Latest Elitmus Aptitude Question SOLUTION: A duster is of dimension 5*5*7 is sinked into a cylindrical vessel of radius 5 cm and height 10 cm .After insertion the level of the water is inccresed by 2.14 cm ..Then by what pe

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(#M40040363) ELITMUS Numerical Ability question Area and Volume Keep an EYE

A duster is of dimension 5*5*7 is sinked into a cylindrical vessel of radius 5 cm and height 10 cm .After insertion the level of the water is inccresed by 2.14 cm ..Then by what percentage the part of the duster is above/below(something like that) the layer?

Asked In Elitmus ASHWINI KUAMR YADAV (10 years ago)

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lalita kesarwani  (10 years ago)

i guess answer is 96% . by relation volume of water displaced we can get the volume of immersed part and we know the toatal volume therefore we can get the answer
vol of water displaced = 3.14*5*5*2.14= 167.99 ==168
this is nothing but the volume of immersed part
now toatal volume of duster=l*b*h= 25*7=175
%below= 16800/175= 96%

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aditya saraswat  (10 years ago)

volume of rise water=pi*5*5*2.14=167.99 ~168
volume of duster=5*5*7=175
%part of duster below the level is=(175-168)*100/175

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rajeev chaubey  (10 years ago)

volume of duster=175cm^3
v of cylnder=3.14*25*10
volume increase by duster=3.14*25*2.14
now suppose x volume of duster is either above or below layer
that much x volume of duster is responsible for gain in volume of cylinder.
so, x=3.14*25*2.14(volume of duster above or below)
tptal volume of duster-x=175-3.14*25*2.14=6.9cm^3
ans-6.9%

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ASHWINI KUAMR YADAV  (10 years ago)

which option was correct..

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kumar  (10 years ago)

i think in this question all available answer were wrong

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Rayagouda   (10 years ago)

Since the level of water level increased is 2.14cm.
volume occupied by duster in water is =pi*r^2*2.14=pi*5^2*2.14=53.50
percentage of part of the duster inside the layer is =53.50/2(lb+bh+hl)
=53.50/190=.28= 28%

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vikrant gupta  (9 years ago)

Dimension of the duster is assumed as 5*5*
Volume of the duster below the water layer
= volume of water increased
= pir2h = 3.14×52×2.14 =168
Total volume of the duster = 5×5×7 = 175

Percentage of the duster part below the layer = 168×100175 = 96%

Percentage of the duster part above the layer = 4%

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ranjan  (7 years ago)

Dimension of the duster is assumed as 5*5*7 cm3 (unit is not given in the question)

Volume of the duster below the water layer
= volume of water increased
= πr2h = 3.14×52×2.14 ≈ 168 cm3

Total volume of the duster = 5×5×7 = 175 cm3

Percentage of the duster part below the layer =
168
×
100
175
168×100175 = 96%

Percentage of the duster part above the layer = 4%

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