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JEE Main · Physics

Properties Of Matter Practice Questions

Generate JEE Main level questions on Properties of Matter. Focus on Elasticity, Viscosity, and Surface Tension.

292 questions · 20 PYQs · 0 AI practice · JEE Main 2027

Q121JEE Main 2023MCQ

A body cools from $80^{\circ} {\text{C}}$ to $60^{\circ} {\text{C}}$ in 5 minutes. The temperature of the surrounding is $20^{\circ} {\text{C}}$ . The time it takes to cool from $60^{\circ} {\text{C}}$ to $40^{\circ} {\text{C}}$ is: [12-Apr-2023 shift 1]

🚀 Solve in Practice Mode

📖 Explanation

Body cools from $80^{\circ} {\text{C}}$ to $60^{\circ} {\text{C}}$ in $5 {\min}$ ${\text{T}}_{\;\text{suroumding }\;}=20^{\circ} {\text{C}}$ Time taken to cool from $60^{\circ} {\text{C}}$ to $40^{\circ} {\text{C}}$ ? we know newton's law of cooling is $-\;{ {\text{d}} \theta }/{ {\text{dt}}}= {\text{b}} (\theta -\theta _0)$ surrounding temp. We take approx value of ${\text{v}}$ as $\;\frac{80+60}{2}=70^{\circ}$ $\;\;\frac{20}{5}= {\text{b}}[70-20]$ . . . (1) $\;\;{20}/{ {\text{t}}}= {\text{b}}[50-20]$ . . . (2)After solving ${\text{t}}=\;\frac{25}{3} {\min}$ In second ${\text{t}}=\;\frac{25}{3} \times 60$ $=500 {\ sec} \;\text{. }\;$

Q122JEE Main 2023MCQ

A fully loaded boeing aircraft has a mass of $5.4 \times 10^5 {\text{kg}}$ . Its total wing area is $500 {\text{m}}^2$ . It is in level flight with a speed of $1080 {\text{km}} / {\text{h}}$ . If the density of air $\rho$ is $1.2 {\text{kg}} {\text{m}}^{-3}$ , the fractional increase in the speed of the air on the upper surface of the wing relative to the lower surface in percentage will be $( {\text{g}}=10 {\text{m}} / {\ s}^2)$ [29-Jan-2023 Shift 2]

🚀 Solve in Practice Mode

📖 Explanation

$\; {\text{P}}_2 {\text{A}}- {\text{P}}_1 {\text{A}}=5.4 \times 10^5 \times {\text{g}}$ ${\text{P}}_2- {\text{P}}_1=\;\frac{5.4 \times 10^6}{500}=5.4 \times 2 \times 10^2 \times 10$ ${\text{P}}_2+0+\;\frac{1}{2} \rho {\text{V}}_2^2= {\text{P}}_1+0+\;\frac{1}{2} \rho {\text{V}}_1^2$ ${\text{P}}_2- {\text{P}}_1=\;\frac{1}{2} \rho ( {\text{V}}_1^{2-} {\text{V}}_2^2) =\;\frac{1}{2} \rho ( {\text{V}}_1- {\text{V}}_2) ( {\text{V}}_1+ {\text{V}}_2)$ $10.8 \times 10^3=\;\frac{1}{2} \times 1.2 ( {\text{V}}_1- {\text{V}}_2) \times 2 \times 3 \times 10^2$ $10.8 \times 10=3.6 ( {\text{V}}_1- {\text{V}}_2)$ ${\text{V}}_1- {\text{V}}_2=30$ $(\;{ {\text{V}}_1- {\text{V}}_2}/{ {\text{V}}}) \times 100=\;\frac{30}{300} \times 100=10 \%$

Q123JEE Main 2023MCQ

An aluminium rod with Young's modulus $Y=7.0 \times 10^{10} {\text{N}} / {\text{m}}^2$ undergoes elastic strain of $0.04 \%$ . The energy per unit volume stored in the rod in SI unit is: [8-Apr-2023 shift 1]

🚀 Solve in Practice Mode

📖 Explanation

Aluminium rod Young's modulus $\; {\text{y}}=7.0 \times 10^{10} \;{ {\text{N}}}/{ {\text{m}}^2}$ $\;\;\text{ strain }\; 0.04 \%$ $\;\;\text{ strain }\;=\;\frac{0.04}{100}$ Energy per unit volume $=\;\frac{1}{2}$ stress $\times$ strain $\;=\;\frac{1}{2} {\text{y}} \;\text{ strain }\; \times \;\text{ strain }\;$ $\;=\;\frac{1}{2} {\text{y}}(\;\text{ strain }\;)^2$ $\;=\;\frac{1}{2} \times 7 \times 10^{10} \times (\;\frac{0.04}{100}) ^2$ Energy per unit volume $=5600 \;{ {\text{J}}}/{ {\text{m}}^3}$

Q124JEE Main 2023MCQ

The height of liquid column raised in a capillary tube of certain radius when dipped in liquid A vertically is, $5 {\text{cm}}$ . If the tube is dipped in a similar manner in another liquid $B$ of surface tension and density double the values of liquid $A$ , the height of liquid column raised in liquid $B$ would be _______ ${\text{m}}$ . [30-Jan-2023 Shift 1]

🚀 Solve in Practice Mode

📖 Explanation

$\;\; {\text{h}}=\;{2 {\text{S}} \cos \theta }/{ {\text{r} \rho g}}$ $\therefore \;\; \;{ {\text{h}}_1}/{ {\text{h}}_2}\;=\;{ {\text{S}}_1}/{ {\text{S}}_2} \;\frac{\rho _2}{\rho _1}$ $\;{5}/{ {\text{h}}_2}\;= [\;\frac{1}{2}] [\;\frac{2}{1}] \Rightarrow {\text{h}}_2=5 {\text{cm}}=0.05 {\text{m}}$ {Info about angle of contact not there so most appropriate is 3 $\}$

Q125JEE Main 2023NAT

A Spherical ball of radius $1 {\text{mm}}$ and density $10.5 {\text{g}} / {\text{cc}}$ is dropped in glycerine of coefficient of viscosity $9.8$ poise and density $1.5 {\text{g}} / {\text{cc}}$ . Viscous force on the ball when it attains constant velocity is $3696 \times 10^{-x} {\text{N}}$ . The value of ${\text{x}}$ is(Given, ${\text{g}}=9.8 {\text{m}} / {\ s}^2$ and $\pi =\;\frac{22}{7}$ ) [24-Jan-2023 Shift 2]

🚀 Solve in Practice Mode

📖 Explanation

When the ball attain terminal velocity $\; {\text{F}}_{ {\text{v}}}= ( {\text{mg}}- {\text{F}}_{ {\text{B}}}) ( \because {\text{a}}=0)$ $\;= {\text{V}} \sigma _{ {\text{b}}} {\text{g}}- {\text{V}} \rho _{ℓ} {\text{g}}$ $\;={\text{Vg}} (\sigma _{ {\text{b}}}-\rho _{ℓ})$ $\;=\;\frac{4}{3} \pi (10^{-3}) ^3 \times 9.8(10.5-1.5) \times 10^3$ $\;=3696 \times 10^{-7} {\text{N}}$ $\;\;\text{ So, }\; {\text{x}}=7$

Q126JEE Main 2023MCQ

A mercury drop of radius $10^{-3} {\text{m}}$ is broken into 125 equal size droplets. Surface tension of mercury is $0.45 {\text{Nm}}^{-1}$ . The gain in surface energy is: [1-Feb-2023 Shift 1]

🚀 Solve in Practice Mode

📖 Explanation

Sol. Initial surface energy $=0.45 \times 4 \pi (10^{-3}) ^2$ $\;\;\frac{4}{3} \pi (10^{-3}) ^3=125 \times \;\frac{4 \pi }{3} R_{\;\text{new }\;}^3$ $\therefore \;10^{-3}=5 R_{\;\text{new }\;}$ $\therefore \;R_{\;\text{new }\;}=\;\frac{10^{-3}}{5} {\text{m}}$ So, final surface energy $=0.45 \times 125 \times 4 \pi (\;\frac{10^{-3}}{5}) ^2$ Increase in energy $=0.45 \times 4 \pi \times (10^{-3}) ^2 [\;\frac{125}{25}-1]$ $=4 \times 0.45 \times 4 \pi \times 10^{-6}$ $=2.26 \times 10^{-5} {\text{J}}$

Q127JEE Main 2023NAT

A spherical drop of liquid splits into 1000 identical spherical drops. If ${\text{u}}_{ {\text{i}}}$ is the surface energy of the original drop and ${\text{u}}_{ {\text{f}}}$ is the total surface energy of the resulting drops, the (ignoring evaporation). $\;{ {\text{u}}_{ {\text{f}}}}/{ {\text{u}}_{ {\text{i}}}}= (\;{10}/{ {\text{x}}})$ . Then value of ${\text{x}}$ is _______: [25-Jan-2023 Shift 2]

🚀 Solve in Practice Mode

📖 Explanation

Surface Tension $=T$ $R$ : Radius of bigger drop $r$ : Radius of smaller dropVolume will remain same $\;\;\frac{4}{3} \pi {\text{R}}^3=1000 \times \;\frac{4}{3} \pi {\text{r}}^3$ $\; {\text{R}}=10 {\text{r}}$ $\; {\text{u}}_{ {\text{i}}}= {\text{T}} \cdot 4 \pi {\text{R}}^2$ $\; {\text{u}}_{ {\text{f}}}= {\text{T}} \cdot 4 \pi {\text{r}}^2 \times 1000$ $\;\;{ {\text{u}}_{ {\text{f}}}}/{ {\text{u}}_{ {\text{i}}}}=\;{1000 {\text{r}}^2}/{ {\text{R}}^2}$ $\;\;{ {\text{u}}_{ {\text{f}}}}/{ {\text{u}}_{ {\text{i}}}}=\;\frac{10}{1}$

Q128JEE Main 2023MCQ

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 ${\text{g}}=10 {\text{m}} / {\ s}^2$ ]$ : [8-Apr-2023 shift 2]

🚀 Solve in Practice Mode

📖 Explanation

From pascal law same $\Delta {\text{P}}$ transmitted through out liquid $\;\Delta {\text{P}}=\;{ {\text{F}}}/{ {\text{A}}}=\;\frac{5000 \times 10}{250 \times 10^{-4}}$ $\;=2 \times 10^6 {\text{Pa}}$

Explanation Figure 1

Q129JEE Main 2023MCQ

Under isothermal condition, the pressure of a gas is given by ${\text{P}}= {\text{aV}}^{-3}$ , where ${\text{a}}$ is a constant and ${\text{V}}$ is the volume of the gas. The bulk modulus at constant temperature is equal to [13-Apr-2023 shift 1]

🚀 Solve in Practice Mode

📖 Explanation

$\; {\text{P}}= {\text{aV}}^{-3}$ $\;\;{ {\text{dP}}}/{ {\text{dV}}}=-3 {\text{aV}}^{-4}$ Bulk modulus, $B=-V \;\frac{d P}{d V}$ $\; {\text{B}}=- {\text{V}} (\;{-3 {\text{a}}}/{ {\text{V}}^4})$ $\; {\text{B}}=3 \;{ {\text{a}}}/{ {\text{V}}^3}=3 {\text{P}}$

Q130JEE Main 2023NAT

The elastic potential energy stored in a steel wire of length $20 {\text{m}}$ stretched through $2 {\text{cm}}$ is $80 {\text{J}}$ . The cross sectional area of the wire is ________ $(.$ Given, $y=2.0 \times 10^{11} {\text{Nm}}^{-2}$ ) ${\text{mm}}^2$ . [13-Apr-2023 shift 1]

🚀 Solve in Practice Mode

📖 Explanation

$\;\;\text{ Energy, }\; {\text{U}}=\;\frac{1}{2} {\text{kx}}^2$ $\;80=\;\frac{1}{2} {\text{k}} (2 \times 10^{-2}) ^2$ $\; {\text{k}}=\;\frac{160}{4 \times 10^{-4}}$ $\; {\text{k}}=4 \times 10^5 {\text{N}} / {\text{m}}$ $\;\;{ {\text{yA}}}/{ℓ}=4 \times 10^5$ $\; {\text{A}}=\;\frac{4 \times 10^5 \times 20}{2 \times 10^{11}}$ $\; {\text{A}}=40 \times 10^{-6} {\text{m}}^2$ $\; {\text{A}}=40 {\text{mm}}^2$

Q131JEE Main 2023MCQ

The frequency $(v)$ of an oscillating liquid drop may depend upon radius (r) of the drop, density $(\rho )$ of liquid and the surface tension (s) of the liquid as : $v= {\text{r}}^{ {\text{a}}} \rho ^{ {\text{b}}} {\ s}^{ {\text{c}}}$ . The values of ${\text{a}}, {\text{b}}$ and ${\text{c}}$ respectively are [24-Jan-2023 Shift 2]

🚀 Solve in Practice Mode

📖 Explanation

$\;{[ {\text{T}}^{-1}]\$ = $[ {\text{L}}^1]$ ^{ {\text{a}}} $[ {\text{M}}^1 {\text{L}}^{-3}]$ ^{ {\text{b}}} $[;{ {\text{MLT}}^{-2}}/{ {\text{L}}}]$ ^{ {\text{c}}}};\Rightarrow {\text{T}}^{-1}= {\text{M}}^{ {\text{b}}+ {\text{c}}} ⋅ {\text{L}}^{ {\text{a}}-3 {\text{b}}} ⋅ {\text{T}}^{-2 {\text{c}}}; {\text{c}}=;\frac{1}{2}, {\text{b}}=-;\frac{1}{2}, ;; {\text{a}}-3 {\text{b}}=0; {\text{a}}+;\frac{3}{2}=0 \Rightarrow {\text{a}}=-;\frac{3}{2}$

Q132JEE Main 2023MCQ

Given below are two statements: one is labelled as Assertion A and the other is labelled as Reason RAssertion A : When you squeeze one end of a tube to get toothpaste out from the other end. Pascal's principle is observed.Reason $\text{bo} {R}$ : 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 [6-Apr-2023 shift 2]

🚀 Solve in Practice Mode

📖 Explanation

As per pascal's law, when we apply pressure to an ideal liquid it is equally distributed in the entire liquid and to the walls as well.Since due to applied pressure, every morning, the tooth paste does not get compressed and we can safely consider it on incompressible liquid.Therefore both statements are true and ${\text{R}}$ is correct explanation of ${\text{A}}$ .

Q133JEE Main 2023NAT

A thin rod having a length of $1 {\text{m}}$ and area of cross-section $3 \times 10^{-6} {\text{m}}^2$ is suspended vertically from one end. The rod is cooled from $210^{\circ} {\text{C}}$ to $160^{\circ} {\text{C}}$ . After cooling, a mass ${\text{M}}$ is attached at the lower end of the rod such that the length of rod again becomes $1 {\text{m}}$ . Young's modulus and coefficient of linear expansion of the rod are $2 \times 10^{11} {\text{Nm}}^{-2}$ and $2 \times 10^{-5} {\text{K}}^{-1}$ , respectively. The value of $M$ is _______ ${\text{kg}}$ . (Take ${\text{g}}=10 {\text{m}} {\ s}^{-2}$ ) [31-Jan-2023 Shift 1]

🚀 Solve in Practice Mode

📖 Explanation

If $\Delta ℓ$ is decease in length of rod due to decease in temperature $\;\Delta ℓ=ℓ \alpha \Delta {\text{T}}$ $\;\alpha =2 \times 10^{-5} {\text{K}}^{-1}, \Delta {\text{T}}=(210-160)$ $\;=50 {\text{K}}$ $\;\Delta ℓ=1 \times 2 \times 10^{-5} \times 50=10^{-3} {\text{m}}$ $\;\;\text{ Young Modulus }\;= {\text{Y}}=\;{ {\text{F}} / {\text{A}}}/{\Delta ℓ / ℓ} \;\; {\text{A}}=3 \times 10^{-6} {\text{m}}^2$ $\;2 \times 10^{11}=\;{ {\text{Mg}} / 3 \times 10^{-6}}/{10^{-3} / 1}$ $\; {\text{Mg}}=2 \times 10^{11} \times 3 \times 10^{-9}=6 \times 10^{-2}$ $\;M=60 {\text{kg}}$ $\;$ Ans is 60 .

Explanation Figure 1

Q134JEE Main 2023NAT

There is an air bubble of radius $1.0 {\text{mm}}$ in a liquid of surface tension $0.075 {\text{nm}}^{-1}$ and density $1000 {\text{kg}} {\text{m}}^{-3}$ at a depth of $10 {\text{cm}}$ below the free surface. The amount by which the pressure inside the bubble is greater than the atmospheric pressure is ________ ${\text{Pa}} ( {\text{g}}=10 {\text{ms}}^{-2})$ [15-Apr-2023 shift 1]

🚀 Solve in Practice Mode

📖 Explanation

$P_{i n}=P_0+\rho g h+\;\frac{2 T}{r}$ $\;P_{\;\text{\in }\;}-P_0=1000 \times 10 \times 0.1+\;\frac{2 \times 0.075}{0.001}$ $\;=1000+150$ $\;P_{\;\text{\in }\;}-P_0=1150 {\text{Pa}}$

Explanation Figure 1

Q135JEE Main 2023MCQ

The figure shows a liquid of given density flowing steadily in horizontal tube of varying cross - section. Cross sectional areas at $A$ is $1.5 {\text{cm}}^2$ , and $B$ is $25 {\text{mm}}^2$ , if the speed of liquid at $B$ is $60 {\text{cm}} / {\ s}$ then $(P_A-P_B)$ is: (Given ${\text{P}}_{ {\text{A}}}$ and ${\text{P}}_{ {\text{B}}}$ are liquid pressures at ${\text{A}}$ and ${\text{B}}$ points)Density $\rho =1000 {\text{kg}} {\text{m}}^{-3}$ $A$ and $B$ are on the axis of tube [13-Apr-2023 shift 1]

🚀 Solve in Practice Mode

📖 Explanation

By equation of continuity, $\; {\text{A}}_1 {\text{V}}_1= {\text{A}}_2 {\text{V}}_2$ $\; (1.5 \times 10^{-4}) {\text{V}}_{ {\text{A}}}= (25 \times 10^{-6}) 60 {\text{cm}} / {\ s}$ $\; {\text{V}}_{ {\text{A}}}=10 {\text{cm}} / {\ s}$ By Bernoulli's theorem. $\; {\text{P}}_{ {\text{A}}}+\;\frac{1}{2} \rho {\text{V}}_{ {\text{A}}}^2= {\text{P}}_{ {\text{B}}}+\;\frac{1}{2} \rho {\text{V}}_{ {\text{B}}}^2$ $\; {\text{P}}_{ {\text{A}}}- {\text{P}}_{ {\text{B}}}=\;\frac{\rho }{2} ( {\text{V}}_{ {\text{B}}}^{2-} {\text{V}}_{ {\text{A}}}^2)$ $\; {\text{P}}_{ {\text{A}}}- {\text{P}}_{ {\text{B}}}=\;\frac{1000}{2} (60^{2-}10^2) \times 10^{-4}$ $\; {\text{P}}_{ {\text{A}}}- {\text{P}}_{ {\text{B}}}=175 {\text{Pa}}$

Q136JEE Main 2023NAT

A body cools from $60^{\circ} {\text{C}}$ to $40^{\circ} {\text{C}}$ in 6 minutes. If, temperature of surroundings is $10^{\circ} {\text{C}}$ . Then, after the next 6 minutes, its temperature will be ${ }^{\circ} {\text{C}}$ . [29-Jan-2023 Shift 1]

🚀 Solve in Practice Mode

📖 Explanation

By average form of Newton's law of cooling $\;\frac{20}{6}= {\text{k}}(50-10)$ . . . (i) $\;\frac{40-T}{6}=K (\;\frac{40+T}{2}-10)$ . . . (ii)From equation (i) and (ii) $\;\;{20}/{40- {\text{T}}}=\;{40}/{10+ {\text{T}} / 2}$ $\;10+\;{ {\text{T}}}/{2}=80-2 {\text{T}}$ $\;\;{5 {\text{T}}}/{2}=70 \Rightarrow {\text{T}}=28^{\circ} {\text{C}}$

Q137JEE Main 2022NAT

In an experiment to verify Newton's law of cooling, a graph is plotted between the temperature difference $(\Delta T)$ of the water and surroundings and time as shown in figure. The initial temperature of water is taken as $80^{\circ} {\text{C}}$ . The value of ${\text{t}}_2$ as mentioned in the graph will be [24-Jun-2022-Shift-2]

🚀 Solve in Practice Mode

📖 Explanation

Temperature of surrounding $=20^{\circ} {\text{C}}$ For $0 \to 6$ minutes, average temp. $=70^{\circ} {\text{C}}$ $\to$ Rate of cooling $\propto 70^{\circ} {\text{C}}-20^{\circ} {\text{C}}=50^{\circ} {\text{C}}$ For $6 \to t_2$ minutes, average temp. $=50^{\circ} {\text{C}}$ $\to$ Rate of cooling $\propto 30^{\circ} {\text{C}}$ $\Rightarrow t_2-6=\;\frac{5}{3}(6$ minutes $)$ $\Rightarrow {\text{t}}_2=16$ minutes

Q138JEE Main 2022MCQ

A steel wire of length $3.2 {\text{m}}( {\text{Y}}_{ {\text{s}}}=2.0 \times 10^{11} {\text{Nm}}^{-2})$ and a copper wire of length $4.4 {\text{m}}( {\text{Y}}_{ {\text{c}}}=1.1 \times 10^{11} {\text{Nm}}^{-2})$ , both of radius $1.4 {\text{mm}}$ are connected end to end. When stretched by a load, the net elongation is found to be $1.4 {\text{mm}}$ . The load applied, in Newton, will be: ( Given $.\pi =\;\frac{22}{7})$ [27-Jul-2022-Shift-2]

🚀 Solve in Practice Mode

📖 Explanation

$\Delta ℓ_1+\Delta ℓ_2=\Delta ℓ$ $\;{ {\text{F}} ℓ_1}/{ {\text{A}}_1 {\text{y}}_1}+\;{ {\text{F}} ℓ_2}/{ {\text{A}}_2 {\text{y}}_2}=\Delta ℓ$ ${\text{F}}=\;{\Delta ℓ}/{{ℓ_1}/{ {\text{A}}_1 {\text{y}}_1}+\;{ℓ_2}/{ {\text{A}}_2 {\text{y}}_2}}=1.54 \times 10^2=154$

Explanation Figure 1

Q139JEE Main 2022MCQ

The velocity of a small ball of mass ' $m$ ' and density $d_1$ , when dropped in a container filled with glycerin, becomes constant after some time. If the density of glycerin is ${\text{d}}_2$ , then the viscous force acting on the ball, will be : [27-Jun-2022-Shift-1]

🚀 Solve in Practice Mode

📖 Explanation

Viscous force acting on the ball will be equal and opposite to net of weight and buoyant force $\Rightarrow F_0=\;\frac{4}{3} \pi r^3 d_1 g-\;\frac{4}{3} \pi r^3 d_2 g$ $=\;\frac{4}{3} \pi r^3 d_1 g (1-\;\frac{d_2}{d_1})$ $=m g (1-\;\frac{d_2}{d_1})$

Q140JEE Main 2022MCQ

Given below are two statements : One is labelled as Assertion (A) and the other is labelled as Reason (R) Assertion (A): Clothes containing oil or grease stains cannot be cleaned by water wash. Reason (R): Because the angle of contact between the oil/grease and water is obtuse. In the light of the above statements, choose the correct answer from the option given below. [29-Jul-2022-Shift-1]

🚀 Solve in Practice Mode

📖 Explanation

Due to obtuse angle of contact the water doesn't wet the oiled surface properly and cannot wash it also. $\Rightarrow$ Assertion is correct and Reason given is a correct explanation.

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