Question

A cylindrical tank of height $3$ units and diameter $2$ units, resting on its base, is partially filled with water up to a height $h$ units. Thereafter it is turned horizontal and placed on a level ground with its lateral surface in contact with the ground. In this position, the maximum depth of water is $\frac{1}{2}$ units. Then $h$ is

• A. $1-\frac{3\sqrt{3}}{4\pi }$
• B. $1-\frac{3}{4\pi }$
• C. $1-\frac{3\sqrt{3}}{2\pi }$
• D. $1-\frac{3}{2\pi }$

Answer 1

Answer: (A)

$1-\frac{3\sqrt{3}}{4\pi }$

A cylinder tank of height $3units$ and radius $1unit$ is placed on the ground with circular base touching the ground.

The height of water in the tank is $hunits$.

The Volume of the water $V=\pi r^{2}h$ ....$\left(1\right)$

Now if the cylinder is placed horizontally on the ground with its lateral surface touching the ground.

The height of the water becomes $\frac{1}{2}units$

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, $QR=x$,

$\to CR=CP=radius=1$

$\to CQ=CR-QR=1-x$

Now in the right triangle $CQP$, $\to C{Q}^2+Q{P}^2=C{P}^2$

$\to (1-x{)}^2+Q{P}^2=1$

$\to QP=\sqrt{2x-x^2}$

The length of the rectangle at height $x$ is $3units$, while the breadth is $2QP$.

The area of rectangle is $A=3\times 2\sqrt{2x-x^2}=6\sqrt{2x-x^2}$

$\Rightarrow V={\int }_0^{\frac{1}{2}}Adx$

$\Rightarrow V={\int }_0^{\frac{1}{2}}6\sqrt{2x-x^2}.dx$

$\Rightarrow V={\int }_0^{\frac{1}{2}}6\sqrt{1-(1-x{)}^2}.dx$

Let's assume $1-x=sin\theta$ or $dx=-cos\theta d\theta$

$\Rightarrow V={\int }_{\frac{\pi }{2}}^{\frac{\pi }{6}}-6cos\theta \sqrt{1-(sin\theta {)}^2}.d\theta$

$\Rightarrow V=-{\int }_{\frac{\pi }{2}}^{\frac{\pi }{6}}6co{s}^2\theta .d\theta$

$\Rightarrow V=-{\int }_{\frac{\pi }{2}}^{\frac{\pi }{6}}6\left(\frac{1+cos2\theta }{2}\right).d\theta$

$\Rightarrow V=-3{\int }_{\frac{\pi }{2}}^{\frac{\pi }{6}}(1+cos2\theta ).d\theta$

$\to V={[-3(\theta )-3(\frac{sin2\theta }{2}\left)\right]}_{\frac{\pi }{2}}^{\frac{\pi }{6}}$

$\Rightarrow V=-3(\frac{\pi }{6}-\frac{\pi }{2})-\frac{3}{2}(sin\frac{\pi }{3}-sin\pi )$

$\Rightarrow V=-\pi -\frac{3}{2}\left(\frac{\sqrt{3}}{2}\right)$

$V=\pi -\frac{3\sqrt{3}}{4}$

From eq. $\left(1\right)$, The volume of the water $V=\pi r^{2}h=\pi (1{)}^{2}h=\pi hunits$

Hence $\pi h=$ $\pi -\frac{3\sqrt{3}}{4}$

So $h=$ $1-\frac{3\sqrt{3}}{4\pi }$

Hence correct option is $A$.