Fluid Statics PYQs: MHT CET, JEE Main, NEET

Q1NEET 2026Easy

Pressure in a Fluid · Pressure at a depth

A submarine is designed to withstand an absolute pressure of 100 atm. How deep can it go below the water surface?

(Consider the density of water $= 1000\text{ kg m}^{-3},\ 1\text{ atm} = 1 \times 10^{5}\text{ Pa}$ and gravitational acceleration $g = 10\text{ m/s}^{2}$ )

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Answer: (a)

Absolute pressure at depth $h$ : $P = P_0 + \rho g h$ .
$100\times10^5 = 1\times10^5 + 1000\times10\times h$
$h = \dfrac{99\times10^5}{10^4} = 990\text{ m}$
So the answer is (a) 990 m.

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Q2JEE Main 2026 · 22 JAN · SHIFT 1Easy

Pressure in a Fluid · Pressure at a depth

Given below are two statements:

Statement I: Pressure of a fluid is exerted only on a solid surface in contact as the fluid-pressure does not exist everywhere in a still fluid.

Statement II: Excess potential energy of the molecules on the surface of a liquid, when compared to interior, results in surface tension.

In the light of the above statements, choose the correct answer from the options given below

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Answer: (a)

Statement I: pressure exists at every point inside a fluid at rest, not only on solid surfaces, so it is false.
Statement II: surface molecules have extra potential energy compared with interior ones; this surface energy per unit area is surface tension, so it is true.
So the answer is (a).

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Q3MHT CET 2026 · 18 APR · SHIFT 1Medium

Pressure in a Fluid · Pressure at a depth

The pressure at half the depth of a lake is equal to two-third pressure at the bottom of the lake. So the depth 'h' of the lake is ( $\rho$ = density of water in the lake, $g$ = acceleration due to gravity, $P_0$ = atmospheric pressure).

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Answer: (a)

$P_0 + \rho g\dfrac{h}{2} = \dfrac{2}{3}(P_0 + \rho g h)$
$3P_0 + \tfrac{3}{2}\rho g h = 2P_0 + 2\rho g h \Rightarrow P_0 = \tfrac{1}{2}\rho g h$
$h = \dfrac{2P_0}{\rho g}$
So the answer is (a).

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Q4MHT CET 2026 · 13 MAY · SHIFT 2Medium

Pressure in a Fluid · Air bubble rising in a lake

When a large bubble rises from bottom of a water lake to its surface, then its radius doubles. If the atmospheric pressure is equal to the pressure of height $H$ of a certain water column then the depth of lake will be

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Answer: (c)

Radius doubles, so volume becomes $8V$ . At constant temperature, $P_1V_1 = P_2V_2$ .
In water-column units: $(H + h)V = H\times8V$
$h = 7H$
So the answer is (c) $7H$ .

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Q5JEE Main 2025 · 3 APR · SHIFT 2Medium

Pressure in a Fluid · U-tube, barometer & connected vessels

Two cylindrical vessels of equal cross sectional area of $2\text{ m}^{2}$ contain water upto heights 10 m and 6 m, respectively. If the vessels are connected at their bottom then the work done by the force of gravity is (Density of water is $10^{3}\text{ kg/m}^{3}$ and $g = 10\text{ m/s}^{2}$ )

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Answer: (c)

Final common height $= \dfrac{10 + 6}{2} = 8\text{ m}$ , so a 2 m layer moves from vessel 1 to vessel 2.
Mass moved: $m = \rho A \times 2 = 1000\times2\times2 = 4000\text{ kg}$ .
Its centre of mass falls from 9 m to 7 m, i.e. 2 m.
$W = mg\Delta h = 4000\times10\times2 = 8\times10^4\text{ J}$
So the answer is (c).

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Q6JEE Main 2025 · 2 APR · SHIFT 1Medium

Pressure in a Fluid · U-tube, barometer & connected vessels

A vessel with square cross-section and height of 6 m is vertically partitioned. A small window of $100\text{ cm}^{2}$ with hinged door is fitted at a depth of 3 m in the partition wall. One part of the vessel is filled completely with water and the other side is filled with the liquid having density $1.5 \times 10^{3}\text{ kg/m}^{3}$ . What force one needs to apply on the hinged door so that it does not get opened? (Acceleration due to gravity $= 10\text{ m/s}^{2}$ )

Numerical answer type. Work it out, then open the answer.

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Answer: 150

Both sides are full (6 m), so the door at depth 3 m feels pressure from each liquid.
Net pressure: $\Delta P = (\rho_2 - \rho_1) g h = (1500 - 1000)\times10\times3 = 1.5\times10^4\text{ Pa}$ .
$F = \Delta P\, A = 1.5\times10^4\times100\times10^{-4} = 150\text{ N}$
So the force needed is 150 N.

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Q7MHT CET 2025 · 20 APR · SHIFT 2Medium

Pressure in a Fluid · Air bubble rising in a lake

The temperature of an air bubble while rising from bottom to surface of a lake remains constant but diameter is doubled. If pressure on the surface is $h$ meter of mercury column and relative density of mercury is ' $\rho$ ' then the depth of the lake is

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Answer: (b)

Diameter doubles, so volume becomes 8 times. Boyle's law: $P_{bottom}V = P_{surface}(8V)$ , so $P_{bottom} = 8P_{surface}$ .
Surface pressure $h$ m of mercury $= h\rho$ m of water.
$h\rho + d = 8h\rho \Rightarrow d = 7h\rho$
So the answer is (b) $7h\rho$ .

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Q8JEE Main 2024 · 4 APR · SHIFT 1Medium

Pressure in a Fluid · Pressure at a depth

Given below are two statements:

Statement I: When speed of liquid is zero everywhere, pressure difference at any two points depends on equation $P_1 - P_2 = \rho g (h_2 - h_1)$ .

Statement II: In ventury tube shown $2gh = v_1^2 - v_2^2$

In the light of the above statements, choose the most appropriate answer from the options given below.

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Answer: (a)

Statement I: for a fluid at rest, $P_1 + \rho g h_1 = P_2 + \rho g h_2$ , so $P_1 - P_2 = \rho g(h_2 - h_1)$ . Correct.
Statement II: Bernoulli at the same level gives $P_1 - P_2 = \tfrac{1}{2}\rho(v_2^2 - v_1^2)$ , and $P_1 - P_2 = \rho g h$ .
So $2gh = v_2^2 - v_1^2$ , not $v_1^2 - v_2^2$ . Statement II is incorrect.
So the answer is (a).

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Q9JEE Main 2024 · 4 APR · SHIFT 2Medium

Pressure in a Fluid · Pressure at a depth

Mercury is filled in a tube of radius $2\text{ cm}$ up to a height of $30\text{ cm}$ . The force exerted by mercury on the bottom of the tube is _________ N.

(Given, atmospheric $= 10^{5}\text{ N m}^{-2},\ g = 10\text{ m s}^{-2},\ \pi = \frac{22}{7})$ pressure, density of mercury

$= 1.36 \times 10^{4}\text{ kg m}^{-3}$

Numerical answer type. Work it out, then open the answer.

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Answer: 177

Pressure at the bottom: $P = P_0 + \rho g h = 10^5 + 1.36\times10^4\times10\times0.30 = 1.408\times10^5\text{ Pa}$ .
Area: $A = \pi r^2 = \dfrac{22}{7}\times(0.02)^2 = 1.257\times10^{-3}\text{ m}^2$ .
$F = PA = 1.408\times10^5\times1.257\times10^{-3} \approx 177\text{ N}$
So the force is 177 N.

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Q10MHT CET 2024 · 9 MAY · SHIFT 1Medium

Pressure in a Fluid · Air bubble rising in a lake

When an air bubble rises from the bottom of lake to the surface, its radius is doubled. The atmospheric pressure is equal to that of a column of water of height ' $H$ '. The depth of the lake is

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Answer: (c)

Radius doubles, so volume becomes 8 times. Boyle's law: $P_1V = P_2(8V)$ .
In water-column units: $H + h = 8H$
$h = 7H$
So the answer is (c) 7 H.

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Q11MHT CET 2024 · 2 MAY · SHIFT 2Difficult

Pressure in a Fluid · Rotating liquids

A drum of radius ' $R$ ' full of liquid of density ' $d$ ' is rotated at angular velocity ' $\omega$ ' $\text{rad/s}$ . The increase in pressure at the centre of the drum will be

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Answer: (a)

In a rotating liquid, $\dfrac{dP}{dr} = d\,\omega^2 r$ .
Integrating from centre to rim: $\Delta P = \int_0^R d\,\omega^2 r\,dr = \dfrac{d\,\omega^2 R^2}{2}$
So the pressure difference between rim and centre is $\dfrac{\omega^2 R^2 d}{2}$ ; the answer is (a).

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Q12JEE Main 2023 · 10 APR · SHIFT 1Easy

Pressure in a Fluid · Pressure at a depth

Given below are two statements:

Statement I: Pressure in a reservoir of water is same at all points at the same level of water. Statement II: The pressure applied to enclosed water is transmitted in all directions equally. In the light of the above statements, choose the correct answer from the options given below:

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Answer: (d)

Statement I: in a fluid at rest $P = P_0 + \rho g h$ , so all points at the same level have the same pressure. True.
Statement II: Pascal's law says pressure applied to an enclosed fluid is transmitted equally in all directions. True.
So the answer is (d).

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Q13JEE Main 2023 · 8 APR · SHIFT 1Medium

Pressure in a Fluid · Air bubble rising in a lake

An air bubble of volume $1\text{ cm}^{3}$ rises from the bottom of a lake $40\text{ m}$ deep to the surface at a temperature of $12^\circ\text{C}$ . The atmospheric pressure is $1 \times 10^{5}\text{ Pa}$ the density of water is $1000\text{ kg/m}^{3}$ and $g = 10\text{ m/s}^{2}$ . There is no difference of the temperature of water at the depth of $40\text{ m}$ and on the surface. The volume of air bubble when it reaches the surface will be:

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Answer: (d)

Pressure at the bottom: $P_1 = 10^5 + 1000\times10\times40 = 5\times10^5\text{ Pa}$ .
Same temperature, so Boyle's law: $P_1V_1 = P_0V_2$ .
$V_2 = \dfrac{5\times10^5\times1}{10^5} = 5\text{ cm}^3$
So the answer is (d).

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Q14NEET 2022 · PHASE 2Medium

Pressure in a Fluid · U-tube, barometer & connected vessels

Two copper vessels A and B have the same base area but of different shapes. A takes twice the volume of water as that B requires to fill upto a particular common height. Then the correct statement among the following is:

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Answer: (b)

Pressure on the base depends only on the liquid height: $P = P_0 + \rho g h$ , not on the vessel shape or volume.
Both are filled to the same height, so the pressure on the bases is the same.
A holds more water, so A (with water) weighs more; (a) and (d) are wrong.
So the answer is (b).

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Q15JEE Main 2022 · 27 JUL · SHIFT 1Medium

Pressure in a Fluid · U-tube, barometer & connected vessels

Two cylindrical vessels of equal cross-sectional area $16\text{ cm}^{2}$ contain water upto heights $100\text{ cm}$ and $150\text{ cm}$ respectively. The vessels are interconnected so that the water levels in them become equal. The work done by the force of gravity during the process, is [Take, density of water $= 10^{3}\text{ kg/m}^{3}$ and $g = 10\text{ m s}^{-2}$ ]:

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Answer: (b)

Final common height $= \dfrac{100 + 150}{2} = 125\text{ cm}$ , so a 25 cm layer moves across.
Mass moved: $m = 10^3\times16\times10^{-4}\times0.25 = 0.4\text{ kg}$ .
Its centre falls from 137.5 cm to 112.5 cm, i.e. 0.25 m.
$W = 0.4\times10\times0.25 = 1\text{ J}$
So the answer is (b) 1 J.

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Q16JEE Main 2021 · 16 MAR · SHIFT 1Medium

Pressure in a Fluid · Pressure at a depth

The pressure acting on a submarine is 3 $\times$ 105 Pa at a certain depth. If the depth is doubled, the percentage increase in the pressure acting on the submarine would be:

(Assume that atmospheric pressure is 1 $\times$ 105 Pa and density of water is 103 kg m ${}^{-3}$ , g = 10 ms ${}^{-2}$ )

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Answer: (b)

$P = P_0 + \rho g h$ : $3\times10^5 = 10^5 + 10^4 h \Rightarrow h = 20\text{ m}$ .
At $2h = 40\text{ m}$ : $P' = 10^5 + 10^4\times40 = 5\times10^5\text{ Pa}$ .
Increase $= \dfrac{5-3}{3}\times100 = \dfrac{200}{3}\%$
So the answer is (b).

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Q17NEET 2020 · PHASE 2Easy

Pressure in a Fluid · U-tube, barometer & connected vessels

A barometer is constructed using a liquid (density = $760\text{ kg/m}^{3}$ ). What would be the height of the liquid column, when a mercury barometer reads 76 cm?

(density of mercury = $13600\text{ kg/m}^{3}$ )

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Answer: (c)

Both barometers balance the same atmospheric pressure: $h_L\rho_L g = h_{Hg}\rho_{Hg} g$ .
$h_L = \dfrac{0.76\times13600}{760} = 13.6\text{ m}$
So the answer is (c) 13.6 m.

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Q18JEE Main 2020 · 4 SEP · SHIFT 2Medium

Pressure in a Fluid · U-tube, barometer & connected vessels

Two identical cylindrical vessels are kept on the ground and each contain the same liquid of density d. The area of the base of both vessels is S but the height of liquid in one vessel is x1 and in the other, x2. When both cylinders are connected through a pipe of negligible volume very close to the bottom, the liquid flows from one vessel to the other until it comes to equilibrium at a new height. The change in energy of the system in the process is:

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Answer: (c)

PE of a liquid column of height $x$ : $U = \tfrac{1}{2} dgSx^2$ .
Initial: $U_i = \tfrac{1}{2}dgS(x_1^2 + x_2^2)$ . Final height $h = \tfrac{x_1+x_2}{2}$ , so $U_f = dgS h^2 = \tfrac{1}{4}dgS(x_1+x_2)^2$ .
$U_i - U_f = \tfrac{1}{4}dgS\left[2x_1^2 + 2x_2^2 - (x_1+x_2)^2\right] = \tfrac{1}{4}dgS(x_2 - x_1)^2$
So the answer is (c).

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Q19JEE Main 2020 · 8 JAN · SHIFT 2Difficult

Pressure in a Fluid · Pressure at a depth

Two liquids of densities $\rho_1$ and $\rho_2$ ( $\rho_2$ = 2 $\rho_1$ ) are filled up behind a square wall of side 10 m as shown in figure. Each liquid has a height of 5 m. The ratio of the forces due to these liquids exerted on upper part MN to that at the lower part NO is (Assume that the liquids are not mixing)

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Answer: (c)

Force on a vertical strip = average pressure $\times$ area; both parts have area $5\times10 = 50\text{ m}^2$ .
MN: average pressure $= \rho_1 g\times2.5$ , so $F_1 = 2.5\rho_1 g\times50$ .
NO: average pressure $= \rho_1 g\times5 + \rho_2 g\times2.5 = 5\rho_1 g + 5\rho_1 g = 10\rho_1 g$ , so $F_2 = 10\rho_1 g\times50$ .
$\dfrac{F_1}{F_2} = \dfrac{2.5}{10} = \dfrac{1}{4}$
So the answer is (c) 1/4.

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Q20JEE Main 2020 · 2 SEP · SHIFT 1Difficult

Pressure in a Fluid · Rotating liquids

A cylindrical vessel containing a liquid is rotated about its axis so that the liquid rises at its sides as shown in the figure. The radius of vessel is 5 cm and the angular speed of rotation is $\omega$ rad s–1. The difference in the height, h (in cm) of liquid at the centre of vessel and at the side will be:

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Answer: (c)

Free surface of a rotating liquid is a paraboloid: $h = \dfrac{\omega^2 r^2}{2g}$ .
Radius $r = 5\text{ cm}$ (diameter 10 cm).
$h = \dfrac{\omega^2 (5)^2}{2g} = \dfrac{25\omega^2}{2g}$
So the answer is (c).

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Q21JEE Main 2019 · 10 APR · SHIFT 2Medium

Pressure in a Fluid · Pressure at a depth

A submarine experiences a pressure of 5.05 × 106 Pa at a depth of d1 in a sea. When it goes further to a depth of d2, it experiences a pressure of 8.08 × 106 Pa. Then d2 –d1 is approximately (density of water = 103 kg/m3 and acceleration due to gravity = 10 ms–2):

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Answer: (c)

$P_2 - P_1 = \rho g (d_2 - d_1)$
$d_2 - d_1 = \dfrac{(8.08 - 5.05)\times10^6}{10^3\times10} = 303\text{ m}$
So the answer is (c) about 300 m.

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Q22NEET 2019Medium

Pressure in a Fluid · U-tube, barometer & connected vessels

In a u-tube as shown in the fig. water and oil are in the left side and right side of the tube respectively. The heights from the bottom for water and oil columns are 15 cm and 20 cm respectively. The density of the oil is

[take $\rho_{\text{water}} = 1000\text{ kg/m}^{3}$ ]

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Answer: (c)

Pressures at the same level in the bottom (both sides open to air) must be equal: $\rho_w g h_w = \rho_o g h_o$ .
$1000\times15 = \rho_o\times20$
$\rho_o = 750\text{ kg/m}^3$
So the answer is (c).

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Q23MHT CET 2019 · 3 MAY · SHIFT 1Medium

Pressure in a Fluid · Air bubble rising in a lake

When a large bubble rises from bottom of a water lake to its surface, then its radius doubles. If the atmospheric pressure is equal to the pressure of height $H$ of a certain water column, then the depth of lake will be

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Answer: (c)

Radius doubles, so volume becomes 8 times. Boyle's law: $P_1V = P_2(8V)$ .
In water-column units: $H + h = 8H$
$h = 7H$
So the answer is (c) 7H.

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Q24JEE Main 2019 · 12 JAN · SHIFT 2Difficult

Pressure in a Fluid · Rotating liquids

A long cylindrical vessel is half filled with a liquid. When the vessel is rotated about its own vertical axis, the liquid rises up near the wall. If the radius of vessel is 5 cm and its rotational speed is 2 rotations per second, then the difference in the heights between the centre and the sides, in cm, will be:

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Answer: (a)

Height difference for a rotating liquid: $h = \dfrac{\omega^2 r^2}{2g}$ .
$\omega = 2\pi f = 4\pi\text{ rad/s}$ , $r = 0.05\text{ m}$ .
$h = \dfrac{16\pi^2\times0.0025}{2\times9.8} \approx 0.020\text{ m} = 2.0\text{ cm}$
So the answer is (a) 2.0.

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Q25JEE Main 2018 · 15 APR · SHIFT 2Difficult

Pressure in a Fluid · Air bubble rising in a lake

When an air bubble of radius r rises from the bottom to the surface of a lake its radius becomes $\frac{5r}{4}$ Taking the atmospheric pressure to be equal to 10 m height of water column, the depth of the lake would approximately be (ignore the surface tension and the effect of temperature):

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Answer: (c)

Volume ratio: $\dfrac{V_2}{V_1} = \left(\dfrac{5}{4}\right)^3 = \dfrac{125}{64}$ .
Boyle's law in water-column units: $(10 + h)\times64 = 10\times125$
$10 + h = 19.53 \Rightarrow h \approx 9.5\text{ m}$
So the answer is (c) 9.5 m.

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Q26NEET 2017Medium

Pressure in a Fluid · U-tube, barometer & connected vessels

A U tube with both ends open to the atmosphere, is partially filled with water. Oil, which is immiscible with water, is poured into one side until it stands at a distance of 10 mm above the water level on the other side. Meanwhile the water rises by 65 mm from its original level (see diagram). The density of the oil is

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Answer: (c)

Water rises 65 mm on one side and falls 65 mm on the other, so take level BC at the oil-water interface.
Water column above BC: $65 + 65 = 130\text{ mm}$ . Oil column above BC: $130 + 10 = 140\text{ mm}$ .
Equal pressure at B and C: $\rho_o\times140 = 1000\times130$
$\rho_o = 928.6 \approx 928\text{ kg m}^{-3}$
So the answer is (c).

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Q27MHT CET 2026 · 19 MAY · SHIFT 2Easy

Density, Compressibility & Pascal's Law · Pascal's law & hydraulic lift

In a hydraulic press, the piston in the small cylinder has a diameter of $d_1$ cm, while the larger piston has a diameter of $d_2$ cm. If a force $F_1$ is applied to the small piston, then the force $F_2$ on the large piston is given by

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Answer: (a)

Pascal's law: same pressure on both pistons, $\dfrac{F_1}{A_1} = \dfrac{F_2}{A_2}$ .
$A \propto d^2$ , so $F_2 = \dfrac{d_2^2}{d_1^2}F_1$ .
So the answer is (a).

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Q28MHT CET 2026 · 12 MAY · SHIFT 1Easy

Density, Compressibility & Pascal's Law · Pascal's law & hydraulic lift

In a hydraulic press, the small cylinder has a diameter of $d_1$ cm, while the large cylinder has a diameter of $d_2$ cm. If a force $F_1$ is applied to a piston of small cylinder, then the force on the piston of large cylinder $F_2$ is

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Answer: (d)

Pascal's law: $\dfrac{F_1}{A_1} = \dfrac{F_2}{A_2}$ with $A \propto d^2$ .
$F_2 = \dfrac{d_2^2}{d_1^2}F_1$
So the answer is (d).

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Q29MHT CET 2026 · 18 APR · SHIFT 1Easy

Density, Compressibility & Pascal's Law · Pascal's law & hydraulic lift

A piston of cross-sectional area $2.5 \times 10^{-2}$ m $^{2}$ is used in a hydraulic lift to exert a force of $250$ N on water. The cross-sectional area of the other piston which supports a car of mass $3000$ kg is ( $g = 9.8$ m/s $^{2}$ )

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Answer: (b)

Pascal's law: $\dfrac{F_1}{A_1} = \dfrac{F_2}{A_2} \Rightarrow A_2 = \dfrac{F_2 A_1}{F_1}$ .
$F_2 = 3000\times9.8 = 29400\text{ N}$
$A_2 = \dfrac{29400\times2.5\times10^{-2}}{250} = 2.94\text{ m}^2$
So the answer is (b).

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Q30JEE Main 2026 · 4 APR · SHIFT 1Medium

Density, Compressibility & Pascal's Law · Density, mixtures & compressibility

The increase in the pressure required to decrease the volume ( $\Delta V$ ) of water is $6.3 \times 10^{7}\text{ N/m}^{2}$ . The percentage decrease in the volume is ______.

(Bulk modulus of water $= 2.1 \times 10^{9}\text{ N/m}^{2}$ .)

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Answer: (b)

Bulk modulus: $B = \dfrac{\Delta P}{\Delta V/V} \Rightarrow \dfrac{\Delta V}{V} = \dfrac{\Delta P}{B}$ .
$\dfrac{\Delta V}{V} = \dfrac{6.3 \times 10^{7}}{2.1 \times 10^{9}} = 3 \times 10^{-2}$
So the answer is (b) $3\%$ .

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Q31JEE Main 2025 · 8 APR · SHIFT 2Easy

Density, Compressibility & Pascal's Law · Density, mixtures & compressibility

A sample of a liquid is kept at 1 atm. It is compressed to 5 atm which leads to a change of volume of 0.8 cm3. If the bulk modulus of the liquid is 2 GPa, the initial volume of the liquid was _______ litre. (Take 1 atm = 105 Pa)

Numerical answer type. Work it out, then open the answer.

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Answer: 4

$B = \dfrac{\Delta P}{\Delta V/V} \Rightarrow V = \dfrac{B\,\Delta V}{\Delta P}$
$\Delta P = 4\text{ atm} = 4\times10^5\text{ Pa}$ , $\Delta V = 0.8\times10^{-6}\text{ m}^3$ .
$V = \dfrac{2\times10^9\times0.8\times10^{-6}}{4\times10^5} = 4\times10^{-3}\text{ m}^3 = 4\text{ L}$
So the initial volume is 4 litre.

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Q32JEE Main 2025 · 24 JAN · SHIFT 2Easy

Density, Compressibility & Pascal's Law · Density, mixtures & compressibility

The increase in pressure required to decrease the volume of a water sample by $0.2\%$ is $P \times 10^{5}\text{ N m}^{-2}$ . Bulk modulus of water is $2.15 \times 10^{9}\text{ N m}^{-2}$ . The value of P is _________.

Numerical answer type. Work it out, then open the answer.

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Answer: 43

$\Delta P = B\dfrac{\Delta V}{V}$
$\Delta P = 2.15\times10^9\times0.002 = 4.3\times10^6 = 43\times10^5\text{ N m}^{-2}$
So $P = 43$ .

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Q33JEE Main 2025 · 29 JAN · SHIFT 1Easy

Density, Compressibility & Pascal's Law · Pascal's law & hydraulic lift

In a hydraulic lift, the surface area of the input piston is 6 cm2 and that of the output piston is 1500 cm2. If 100 N force is applied to the input piston to raise the output piston by 20 cm, then the work done is _______ kJ.

Numerical answer type. Work it out, then open the answer.

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Answer: 5

Output force: $F_2 = F_1\dfrac{A_2}{A_1} = 100\times\dfrac{1500}{6} = 25000\text{ N}$ .
Work done on the load: $W = F_2 d = 25000\times0.20 = 5000\text{ J}$
So the work done is 5 kJ.

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Q34JEE Main 2025 · 29 JAN · SHIFT 1Medium

Density, Compressibility & Pascal's Law · Density, mixtures & compressibility

The fractional compression $\left(\frac{\Delta V}{V}\right)$ of water at the depth of 2.5 km below the sea level is __________ %. Given, the Bulk modulus of water = $2 \times 10^{9}$ N m $^{-2}$ , density of water = $10^{3}$ kg m $^{-3}$ , acceleration due to gravity $g = 10$ m s $^{-2}$ .

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Answer: (b)

Extra pressure at depth $h$ : $P = \rho g h = 10^{3} \times 10 \times 2500 = 2.5 \times 10^{7}$ Pa.
$\dfrac{\Delta V}{V} = \dfrac{P}{B} = \dfrac{2.5 \times 10^{7}}{2 \times 10^{9}} = 1.25 \times 10^{-2}$
So the answer is (b) $1.25\%$ .

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Q35JEE Main 2024 · 5 APR · SHIFT 2Easy

Density, Compressibility & Pascal's Law · Pascal's law & hydraulic lift

A hydraulic press containing water has two arms with diameters as mentioned in the figure. A force of $10\text{ N}$ is applied on the surface of water in the thinner arm. The force required to be applied on the surface of water in the thicker arm to maintain equilibrium of water is _________ N.

Numerical answer type. Work it out, then open the answer.

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Answer: 1000

Pascal's law: $\dfrac{F_1}{A_1} = \dfrac{F_2}{A_2}$ , and $A \propto d^2$ .
$F_2 = 10\times\left(\dfrac{14}{1.4}\right)^2 = 10\times100$
So the force is 1000 N.

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Q36JEE Main 2024 · 31 JAN · SHIFT 1Medium

Density, Compressibility & Pascal's Law · Density, mixtures & compressibility

The depth below the surface of sea to which a rubber ball be taken so as to decrease its volume by $0.02\%$ is _______ $m$ .

(Take density of sea water $= 10^{3}\text{ kg m}^{-3}$ , Bulk modulus of rubber $= 9 \times 10^{8}\text{ N m}^{-2}$ , and $g = 10\text{ m s}^{-2}$ )

Numerical answer type. Work it out, then open the answer.

Show answer and solution

Answer: 18

$\dfrac{\Delta V}{V} = \dfrac{\rho g h}{B} \Rightarrow h = \dfrac{B\,(\Delta V/V)}{\rho g}$
$h = \dfrac{9 \times 10^{8} \times 2 \times 10^{-4}}{10^{3} \times 10} = \dfrac{1.8 \times 10^{5}}{10^{4}}$
$h = 18$ m

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Q37JEE Main 2024 · 27 JAN · SHIFT 1Medium

Density, Compressibility & Pascal's Law · Density, mixtures & compressibility

If average depth of an ocean is $4000\text{ m}$ and the bulk modulus of water is $2 \times 10^{9}\text{ N m}^{-2}$ , then fractional compression $\frac{\Delta V}{V}$ of water at the bottom of ocean is $\alpha \times 10^{-2}$ . The value of $\alpha$ is _______ (Given, $g = 10\text{ m s}^{-2},\ \rho = 1000\text{ kg m}^{-3}$ )

Numerical answer type. Work it out, then open the answer.

Show answer and solution

Answer: 2

Pressure at the bottom: $P = \rho g h = 1000 \times 10 \times 4000 = 4 \times 10^{7}$ Pa.
$\dfrac{\Delta V}{V} = \dfrac{P}{B} = \dfrac{4 \times 10^{7}}{2 \times 10^{9}} = 2 \times 10^{-2}$
So $\alpha = 2$ .

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Q38JEE Main 2023 · 1 FEB · SHIFT 1Easy

Density, Compressibility & Pascal's Law · Density, mixtures & compressibility

A certain pressure ' $P$ ' is applied to 1 litre of water and 2 litre of a liquid separately. Water gets compressed to $0.01\%$ whereas the liquid gets compressed to $0.03\%$ . The ratio of Bulk modulus of water to that of the liquid is $\frac{3}{x}$ . The value of $x$ is ____________.

Numerical answer type. Work it out, then open the answer.

Show answer and solution

Answer: 1

$B = \dfrac{P}{\Delta V/V}$ ; with the same $P$ , $B \propto \dfrac{1}{\Delta V/V}$ (original volume does not matter).
$\dfrac{B_w}{B_l} = \dfrac{0.03\%}{0.01\%} = 3 = \dfrac{3}{x}$
So $x = 1$ .

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Q39JEE Main 2023 · 8 APR · SHIFT 2Easy

Density, Compressibility & Pascal's Law · Pascal's law & hydraulic lift

A hydraulic automobile lift is designed to lift vehicles of mass $5000\text{ kg}$ . The area of cross section of the cylinder carrying the load is $250\text{ cm}^{2}$ . The maximum pressure the smaller piston would have to bear is $[$ Assume $g = 10\text{ m/s}^{2}]$

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Answer: (d)

Pressure is the same throughout the fluid (Pascal's law), so the small piston bears the same pressure as the load cylinder.
$P = \dfrac{mg}{A} = \dfrac{5000\times10}{250\times10^{-4}} = 2\times10^6\text{ Pa}$
So the answer is (d).

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Q40JEE Main 2023 · 6 APR · SHIFT 2Easy

Density, Compressibility & Pascal's Law · Pascal's law & hydraulic lift

Given below are two statements: one is labelled as Assertion $\mathbf{A}$ and the other is labelled as Reason $\mathbf{R}$

Assertion A: When you squeeze one end of a tube to get toothpaste out from the other end, Pascal's principle is observed.

Reason R: A change in the pressure applied to an enclosed incompressible fluid is transmitted undiminished to every portion of the fluid and to the walls of its container.

In the light of the above statements, choose the most appropriate answer from the options given below

Show answer and solution

Answer: (d)

Squeezing one end raises the pressure, and that pressure is transmitted through the paste to the open end, pushing it out: Pascal's principle. A is correct.
R is the statement of Pascal's law, which is correct and explains A.
So the answer is (d).

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Q41JEE Main 2022 · 24 JUN · SHIFT 1Easy

Density, Compressibility & Pascal's Law · Density, mixtures & compressibility

The bulk modulus of a liquid is 3 $\times$ 1010 Nm ${}^{-2}$ . The pressure required to reduce the volume of liquid by 2% is:

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Answer: (c)

$B = \dfrac{P}{\Delta V/V} \Rightarrow P = B\dfrac{\Delta V}{V}$
$P = 3\times10^{10}\times0.02 = 6\times10^8\text{ N m}^{-2}$
So the answer is (c).

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Q42JEE Main 2021 · 24 FEB · SHIFT 1Easy

Density, Compressibility & Pascal's Law · Pascal's law & hydraulic lift

A hydraulic press can lift 100 kg when a mass 'm' is placed on the smaller piston. It can lift ___________ kg when the diameter of the larger piston is increased by 4 times and that of the smaller piston is decreased by 4 times keeping the same mass 'm' on the smaller piston.

Numerical answer type. Work it out, then open the answer.

Show answer and solution

Answer: 25600

Load lifted $M = m\dfrac{A_2}{A_1} = m\left(\dfrac{D}{d}\right)^2$ .
New ratio: $\dfrac{4D}{d/4} = 16\dfrac{D}{d}$ , so area ratio becomes $16^2 = 256$ times larger.
$M' = 100\times256 = 25600\text{ kg}$
So it can lift 25600 kg.

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Q43JEE Main 2021 · 17 MAR · SHIFT 2Medium

Density, Compressibility & Pascal's Law · Density, mixtures & compressibility

An object is located at 2 km beneath the surface of the water. If the fractional compression $\frac{\Delta V}{V}$ is 1.36%, the ratio of hydraulic stress to the corresponding hydraulic strain will be ____________. [Given: density of water is 1000 kgm ${}^{-3}$ and g = 9.8 ms ${}^{-2}$ ]

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Answer: (b)

Hydraulic stress $= \rho g h = 1000 \times 9.8 \times 2000 = 1.96 \times 10^{7}$ Pa.
Stress/strain is the bulk modulus: $B = \dfrac{\rho g h}{\Delta V/V} = \dfrac{1.96 \times 10^{7}}{0.0136}$
$B \approx 1.44 \times 10^{9}\ \text{N m}^{-2}$
So the answer is (b) $1.44 \times 10^{9}\ \text{N m}^{-2}$ .

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Q44JEE Main 2021 · 26 FEB · SHIFT 1Medium

Density, Compressibility & Pascal's Law · Density, mixtures & compressibility

The normal density of a material is $\rho$ and its bulk modulus of elasticity is K. The magnitude of increase in density of material, when a pressure P is applied uniformly on all sides, will be:

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Answer: (c)

Mass is fixed, so $\rho V =$ constant $\Rightarrow \dfrac{\Delta \rho}{\rho} = \dfrac{|\Delta V|}{V}$ .
From bulk modulus: $\dfrac{|\Delta V|}{V} = \dfrac{P}{K}$ .
$\Delta \rho = \dfrac{\rho P}{K}$
So the answer is (c) $\dfrac{\rho P}{K}$ .

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Q45JEE Main 2021 · 31 AUG · SHIFT 1Medium

Density, Compressibility & Pascal's Law · Density, mixtures & compressibility

When a rubber ball is taken to a depth of __________m in deep sea, its volume decreases by 0.5%. (The bulk modulus of rubber = 9.8 $\times$ 108 Nm ${}^{-2}$ , Density of sea water = 103 kgm ${}^{-3}$ , g = 9.8 m/s2)

Numerical answer type. Work it out, then open the answer.

Show answer and solution

Answer: 500

$\dfrac{\Delta V}{V} = \dfrac{\rho g h}{B} \Rightarrow h = \dfrac{B\,(\Delta V/V)}{\rho g}$
$h = \dfrac{9.8 \times 10^{8} \times 0.005}{10^{3} \times 9.8} = \dfrac{4.9 \times 10^{6}}{9.8 \times 10^{3}}$
$h = 500$ m

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Q46JEE Main 2018Difficult

Density, Compressibility & Pascal's Law · Density, mixtures & compressibility

A solid sphere of radius r made of a soft material of bulk modulus K is surrounded by a liquid in a cylindrical container. A massless piston of area a floats on the surface of the liquid, covering entire cross section of cylindrical container. When a mass m is placed on the surface of the piston to compress the liquid, the fractional decrement in the radius of the sphere, $\left(\frac{dr}{r}\right)$ is:

Show answer and solution

Answer: (d)

Extra pressure on the liquid (Pascal's law): $P = \dfrac{mg}{a}$ , transmitted to the sphere.
$\dfrac{\Delta V}{V} = \dfrac{P}{K} = \dfrac{mg}{Ka}$
$V \propto r^{3} \Rightarrow \dfrac{\Delta V}{V} = 3\dfrac{dr}{r} \Rightarrow \dfrac{dr}{r} = \dfrac{mg}{3Ka}$
So the answer is (d) $\dfrac{mg}{3Ka}$ .

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Q47NEET 2015 · AIPMT · CANCELLED PAPERMedium

Density, Compressibility & Pascal's Law · Density, mixtures & compressibility

The approximate depth of an ocean is 2700 m. The compressiblity of water is 45.4 $\times$ 10 $-$ 11 Pa ${}^{-1}$ and density of water is 103 kg/m3. What fractional compression of water will be obtained at the bottom of the ocean?

Show answer and solution

Answer: (a)

Pressure at the bottom (gauge): $P = \rho g h = 10^3\times9.8\times2700 = 2.65\times10^7\text{ Pa}$ .
Fractional compression $= kP = 45.4\times10^{-11}\times2.65\times10^7$
$\dfrac{\Delta V}{V} \approx 1.2\times10^{-2}$
So the answer is (a).

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Q48JEE Main 2014 · 11 APR · SHIFT 1Easy

Density, Compressibility & Pascal's Law · Density, mixtures & compressibility

The Bulk moduli of Ethanol, Mercury and water are given as 0.9, 25 and 2.2 respectively in units of 109 Nm ${}^{-2}$ . For a given value of pressure, the fractional compression in volume is $\frac{\Delta V}{V}$ . Which of the following statements about $\frac{\Delta V}{V}$ for these three liquids is correct?

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Answer: (a)

$\dfrac{\Delta V}{V} = \dfrac{P}{B}$ , so for the same pressure the smaller $B$ gives larger compression.
$B$ : ethanol 0.9 < water 2.2 < mercury 25.
So ethanol > water > mercury; the answer is (a).

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Q49JEE Main 2010 · AIEEEMedium

Density, Compressibility & Pascal's Law · Density, mixtures & compressibility

A ball is made of a material of density $\rho$ where $\rho_{oil} < \rho < \rho_{water}$ with $\rho_{oil}$ and $\rho_{water}$ representing the densities of oil and water, respectively. The oil and water are immiscible. If the above ball is in equilibrium in a mixture of this oil and water, which of the following pictures represents its equilibrium position?

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Answer: (b)

Oil is lighter than water, so oil forms the upper layer and water the lower layer.
$\rho > \rho_{oil}$ : the ball sinks through oil. $\rho < \rho_{water}$ : it cannot sink into water fully.
So it rests at the oil–water interface, partly in each liquid.
So the answer is (b).

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Q50MHT CET 2026 · 12 MAY · SHIFT 1Easy

Archimedes' Principle & Floatation · Floatation

A vessel contains oil (density $= 0.6$ g/cm $^{3}$ ) over mercury (density $= 13.6$ g/cm $^{3}$ ). A homogeneous sphere floats with half its volume immersed in mercury and the remaining half in oil. The density of the material of the sphere in g/cm $^{3}$ is

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Answer: (c)

Weight = total upthrust: $V\rho g = \dfrac{V}{2}\rho_{Hg} g + \dfrac{V}{2}\rho_{oil} g$ .
$\rho = \dfrac{13.6 + 0.6}{2} = 7.1$ g/cm $^{3}$
So the answer is (c) $7.1$ .

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Q51MHT CET 2026 · 11 APR · SHIFT 1Easy

Archimedes' Principle & Floatation · Floatation

A vessel contains oil (density $= 0.8$ gm $/\text{cm}^{3}$ ) over mercury (density $= 13.6$ gm $/\text{cm}^{3}$ ). A homogeneous sphere floats with half of its volume immersed in mercury and the other half in oil. The density of the material of the sphere in gm $/\text{cm}^{3}$ is

Show answer and solution

Answer: (b)

Weight = total upthrust: $V\rho g = \dfrac{V}{2}\rho_{Hg} g + \dfrac{V}{2}\rho_{oil} g$ .
$\rho = \dfrac{13.6 + 0.8}{2} = 7.2$ g/cm $^{3}$
So the answer is (b) $7.2$ .

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Q52JEE Main 2026 · 24 JAN · SHIFT 2Easy

Archimedes' Principle & Floatation · Floatation

A cubical block of density $\rho_b = 600\text{ kg/m}^{3}$ floats in a liquid of density $\rho_{\text{e}} = 900\text{ kg/m}^{3}$ . If the height of block is $H = 8.0\text{ cm}$ then height of the submerged part is

______ cm.

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Answer: (d)

Floating: $\rho_b A H g = \rho_e A h g \Rightarrow h = \dfrac{\rho_b}{\rho_e} H$
$h = \dfrac{600}{900} \times 8.0 = 5.33$ cm
So the answer is (d) 5.3 cm.

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Q53JEE Main 2026 · 2 APR · SHIFT 1Medium

Archimedes' Principle & Floatation · Floatation

A tub is filled with water and a wooden cube $10$ cm $\times 10$ cm $\times 10$ cm is placed in the water. The wooden cube is found to float on the water with a part of it submerged in water. When a metal coin is placed on the wooden cube, the submerged part is increased by $3.87$ cm. The mass of the metal coin is ________ gram.

(Take water density as $1$ g/cm $^{3}$ and density of wood as $0.4$ g/cm $^{3}$ )

Numerical answer type. Work it out, then open the answer.

Show answer and solution

Answer: 387

Coin weight = extra upthrust = weight of extra water displaced.
Extra displaced volume $= 10 \times 10 \times 3.87 = 387$ cm $^{3}$ .
Mass of coin $= 387 \times 1 = 387$ g

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Q54JEE Main 2025 · 28 JAN · SHIFT 2Easy

Archimedes' Principle & Floatation · Floatation

A 400 g solid cube having an edge of length 10 cm floats in water. How much volume of the cube is outside the water?

(Given: density of water = 1000 kg m-3)

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Answer: (b)

Floating: displaced water mass = cube mass = 400 g, so submerged volume $= 400$ cm $^{3}$ (water is 1 g/cm $^{3}$ ).
Total volume $= 10^{3} = 1000$ cm $^{3}$ .
Volume outside $= 1000 - 400 = 600$ cm $^{3}$
So the answer is (b) 600 cm $^{3}$ .

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Q55MHT CET 2025 · 23 APR · SHIFT 2Medium

Archimedes' Principle & Floatation · Floatation

A solid cylinder of length $l$ and cross-sectional area $\frac{a}{5}$ is immersed such that it floats with its axis vertical at the liquid-liquid interface with length $l/4$ in the denser liquid as shown in figure. The lower density liquid ( $\rho$ ) is open to atmosphere having pressure $P_0$ . The density d of solid cylinder is

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Answer: (b)

From the figure, the lower liquid has density $3\rho$ and the upper liquid $\rho$ ; the cylinder has $l/4$ in $3\rho$ and $3l/4$ in $\rho$ .
Weight = upthrust: $d\,A l g = 3\rho\,A\dfrac{l}{4} g + \rho\,A\dfrac{3l}{4} g$
$d = \dfrac{3\rho}{4} + \dfrac{3\rho}{4} = \dfrac{3}{2}\rho$
So the answer is (b) $\dfrac{3}{2}\rho$ .

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Q56MHT CET 2025 · 19 APR · SHIFT 2Difficult

Archimedes' Principle & Floatation · Archimedes' principle & upthrust

A small metal sphere of density $\rho$ is dropped from height $h$ into a jar containing liquid of density $\sigma (\sigma > \rho)$ . The maximum depth up to which the sphere sinks is (Neglect damping forces)

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Answer: (d)

Speed on reaching the liquid: $v^2 = 2gh$ .
In the liquid, net upward force $= V(\sigma - \rho)g$ , so retardation $a = \dfrac{(\sigma - \rho)g}{\rho}$ .
Maximum depth $= \dfrac{v^2}{2a} = \dfrac{2gh\,\rho}{2(\sigma - \rho)g} = \dfrac{h\rho}{\sigma - \rho}$
So the answer is (d).

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Q57JEE Main 2024 · 9 APR · SHIFT 1Medium

Archimedes' Principle & Floatation · Floatation

A sphere of relative density $\sigma$ and diameter $D$ has concentric cavity of diameter $d$ . The ratio of $\frac{D}{d}$ , if it just floats on water in a tank is:

Show answer and solution

Answer: (d)

Just floats: weight of shell = upthrust on the whole sphere (fully submerged).
$\sigma \rho_w \dfrac{\pi}{6}(D^3 - d^3) g = \rho_w \dfrac{\pi}{6} D^3 g$
$\sigma D^3 - \sigma d^3 = D^3 \Rightarrow D^3(\sigma - 1) = \sigma d^3$
$\dfrac{D}{d} = \left(\dfrac{\sigma}{\sigma - 1}\right)^{1/3}$ , so the answer is (d).

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Q58JEE Main 2024 · 8 APR · SHIFT 2Medium

Archimedes' Principle & Floatation · Floatation

A cube of ice floats partly in water and partly in kerosene oil. The ratio of volume of ice immersed in water to that in kerosene oil (specific gravity of Kerosene oil = 0.8, specific gravity of ice = 0.9):

Show answer and solution

Answer: (d)

Let $V_w$ , $V_k$ be the volumes in water and kerosene, $V = V_w + V_k$ .
Weight = upthrust: $0.9\,V = 1 \cdot V_w + 0.8\,V_k$
$0.9(V_w + V_k) = V_w + 0.8 V_k \Rightarrow 0.1 V_k = 0.1 V_w$
$V_w : V_k = 1 : 1$ , so the answer is (d).

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Q59MHT CET 2024 · 4 MAY · SHIFT 2Difficult

Archimedes' Principle & Floatation · Archimedes' principle & upthrust

A hemispherical portion of radius ' $R$ ' is removed from the bottom of a cylinder of radius ' R '. The volume of the remaining cylinder is ' V ' and its mass is ' M '. It is suspended by a string in a liquid of density ' $\rho$ ', where it stays vertical. The upper surface of the cylinder is at a depth ' $h$ ' below the liquid surface. The force on the bottom of the cylinder by the liquid is

Show answer and solution

Answer: (d)

Upthrust = (force on bottom, upward) − (force on top, downward) $= V\rho g$ .
Force on top surface (gauge): $F_{top} = \rho g h \cdot \pi R^2$ .
$F_{bottom} = V\rho g + \pi R^2 h\rho g = \rho g(V + \pi R^2 h)$
So the answer is (d).

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Q60MHT CET 2023 · 13 MAY · SHIFT 1Difficult

Archimedes' Principle & Floatation · Archimedes' principle & upthrust

The fundamental frequency of a sonometer wire carrying a block of mass ' $M$ ' and density ' $\rho$ ' is ' $n$ ' Hz. When the block is completely immersed in a liquid of density ' $\sigma$ ' then the new frequency will be

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Answer: (a)

$n = \dfrac{1}{2l}\sqrt{\dfrac{T}{\mu}} \Rightarrow n \propto \sqrt{T}$ .
In air $T = Mg$ . Immersed: $T' = Mg - \dfrac{M}{\rho}\sigma g = Mg\,\dfrac{\rho - \sigma}{\rho}$ .
$n' = n\sqrt{\dfrac{T'}{T}} = n\left[\dfrac{\rho - \sigma}{\rho}\right]^{1/2}$
So the answer is (a).

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Q61MHT CET 2023 · 11 MAY · SHIFT 1Difficult

Archimedes' Principle & Floatation · Archimedes' principle & upthrust

A body of density ' $\rho$ ' is dropped from rest at a height ' $h$ ' into a lake of density ' $\sigma (\sigma > \rho)$ . The maximum depth to which the body sinks before returning to float on the surface is (neglect air dissipative forces)

Show answer and solution

Answer: (a)

Speed at the lake surface: $v^2 = 2gh$ .
In water, net upward force $= V(\sigma - \rho)g$ , so retardation $a = \dfrac{(\sigma - \rho)g}{\rho}$ .
Maximum depth $= \dfrac{v^2}{2a} = \dfrac{h\rho}{\sigma - \rho}$
So the answer is (a).

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Q62MHT CET 2021 · 20 SEP · SHIFT 2Difficult

Archimedes' Principle & Floatation · Archimedes' principle & upthrust

A body of density $V$ is dropped from (at rest) height ' $h$ ' into a lake of density ' $\delta$ ' $(\delta > \rho)$ . The maximum depth to which the body sinks before returning to float on the surface is [Neglect all dissipative forces]

Show answer and solution

Answer: (d)

Speed at the lake surface: $v^2 = 2gh$ .
In water, net upward force $= V(\delta - \rho)g$ , so retardation $a = \dfrac{(\delta - \rho)g}{\rho}$ .
Maximum depth $= \dfrac{v^2}{2a} = \dfrac{h\rho}{\delta - \rho}$
So the answer is (d).

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Q63JEE Main 2020 · 8 JAN · SHIFT 1Easy

Archimedes' Principle & Floatation · Archimedes' principle & upthrust

Consider a solid sphere of radius R and mass density

$\rho(r) = \rho_0 \left(1 - \frac{r^2}{R^2}\right),\ 0 < r \leq R$

The minimum density of a liquid in which it will float is:

Show answer and solution

Answer: (b)

It floats if the liquid density is at least the average density of the sphere.
$M = \int_0^R \rho_0\left(1 - \dfrac{r^2}{R^2}\right) 4\pi r^2\,dr = 4\pi\rho_0\left(\dfrac{R^3}{3} - \dfrac{R^3}{5}\right) = \dfrac{8\pi\rho_0 R^3}{15}$
$\rho_{avg} = \dfrac{M}{\frac{4}{3}\pi R^3} = \dfrac{8\rho_0}{15} \times \dfrac{3}{4} = \dfrac{2\rho_0}{5}$
So the answer is (b) $\dfrac{2\rho_0}{5}$ .

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Q64JEE Main 2020 · 5 SEP · SHIFT 1Medium

Archimedes' Principle & Floatation · Floatation

A hollow spherical shell of outer radius R floats just submerged under the water surface. The inner radius of the shell is r. If the specific gravity of the shell material is $\frac{27}{8}$ w.r.t water, the value of r is:

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Answer: (d)

Just submerged: weight of shell = upthrust on full outer volume.
$\dfrac{27}{8}\cdot\dfrac{4}{3}\pi(R^3 - r^3) = \dfrac{4}{3}\pi R^3 \Rightarrow R^3 - r^3 = \dfrac{8}{27}R^3$
$r^3 = \dfrac{19}{27}R^3 \Rightarrow r = (0.704)^{1/3} R \approx 0.89\,R$
$\dfrac{8}{9} = 0.889$ , so the answer is (d) $\dfrac{8}{9}R$ .

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Q65JEE Main 2020 · 8 JAN · SHIFT 1Medium

Archimedes' Principle & Floatation · Floatation

A leak proof cylinder of length 1m, made of a metal which has very low coefficient of expansion is floating vertically in water at 0°C such that its height above the water surface is 20 cm. When the temperature of water is increased to 4°C, the height of the cylinder above the water surface becomes 21 cm. The density of water at T = 4°C, relative to the density at T = 0°C is close to:

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Answer: (c)

Weight is constant, so $\rho\,h_{sub}$ = constant (same cross-section).
Submerged length: $100 - 20 = 80$ cm at $0^\circ$ C, $100 - 21 = 79$ cm at $4^\circ$ C.
$\rho_0 \times 80 = \rho_4 \times 79 \Rightarrow \dfrac{\rho_4}{\rho_0} = \dfrac{80}{79} \approx 1.013$
So the answer is (c) 1.01.

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Q66JEE Main 2020 · 4 SEP · SHIFT 1Difficult

Archimedes' Principle & Floatation · Archimedes' principle & upthrust

A air bubble of radius 1 cm in water has an upward acceleration 9.8 cm s–2. The density of water is 1 gm cm–3 and water offers negligible drag force on the bubble. The mass of the bubble is (g = 980 cm/s2).

Show answer and solution

Answer: (d)

Newton's law for the bubble: $V\rho_w g - mg = ma \Rightarrow m = \dfrac{V\rho_w g}{g + a}$ .
$V = \dfrac{4}{3}\pi(1)^3 = 4.19$ cm $^{3}$
$m = \dfrac{4.19 \times 1 \times 980}{980 + 9.8} = \dfrac{4.19}{1.01} \approx 4.15$ g
So the answer is (d) 4.15 gm.

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Q67JEE Main 2019 · 10 APR · SHIFT 2Medium

Archimedes' Principle & Floatation · Floatation

A cubical block of side 0.5 m floats on water with 30% of its volume under water. What is the maximum weight that can be put on the block without fully submerging it under water? [Take, density of water = 103 kg/m3]

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Answer: (b)

Volume $V = 0.5^3 = 0.125$ m $^{3}$ . Block mass = displaced water mass $= 0.3 \times 0.125 \times 10^{3} = 37.5$ kg.
Fully submerged, upthrust supports $0.125 \times 10^{3} = 125$ kg.
Extra load $= 125 - 37.5 = 87.5$ kg
So the answer is (b) 87.5 kg.

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Q68JEE Main 2019 · 9 APR · SHIFT 2Medium

Archimedes' Principle & Floatation · Floatation

A wooden block floating in a bucket of water has 4/5 of its volume submerged. When certain amount of an oil is poured into the bucket, it is found that the block is just under the oil surface with half of its volume under water and half in oil. The density of oil relative to that of water is :-

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Answer: (c)

In water alone: $\rho_b = \dfrac{4}{5}\rho_w = 0.8\rho_w$ .
With oil: $\rho_b V = \dfrac{V}{2}\rho_w + \dfrac{V}{2}\rho_{oil}$
$0.8 = 0.5 + 0.5\,\dfrac{\rho_{oil}}{\rho_w} \Rightarrow \dfrac{\rho_{oil}}{\rho_w} = 0.6$
So the answer is (c) 0.6.

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Q69JEE Main 2018 · 15 APR · SHIFT 1Difficult

Archimedes' Principle & Floatation · Floatation

A thin uniform tube is bent into a circle of radius $r$ in the vertical plane. Equal volumes of two immiscible liquids, whose densities are $\rho_1$ and $\rho_2\ (\rho_1 > \rho_2),$ fill half the circle. The angle $\theta$ between the radius vector passing through the common interface and the vertical is:

Show answer and solution

Answer: (c)

Each liquid fills a quarter circle ( $90^\circ$ ). With the interface at angle $\theta$ from the vertical, the free surfaces are at angles $\theta - 90^\circ$ and $\theta + 90^\circ$ .
Height of heavy liquid ( $\rho_1$ ) above the interface: $r(\cos\theta - \sin\theta)$ ; of light liquid ( $\rho_2$ ): $r(\cos\theta + \sin\theta)$ .
Equal pressure at the interface: $\rho_1 g r(\cos\theta - \sin\theta) = \rho_2 g r(\cos\theta + \sin\theta)$
$\tan\theta = \dfrac{\rho_1 - \rho_2}{\rho_1 + \rho_2}$ , so the answer is (c).

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Q70NEET 2016 · PHASE 1Medium

Archimedes' Principle & Floatation · Floatation

Two non-mixing liquids of densities $\rho$ and n $\rho$ (n > 1) are put in a container. The height of each liquid is h. A solid cylinder of length L and density d is put in this container. The cylinder floats with its axis vertical and length pL (p < 1) in the denser liquid. The density d is equal to

Show answer and solution

Answer: (b)

Length $pL$ is in liquid $n\rho$ and $(1-p)L$ in liquid $\rho$ .
Weight = upthrust: $d A L g = n\rho A pL g + \rho A (1-p) L g$
$d = \rho\,[np + 1 - p] = \{1 + (n-1)p\}\rho$
So the answer is (b).

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Q71JEE Main 2014Difficult

Archimedes' Principle & Floatation · Floatation

An open glass tube is immersed in mercury in such a way that a length of $8\text{ cm}$ extends above the mercury level. The open end of the tube is then closed and sealed and the tube is raised vertically up by additional $46\text{ cm}$ . What will be length of the air column above mercury in the tube now? (Atmospheric pressure $= 76\text{ cm}$ of $Hg$ )

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Answer: (a)

Initially air column 8 cm at 76 cm Hg. After raising, the tube top is $8 + 46 = 54$ cm above the outside mercury level.
Let air length be $x$ ; mercury column inside above outside level $= 54 - x$ , so air pressure $= 76 - (54 - x) = 22 + x$ cm Hg.
Boyle's law: $76 \times 8 = x(22 + x) \Rightarrow x^2 + 22x - 608 = 0$
$x = \dfrac{-22 + \sqrt{484 + 2432}}{2} = \dfrac{-22 + 54}{2} = 16$ cm
So the answer is (a) 16 cm.

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Q72JEE Main 2014Difficult

Archimedes' Principle & Floatation · Floatation

There is a circular tube in a vertical plane. Two liquids which do not mix and of densities $d_1$ and $d_2$ are filled in the tube. Each liquid subtends $90^\circ$ angle at center. Radius joining their interface makes an angle $\alpha$ with vertical. Ratio $\frac{d_1}{d_2}$ is:

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Answer: (c)

Same as a quarter-circle each: heavy liquid $d_1$ rises $r(\cos\alpha - \sin\alpha)$ above the interface, light liquid $d_2$ rises $r(\cos\alpha + \sin\alpha)$ .
Equal pressure at the interface: $d_1 g r(\cos\alpha - \sin\alpha) = d_2 g r(\cos\alpha + \sin\alpha)$
$\dfrac{d_1}{d_2} = \dfrac{\cos\alpha + \sin\alpha}{\cos\alpha - \sin\alpha} = \dfrac{1 + \tan\alpha}{1 - \tan\alpha}$
So the answer is (c).

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Q73JEE Main 2008 · AIEEEMedium

Archimedes' Principle & Floatation · Floatation

A jar is filled with two non-mixing liquids $1$ and $2$ having densities $\rho_1$ and $\rho_2$ respectively. A solid ball, made of a material of density $\rho_3$ , is dropped in the jar. It comes to equilibrium in the position shown in the figure. Which of the following is true for $\rho_1,\ \rho_1$ and $\rho_3$ ?

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Answer: (d)

Liquid 1 is on top of liquid 2, so $\rho_1 < \rho_2$ .
The ball sits at the interface: it sinks through liquid 1 ( $\rho_3 > \rho_1$ ) but floats on liquid 2 ( $\rho_3 < \rho_2$ ).
So the answer is (d) $\rho_1 < \rho_3 < \rho_2$ .

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