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JEE Advanced 1995 · Mechanical Properties of Fluids

A container of large uniform cross-sectional area AA resting on a horizontal surface, holds two immiscible, non-viscous and incompressible liquids of densities dd and 2d2d, each of height H/2H/2 as shown in the figure. The lower density liquid is open to the atmosphere having pressure P0P_0. (a) A homogeneous solid cylinder of length L(L<H/2)L(L < H/2), cross-sectional area A/5A/5 is immersed such that it floats with its axis vertical at the liquid-liquid interface with length L/4L/4 in the denser liquid. Determine: (i) the density DD of the solid and (ii) the total pressure at the bottom of the container. (b) The cylinder is removed and the original arrangement is restored. A tiny hole of area s(s≪A)s(s \ll A) is punched on the vertical side of the container at a height h(h<H/2)h(h < H/2). Determine : (i) the initial speed of efflux of the liquid at the hole, (ii) the horizontal distance xx travelled by the liquid initially, and (iii) the height hmh_m at which the hole should be punched so that the liquid travels the maximum distance xmx_m initially. Also calculate xmx_m. (Neglect the air resistance in these calculations.)

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Written solution

Step 1: Determine the density DD of the solid cylinder.

The cylinder floats in equilibrium at the interface, meaning its weight is balanced by the total buoyant force.

W=D(A5)LgW = D \left(\frac{A}{5}\right) L g FB=d(A5)(3L4)g+2d(A5)(L4)gF_B = d \left(\frac{A}{5}\right) \left(\frac{3L}{4}\right) g + 2d \left(\frac{A}{5}\right) \left(\frac{L}{4}\right) g FB=5d4(A5)LgF_B = \frac{5d}{4} \left(\frac{A}{5}\right) L g D(A5)Lg=5d4(A5)LgD \left(\frac{A}{5}\right) L g = \frac{5d}{4} \left(\frac{A}{5}\right) L g D=5d4D = \frac{5d}{4}

Step 2: Determine the total pressure at the bottom of the container.

The total pressure is the sum of atmospheric pressure and the pressure due to the weight of the liquids and the cylinder.

Pbottom=P0+Wliq+WcylAP_{\text{bottom}} = P_0 + \frac{W_{\text{liq}} + W_{\text{cyl}}}{A} Wliq=dA(H2)g+2dA(H2)gW_{\text{liq}} = d A \left(\frac{H}{2}\right) g + 2d A \left(\frac{H}{2}\right) g Wcyl=D(A5)LgW_{\text{cyl}} = D \left(\frac{A}{5}\right) L g Pbottom=P0+32dAHg+14dALgAP_{\text{bottom}} = P_0 + \frac{\frac{3}{2} d A H g + \frac{1}{4} d A L g}{A} Pbottom=P0+dg(3H2+L4)P_{\text{bottom}} = P_0 + dg\left(\frac{3H}{2} + \frac{L}{4}\right)

Step 3: Determine the initial speed of efflux vv at the hole.

The pressure difference between the inside of the container at height hh and the atmosphere is calculated.

ΔP=dg(H2)+2dg(H2−h)\Delta P = dg\left(\frac{H}{2}\right) + 2dg\left(\frac{H}{2} - h\right) v=2 ΔP2dv = \sqrt{\frac{2\,\Delta P}{2d}} v=g(3H2−2h)v = \sqrt{g\left(\frac{3H}{2} - 2h\right)}

Step 4: Determine the horizontal distance xx travelled by the liquid.

The time of flight is determined by the vertical motion from height hh.

t=2hgt = \sqrt{\frac{2h}{g}} x=vtx = v t x=g(3H2−2h)2hgx = \sqrt{g\left(\frac{3H}{2} - 2h\right)} \sqrt{\frac{2h}{g}} x=h(3H−4h)x = \sqrt{h(3H - 4h)}

Step 5: Determine the height hmh_m for maximum distance and calculate xmx_m.

Maximize the square of the distance with respect to hh.

x2=h(3H−4h)x^2 = h(3H - 4h) dx2dh=3H−8h=0\frac{dx^2}{dh} = 3H - 8h = 0 hm=3H8h_m = \frac{3H}{8} xm=3H8(3H−4(3H8))x_m = \sqrt{\frac{3H}{8}\left(3H - 4\left(\frac{3H}{8}\right)\right)} xm=3H4x_m = \frac{3H}{4}

Answer:

(a) (i) D=5d4D = \dfrac{5d}{4}

(ii) Pbottom=P0+dg(3H2+L4)P_{\text{bottom}} = P_0 + dg\left(\dfrac{3H}{2} + \dfrac{L}{4}\right)

(b) (i) v=g(3H2−2h)v = \sqrt{g\left(\dfrac{3H}{2} - 2h\right)}

(ii) x=h(3H−4h)x = \sqrt{h(3H - 4h)}

(iii) hm=3H8h_m = \dfrac{3H}{8}, xm=3H4x_m = \dfrac{3H}{4}

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