Q161JEE Main 2022MCQ

An air bubble of negligible weight having radius $r$ rises steadily through a solution of density $\sigma$ at speed $v$ . The coefficient of viscosity of the solution is given by : [30-Jun-2022-Shift-1]

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📖 Explanation

Air bubble moves with constant speed $v$ . So net force $=0$ . $\therefore$ Buoyant Force $=$ Viscous force $\Rightarrow F_b=F_v$ $\Rightarrow \sigma \times \;\frac{4}{3} \pi r^3 g=6 \pi n r v$ $\Rightarrow n=\;\frac{2 \sigma r^2 g}{9 v}$

Q162JEE Main 2022NAT

A string of area of cross-section $4 {\text{mm}}^2$ and length $0.5 {\text{m}}$ is connected with a rigid body of mass $2 {\text{kg}}$ . The body is rotated in a vertical circular path of radius $0.5 {\text{m}}$ . The body acquires a speed of $5 {\text{m}} / {\ s}$ at the bottom of the circular path. Strain produced in the string when the body is at the bottom of the circle is ____ $\times 10^{-5}$ (use Young's modulus $10^{11} {\text{N}} / {\text{m}}^2$ and ${\text{g}}=10 {\text{m}} / {\ s}^2$ ) [28-Jul-2022-Shift-2]

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📖 Explanation

Strain $=$ F/AY $=\;\frac{m g+\frac{m v^2}{R}}{A Y}$ $=\;\frac{20+\frac{2(5)^2}{0.5}}{3 \times 10^{-6} \times 10^{11}}=30 \times 10^{-5}$

Q163JEE Main 2022NAT

The diameter of an air bubble which was initially $2 {\text{mm}}$ , rises steadily through a solution of density $1750 {\text{kg}} {\text{m}}^{-3}$ at the rate of $0.35 {\text{cms}}^{-1}$ . The coefficient of viscosity of the solution is _______ poise (in nearest integer). (the density of air is negligible). [28-Jul-2022-Shift-1]

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📖 Explanation

As the bubble is rising steadily the net force acting on it will be zero (Because of density of air the value of ${\text{mg}}$ can be neglected)So $B=F \Rightarrow \;\frac{4 \pi }{3} R^3 \rho g=6 \pi η R v$ Putting $\;\; {\text{R}}=1 {\text{mm}}=10^{-3} {\text{m}}$ ${c} \rho =1.75 \times 10^3 {\text{kg}} / {\text{m}}^3$ ${\text{g}}=10 {\text{m}} / {\ s}^2$ ${\text{v}}=0.35 \times 10^{-2} {\text{m}} / {\ s}$ $η=\;\frac{10}{9} ≃ 1 \cdot 11 \;\text{ SI unit }\;=11 \;\text{ poise }\;( {\text{CGS}})$

Q164JEE Main 2022MCQ

Given below are two statements : One is labelled as Assertion A and the other is labelled as Reason R.Assertion A : Product of Pressure (P) and time (t) has the same dimension as that of coefficient of viscosity.Reason R : Coefficient of viscosity $=\;\frac{\;\text{ Force }\;}{\;\text{ Velocity gradient }\;}$ Choose the correct answer from the options given below : [28-Jun-2022-Shift-1]

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📖 Explanation

$[$ Pressure $][$ Time $]= [\;\frac{\;\text{ Force }\;}{\;\text{ Area }\;}] [\;\frac{\;\text{ distance }\;}{\;\text{ Area }\;}]$ [Coefficient of viscosity $]= [\;\frac{\;\text{ Force }\;}{\;\text{ Area }\;}] [\;\frac{\;\text{ distance }\;}{\;\text{ Area }\;}]$ Statement ' $A$ ' is trueBut Statement ' $R$ ' is false are coefficient of viscosity $=\;\frac{\;\text{ Force }\;}{\;\text{ Area }\; \times \;\text{ Velocity gradient }\;}$

Q165JEE Main 2022NAT

The area of cross-section of a large tank is $0.5 {\text{m}}^2$ . It has a narrow opening near the bottom having area of cross-section $1 {\text{cm}}$ . A load of $25 {\text{kg}}$ is applied on the water at the top in the tank. Neglecting the speed of water in the tank, the velocity of the water, coming out of the opening at the time when the height of water level in the tank is $40 {\text{cm}}$ above the bottom, will be ____ ${\text{cms}}^{-1} \cdot \$ [. $Take$ {\text{g}}=10 {\text{ms}}^{-2} $]\$ [27-Jun-2022-Shift-1]

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📖 Explanation

By Bernoulli's theorem: $\;\frac{250}{0.5}+\rho g h=\;\frac{1}{2} \rho v^2$ $\Rightarrow {\text{v}}=3 {\text{m}} / {\ s}$ $\Rightarrow {\text{v}}=300 {\text{cm}} / {\ s}$

Q166JEE Main 2022NAT

A tube of length $50 {\text{cm}}$ is filled completely with an incompressible liquid of mass $250 {\text{g}}$ and closed at both ends. The tube is then rotated in horizontal plane about one of its ends with a uniform angular velocity $x \sqrt{F} {\text{rads}}^{-1}$ . If ${\text{F}}$ be the force exerted by the liquid at the other end then the value of $x$ will be ________. [29-Jul-2022-Shift-2]

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📖 Explanation

${\text{F}}=∫( {\text{dm}}) \omega ^2 {\text{x}}$ $=∫_{0}^{ {\text{L}}} (\;{ {\text{m}}}/{ {\text{L}}} {\text{dx}}) \omega ^2 {\text{x}}$ $=\;{ {\text{m}}}/{ {\text{L}}} \omega ^2 \;{ {\text{L}}^2}/{2}$ $=\;{ {\text{m}} \omega ^2 {\text{L}}}/{2}$ $\omega =\sqrt{\;{2}/{ {\text{mL}}}} \sqrt{ {\text{F}}}$ $=\sqrt{\;\frac{2}{0.25 \times 0.5}} \sqrt{ {\text{F}}}$ $={\sqrt{16}} \sqrt{ {\text{F}}}$ $=4 \sqrt{ {\text{F}}}$

Q167JEE Main 2022NAT

A uniform heavy rod of mass $20 {\text{kg}}$ , cross sectional area $0.4 {\text{m}}^2$ and length $20 {\text{m}}$ is hanging from a fixed support. Neglecting the lateral contraction, the elongation in the rod due to its own weight is $x \times 10^{-9} {\text{m}}$ . The value of $x$ is ________. (Given, young modulus $Y=2 \times 10^{11} {\text{Nm}}^{-2}$ and ${\text{g}}=10 {\text{ms}}^{-2}$ ) [26-Jul-2022-Shift-2]

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📖 Explanation

$Y=\;\frac{T}{A} \;\frac{d x}{d y}$ ${\text{m}}=20 {\text{kg}}$ $A=\mathbf{0 . 4 \mathrm { m } ^ { 2 }}$ $1=20 {\text{m}}$ ${\text{Y}}=\;\frac{\;\text{ stress }\;}{\;\text{ strain }\;}$ $Y=\;{\;{ {\text{T}}}/{ {\text{A}}}}/{{ {\text{dy}}}/{ {\text{dx}}}}=\;{ {\text{T}}}/{ {\text{A}}} \cdot \;{ {\text{dx}}}/{ {\text{dy}}}$ $d y=\;\frac{T d x}{A Y}$ Tension at a distance ${\text{x}}$ from lower end $=\;{ {\text{mg}}}/{ℓ} {\text{x}}$ So. $∫_0^{\Delta l} d {\text{y}}=∫_0^{ℓ} \;{ {\text{mg}}}/{ℓ} {\text{x}} \;{ {\text{dx}}}/{ {\text{AY}}}$ $\Delta ℓ=\;{ {\text{mg}}}/{ℓ {\text{AY}}}\$ [;{ {\text{x}}^2}/{2}]_0^{ℓ}∆ ℓ=;{ {\text{mg}} ℓ}/{2 {\text{AY}}}∆ ℓ=;\frac{20 × 10 × 20}{2 × 0.4 × 2 × 10^{11}}2500 × 10^{-11}∆ ℓ=25 × 10^{-9}= {\text{x}} × 10^{-9}{\text{x}}=25$

Q168JEE Main 2022NAT

The velocity of upper layer of water in a river is $36 {\text{kmh}}^{-1}$ . Shearing stress between horizontal layers of water is $10^{-3} {\text{Nm}}^{-2}$ . Depth of the river is m. (Co-efficient of viscosity of water is _____ $10^{-2} {\text{Pa}} . {\text{s}}$ ) [25-Jun-2022-Shift-1]

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📖 Explanation

$F=-η A \;\frac{d u}{d x}$ $\Rightarrow 10^{-3}=10^{-2} \times \;\frac{10}{h}$ $\Rightarrow h=\;\frac{10^{-1}}{10^{-3}} {\text{m}}=100 {\text{m}}$ $\Rightarrow (100)$

Q169JEE Main 2022MCQ

A water drop of radius $1 {\text{cm}}$ is broken into 729 equal droplets. If surface tension of water is 75 dyne/ ${\text{cm}}$ , then the gain in surface energy upto first decimal place will be : (Given $\pi =3.14$ ) [26-Jul-2022-Shift-1]

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📖 Explanation

$729 \times \;\frac{4}{3} \pi r^3=\;\frac{4}{3} \pi R^3$ $\Rightarrow R=9 r . . . . . .(1)$ $\Delta U=S \times \Delta A . . . . . .(2)$ $\Rightarrow \Delta U=S \times \{-4 \pi R^{2+}729 \times 4 \pi r^2\}$ $=S \times 4 \pi \{729 r^{2-}81 r^2\}$ $=7.5 \times 10^{-4} J$

Q170JEE Main 2022NAT

The elastic behaviour of material for linear stress and linear strain, is shown in the figure. The energy density for a linear strain of $5 \times 10^{-4}$ is_____ ${\text{kJ}} / {\text{m}}^3$ . Assume that material is elastic upto the linear strain of $5 \times 10^{-4}$ . [26-Jun-2022-Shift-1]

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📖 Explanation

slope of strain - stress curve given by $=\;\frac{10^{-10}}{20}$ for strain of $5 \times 10^{-4}$ stress is given by $5 \times 10^{-4}=\;\frac{10^{-10}}{20} \times$ stressstress $=10^8 {\text{N}} / {\text{m}}^2$ Energy density $=\;\frac{1}{2} \times$ strain $\times$ stress $=\;\frac{1}{2} \times 5 \times 10^{-4} \times 10^8$ $=25000 {\text{J}} / {\text{m}}^3$ $=25 {\text{kJ}} / {\text{m}}^3$

Q171JEE Main 2022NAT

The speed of a transverse wave passing through a string of length $50 {\text{cm}}$ and mass $10 {\text{g}}$ is $60 {\text{ms}}^{-1}$ . The area of cross-section of the wire is $2.0 {\text{mm}}^2$ and its Young's modulus is $1.2 \times 10^{11} {\text{Nm}}^{-2}$ . The extension of the wire over its natural length due to its tension will be $x \times 10^{-5} {\text{m}}$ . The value of $x$ is _______. [29-Jul-2022-Shift-2]

🚀 Solve in Practice Mode

📖 Explanation

${\text{V}}_{ {\text{w}}}=\sqrt{\;{ {\text{T}}}/{\mu }}$ $60=\sqrt{\;{ {\text{T}}}/{10 \times 10^{-3}} \times 0.5}$ ${\text{T}}=\;\frac{(60)^2 \times 10^{-2}}{0.5}=72 {\text{N}}$ $\Delta ℓ=\;{ {\text{F}} ℓ}/{ {\text{AY}}}=\;\frac{72 \times 0.5}{2 \times 10^{-6} \times 1.2 \times 10^{11}}$ $=\;\frac{72 \times 5}{24} \times 10^{-5}=15 \times 10^{-5}$

Q172JEE Main 2022MCQ

A drop of liquid of density $\rho$ is floating half immersed in a liquid of density $\boldsymbol{\sigma }$ and surface tension $7.5 \times 10^{-4} {Ncm}^{-1}$ . The radius of drop in ${cm}$ will be : $(g=10 m s^{-2})$ [25-Jul-2022-Shift-2]

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📖 Explanation

Balancing the forces on drop $2 \pi R T+\;\frac{4}{3} \pi R^{3} \rho g=\;\frac{2}{3} \pi R^{3} \sigma g$ $\Rightarrow 2 T=\;\frac{2 R^{2}}{3}(\sigma -2 \rho ) \times 10$ $\Rightarrow \;\frac{15 \times 10^{-2} \times 3}{10(\sigma -2 \rho ) 2}=R^{2}$ $R=\;\frac{3}{2 \times 10} \sqrt{\;\frac{1}{(\sigma -2 \rho )}}$ $=\;\frac{3}{20} \sqrt{\;\frac{1}{\sigma -2 \rho }}(\;\text{ \in m }\;)$ $( {R})$ in ${cm}=\;\frac{3 \times 100}{20} \sqrt{\;\frac{1}{\sigma -2 \rho }}=15 \times \sqrt{\;\frac{1}{\sigma -2 \rho }}$ Now if $2 \rho >\sigma (R_{\;\text{\in }\; c m})=\;\frac{15}{\sqrt{2 \rho -\sigma }}$

Q173JEE Main 2022NAT

A square aluminum (shear modulus is $25 \times 10^9 {\text{Nm}}^{-2}$ ) slab of side $60 {\text{cm}}$ and thickness $15 {\text{cm}}$ is subjected to a shearing force (on its narrow face) of $18.0 \times 10^4 {\text{N}}$ . The lower edge is riveted to the floor. The displacement of the upper edge is _______ $\mu {\text{m}}$ . [27-Jul-2022-Shift-1]

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📖 Explanation

$\;{ {\text{F}}}/{ {\text{A}}}=η \;{ {\text{x}}}/{ℓ} \Rightarrow \;{ {\text{F}} ℓ}/{ {\text{A}} η}= {\text{x}}$ $\Rightarrow {\text{x}}=\;\frac{18 \times 10^4 \times 60 \times 10^{-2}}{60 \times 10^{-2} \times 15 \times 10^{-2} \times 25 \times 10^9}$ $=48 \times 10^{-6} {\text{m}}=48 \mu {\text{m}}$

Q174JEE Main 2022MCQ

A wire of length $L$ is hanging from a fixed support. The length changes to $L_1$ and $L_2$ when masses $1 {\text{kg}}$ and $2 {\text{kg}}$ are suspended respectively from its free end. Then the value of $L$ is equal to : [29-Jun-2022-Shift-1]

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📖 Explanation

By Hooke's Lawso ${\text{F}} \alpha \Delta {\text{L}}$ $\;{ {\text{F}}_1}/{ {\text{F}}_2}=\;{\Delta {\text{L}}_1}/{\Delta {\text{L}}_2}$ $\;\frac{10}{20}=\;{ ( {\text{L}}_1- {\text{L}}) }/{ ( {\text{L}}_2- {\text{L}}) }$ ${\text{L}}=2 {\text{L}}_1- {\text{L}}_2$

Q175JEE Main 2022NAT

A small spherical ball of radius $0.1 {\text{mm}}$ and density $10^4 {\text{kg}} {\text{m}}^{-3}$ falls freely under gravity through a distance h before entering a tank of water. If, after entering the water the velocity of ball does not change and it continue to fall with same constant velocity inside water, then the value of h will be _____m.(Given ${\text{g}}=10 {\text{ms}}^{-2}$ , viscosity of water $=1.0 \times 10^{-5} {\text{N}}- {\text{sm}}^{-2}$ ). [29-Jun-2022-Shift-2]

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📖 Explanation

${\sqrt{2 g h}}=\;\text{ terminal speed }\;$ $\Rightarrow {\sqrt{2 g h}}=\;\frac{2}{9} \;\frac{r^2 g (\rho -\rho ^{'}) }{η}$ $=\;\frac{2}{9} \times \;\frac{10^{-8} \times 10 \times 9000}{10^{-5}}$ $\Rightarrow h=\;\frac{400}{2 g}$ $\Rightarrow h=20 {\text{m}}$

Q176JEE Main 2022NAT

The excess pressure inside a liquid drop is $500 {\text{Nm}}^{-2}$ . If the radius of the drop is $2 {\text{mm}}$ , the surface tension of liquid is $x \times$ $10^{-3} {\text{Nm}}^{-1}$ . The value of $x$ is____ [30-Jun-2022-Shift-1]

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📖 Explanation

${\text{P}}= {\text{P}}_0+\;\frac{2 T}{R}$ $\Rightarrow P-P_0=\;\frac{2 T}{R}$ $500=\;\frac{2 \times T}{2 \times 10^{-3}}$ $T=500 \times 10^{-3}$ So, $x=500$

Q177JEE Main 2022MCQ

Potential energy as a function of ${\text{r}}$ is given by $U=\;\frac{A}{r^{10}}-\;\frac{B}{r^5}$ , where $r$ is the interatomic distance, A and ${\text{B}}$ are positive constants. The equilibrium distance between the two atoms will be : [24-Jun-2022-Shift-2]

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📖 Explanation

For equilibrium $-\;\frac{d U}{d r}=0=\;\frac{10 A}{r^{11}}-\;\frac{5 B}{r^6}$ $\Rightarrow r^5=\;\frac{2 A}{B}$ And $r=(\;\frac{2 A}{B})^{1 / 5}$

Q178JEE Main 2022NAT

A wire of length ${L}$ and radius ${r}$ is clamped rigidly at one end. When the other end of the wire is pulled by a force ${F}$ , its length increases by $5 {cm}$ . Another wire of the same material of length $4 {L}$ and radius $4 {r}$ is pulled by a force $4 {F}$ under same conditions. The increase in length of this wire is ${cm}$ [25-Jul-2022-Shift-1]

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📖 Explanation

$\;\frac{F / A}{\Delta L / L}=Y$ $\Rightarrow \Delta L=\;\frac{F L}{A Y}$ $\;\frac{\Delta L_{2}}{\Delta L_{1}}=(\;\frac{F_{2}}{F_{1}}) \times (\;\frac{L_{2}}{L_{1}}) \times (\;\frac{A_{1}}{A_{2}})$ $=4 \times 4 \times \;\frac{1}{16}=1$ $\Delta L_{2}=\Delta L_{1}=5 {cm} .$

Q179JEE Main 2022NAT

The elongation of a wire on the surface of the earth is $10^{-4} {\text{m}}$ . The same wire of same dimensions is elongated by $6 \times 10^{-5} {\text{m}}$ on another planet. The acceleration due to gravity on the planet will be ___ ${\text{ms}}^{-2}$ . (Take acceleration due to gravity on the surface of earth $=10 {\text{ms}}^{-2}$ ) [26-Jun-2022-Shift-1]

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📖 Explanation

on earth, $\Delta l=10^{-4} {\text{m}}$ on other planet $\Delta l^{'}=6 \times 10^{-5} {\text{m}}$ $\Delta l=\;\frac{F l}{A y} \Rightarrow \;\frac{\Delta l^{'}}{\Delta l}=\;{\;\frac{F^{'} l}{A y}}/{\frac{F l}{A y}}=\;\frac{m g^{'}}{m g}$ $\Rightarrow g^{'}=\;\frac{\Delta l^{'}}{\Delta l} \times g$ $=\;\frac{6 \times 10^{-5}}{10^{-4}} \times 10$ $\Rightarrow g^{'}=6 {\text{m}} / {\text{s}}^2$

Q180JEE Main 2022MCQ

A pressure-pump has a horizontal tube of cross sectional area $10 {\text{cm}}^2$ for the outflow of water at a speed of $20 {\text{m}} / {\ s}$ . The force exerted on the vertical wall just in front of the tube which stops water horizontally flowing out of the tube, is :$[given: density of water $=1000 {\text{kg}} / {\text{m}}^3$ ]$ [28-Jul-2022-Shift-2]

🚀 Solve in Practice Mode

📖 Explanation

$F_w=\rho A v^2$ $=10^3 \times 10 \times 10^{-4} \times 20 \times 20$ $=400 N$