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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. |
A cone of radius 8 cm and height 12 cm is divided into two parts by a plane through the mid-point of its axis parallel to its base Find the ratio of the volumes of two parts |
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Answer» `AD=12/2=6 cm =h/2` `tantheta=r/h=(DE)/(h/2)` `DE=r/2` Volume of cone AEF=`1/3*pi(r/2)^2(h/2)` `=1/3pir^2h/8` Volume of cone=`1/3*pir^2h` Volume of `2^(nd)` part=`1/3pir^2h-1/3pir^2h/8` `=7/8pir^2h/3` `V_1/V_2=(1/3pir^2h/8)/(7/8pir^2h/3)=1/7`. |
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| 2. |
If volume of two spheres are in the ratio 64: 27, then the ration of their surface area isA. `3:4`B. `4:3`C. `9:16`D. `16:9` |
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Answer» Correct Answer - D Let the radii of two spheres are `r_(1) ` and `r_(2)` respectively . `therefore ` Volume of the sphere of radius , `r_(1) = V_(1) = (4)/(3) pir_(1)^(3)" "……..(i) " "[therefore "Volume of sphere "= (4)/(3) pi ("radius")^(3)]` and volume of the sphere of radius , `r_(2) = v_(1) = (4)/( 3) pi r_(2)^(3)" "....(ii)` Given, ratio of volume `= V_(1): V_(2)= (64)/(27) rArr ((3)/(4) pir_(1)^(3))/((4)/(3)pir_(2)^(3))" "["Using Eqs, (i) and (ii)"]` `rArr " "(r_(1)^(3))/(r_(2)^(3)) =(64)/(27)rArr (r_(1))/(r_(2)) = (4)/(3)" "......(i)` Now, ratio of surface area `= (4pir_(1)^(2))/(4pir_(2))" "[therefore "Surface area of a sphere"= 4 pi ("radius")^(2)]` `" "(r_(1)^(2))/(r_(2)^(2))` `" "= ((r_(1))/(r_(2)))^(2) = ((4)/(3))^(2) " "["Using Eq. (iii)"]` Hence, the required ratio of their surface area is `16:9` |
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| 3. |
If volumes of two spheres are in the ratio 64:27 then the ratio of their surface areas is |
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Answer» `V=4/3pir^3` `V_1/V_2=(4/3pir_1^3)/(4/3pir_2^3)=64/27` `r_1^3/r_2^3=4^3/3^3` `r_1/r_2=4/3` `SA=4pir^2` `(SA_1)/(SA_2)=(4pir_1^2)/(4pir_2^2)=(r_1/r_2)^2` `(4/3)^2=16/9` option d is correct |
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| 4. |
The radius of a spherical balloon increases from 7 cm to 14 cm as air is being pumped into it. Find the ratio of surface areas of the balloon in the two cases. |
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Answer» case 1`r= 7cm` TSA=`4 pi r^2` `=4 *22/7*7*7= 616cm^2` case 2`r=14cm` TSA=`4*22/7*14*14= 2464cm^2` ratio of surface areas= `(TSA^2)/(TSA_1)` `=2464/616` `=4:1` |
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| 5. |
A hemispherical bowl made of brass has inner diameter 10.5 cm. Find the cost of tin-plating it on the inside at the rate of Rs 16 per`100 cm^2`. |
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Answer» Curved surface area=`2pir^2` =`2*22/7*(10.5/2)^2` =`173.25cm^2` Total cost=curved surface area *cost =173.25*0.16 =27.72 Rs. |
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| 6. |
Rasheed got a playing top (lattu) as his birthday present, which surprisingly had no colour on it. He wanted to colour it with his crayons. The top is shaped like a cone surmounted by a hemisphere. The entire top is 5 cm in height and the diameter of the top is 3.5 cm. Find the area he has to colour. |
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Answer» CSA of hemisphere=`2 pi r^2` =`2*22/7*(3.5/2)^2` =19.2775` cm^2` Surface Area of cone= `pi r l` =`22/7*(3.5/2)*sqrt((3.5/2)^2+(6.5/2)^2)` =20.35`cm^2` Total area=CSA of hemisphere+ SA of Cone =19.2775+20.35 =39.6275 `cm^2`. |
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| 7. |
Savitri had to make a model of a cylindrical kaleidoscope for her science project. She wanted to use chart paper to make the curved surface of the kaleidoscope. (see Fig 13.10). What would be the area of chart paper required by her, if she wanted |
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Answer» Required area can be given as the curved surface area of the cyllinder. We know, curved surface area of a cyllinder, `A = 2pirh` Here, `h = 25cm and r = 3.5cm` So, `A = 2**22/7**3.5**25 = 22**25 = 550cm^2` So, required area will be `550cm^2.` |
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| 8. |
Mary wants to decorate her Christmas tree. She wants to place the tree on a wooden box covered with coloured paper with picture of Santa Claus on it (see Fig. 13.4). She must know the exact quantity of paper to buy for this purpose. If the box has length, breadth and height as 80 cm, 40 cm and 20 cm respectively how many square sheets of paper of side 40 cm would she require? |
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Answer» Area of sheet Required=TSA of cuboid =2(lb+bh+lh) =2(80*40+40*20+80*20) =11200`cm^2` No. of sheets required=(total area)/(area of one sheet) =`(11,200)/1600=7` |
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| 9. |
A solid cylinder of radius r and height h is palced over other cylinder same height and radius . The total surface area of the shape so formed is `4pirh + 4 pir^(2)`. |
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Answer» False Since , the total surface area of cylinder of radius , r and height , `h = 2pirh + 2 pir^(2)` When one cylinder is placed over the other cylinder of same height and radius , then height of the new cylinder= 2h and radius of the new cylinder = r `therefore` Total surface area the new cylinder = `2pir (2h) + 2pir^(2)= 4 pih + 2pir^(2)` |
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| 10. |
During conversion of a solid from one shape to another, the volume the new shape willA. increaseB. decreaseC. remainD. be doubled |
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Answer» Correct Answer - C During conversion of a solid form one shape to another, the volume of the new shape will remain unaltered. |
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| 11. |
To fill swimming pool two pipes are to be used. If the pipe of larger diameter is used for 4 hours and the pipe of smaller diameter for 9 hours, only half the pool can be filled. Find, how long it would take for each pipe to fill the pool separately, if the pipe of smaller diameter takes 10 hours more than the pipe of larger diameter to fill the pool |
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Answer» small`=x+10`,larger `=x` `1/x xx4+1/x+10xx9=1/2` `4/x+9/(x+10)=1/2` `(4(x+10)+9x)/(x(x+10))=1/2` `(4x+40+9x)/(x^2+10)=1/2` `26x+80=x^2+10x` `x^2-16x-80=0` `x^2-20x+4x-80=0` `x(x-20)+4(x-20)=0` `x=20` larger`=xhr=20hr` smaller`=(x+10)hr=30hr` `` `` |
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| 12. |
A fez, the cap used by the Turks, is shaped like the frustum of a cone. If its radius on the open side is 10 cm, radius at the upper base is 4 cm and its slant height is 15 cm, find the area of material used for making it. |
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Answer» Please refer to video for the figure. Area of material used for making fez, `A` = Curved Surface Area of frustum + Area of circular part at the top of fez So, `A = pi**(r_1+r_2)**l +pir_2^2` Here, `l = 15, r_1 = 10 and r_2 = 4` Putting these values, `A = pi(10+4)**15+pi**16=226**22/7 ~= 710.6cm^2` |
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| 13. |
A vessel is in the form of an inverted cone. Its height is 8 cm and the radius of its 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. |
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Answer» Volume of vessel=`1/3pir^2h` =`1/3pi5^2*8` =`200pi/3` Volume of lead shots=`4/3pir^3` `n4/3pi(0.5)^3=1/4pi/3(200)` `n=100` total no. of lead shots=100 |
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| 14. |
A heap of wheat is in the form of a cone ofdiameter 9m and height 3.5m. Find its volume. How much canvas cloth isrequired to just cover the heap? `(U s e pi=3. 14)` |
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Answer» Given that, a heap of rice is in the form of a cone. Height of a heap of rice. i.e., cone (h) = 3.5 m and diameter of a heap of rice i.e., cone = 9 cm Radius of a heap of rice i.e., cone (r) `= (9)/(2)m` So, " "volume of rice `= (1)/(3) pi xx r^(2)h` ` = (1)/(3) xx (22)/(7) xx (9)/(2) xx (9)/(2) xx 3.5` `= (6237)/(84) = 74.25 m^(3)` Now, canvas cloth required to just cover heap of rice = Surface area of a heap of rice `= pirl` `= (22)/(7) xx (9)/(2) xx sqrt(((9)/(2))^(2) + (3.5)^(2))` `=(11 xx 9)/(7) xx sqrt((81)/(4) + 12.25)` `= (99)/(7) xxsqrt((130)/(4)) = (99)/(7) xx sqrt(32.5)` `14.142 xx 5.7` `= 80.61 m^(2)` Hence,`80.61 m^(2)` canvas cloth is required to just cover heap. |
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| 15. |
A bus stop is barricaded from the remaining part of the road, by using 50 hollowcones made of recycled cardboard. Each cone has a base diameter of 40 cm and height1 m. If the outer side of each of the cones is to be painted and the cost of pain |
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Answer» 1)`CSA=pirl` =`3.14*20*10^(-2)*sqrt(1^2+(0.2)^2)` =`0.64056m^2` TSA=50*0.64056 =32.28`m^2` `1m^2=12Rs` `32.28m^2=38.736. |
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| 16. |
Find (i) the curved surface area and (ii) the total surface area of a hemisphere of radius 21 cm. |
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Answer» (i) The curved surface area of a hemisphere can be given as `A_C = 2pir^2` Here, `r = 21 cm` So, `A_C = 2**22/7**21**21 = 22**26 = 2772cm^2` (ii) Total Surface Area of a hemisphere, `A_T = `Curved surface area + area of base So, `A_T = A_C+pir^2` `A_T = 2272+22/7**21**21 = 2272+22**63` `A_T = 2272+1386 = 4158 cm^2` |
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| 17. |
A right triangle ABC with sides 5 cm, 12 cm and 13 cm is revolved about the side 12 cm.Find the volume of the solid so obtained. |
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Answer» When we revolve a right angle triangle, we get a conical shape. Please refer to video for the figure of cone. So, here, height of cone, `h = 12cm` Radius of cone, `r = 5cm` Volume of cone, `V = 1/3pir^2h` `V = 1/3**22/7**5**5**12=2200/7 = 314.21cm^3` So, volume of obtained solid will be `314.21cm^3`. |
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| 18. |
Find the curved surface area of a right circular cone whose slant height is 10 cm and base radius is 7 cm |
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Answer» CSA=`pirl` =`22/7*7*10` =`220cm^2`. |
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| 19. |
The volume of a right circular cone is`9856 cm^3`. If the diameter of the base is 28 cm,Find (i) height of the cone (ii) slant height of the cone (iii) curved surface area of the cone |
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Answer» (i)Here, Volume of cone, `V = 9856cm^3` Diameter of cone, `d = 28cm` So, radius of cone, `r = d/2 = 28/2 = 14cm` We know, volume of a cone, `V = 1/3pir^2h` So, `9856 = 1/3**22/7**14**14**h` `h = (9856**3)/(22**28) = (64**3)/4 = 48cm` So, height of cone, `h = 48cm` (ii)Now, Slant height, `l = sqrt(h^2+r^2)` `l = sqrt(48^2+14^2) = sqrt(2^2**24^2+2^2**7^2` `l = 2sqrt(24^2+7^2) = 2sqrt(576+49) = 2**25 = 50cm` (iii) Curved surface area of a cone, `A = pirl` So, `A = 22/7**14**50 = 2200cm^2` |
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| 20. |
Diameter of the base of a cone is 10.5 cm and its slant height is 10 cm. Find its curvedsurface area. |
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Answer» The curved surface area of a cone can be given as, `A_C = pirl` Here, we are given, `r = d/2 =10.5/2cm and l = 10cm` So, `A_C = 22/7**10.5/2**10 = 660/4 = 165cm^2` |
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| 21. |
Find the total surface area of a cone, if its slant height is 21 m and diameter of its baseis 24 m. |
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Answer» TSA=CSA+BA =`pirl+pir^2` =`pir(l+r)` =`22/7*12(21+12)` =1244.7`m^2` |
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| 22. |
A cylinder with height and radius `2:1` is filled with soft drink and then it is tilted so as to allow some soft drink to flow off to an extent where the level of soft drink just touches the lowest point of the upper mouth. If `2.1` L of soft drink is retained in the cylinder, what is the capacity of the cylinder ? |
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Answer» Let radius of cylinder ` = r` `:.` Height of cylinder `= 2r` When cylinder is tilted, then the liquid retained in cylinder ` = (pir^2h)/2 = pir^3 = 2.1` Now, Capacity of cylinder `V = pir^2h = 2pir^3` `:.V = 2*pir^3 = 2**2.1 = 4.2 L` |
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| 23. |
The front compound wall of a house is decorated by wooden spheres of diameter 21 cm, placed on small supports as shown in Fig 13.32. Eight such spheres are used for this purpose, and are to be painted silver. Each support is a cylinder of radius 5cm and height 7cm and is painted black.Find the cost of paint required if silver paint cost 25 paisa per `cm^2` and black paint cost 5 paisa per `cm^2` |
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Answer» Area to be painted silver=8*Surface area of sphere. =`8*4pir^2` =`8*4*22/7*(21/2)^2` =11088`cm^2` Area to be painted black=8*Curved surface area of cylinder =`8*2pirh` =`8*2*22/7*15*7` =`528cm^2` Total cost=Silver are*cost+black area*cost =11088*0.25+528*0.05 =2798.4Rs |
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| 24. |
The diameter of a sphere is decreased by 25%. By what per cent does its curved surface area decrease? |
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Answer» `d_2=d_1-1/4d_1` `d_2=3/4d_1` =`(d_2-d_1)/d_1*100` =`(4pi(r_2^2-r_1^2))/(4pir_1^2)*100` =`(((3d_1)/4)-d_1^2)/d_1^2 *100` =43.75. |
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| 25. |
The volume of the frustum of a cone is `(1)/(3) pih [r_(1)^(2) +r_(2)^(2)- r_(1)r_(2)]`, where h is vertical height of the frustum and `r_(1), r_(2)` are the radii of the ends. |
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Answer» False Since, the volume of the frustum of a come is `(1)/(3) pi h [r_(1)^(2) +r_(2)^(2) + r_(1)r_(2)]` where h is vertical height of the frustum and `r_(1),r_(2)` are the radii of the ends. |
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| 26. |
The volume of a sphere is `905(1)/7 cm^3`. Determine its diameter, and its surface area. |
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Answer» Volume of the sphere`(V) = 905 1/7 cm^3 = 6336/7 cm^3` We know, `V = 4/3pir^3` `:. 4/3pir^3 = 6336/7` `=>r^3 = 6336/7**3/4**7/22 = 216` `=>r = 6cm` So, radius of the sphere is `6` cm. `:.` Diameter of the sphere ` = 2r = 12cm` Now, surface area of the sphere, `(S) = 4pir^2 = 4**22/7**6**6 = 3168/7 cm^2 = 452.57cm^2` |
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| 27. |
The decorative block is made of two solids — a cube and a hemisphere. The base of the block is a cube with edge 5 cm, and the hemispherefixed on the top has a diameter of 4.2 cm. Find the total surface area of the block. |
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Answer» Total Area=`6a^2`+CSA of hemisphere-Base area of hemishere) =`6a^2+4/2pir^2=pir^2` =`6a^2+pir^2` =`6(5)^2+22/7(4.2/2)^2` =`258.6 cm^2` |
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| 28. |
A solid ball is exactly fitted inside the cubical box of side `a`. The volume of the ball is `(4)/(3) pi a^(3)`. |
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Answer» False Because solid ball is exctly fitted inside the cubical box of side a. So, os the diameter for the solid ball. `therefore`" " Radius of the ball `= (a)/(2)` So, " " volume of the ball `= (4)/(3) pi ((a)/(2))^(3) = (1)/(6) pia^(3)` |
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| 29. |
Determine the ratio of the volume of a cube to that of a sphere which will exactly fit inside the cube. |
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Answer» Let side of cube is `a` and radius of sphere is `r`. As sphere exactly fit inside the cube, `:. a = 2r` Ratio of volume of cube to that of sphere,`V_c/V_s = a^3/(4/3pir^3)` `V_c/V_s = (2r)^3/(4/3pir^3)` `=>V_c/V_s = (8r^3)/(4/3pir^3) = 6/pi` So, the required ratio is `6:pi`. |
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| 30. |
Metal sphere, each of radius 2 cm, are packed into a rectangular box of internal dimension `16 cm xx 8 cm xx 8 cm`. When 16 spheres are packed the box is filled with presrvative liquied .Find the volume of this liquid. Give your answer to the nerest interger. [ use `pi = 3.14`]. |
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Answer» Given, dimension of the cuboidal `= 16 cm xx 8 cm xx 8 cm ` ` therefore ` Volume of the cuboid `= 16 xx 8 xx = 1024 cm^(3)` Also, given radius of one glass sphere = 2 cm `therefore ` Volume of one glass sphere `= (4)/(3) pir^(3) = (4)/(3) xx (22)/(7) xx (2)^(3)` `" "= (704)/(21) = 33.523 cm^(3)` Now, volume of 16 glass sphere `= 16 xx 33.523 cm^(3)` `therefore ` Required volume of water = Volume of cuboidal - Volume of 16 galss spheres `" "= 1024- 536.37 = 487.6 cm^(3)` |
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| 31. |
Find the volume of a sphere of radius 11.2 cm |
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Answer» given that`r= 11.2`cm volume of sphere`= 4/3 pi r^3` =4/3*22/7*11.2*11.2*11.2`` `= 5887.32cm^2` |
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| 32. |
At a Ramzan Mela, a stall keeper in one of the food stalls has a large cylindrical vessel of base radius 15 cm filled up to a height of 32 cm with orange juice. The juice is filled in small cylindrical glasses (see Fig. 13.27) of radius 3 cm up to a height of 8 cm, and sold for Rs 3 each. How much money does the stall keeper receive by selling the juice completely? |
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Answer» Number of glass*Volume of juice in each glass=Volume of orange `n*pi*r^2*h=piR^2H` `n(3)^2(8)=15^2*32` `n=100 glasses` Total Money=No. of glasses*cost =100*3 =300Rs |
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| 33. |
A metallic hemisphere is melted and recast in the shape of a cone withthe same base radius R as that of the hemisphere. If `H`is the height of the cone, then write the value of `H//Rdot` |
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Answer» Note the volume will remain the same. volume of hemisphere=`2/3pir^3` volume of cone=`1/3pir^2h` `1/3pir^2h=2/3pir^3` =>`h/r=2/1` |
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| 34. |
A farmer connects a pipe of internal diameter 20 cm from a canal into a cylindrical tank in her field, which is 10 m in diameter and 2 m deep. If water flows through the pipe at the rate of 3 km/h, in how much time will the tank be filled? |
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Answer» Volume rate of water flow=A*3km/h =`pi(10)^2*3*10^3m^3/h` =`3pim^3/h` time *Volume rate=Volume filled in that time `t*3pim^3/h=50pim^3` `t=16hrs` |
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| 35. |
A rectangular water tank of base `11 m xx 6m` contains water upto a height of 5 m. If the water in the tank is transferred to a cylindrical tank of radius 3.5 m, find the height of the water level in the tank. |
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Answer» Given,dimenison of base of rectaangular tank `= 11 m xx 6m` and height of water g= 5 m Volume of the water in rectangular tank `= 11 xx 6 xx 5 = 330 m^(3)` Also given radius of the cylindrical tank = 3.5 m Let height of water level in cylinder tank be h. Then, volume of the water is cylindrical `= pir^(2)h = pi (3.5)^(2) xxh` `" " =(22)/(7) xx 3.5 xx 3.5 xx h` According to the question, `330= 28.5 h " "["since, volume of water is same in both tanks "]` `therefore" "h = (330)/(38.5)= (3300)/(385)` ` therefore " " = 8.75 m or 8.6 m` Hence, the height of water level in cylindrical tank is 8.6 m. |
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| 36. |
The height of a cone is 40 cm. A small cone is cut off at the top by a plane parallel to its base. If the volume of a small cone is `1/64` of the volume of the given cone, at what height above the base is the section made? |
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Answer» `h/H = r/R = l/L` volume of small cone= `1/3 pi r^2 h` volume of bigger cone =`1/3 pi R^2 H` Now, `1/3 pi R^2 H xx 1/64 = 1/3 pi r^2 h` `(r^2H)/64 = r^2 h` `h/40 = r/R` `r= (hR)/40` now, `(R^2H)/64 = (hR/40)^2 xx h` `r^2 xx 40/64 = h^2R^2/(40)^2 xx h` `h^3 = (40)^3/64` `h= 40/4 = 10cm` from the height 30cm from base of cone, the section made Ans`h=10cm` Answer |
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| 37. |
A metallic right circular cone 20 cm high and whose vertical angle is `60o`is cut into two parts at the middle of its height by a plane parallel to its base. If the frustum so obtained be drawn into a wire of diameter `1/(16)c m` find the le |
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Answer» Volume of wire=Volume of frustration =`1/3pi(r_1^2+r_2^2+r_1r_2)h` =`1/3(400*3+100*3+200*3)/((1/32)(1/32))*10` =71.68km |
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| 38. |
A solid iron rectangular block of dimensions 4.4 m , 2.6 m and 1 m is cast into a hollow cylindrical pipe of internal radius 30 cm and thickness 5 cm.Find the length of the pipe. |
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Answer» volume of cylindrical pipe=volume of rectangular block. volume of external cylinder-volume of interal cylinder=l*b*h `piR^2h-pir^2h=l*b*h` `pih(R^2-r^2)=l*b*h` `pi*h(35+30)(35-30)=4.4*2.6*1` `h=112m.` |
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| 39. |
It is required to make a closed cylindrical tank of height 1 m and base diameter 140 cmfrom a metal sheet. How many square metres of the sheet are required for the same? |
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Answer» Metal sheet required to make a closed cylindrical tank = Curved surface area of cylinder + 2*Base area of cylinder `= 2pirh+2pir^2=2pir(r+h)` Here, `r = d/2 = 140/2 = 70cm = 0.7m` `h = 1m` So, required area, `A = 2**22/7**0.7(1+0.7)` `A = 4.4**1.7 = 7.48m^2` So, `7.48m` square metres of the sheet are required to make the given cylinder. |
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| 40. |
A solid right circular cone is cut into two parts at the middle of itsheight by a plane parallel to its base. The ratio of the volume of thesmaller cone to the whole cone is: 1:2 (b) 1:4 (c) 1:6 (d) 1:8 |
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Answer» For larger cone,`tan(theta)= R/h`For smaller cone,`tan(theta)=r/(h/2)=2r/h`Hence, `R=2r` Required ratio= `(1/3*r^2*h/2)/(1/3*R^2*h)`Therefore, Requied Ratio=1/8. |
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| 41. |
A copper rod of diameter 1 cm and length 8 cm is drawn into a wire of length 18 m of uniform thickness. Find the thickness of the wire. |
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Answer» Note the volume of the material in both the shapes will be the same only. Volume of rod1=volume of rod2 `pi(r_1)^2h_1=pi(r_2)^2h_2` substituting values we get, `1/2 xx1/2xx8=1800xxr_2^2` =>`(r_2)=1/30 cm`thickness of wire=`2/30 cm` |
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| 42. |
The capacity of a closed cylindrical vessel of height 1 m is 15.4 litres. How manysquare metres of metal sheet would be needed to make it? |
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Answer» Volume=15.4 liters `pir^2h=15.4*10^(-3)m^3` `r^2=49*10^(-4)m^2` `4=7*10^(-2)m` TSA of cylinder=CSA +2BA =`2piRh+2pir^2` =`2pir(r+h)` =`2*22/7*7(1+0.07)m` =`47.08*10^(-2)m^2` |
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| 43. |
A metal pipe is 77 cm long. The inner diameter of a cross section is 4 cm, the outer diameter being 4.4 cm. Find its(i) inner curved surface area,(ii) outer curved surface area,(iii) total surface area |
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Answer» 1) Inner surface area=`2pirh` `=2*22/7*77*2` =968`cm^2` 2) Outer surfacce area=`2pirh` `=2*22/7*2.2*77` =10647.8`cm^2` 3) Total surface area=outer surface area+inner surface area+ring. =968+1064.8+`pi(R^2+r^2)` =2032.8+22/7(4.4^2-4^2) =2034.36`cm^2` |
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| 44. |
A hemispherical tank full of water is emptied by a pipe at the rate of `(25)/7`litres per second. How much time will it take to empty half the tank, if it is 3m in diameter? |
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Answer» Volume of hemispeherical tank, `V = 2/3pir^3` Here, diameter,`d= 3m` So, radius, `r = 1.5m` So, `V = 2/3**22/7**(1.5)^3 = 2/3**22/7**27/8=99/14` `V/2 = 99/28` Let `t` is the time taken in seconds to empty half of the tank, As, the tank is emptied at the rate of `25/7` litres per second. then, `25/7**t = 99**1000/28`(Multiplied by 1000 to convert it into litres) `=>t = 99000/28**7/25 = 990 ` seconds` = 990/60 = 16.5` minutes |
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| 45. |
In a hot water heating system, there is a cylindrical pipe of length 28 m and diameter5 cm. Find the total radiating surface in the system. |
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Answer» Radiating Surface=CSA of given cylinder =`2pirh` =`2*22/h*0.05*28` =`4.4m^2`. |
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| 46. |
The circumference of the base of a cylindrical vessel is 132 cm and its height is 25 cm.How many litres of water can it hold? (`1000 cm^3 =1l`) |
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Answer» amount of water this can hold=Volume of vessel =`pir^2h` =`(pir)^2/pih` =`66^2/22*7*25` =`34650cm^3` liters=34.650l. |
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| 47. |
The inner diameter of a cylindrical wooden pipe is 24 cm and its outer diameter is 28 cm. The length of the pipe is 35 cm. Find the mass of the pipe, if 1 cm3 of wood has a mass of 0.6 g. |
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Answer» Volume=Volume of outer cylinder-Volume of inner cylinder =`piR^2h-pir^2h` =`pih(R^2-r^2)` =`22/7*35(14^2-12^2)` =`5720cc` Mass=0.69*5720 =3432g. |
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| 48. |
A water flows through a cylindrical pipe, whose inner radius is 1 cm, at the rate of `80 cm^(-1)` in an empty cylindrical tank, the radius of whose base is 40 cm. What is the rise of water level in tank in half an hour? |
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Answer» Given , radius of tank , `r_(1) = 40 cm` Let hight of water level in tank in half an hour =` h_(1)` Also , given internal radius of cylindrical pipe, `r_(2) = 1cm` and speed of water `= 80 cm//s ` i.e., in 1 water flow = 80 cm `therefore` In 30 (min) water flow `= 80 xx 60 xx 30 = 144000 cm` According to the question, Volume of water in cylindrical tank = Volume of water flow from the circular pipe in half an hour `rArr " " pir_(1)^(2) h_(1) = pi r_(2)^(2) h_(2)` `rArr " " 40 xx 40 xx h_(1) = 1 xx 1xx 144000` `therefore " " h_(1) = (144000)/(40 xx 40) = 90` cm in half an hour. |
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| 49. |
There are 3 stair-steps as shown in the figure. Each stair step has width 25 cm, height 12 cm and length 50 cm. How many bricks have been used in it if each brick is `12.5cm xx 6.25cm xx 4 cm` |
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Answer» Volume of stain(1+2+3)=N*Volume of 1 brick `50*12*225+51*24*25+50*36*25=N*12.5*6.5*4` `N=72/4` `N=28`. |
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| 50. |
A fish tank is in the form of a cuboid, external measures of that cuboid are 80 cm x 40 cm x 30 cm. The base, side faces and back face are to be covered with a coloured paper. Find the area of the paper needed |
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Answer» Total area of cuboide=`2(80*30+30*40+40*80)` `13600cm^2` Area of paper=(Total area)-area of top+area of front. `=13600-(80*40+80*30)` `=8000cm^2`. |
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