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InterviewSolution
This section includes InterviewSolutions, each offering curated multiple-choice questions to sharpen your knowledge and support exam preparation. Choose a topic below to get started.
| 1. |
If the wole surface area of a sphere be 5544 sq-cm, then the volume of the sphere will beA. 38808 ccB. 42304 ccC. 22176 ccD. 33951 cc |
| Answer» Correct Answer - A | |
| 2. |
`127(2)/(7)` sq. cm of sheet is required to make a hemispherical bowl. Calculate the length of diameter of the forepart of the bowl. |
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Answer» Let the radius of the forepart of the bowl be r cm . `therefore` the quantity of sheet required to make the bowl `=2pir^(2)` sq.cm. As per question `2pir^(2)=172(2)/(7)rArr 2xx(22)/(7)xxr^(2)=(891)/(7)rArr r^(2)=(891xx7)/(7xx2xx22)rArrr^(2)=(81)/(4) rArr r=(9)/(2).` So, the diameter of ot the bowl =`2xx(9)/(2)`cm=9 cm. |
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| 3. |
If the cost of making a leather ball is Rs.431.20 at the rate of Rs.17.50 per square cm. Calculate the length of diameter of the ball. |
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Answer» Let the radius of the ball =r cm. `therefore` the whole surface area of the ball `=4pir^(2)` sq-cm. Rs. 17.50 is required to make a ball of leather of 1 sq. cm . `therefore ` Rs. 431.20 is requied to make a ball of leather of `(431.20)/(17.50) sq. cm.` `therefore 4 pir^(2) =(431.20)/(17.50)` or ` 4 xx(22)/(7)xxr^(2) =(4312)/(175)` or, `r^(2)=(4312)/(175xx4xx22) =(49)/(25)` or , `r=sqrt((49)/(25))=7/5=1.4` `therefore ` the diameter of the ball `=2xx 1.4 cm =2.8 cm` |
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| 4. |
The length of radius of a spherical gas ballon increases from 7 cm to 21 cm as air is being pumped into it, then find the ratio of surface areas of the ballon in two cases . |
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Answer» The previous radius of the balloon `=4xx(22)/(7)xx7^(2)` sq.cm `therefore` the ratio of the whole surface areas of the balloon in two cases `=4xx(22)/(7)xx7^(2):4xx(22)/(7)xx21^(2)=(7xx7)/(21xx21)=(1)/(9)=1:9`. Hence the required ratio=1:9. |
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| 5. |
A large new sphere is made by melting two metal sphere of radii `r_(1) "unit and" r_(2)` unit respectively. Find the radius of the large sphere. |
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Answer» The total volume of the sphere `=(4)/(3)pi(r_(1)^(3)+r_(2)^(3))` cubic,unit. Now, let the radius of the large be r unit, `therefore` the volume of the large sphere `=(4)/(3)pir^(3)` cu,unit. As per condition `(4)/(3)pir^(3)=(4)/(3)pi(r_(1)^(3)+r_(2)^(3)) rArr r=root(3)(r_1^3+r_2^3).` Hence the required of the large sphere `=root(3)(r_1^3+r_2^3).` unit. |
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| 6. |
If the length of diameter of a soild sphere is 28 cm and it is completely immersed into the water then calculate the volume of water displaced by the sphere |
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Answer» The diameter of the sphere =28 cm `therefore` radius of the sphere `=(28)/(2)`cm=14cm `therefore` the volume of the sphere `=(4)/(3)xx(22)/(7)xx(14)^(3)cc=(34496)/(3)cc=11498(2)/(3)`cc. Hence the required volume of water =`11498(2)/(3)`cc. |
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| 7. |
A circular iron-plate of thickness `(2)/(3)` cm is made by beating an iron-sphere of diameter 4 cm. What will be the radius of the iron-plate? |
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Answer» The radius of the iron-sphere `-(4)/(2)`cm=2cm. `therefore` the volume of the iron-sphere `=(4)/(3)pixx2^(3)cc.=(32pi)/(3)cc.` Let the radius of the iron-plate be r cm. `therefore` the area of the iron-plate `=pir^(2)` sq-cm. Since the iron-plate is of thickness `(2)/(3)` cm, its volume =`(2)/(3)pir^(2)` cu.cm. As per question `(2)/(3)pir^(2)=(32pi)/(3)rArr r^(2)=16rArr r=sqrt(16)=4` Hence the radius of the circular iron-plate=4 cm. |
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| 8. |
on the curved surface of the axis of a globe with the length of 14 cm radius, two circular holes area made each of which has the length of radius 0.7 cm.Calculate the area of metal sheet surrounding its curved surface. |
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Answer» The whole surface area of the globe =`4pixx(14)^(2)` sq-cm. The areas of the holes `=2pixx(0.7)^(2)` sq-cm. So the required surface area of the two circular holes =`{4pixx(14)^(2)-2pixx(0.7)^(2)} "sq-cm."=2pi[2xx(14)^(2)-(0.7)^(2)]`sq-cm `=2xx(22)/(7)xx(392-0.49) "sq.cm="2xx(22)/(7)xx391.51 "sq.cm"=2660.92`sq.cm Hence the required surface area of the cicular metalic sheet =2660.92 sq.cm. |
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| 9. |
If by melting a solid hemisphere be made into a sphere, then what will be the ratio of their radii? |
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Answer» Let the radii of the hemisphere and the sphere be `r_(1)` unit and r_(2) uint respectively. ltbgt So, the volume of the hemisphere =`(2)/(3)pi_(1)^(3)` cubic.unit and the volume of the original sphere`(4)/(3)pir_(2)^(3)` cu.unit. As per question, `(2)/(3)pir_(1)^(3)=(4)/(3)pir_(2)^(3) "or," r_(1)^(3)/r_(1)^(3)=(4)/(2)=2 "or," (r_(1)/r_(2))^(3)=2 or, r_(1)/(r_(2))=root(3)(2)` `thereforer_(1):r_(2)=root(3)(2):1` Hence the required ratio =`root(3)(2):1` |
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| 10. |
A sphere of greates volume is cut off from a metal hemisphere of radius 6cm. If the cost of 1 cc metal be ₹42, then what will be the cost of the remaining hemisphere? |
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Answer» The radius of the metal hemisphere =6 cm. So, the volume of the hemisphere `=(2)/(3)xx6^(3)`cc. Again, the diameter of the greates sphere that can be cut off from the hemisphere will be 6 cm . So, its radius will be `(6)/(2)`cm =3 cm. `:.` the volume of the greatest sphere cut off frome the hemisphere `=(4)/(3)pi3^(3)`cc. `:.` The volume of the remaining part of the hemisphere `=((2)/(3)pixx6^(3)-(4)/(3)pi3^(3)) "cc" =(2)/(3)pi(216-54)"cc"=(2)/(3)pixx162"cc"=108` So, the cost of the remaining part of the metal hemishere =₹`108xx(22)/(7)xx42=₹14256.` Hence the required cost=₹14256. |
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| 11. |
If the diameter of a sphere be 6 cm, then the volume of the sphere will beA. 18 `pi` ccB. 27 `pi` ccC. 36 `pi` ccD. 45 `pi` cc |
| Answer» Correct Answer - C | |
| 12. |
The curved surface of a solid metalic sphere is cut in such a way that the curved surface area of the new sphere new sphere is half of that previous one. Calculate the ratio of the volumes of the portion cut off and remaining portion of the shere. |
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Answer» Let the radius of the metalic sphere be R unit and that of the new sphere produced be r unit. As per question, `4pir^(2)=(1)/(2)xx4piR^(2) rArrR^(2)=2r^(2)rArrR=sqrt(2r).` `:.` the volume of the new sphere `=(4)/(3)pir^(3)` cubic.units. Also, the volume of the portion cut off `=((4)/(3)piRr^(3)=(4)/(3)pir^(2))` =`(4)/(3)pi(R^(3)-r^(3))` cubic.units. `=(4)/(3)xx(22)/(7){(sqrt(2)r)^(3)-r^(3)}` cubic.units `=(4)/(3)xx(22)/(7)(2sqrt(2)r^(3)-r^(3))` cubic.units `=(4)/(3)xx(22)(2sqrt(2)-1)r^(3)` cubic.units `:.` ratio of the volumes of the cut off the large sphere and the volume of the remaining part. `=(4)/(3)xx(22)/(7)xx(2sqrt(2)-1)r^(3):(4)/(3)pir^(3)=(2sqrt(2)-1):1` Hence the required ratio=`=(2sqrt(2)-1):1` |
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| 13. |
What will be the ratio of diameter of a hemisphere and the length of its circumference? |
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Answer» Let the radius of the hemisphere be r unit `therefore` its diameter =2r Also the circumference of the hemisphere=`(pir+2r)` unit. As per question , 2r:`(pir+2r)=2r:r(pi+2)=2:(pi+2)` Hence the ratio=2:`(pi+2)`. |
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| 14. |
If radius of a sphere be increased by 3 cm, the volume of the sphere is increased by 264 cc. Find the radius of the sphere. |
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Answer» Let the radius of the sphere be r cm. `:.` the volume of th sphere `=(4)/(3)pir^(3)` cc. If the radiuys of the sphere be incereased by 3 cm, then its volume will be `(4)/(3) pi(r+3)^(3)` cc. As per question `(4)/(3)pi (r+3)^(3)- (4)/(3)pir^(3)=264`. or, `(4)/(3)pi[(r+3)^(3)-r^(3)]=264 "or" (4)/(3)xx(22)/(7)(r^(3)+9r^(2)+27r+27-r^(3))=264` or, `(4)/(3)xx(22)/(7)(9r^(2)+27r+27)=264 "or," 9r^(2)+27r+27=(264xx3xx7)/(4xx22)` or,`9(r^(2)+3r+3)=63 "or,"r^(2)+3r+3=7 "or" r^(2)+3r+3-7=0 "or" r^(2)+3r-4=0` or, `r^(2)`+4r-r-4=0 or, r(r+4)-1(r+4)=0 or, (r+4) (r-1)=0 `:.` either r+4=0, or, r-1 =0 `rArr` r=-4 or, r=1 But the value of r cannot be negative, `:.` r=1. Hence the requied radius of the sphere was 1 cm. |
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| 15. |
The diameter of a sphere is double of the diameter of another sphere. Then how much times will be the volume of smaller sphere than the volume of the greater sphere? |
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Answer» Let the radius of the smaller sphere be r unit. Then the radius of the greater sphere will be 2r unit, since the diameter of the greater sphere is double of the smaller one. Now, volume of the smaller sphere, `V=(4)/(3)pir^(3)` cu.unit and the volume of the greater sphere `=(4)/(3)pixx(2r)^(3)"cu.unit"=(4)/(3)pixx8r^(3) "cu.unit"8xx(4)/(3)pir^(3)` cu.unit. i.e, 8 times of the smaller sphere. Hence the volume of the greater sphere is 8 times of the volume of the smaller sphere. |
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| 16. |
If the length of radius of a sphere is increased by 50%, how much percent will be increased of its curved surface area? |
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Answer» Let the radius of the sphere be r unit. `therefore` the of the curved surface of the sphere=`4pir^(2)` sq-unit. If the radius of the sphere be increased by 50% then the new radius will be `(r+rxx(50)/(100)) "unit"=(3r)/(2)` unit. Then the curved surface area of the sphere will be `4pi((3r)/(2))^(2)"sq.units"=4pi.(9r^(2))/(4)` sq.units `=9pir^(2)` sq.units. `therefore` increase of corved surface area `=(9pir^(2)-4pir^(2)) "sq.unit"=5pir^(2)` sq.unit `therefore` the percent of increment of the curved surface area of the sphere =`(5pir^(2))/(4pir^(2))xx100%=125%` Hence the required percent=125% |
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| 17. |
If the radius if a sphere is decreased by `(1)/(2)` times, then the curved surface area of the sphere will be changed byA. `(1)/(2)` timesB. `(1)/(4)` timesC. `(1)/(8)` timesD. `(1)/(16)` times |
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Answer» If the radius of a sphere be r units, the curved surface area of it will be =S= `4pir^(2)` sq-units. If the radius is decreased by `(1)/(2)` times, then the curved surface area `=4pixx((r)/(2))^(2) "sq-units" =pir^(2) "sq-unit"=(1)/(4)xx4pir^(2) "sq-unit"=(1)/(4)xx "S sq-unte" =(S)/(4) "sq-units"` `therefore` the new curved surface area of the sphere will be `(1)/(4)` times of the previous curved surface area. |
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| 18. |
If the voume of a sphere is incereased by 8 times, then its radius will be increased byA. 2 timesB. 4 timesC. 6 timesD. 8 times |
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Answer» Let the radius of the sphere be r units and its volume be V cubic-units. `therefore V=(4)/(3)pir^(3)` (by formula) Now,if the volume of the sphere be increased by 8 times, i.e., its volume be 8V cubic-units, then let its radius be `r_(1)` units. `therefore 8V =(4)/(3)pir_(1)^(3) rArr 8xx(4)/(3)pir^(3) =(4)/(3)pir_(1)^(3) rArr r_(1)^(3)=8r^(3)` `rArr r_(1)^(3)=(2r)^(3) rArr r_(1)=2r,` i.e., the radius will be increased by 2 times. Hence (a) is correct. |
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| 19. |
The length of diameter of base of a hemispherical tomb is 42 dcm. Calculate the cost of colouring the upper surface of the tomb at the rate of ₹ 35 per square metre. |
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Answer» The diameter of base of the hemispherical tomb = 42 dcm `therefore ` its radius ` =42/2 ` dcm =21 dcm ` therefore ` the surface area of the upper surface of the tomb ` =2 xx 22/7 xx 21^2` sq. dcm = 2772 sq. dcm= 27. 72 sq. metres So the cost of coluring the tomb `= ₹ 27.72 xx 35 = ₹ 970.2 ` Hence the required cost = ₹ 970.2 |
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| 20. |
The length of diameter of a solid sphere of lead si 14 cm .if the sphere is melted, then calculate how many sphers with length of 3.5 cm radius can be made ? |
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Answer» The radius of the large solid sphere`=(14)/(2)` cm=7 cm ` therefore` volume of it `(4)/(3)pi xx7^(3)c c` The radius of each samller sphere `=.5 m` `:.` volume of each of it `=(4)/(3) pi xx (3.5)^(3)c c` Let the number of smaller spheres which can be made is x. `:.` the volume of x smaller sphere `x xx(4)/(3) pi xx (3.5)^(3) c c ` As per condition, `x xx (4)/(3) pi xx (3.5) =(4)/(3) pi xx 7^(3) rArr x =( 7 xx 7 xx 7)/( 3.5 xx 3.5xx 3.5) rArr x=8` Hence the required number of smaller sphere 8 |
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| 21. |
The external deameter of a hollow gold-sphere is 12 cm and it is made of a plate of thickness 1cm. If the mass of 1cc gold be 19.5 gm and the price of 1 gm gold be ₹ 1280, then what is the price of the whole solid sphere? |
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Answer» The external deameter of a hollow gold-sphere is 12 cm and the thickness of it is 1 cm. `:.` the external radius of the gold-radius is `(12)/(2) `cm =6 cm and its internal radius =(6-1) cm =5 cm. So the volume of the sphere `=((4)/(3)pixx6^(3)-(4)/(3)pixx5^(3))cc =(4)/(3)pi(216-125) cc` `=(4)/(3)xx(22)/(7)xx91 "cc" =(88xx13)/(3) ` cc. The mass of 1 cc gold =19.5 gm. `:.` the mass of `(88xx13)/(3) "cc gold" =(88xx13)/(3)xx19.5` gm `:.` the price of the gold-sphere `=₹(88xx13)/(3)xx19.5xx1280`=₹9518080. Hence the required cost of the hollow gold-sphere =₹9618080. |
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| 22. |
The ratio of the volumes of two spherers is 216 : 125. If the sum of the radii of the two sphere be 22 cm , then find the radii pf the spheres. |
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Answer» Let the radii of the two spheres be `r_(1)`cm and `r_(2)` cm respectively. As per question ,`(4)/(3)pir_(1)^(3):(4)/(3)pir_(1)^(3)=216:125rArr(r_(1)/(r_(2))^(3))=(216)/(125)=((6)/(5))^(3)` `rArr r_(1)/r_(2)=(6)/(5)rArr r_(1)=(6r_(2))/(5)………..`(1) Again, by question, `r_(1)+r_(2)=22` `rArr (6r_(2))/(5)+r_(2)=22 [by (1)]rArr 11 r_(2)=22xx5 rArr r_(2)=(22xx5)/(11)=10` `:.` from (1) we get, `r_(1)=(6xx10)/(5)`=12. Hence the radii of the two sphere were 10 cm and 12 cm respectively. |
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| 23. |
If the external and internal radii of a hollow sphere be 6 cm and 3 cm respectively, then the volume (i.e, the material) of the sphere will beA. 629 cu-unitsB. 792 cu-unitsC. 829 cu-unitsD. 929 cu-units |
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Answer» The required volume of the sphere `=((4)/(3)pixx6^(3)-(4)/(3)pixx3^(3))` cu. units `(4)/(3)xx(22)/(7)(216-27)` cu.units =`(4)/(3)xx(22)/(7)xx189` cubic.units= 792 cubi.units `therefore` (b) is correct |
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| 24. |
If curved surface area of a solid sphere is S and volume is V, then find the value of `(S^(3))/V^(2)` [not putting the value of `pi`]. |
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Answer» Let the radius of the solid sphere be r unit. Then the curved surface area of the sphere=`4pir^(2)` sq.units. As per question, S =`4our^(2)……….` (1) Also, the volume of the sphere`=(4)/(3)pir^(3)` cu.units. As per question `V=(4)/(3)pir^(3) ………….`(2) `therefore (S^(3))/(V^(3))=(4pir^(2))^(3)/((4/3pir^(3))^(2))` [Dividing cube of (1) by square of (2)]=`(64pir^(3)r^(3))/(16/3pir^(2)r^6)=(64xx9)/(16)pi=36pi` Hence the required value of `S^(3)/(V^(2))=36pi.` |
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| 25. |
The curved surface area of a solid sphere is equal to the surface area of a solid right circular cylinder. The lengths of both height and diameter of cylinder are 12 cm . Find the length of radius of the sphere. |
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Answer» Let the radius of the sphere be r cm. `therefore` curved surface area of the solid sphere =`4pir^(2)` sq-cm The radius of the right circular cylinder`=(12)/(2)cm=6cm.` and heght of it =12 cm. `therefore` surface area of the right circular cylinder =`2pixx6xx12` sq-cm. As per question, `4pir^(2)=2pixx6xx12 or, r^(2)`=36 or , r=6 Hence the required radius of the sphere=6cm. |
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| 26. |
The ratio of the surface area of a sphere and whole surface area of a hollow hemisphere of same measurement will beA. `1:1`B. `2:1`C. `3:1`D. `4:3` |
| Answer» Correct Answer - B | |
| 27. |
If the ratio of curved surface areas of two hemisphere is 4:9, then the ratio of their lenghts of radii is 2:3 |
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Answer» Let the radii of the rwo spheres be `r_(1) "and" r_(2)` units.As per question, `2pir_(1)^(2):2pir_(2)^(2)=4:9 rArr (2pir_(1)^(2))/(2pir_(1)^(2))=(4)/(9)rArr(r_(1)/(r_(2)))^(2)=((2)/(3))^(2)rArrr_(1)/r_(2)=(2)/(3)rArrr_(1):r_(2)=2:3` Hence the given atatement is true. |
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| 28. |
Whole surface area of a solid hemisphere is equal to the curved surface area of a solid sphere. Find the ratio of lenghts of radius of hemisphere and sphere. |
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Answer» Let the radius of the solid hemisphere be `r_(1)` unit and radius of the solid sphere be `r_(2)` unit. `therefore` the whole surface area of the solid hemisphere `=3pir_(1)^(2)` sq-unit and the curved surface area of the solid sphere=`4pir_(2)^(2)` sq-unit As per question, `3pir_(1)^(3)=4pir_(2)^(2) rArrr_(1)^(2)/(r_(2)^(2))=(4)/(3)rArr((r)/(r_(2)))^(2)=((2)/(sqrt3))=2:sqrt(3).` Hence the required ratio=`2:sqrt(3).` |
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| 29. |
Two hollow sphere with the lenghts of diameter 21 cm and 17.5 cm respectively are made from the sheets of the same metal, Calculate the ratio of the curved surface areas of the two hollow sphere. |
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Answer» The lenghts of diameters of the two hollow spheres are 21 cm and 17.5 cm. `therefore` their radii are `(21)/(2) "cm and" (17.5)/(2)` cm So, the ratio of the curved surface areas of the two hollow sphere `=4pixx((21)/(2)):4pixx((17.5)/(2))^(2)= (21^(2))/(4):(17.5)^(2)/(4)=21xx21:17.5xx17.5=(21xx21xx100)/(175xx175)=(36)/(25)=36:25` |
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| 30. |
Calculate how many marbles with lenghts of 1 cm radius may be formed by melting a solid sphere of iron having 8 cm length of radius. |
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Answer» The radius of solid iron-sphere =8cm `:.` the volume of the iron-sphere `=(4)/(3)xxpixx8^(3)`cc The radius of each marble =1 cm Math (X) -34 `:.` volume of each marble `=(4)/(3)pixx1^(3)`cc Let x marbles can be made. `:.` as per condition `x xx(4)/(3)pixx1^(3)=(4)/(3)pi8^(3) rArr x=(8^(3))/(1^(3))=512` Hence 512 mardition can be made. |
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| 31. |
If the numerical values of the volume and curved surface area of a sphere ve equal, then the radius of the sphere will beA. 1 unitB. 2 unitC. 3 unitD. 3 units |
| Answer» Correct Answer - C | |
| 32. |
If the curved surface area of a globe be `(22)/(7)` sq-cm, then the radius of the sphere will beA. 0.25 metreB. 0.5 metreC. 1 metreD. 2 metre |
| Answer» Correct Answer - B | |
| 33. |
The whole surface area of a hollow hemisphere will beA. 308 sq-cmB. 462 sq-cmC. 506 sq-cmD. 612 sq-cm |
| Answer» Correct Answer - A | |
| 34. |
If the ratio of volumes of two solid spheres is 1:8, the ratio of their curved surface areas isA. `1:2`B. `1:4`C. `1:8`D. `1:16` |
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Answer» Let the radii of the two sphere be `r_(1) "and" r_(2)` units. So the volumes of two spheres are `(4)/(3)pir_(1)^(3)` cubic-units and `(4)/(3)pir_(1)^(3)` cubic-units. As per questions , `(4)/(3)pir_(1)^(3):(4)/(3)pir_(1)^(3) =1:8 rArr (4/3pir_(1)^(3))/(4/3pir_(2)^(3))=(1)/(8) or,(r_(1)^(3))/(r_(2)^(3)) =(1)/(8)or,(r_(1)/(r_(2)))^(3)=((1)/(2))^(3) or, r^(1)/r^(2)=(1)/(2)` So, the ratio of their curved surface areas `=4pir_(1)^(2):4pir_(1)^(2)=(4pir_(1)^(2))/(4pir_(2)^(2))=(r_(1)/(r_(2)))^(2)=((1)/(2))^(2)=(1)/(4)=1:4` Hence (b) is correct. |
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| 35. |
The external radius of a hollow hemisphere is 6 cm and it is of thickness 2 cm.What will be the whole surface area of the hemisphere? |
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Answer» The external radius of the hemisphere is 6 cm and the thickness of it is 2cm. `therefore` internal radius of the hemisphere =(6-2) cm =4 cm. `therefore` the whole surface area of the hemisphere lt brgt =`(2pixx6^(2)+2pixx4^(2)+pixx6^(2)-pixx4^(2))` sq.cm `= pi(2xx36+2xx16+36-16) ` sq.cm `=(22)/(7)xx(72+32+20) "sq.cm" =(22)/(7)xx124 "sq.cm"=(1728)/(7) "sq.cm"=389(5)/(7)`sq.cm. Hence the whole surface area of the hemisphere `=389(5)/(7)` sq,cm. |
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| 36. |
If numerical value of curved surface area of a solid sphere is three times of its volume, the length of its radius isA. 1 unitB. 2 unitC. 3 unitD. 4 unit |
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Answer» Let the radius of the sphere = r units `therefore` curved surface area of the sphere `=4pir^(2)` sq-units `therefore` the volume of the sphere `=(4)/(3)pir^(3)` cubic-units As per questions, `4pir^(2)=3xx(4)/(3)pir^(3)` or, `4pir^(2)=4pir^(3) rArr1=r rArrr=1` `therefore` radius =1unit `therefore` radius =1 unit `therefore` (a) is correct. |
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| 37. |
If the volume of a sphere be V cubic-units, then the radius of the sphere will beA. `((3V)/(4pi))^(3)` unitB. `(3V)/(4pi)` unitC. `sqrt((3V)/(4pi))` unitsD. `root(3)((3V)/(4pi)` units |
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Answer» Let the radius of the sphere be r units `therefore` the volume of the sphere `=(4)/(3)pir^(3)` cubic-units As per question ,`(4)/(3)pir^(3)=V "or ," r^(3)=(3V)/(4pi) "or" , r=root(3)((3V)/(4pi)` `therefore` (d) is correct. |
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| 38. |
If the whole surface area of a sphere be `pir^(2)` sq-uints, then the volume of the sphere will beA. `pix^(3)` cubic-unitsB. `(pix^(3))/(6)` cubic-unitsC. `(pix^(3))/(2)` cubic-unitsD. `(4)/(3)pix^(3)` cubic-units |
| Answer» Correct Answer - B | |
| 39. |
If the radius of a solid sphere be decreased by 1 cm, then the curved surface area of it is decreased by 88 sq-cm. What was the radius of the square? |
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Answer» Let the radius of the solid sphere be r cm . `:.` the curved surface area of the sphere `=4pir^(2)` sq-cm. Now, if the radius of the sphere is decreased by 1 cm, then the curved surface of it will be `4pi(r-1)^(2)` sq-cm. As per question , `4pir^(2)-4pi(r-1)^(2)`=88 or, `4pi[r^(2)-(r-1)^(2)]`=88 or, `4xx(22)/(7)[(r+r-r)]=88 "or" 4xx(22)/(7)[(2r-1)]=88` or, `2r-1=(88xx7)/(4xx22)` or,2r-1=7 or, 2r=8 or, r=4. Hence the radius of the solid sphere was 4 cm. |
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| 40. |
If the length of radius of a solid sphere be doubled, the volume of sphere will be doubled. |
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Answer» Let the radius of the sphere be r units, then the volume of the sphere will be `(4)/(3)pir^(3)` cu-units. Now, if the radius be doubed, the volume of the sphere will be `(4)/(3)pixx(2r)^(3)` cu-units =`8xx(4)/(3)pir^(3)` cubic-units. ltbrlt I,e, the volume of the sphere will be 8 times of its previous volume. Hence the given statement is false. |
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| 41. |
If the volume of a solid hemisphere be 19404 cc, then the radius of the hemisphere will beA. 7 cmB. 14 cmC. 21 cmD. 28 cm |
| Answer» Correct Answer - C | |
| 42. |
The external and interenal diameter of a hollow copper-sphere are 20 cm and 16 cm respectively . By melting this sphere, how many solid bullets of diameter 4 cm each can be made? |
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Answer» The external radius of the hollow sphere `=(20)/(2)`cm=10. and the internal radius of the hollow-sphere `=(16)/(2) `cm =8 cm. `therefore` the volume of materials of the hollow sphere `=[(4)/(3)pixx10^(3)-(4)/(3)pixx8^(3)]cc.=(4)/(3)pi(1000-512)cc=(4)/(3)pixx488cc` The radius solid bullets `=(4)/(2)` cm =2 cm `:.` volume of solid bullets `=(4)/(3)pixx2^(3)` cc. Let the number of solid bullets that can be made be x. `:. x xx (4)/(3)pixx2^(3)=(4)/(3)pixx(488)/(8) = 61.` Hence 61 solid bullets can be made. |
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| 43. |
If the ratio of the curved surface area of two spheres b 4:9, then the ratio of the volumes of the spheres will beA. `2:3`B. `4:9`C. `8:27`D. `16:81` |
| Answer» Correct Answer - C | |
| 44. |
If the volume of a sphere be 8 times the volume of another sphere, then the ratio of the curved surface areas of the two spheres will beA. `2:3`B. `1:8`C. 3:16`D. `1:4` |
| Answer» Correct Answer - D | |
| 45. |
If the ratio of curved surface areas of two solid spheres is 16:9, the ratio of their volumes isA. `64:27`B. `4:3`C. `27:64`D. `3:4` |
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Answer» Let the radii of two solid spheres be `r_(1)` units and `r_(2)` units respectively. `therefore` the curved surface areas of the spheres are `4pi_(1)^(2)" ""sq-units and"" "4pi_(2)^(2) `sq-units respectively As per question, `4pir_(1)^(2):4pir_(2)^(2)=16:9rArr(4pir_(1)^(2))/(4pir_(1)^(2))=(16)/(9)rArr(r_(1)/(r))^(2)=((4)/3)^(2)rArrr_(1)/r_(2)=(4)/(3)` `therefore` The ratio of their volumes =`=(4)/(3)pir_(1)^(3):=(4)/(3)pir_(2)^(3)=(4/3pir_(1)^(3))/(4/3pir_(2)^(3))=(r_(1)/(r_(2)))^(3)=((4)/(3))^(3)=(64)/(27)=64:27` `therefore` is correct. |
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| 46. |
The name of solid which is composed of only one surface is __________. |
| Answer» Answer: sphere | |
| 47. |
The whole surface area of a solid hemisphere with length of 7 cm radius isA. 588`pi` sq-cmB. 392 `pi` sq-cmC. 147`pi` sq-cmD. 98 `pi` sq-cm |
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Answer» The radius of the solid hemisphere =7. `therefore` the whole surface area of it =`3pixx(7)^(2) sq-cm =3pixx49=sq-cm=147pi sq-cm.` `therefore (c) is correct. |
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| 48. |
The radius of a copper-sphere is twice the radius of an iron-sphere. The numerical value of the whole surface area of the copper-sphere is equal to the numerical value of the volume of the iron-sphere. Then the radius of the copper-sphere will beA. 6 unitB. 12 unitC. 18 unitD. 24 unit |
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Answer» Let the radius of the iron-sphere be r units. `therefore` the radius of the copper-sphere is 2r units. `therefore` the whole surface area of the copper-sphere `4pixx(2r)^(2) "sq-units" =16 pir^(2)` sq-units Also, the volume of the iron sphere`=(4)/(3)pir^(3)` cubic-units As per question, `(4)/(3)pir^(3) =16pir^(2) rArr r =12` `therefore` the radius of the copper-sphere=2xx12 units 24 units. `therefore` (d) is correct. |
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| 49. |
Two solid sphere of radii 3 cm and 4 cm are melted to make a hollow sphere of external radius of 6 cm, then wahat will be the thickness of the new hollow sphere? |
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Answer» The radii of the twoo solid spheres are 3cm and 4 cm. So, the total volume of the two spheres `=((4)/(3)pixx3^(3)+(4)/(3)pixx4^(3)) "cc" =(4)/(3)pi(27+64) "cc"=(4)/(3)pixx91` cc. The external radius of the hollow sphere =6 cm Let the internal radius of it =r cm. `:.` the volume of material of the hollow sphere` =(4)/(3)pi(6^(3)-r^(3)) "cc" (4)/(3)pi(216-r^(3))` cc `:. (4)/(3)pi(216-r^(3))=(4)/(3)pixx91 "or," 216-r^(3)=91 "or", =r^(3)=125 "or" r^(3)=5^(3)rArrr=5`. `:.` thickness of the hollow sphere =(6-5)cm =1cm Hence the thickness of the hollow sphere =1 cm . |
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| 50. |
Three spheres made of copper having of 3 cm, 4cm and 5cm radii are melted and a large sphere is made.Calculate the lenghth of the large sphere. |
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Answer» The radii of three smaller spheres are 3 cm, 4cm and 5cm. `:.` the total volume of the three smaller spheres `=(4)/(3)pi(3^(3)+4^(3)+5^(3))cc.=(4)/(3)pi(27+64+125)cc.=(4)/(3)xx(22)/(7)xx216`cc. Let the radius of the large sphere be R cm `:.` volume of the large sphere `=(4)/(3)xx(22)/(7)xxR^(3)`cc. As per question,`(4)/(3)xx(22)/(7)xxR^(3)= 4/3 xx 22/7 xx 216 rArr R^3= 216 rArr R^3 = 6^3 rArr R= 6 ` Hence the radius of the large sphere = 6 cm |
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