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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.
| 101. |
How many metallic balls of radius 1 cm can be recast by melting a metallic sphere of radius 8 cm? |
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Answer» Correct Answer - 3.14 cm Radius of metallic ball = 1cm Volume of one ball = `(4)/(3)pi(1^(3))=(4)/(3)picm^(3)` Radius of sphere=8cm Volume of sphere=`(4)/(3)pi(8)^(3)cm^(3)` Now the numebr of metallic balls =`("volume of sphere")/("volume of one ball")=((4)/(3)pi(8)^(3))/((4)/(3)pi)=(8)^(3)=512` |
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| 102. |
Ashot-putt is a metallic sphere of radius 4.9 cm. If the density of the metalis `7. 8" "g" "p e r" "c m^3`, find the mass of the shot-putt |
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Answer» A shot-put is a solid sphere of metal and its mass is equal to the product of its volume and density Volume of the shot-put `= (4)/(3)pi r^(3)` `= ((4)/(3) xx (22)/(7) xx (49)/(10) xx (49)/(10) xx (49)/(10)) cm^(3)` `= (1479016)/(3000) cm^(3) = 493 cm^(3)` (nearly) Mass of `1 cm^(3)` of the shot-put = 7.8g Mass of the shot-put `= (7.8 xx 493)g` `= ((78)/(10) xx (493)/(1000)) kg = (38454)/(10000) kg = 3.85kg` (nearly) Hence, the mass of the given shot-put is 3.85 kg |
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| 103. |
Find the length of 13.2 kg of copper wire of diameter 4 mm, when 1 cubic centimetre of copper weights 8.4 g. |
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Answer» Correct Answer - 125 m Let the required length of wire be x cm Then, its volume `= ((22)/(7) xx (2)/(10) xx (2)/(10) xx x) cm^(3) = ((22x)/(175)) cm^(3)` `:.` its weight `= ((22x)/(175) xx 8.4) g = ((22x xx 8.4)/(175 xx 1000)) kg` `:. (22x xx 8.4)/(175 xx 1000) = 13.2` `rArr x = ((13.2 xx 175 xx 1000)/(22 xx 8.4)) cm = ((132 xx 175 xx 1000)/(22 xx 84 xx 100))m = 125m` |
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| 104. |
Shanti Sweets Stall was placingan order for making cardboard boxes for packing their sweets. Two sizes ofboxes were required. The bigger of dimensions `25" "c m" "xx" "20" "c m" "xx" "5c m`and the smaller of dimensions `15" "c m" "xx" " |
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Answer» Surface area of one bigger box =`2(25xx20xx20xx5+5xx25)` 1450 `cm^(2)` By adding 5% overlaps =1450+72.5=1522.5 `cm^(2)` Area for 250 bigger boxes =1522.5`xx`250 `cm^(2)` cost for 250 bigger boxes =`1522.5xx250xx(4)/(1000)`=Rs 1522.5 Surface area of one smaller box =`2(15xx12+12xx5+xx15)=630 cm^(2)` By adding 5% for overlaps =630+31.5=661.5 `cm^(2)` Area for 250 small boxes =`661.5xx250xx(4)/(1000)=Rs 661.5` Total cost =1522.5+661.5=Rs 2184 |
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| 105. |
How manylitres of water flow out of a pipe having an area of cross-section of `5 c m^2`in oneminute, if the speed of water in the pipe is `30 c m//s e c`? |
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Answer» Correct Answer - 9 lites Speed of water = 30m/s Volume of water that flows out of the pipe in 1s = area of cross section `xx` length of water flows in 1s `= (5xx 30) cm^(3) = 150 cm^(3)` Volume of water that flows in 1 minute `= (150 xx 60) cm^(3) = ((150 xx 60)/(1000))` litres = 9 litres |
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| 106. |
A well with10m inside diameter is dug 8.4m deep. Earth taken out of it is spread allaround it to a width of 7.5m to form an embankment. Find the height of theembankment. |
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Answer» Correct Answer - 1.6 m Volume of earth dug out = volume of earth of earth in embankment `pi xx 5 xx 5 xx 8.4 = pi xx {(12.5)^(2) - 5^(2)} xx h`. Find h |
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| 107. |
The capacity of a cuboidal tank is 50000 litres ofwater. Find the breadth of the tank,if its length and depth are respectively 2.5 m and 10 m |
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Answer» Correct Answer - 2 m Volume of the tank = 50000 litres `= (50000)/(1000) m^(3) = 50 m^(3)` `:. 10 xx x xx 2.5 = 50`. Find x |
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| 108. |
The diameter of a roller, 120 cm long, is 84 cm. If it takes 500 complete revolutions to level a playground, find the cost of levelling it at Rs 5 per square metre. |
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Answer» Radius of the roller, r = 42 cm, and its length, h = 120 cm Area covered by the roller in 1 revolution = curved surface area of the roller `= (2pi rh)` sq units `= (2 xx (22)/(7) xx 42 xx 120) cm^(2) = 31680 cm^(2)` Area covered by the roller in 500 revolutions `(31680 xx 500) cm^(2) = ((31680 xx 500)/(100 xx 100)) m^(2) = 1584 m^(2)` `:.` area of the playground `= 1584 m^(2)` Cost of levelling the playground `= Rs (1584 xx 5) = Rs 7920` |
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| 109. |
Find the volume of a sphere whose surface area is `154 cm^(2)` |
| Answer» Correct Answer - `179.67 cm^(3)` | |
| 110. |
How many lead balls, each of radius 1 cm, can be made from a sphere of radius 8 cm ? |
| Answer» Correct Answer - 512 | |
| 111. |
The volume of a sphere of radius `2r` isA. `(32 pi r^(3))/(3)`B. `(16 pi r^(3))/(3)`C. `(8pi r^(3))/(3)`D. `(64 pi r^(3))/(3)` |
| Answer» Correct Answer - A | |
| 112. |
The surface area of sphere is `(576pi) cm^(2)`. Find its volume |
| Answer» Correct Answer - `(2304 pi) cm^(3)` | |
| 113. |
The volume of a right circular cone is 9856 `c m^3`. Ifthe diameter of the base is 28cm, find:(i)height of the cone (ii) slant height of the cone(iii)curved surface area of thecone. |
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Answer» We have, `V = 9856 cm^(3) and r = 14 cm` (i) Let the height of the cone be h cm Then, volume `= (1)/(3) pi r^(2)h` `rArr 9856 = (1)/(3) xx (22)/(7) xx 14 xx 14 xx h` `rArr h = ((9856 xx 3 xx 7)/(22 xx 14 xx 14)) = 48` `:.` height of the cone is 48 cm (ii) Let the slant height of the cone be l cm. Then, `l^(2) = r^(2) + h^(2)` `rArr l^(2) = (14)^(2) + (48)^(2) = 196 + 2304 = 2500` `rArr l = sqrt(2500)= 50` `:.` height of the cone is 50 cm (iii) Curved surface area of the cone `= pi rl` `= ((22)/(7) xx 14 xx 50) cm^(2) = 2200 cm^(2)` `:.` curved surface area of the cone is `2200 cm^(2)` |
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| 114. |
If the surface area of a sphere is `(144 pi)m^(2)` then its volume isA. `(288pi)m^(3)`B. `(188 pi)m^(3)`C. `(300 pi)m^(3)`D. `(316pi) m^(3)` |
| Answer» Correct Answer - A | |
| 115. |
The surface area of a sphere is `1386 cm^(2)`. Its volume isA. `1617 cm^(3)`B. `3234 cm^(3)`C. `4158 cm^(2)`D. `9702 cm^(2)` |
| Answer» Correct Answer - C | |
| 116. |
The volume of a sphere of radius 10.5 cm isA. `9702 cm^(3)`B. `4851 cm^(3)`C. `19494 cm^(3)`D. `14553 cm^(3)` |
| Answer» Correct Answer - B | |
| 117. |
The height of a cone is 21 cm and its slant height is 28 cm. The volume of the cone is |
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Answer» We have, `h = 21 cm and l = 28 cm` `:. l^(2) = r^(2) + h^(2) rArr r = sqrt(l^(2) - h^(2)) = sqrt((28)^(2) - (21)^(2)) cm` `rArr r = sqrt((28 + 21) xx (28 - 21)) cm = sqrt(49 xx 7) cm` `= 7 sqrt7 cm` Volume of the cone `= (1)/(3) pi r^(2) h` `= {(1)/(3) xx (22)/(7) xx (7 sqrt7)^(2) xx 21} cm^(3)` `= (22 xx 343) cm^(3) = 7546 cm^(3)` |
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| 118. |
The radiiof the ends of a bucket 30 cm high are 21 cm and 7 cm. Find its capacity inlitres and the amount of sheet required to make this bucket. |
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Answer» Although we can solve this problem by suing the similar trianlges, but here we are using the direct formula. Volume of bucket=`(1)/(3)pih(r_(1^(2))+r_(1)r_(2)+r_(2^(2))` `=(1)/(3)xx(22)/(7)xx30(21^(2)+21xx7+7^(2))` `=(1)/(3)xx(22)/(7)xx30x637=20020 cm^(3)=(20020)/(1000)lit` Capacity =20.2 lit Area of sheeet = C.S.A of bucket +Area of bse ,=`pil(r_(1)+r_(2))+pir_(2)^(2)=pi[l(r_(1)+r_(2))+r_(2)^(2))]` `=(22)/(7)[33.10^(28)+49]` 3067.32=3067 `cm^(2)` (approx) |
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| 119. |
The surface area of a sphere of radius 21 cm isA. `2772 cm^(2)`B. `1386 cm^(2)`C. `4158 cm^(2)`D. `5544 cm^(2)` |
| Answer» Correct Answer - D | |
| 120. |
Find the volume and surface area of a sphere of radius 21 cm. |
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Answer» Radius of the sphere = 21 cm Volume of the sphere `= ((4)/(3) pi r^(3))` cubic units `= ((4)/(3) xx (22)/(7) xx 21 xx 21 xx 21) cm^(3)` `= 38808 cm^(3)` Surface area of the sphere `= (4pir^(2))` sq units `= (4 xx (22)/(7) xx 21 xx 21) cm^(2) = 5544 cm^(2)` |
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| 121. |
Find the surface area of a sphere whose volume is `4851 cm^(3)` |
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Answer» Let the radius of the sphere by r cm Then, its volume `= ((4)/(3) pi r^(3)) cm^(3)` `:. (4)/(3) pi r^(3) = 4851 rArr (4)/(3) xx (22)/(7) xx r^(3) = 4851` `rArr r^(3) = (4851 xx (3)/(4) xx (7)/(22)) = ((441 xx 21)/(8)) = ((21)/(2))^(3)` `rArr r = (21)/(2) = 10.5` Thus, the radius of the sphere is 10.5 cm Surface area of the sphere `= (4pi r^(2))` sq units `= (4 xx (22)/(7) xx (21)/(2) xx (21)/(2)) cm^(2)` `= 1386 cm^(2)` Hence, the surface area of the given sphere is `1386 cm^(2)` |
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