a conical pit of top diameter 3.5m is 12m deep. what is its capacity in kiloliters? (assume π = 22/7) - maths q&a
given radius of cone (r) = 3.5/2 m height of cone (h) = 12 m find out we have to find the capacity in kiloleters solution volume of cone = (1/3) × πr2h on taking π=22/7 = (1/3) × (22/7) × (3.5/2) × (3.5/2) × 12 = (22 × 3.5 × 3.5 × 12) / (7 × 3 × 2 × 2) = 38.5 m3 capacity of conical pit in kilo liters: 1000 m3 = 1 liter 38.5 m3 = 1000 × 38.5 liters = 38,500 liters = 38.5 kilo liters answer the capacity of the container is 38.5 kilo liters
radius = 1.75 m
height (h) = 12 m
volume = 
= ×  × 1.75 × 1.75 × 12
= 38.5 m3
since, 1 m3 = 1 kiloliter
capacity of the pit = (38.5 × 1)
= 38.5 kilolitres given: volume of cylinder = 1617 cm3let r be the radius of base and h be the height of coner:h = 2:3∴ r/h = 2/33r = 2hh = 3r/2 …(i)formula: volume of cylinder = πr2h∴ 1617 = (22/7) r2hr2h = 514.5using (i) we have∴ r2 × (3r/2) = 514.53r3 = 1029r3 = 343r = 7 cmh = 21/2 cmtotal surface area of cylinder = 2πr2 + 2πrh= 2 × (22/7) × 72 + 2 × (22/7) × 7 × (21/2)= 308 + 462= 770 cm2therefore total surface area of cylinder = 770 cm2 32 lit Correct answer is 26 mm. Mass of powder is constant. mi=mf  ⇒ ρiVi = ρfVf ρihi = ρf hf ⇒ hi = (ρf hf)/ρi On substitution, (1755×12)/ 810= 26 mm  


Correct answer is 26 mm.


Mass of powder is constant.


mi=mf ⇒ ρiVi= ρfVf


ρihi = ρf hf⇒ hi = (ρf hf)/ρi


On substitution, (1755×12)/ 810= 26 mm


 


[See less](#) mass of titanium that can... Correct option is A) 9.378 m2


![](https://lh3.googleusercontent.com/-UueuqyijXPQ/YLtK0v0m9CI/AAAAAAAABro/xFcfQOQAOx8jHAtpqiORvW_8DRFX1yK3ACLcBGAsYHQ/image.png)


result must have significant figures equal to least figures being multiplid do area of metal sheet 9.378 m2  (four significant digits) the dimensions of base of a rectangular box of greatest volume that can be constructed from 100 cm square of hard board sheet, if the box is three times as long as it is wide, are \_\_\_\_\_\_. [readmore]typed answerh2 + i2 ⇌ 2hi      kc = 4.6i            0.5   0.5     0.5c          -x      -x       +2xe        0.5-x 0.5-x   0.5+2xkc=hi2hii2kc=0.5+2x20.5-x0.5-xkc=0.5+2x20.5-x24.6=0.5+2x0.5-xx=0.138hi=0.5+2x=0.776 m v = (4/3)πr³ 

dv/dt = 4πr².dr/dt

so, when the diameter is 20 cm we have:

-4π = 4πr².dr/dt

hence, dr/dt = -1/r²

i.e. dr/dt = -1/10² => -1/100

now, d = 2r

so, d(d)/dt = 2.dr/dt

=> d(d)/dt = -1/50 cm/hour

or, -0.2 mm/hour

:)> V(NaOH) = 2×9×0.122/0.141 = 15.57 mL