The correct option is
B 1−3√34πA cylinder tank of height 3 units and radius 1 unit is placed on the ground with circular base touching the ground.
The height of water in the tank is h units.
The Volume of the water V=πr2h ....(1)
Now if the cylinder is placed horizontally on the ground with its lateral surface touching the ground.
The height of the water becomes 12 units
The figure shows the cross-section of the cylinder when placed horizontally
The cross-section of the cylinder when placed horizontally is rectangular. Hence Area of cross-section will be Area of the rectangle. The length of the rectangle is same as the length of the cylinder in horizontal position and breadth is along the diameter of the circular cross-section area.
Let's take a general rectangular cross-section at the height x from the bottom.
From the figure, QR=x,
→CR=CP=radius=1
→CQ=CR−QR=1−x
Now in the right triangle CQP, →CQ2+QP2=CP2
→(1−x)2+QP2=1
→QP=√2x−x2
The length of the rectangle at height x is 3 units, while the breadth is 2QP.
The area of rectangle is A=3×2√2x−x2=6√2x−x2
⇒V=∫120Adx
⇒V=∫1206√2x−x2.dx
⇒V=∫1206√1−(1−x)2.dx
Let's assume 1−x=sinθ or dx=− cos θ dθ
⇒V=∫π6π2−6 cosθ√1−(sinθ)2.dθ
⇒V=−∫π6π26 cos2θ.dθ
⇒V=−∫π6π26(1+cos2θ2).dθ
⇒V=−3∫π6π2(1+cos2θ).dθ
→V=[−3(θ)−3(sin2θ2)]π6π2
⇒V=−3(π6−π2)−32(sinπ3−sinπ)
⇒V=−π−32(√32)
V=π−3√34
From eq. (1), The volume of the water V=πr2h=π(1)2h=πh units
Hence π h= π−3√34
So h= 1−3√34π
Hence correct option is A.
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