JEE Advance - Physics (2020 - Paper 1 Offline - No. 14)
An open-ended U-tube of uniform cross-sectional area contains water (density 103 kg m−3
). Initially the
water level stands at 0.29 m from the bottom in each arm. Kerosene oil (a water-immiscible liquid) of
density 800 kg m−3
is added to the left arm until its length is 0.1 m, as shown in the schematic figure
below. The ratio $$\left( {{{{h_1}} \over {{h_2}}}} \right)$$ of the heights of the liquid in the two arms is :
$${{15} \over {14}}$$
$${{35} \over {33}}$$
$${7 \over 6}$$
$${5 \over 4}$$
Explanation
We have, h1 + h2 = 0.29 $$ \times $$ 2 + 0.1
h1 + h2 = 0.68 .......(i)
$$ \Rightarrow $$ p0 + $$\rho $$kg(0.1) +$$\rho $$wg(h1 $$-$$ 0.1) $$-$$ $$\rho $$wgh2 = p0
where, $$\rho $$k = density of kerosene and $$\rho $$w = density of water.
$$ \Rightarrow $$ $$\rho $$kg(0.1) + $$\rho $$wgh1 $$-$$ $$\rho $$wg $$ \times $$ (0.1) = $$\rho $$wgh2
$$ \Rightarrow $$ 800 $$ \times $$ 10 $$ \times $$ 0.1 + 1000 $$ \times $$ 10 $$ \times $$ h1 $$-$$ 1000 $$ \times $$ 10 $$ \times $$ 0.1 = 1000 $$ \times $$ 10 $$ \times $$ h2
$$ \Rightarrow $$ 10000(h1 $$-$$ h2) = 200
$$ \Rightarrow $$ h1 $$-$$ h2 = 0.01 .......(ii)
Solving Eqs. (i) and (ii), we get
h1 = 0.35
and h2 = 0.33
So, $${{{h_1}} \over {{h_2}}} = {{35} \over {33}}$$
h1 + h2 = 0.68 .......(i)
$$ \Rightarrow $$ p0 + $$\rho $$kg(0.1) +$$\rho $$wg(h1 $$-$$ 0.1) $$-$$ $$\rho $$wgh2 = p0
where, $$\rho $$k = density of kerosene and $$\rho $$w = density of water.
$$ \Rightarrow $$ $$\rho $$kg(0.1) + $$\rho $$wgh1 $$-$$ $$\rho $$wg $$ \times $$ (0.1) = $$\rho $$wgh2
$$ \Rightarrow $$ 800 $$ \times $$ 10 $$ \times $$ 0.1 + 1000 $$ \times $$ 10 $$ \times $$ h1 $$-$$ 1000 $$ \times $$ 10 $$ \times $$ 0.1 = 1000 $$ \times $$ 10 $$ \times $$ h2
$$ \Rightarrow $$ 10000(h1 $$-$$ h2) = 200
$$ \Rightarrow $$ h1 $$-$$ h2 = 0.01 .......(ii)
Solving Eqs. (i) and (ii), we get
h1 = 0.35
and h2 = 0.33
So, $${{{h_1}} \over {{h_2}}} = {{35} \over {33}}$$
Comments (0)
