533x261":{"file":"fig-18-03-2019_13-08-06-533x261.jpg","width":533,"height":261,"mime-type":"image\/jpeg","source_url":"https:\/\/cracku.in\/blog\/wp-content\/uploads\/2019\/03\/fig-18-03-2019_13-08-06-533x261.jpg"},"td_534x462":{"file":"fig-18-03-2019_13-08-06-534x462.jpg","width":534,"height":462,"mime-type":"image\/jpeg","source_url":"https:\/\/cracku.in\/blog\/wp-content\/uploads\/2019\/03\/fig-18-03-2019_13-08-06-534x462.jpg"},"td_696x0":{"file":"fig-18-03-2019_13-08-06-696x365.jpg","width":696,"height":365,"mime-type":"image\/jpeg","source_url":"https:\/\/cracku.in\/blog\/wp-content\/uploads\/2019\/03\/fig-18-03-2019_13-08-06-696x365.jpg"},"td_696x385":{"file":"fig-18-03-2019_13-08-06-696x385.jpg","width":696,"height":385,"mime-type":"image\/jpeg","source_url":"https:\/\/cracku.in\/blog\/wp-content\/uploads\/2019\/03\/fig-18-03-2019_13-08-06-696x385.jpg"},"td_741x486":{"file":"fig-18-03-2019_13-08-06-741x472.jpg","width":741,"height":472,"mime-type":"image\/jpeg","source_url":"https:\/\/cracku.in\/blog\/wp-content\/uploads\/2019\/03\/fig-18-03-2019_13-08-06-741x472.jpg"},"td_0x420":{"file":"fig-18-03-2019_13-08-06-801x420.jpg","width":801,"height":420,"mime-type":"image\/jpeg","source_url":"https:\/\/cracku.in\/blog\/wp-content\/uploads\/2019\/03\/fig-18-03-2019_13-08-06-801x420.jpg"}},"image_meta":{"aperture":"0","credit":"","camera":"","caption":"","created_timestamp":"0","copyright":"","focal_length":"0","iso":"0","shutter_speed":"0","title":"","orientation":"0"}},"post":null,"source_url":"https:\/\/cracku.in\/blog\/wp-content\/uploads\/2019\/03\/fig-18-03-2019_13-08-06.jpg"},"yoast_head":"\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\t\n\t\n\n\n\n