Q1JEE Main 2024NAT

Four particles each of mass $1 {\text{kg}}$ are placed at four corners of a square of side $2 {\text{m}}$ . Moment of inertia of system about an axis perpendicular to its plane and passing through one of its vertex is____ ${\text{kgm}}^2$ . [27-Jan-2024 Shift 1]

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📖 Explanation

$\; {\text{I}}= {\text{ma}}^{2+} {\text{ma}}^{2+} {\text{m}}(\sqrt{2 {\text{a}}})^2$ $\;=4 {\text{ma}}^2$ $\;=4 \times 1 \times (2)^2=16$

Q2JEE Main 2024NAT

A body of mass ' $m$ ' is projected with a speed ' $u$ ' making an angle of $45^{\circ}$ with the ground. The angular momentum of the body about the point of projection, at the highest point is expressed as $\;{{\sqrt{2}} {\mu}^3}/{ {\text{Xg}}}$ . The value of ' ${\text{X}}$ ' is______ [31-Jan-2024 Shift 2]

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📖 Explanation

${\text{L}}= {\mu} \cos \theta \;{ {\text{u}}^2 sin \;^2 \theta }/{2 {\text{g}}}$ $=m u^3 \;{1}/{4 {\sqrt{2}} g} \Rightarrow x=8$

Q3JEE Main 2024MCQ

A heavy iron bar of weight $12 {\text{kg}}$ is having its one end on the ground and the other on the shoulder of a man. The rod makes an angle $60^{\circ}$ with the horizontal, the weight experienced by the man is : [27-Jan-2024 Shift 2]

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📖 Explanation

Torque about ${\text{O}}=0$ $\;120 (\;{ {\text{L}}}/{2} \cos 60^{\circ}) - {\text{N}}_2 {\text{L}}=0$ $\; {\text{N}}_2=30 {\text{N}}$

Q4JEE Main 2024MCQ

A particle of mass $m$ projected with a velocity ' $u$ ' making an angle of $30^{\circ}$ with the horizontal. The magnitude of angular momentum of the projectile about the point of projection when the particle is at its maximum height $h$ is : [30-Jan-2024 Shift 1]

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📖 Explanation

$\; {\text{L}}=m u \cos \theta H$ $\;=m u \cos \theta \times \;\frac{u^2 sin \;^2 \theta }{2 g}$ $\;=\;\frac{m u^3}{2 g} \times \;{{\sqrt{3}}}/{2} \times (\;\frac{1}{2}) ^2=\;{{\sqrt{3}} m u^3}/{16 g}$

Q5JEE Main 2024NAT

A body of mass $5 {\text{kg}}$ moving with a uniform speed $3 {\sqrt{2}} {\text{ms}}^{-1}$ in ${\text{X}}- {\text{Y}}$ plane along the line ${\text{y}}= {\text{x}}+4$ . The angular momentum of the particle about the origin will be ____ ${\text{kg}} {\text{m}}^2 {\ s}^{-1}$ . [29-Jan-2024 Shift 2]

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📖 Explanation

$y-x-4=0$ $d_1$ is perpendicular distance of given line from origin. ${\text{d}}_1={| \;{-4}/{{\sqrt{1^{2+}1^2}}} |} \Rightarrow 2 {\sqrt{2}} {\text{m}}$ So, $|{ {\text{L}}}^{\to }|={\ mvd}_1\;=5 \times 3 {\sqrt{2}} \times 2 {\sqrt{2}} {\text{kg}} {\text{m}}^2 / {\ s}$ $\;=60 {\text{kg}} {\text{m}}^2 / {\ s}$

Q6JEE Main 2024NAT

A ring and a solid sphere roll down the same inclined plane without slipping. They start from rest. The radii of both bodies are identical and the ratio of their kinetic energies is $\;{7}/{ {\text{x}}}$ where ${\text{x}}$ is [27-Jan-2024 Shift 2]

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📖 Explanation

In pure rolling work done by friction is zero. Hence potential energy is converted into kinetic energy. Since initially the ring and the sphere have same potential energy, finally they will have same kinetic energy too. $\therefore$ Ratio of kinetic energies $=1$ $\Rightarrow \;\frac{7}{x}=1 \Rightarrow x=7$

Q7JEE Main 2024NAT

A cylinder is rolling down on an inclined plane of inclination $60^{\circ}$ . It's acceleration during rolling down will be $\;{ {\text{x}}}/{{\sqrt{3}}} {\text{m}} / {\ s}^2$ , where ${\text{x}}=$ ______ (use ${\text{g}}=10 {\text{m}} / {\ s}^2$ ). [29-Jan-2024 Shift 1]

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📖 Explanation

For rolling motion, ${\text{a}}=\;{ {\text{g}} sin \;\theta }/{1+\;{ {\text{I}}_{ {\text{cm}}}}/{ {\text{MR}}^2}}$ $\;a=\;\frac{g sin \;\theta }{1+\;\frac{1}{2}}$ $\;=\;{2 \times 10 \times {{\sqrt{3}}}/{2}}/{3}$ $\;=\;{10}/{{\sqrt{3}}}$ Therefore ${\text{x}}=10$

Q8JEE Main 2024NAT

Two identical spheres each of mass $2 {\text{kg}}$ and radius $50 {\text{cm}}$ are fixed at the ends of a light rod so that the separation between the centers is $150 {\text{cm}}$ . Then, moment of inertia of the system about an axis perpendicular to the rod and passing through its middle point is $\;\frac{x}{20} {\text{kg}} {\text{m}}^2$ , where the value of ${\text{x}}$ is___ [31-Jan-2024 Shift 2]

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📖 Explanation

$\; {\text{I}}= (\;\frac{2}{5} {\text{mR}}^{2+} {\text{md}}^2) \times 2$ $\; {\text{I}}=2 (\;\frac{2}{5} \times 2 \times (\;\frac{1}{2}) ^{2+}2 \times (\;\frac{3}{4}) ^2) =\;\frac{53}{20} {\text{kg}}- {\text{m}}^2$ $\; {\text{X}}=53$

Q9JEE Main 2024NAT

Two discs of moment of inertia $I_1=4 {\text{kg}} {\text{m}}^2$ and ${\text{I}}_2=2 {\text{kg}} {\text{m}}^2$ about their central axes & normal to their planes, rotating with angular speeds $10 {\text{rad}} / {\ s}$ & $4 {\text{rad}} / {\ s}$ respectively are brought into contact face to face with their axe of rotation coincident. The loss in kinetic energy of the system in the process is ______J. [30-Jan-2024 Shift 2]

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📖 Explanation

$\; {\text{I}}_1 \omega _1+ {\text{I}}_2 \omega _2= ( {\text{I}}_1+ {\text{I}}_2) \omega _0 \;\text{ (C.O.A.M.) }\;$ $\;\;\text{ gives }\; \omega _0=8 {\text{rad}} / {\ s}$ $\; {\text{E}}_1=\;\frac{1}{2} {\text{I}}_1 \omega _1^{2+}\;\frac{1}{2} {\text{I}}_2 \omega _2^2=216 {\text{J}}$ $\; {\text{E}}_2=\;\frac{1}{2} ( {\text{I}}_1+ {\text{I}}_2) \omega _0^2=192 {\text{J}}$ $\; \therefore \Delta {\text{E}}=24 {\text{J}}$

Q10JEE Main 2024NAT

Consider a Disc of mass $5 {\text{kg}}$ , radius $2 {\text{m}}$ , rotating with angular velocity of $10 {\text{rad}} / {\ s}$ about an axis perpendicular to the plane of rotation. An identical disc is kept gently over the rotating disc along the same axis. The energy dissipated so that both the discs continue to rotate together without slipping is______ J. [30-Jan-2024 Shift 1]

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📖 Explanation

$\;\vec{L}_i=I \omega _i=\;\frac{M R^2}{2} \cdot \omega =100 {\text{kgm}}^2 / {\ s}$ $\;E_i=\;\frac{1}{2} \cdot \;\frac{M R^2}{2} \cdot \omega ^2=500 {\text{J}}$ $\;\vec{L}_i=\vec{L}_f \Rightarrow 100=2 I \omega _f$ $\;\omega _{ {\text{f}}}=5 {\text{rad}} / {\ sec}$ $\;E_f=2 \times \;\frac{1}{2} \cdot \;\frac{5(2)^2}{2} \cdot (5)^2=250 {\text{J}}$ $\;\Delta E=250 {\text{J}}$

Q11JEE Main 2023MCQ

A disc is rolling without slipping on a surface. The radius of the disc is ${\text{R}}$ . At ${\text{t}}=0$ , the top most point on the disc is ${\text{A}}$ as shown in figure. When the disc completes half of its rotation, the displacement of point ${\text{A}}$ from its initial position is [13-Apr-2023 shift 1]

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📖 Explanation

$\;\;\text{ Displacement }\;= {\text{A}}^{'} {\text{A}}$ $\; {\text{A}}^{'} {\text{A}}=\sqrt{(\pi {\text{R}})^{2+}(2 {\text{R}})^2}$ $\; {\text{A}}^{'} {\text{A}}= {\text{R}} {\sqrt{\pi ^{2+}4}}$

Q12JEE Main 2023NAT

If a solid sphere of mass $5 {\text{kg}}$ and a disc of mass 4 ${\text{kg}}$ have the same radius. Then the ratio of moment of inertia of the disc about a tangent in its plane to the moment of inertia of the sphere about its tangent will be $\;\frac{x}{7}$ . The value of $x$ is ______. [25-Jan-2023 Shift 2]

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📖 Explanation

$\;I_1=\;\frac{2}{5} m_1 R^{2+}m_1 R^2$ $\;I_1=m_1 R^2 (\;\frac{7}{5})$ $\;I_1=7 R^2$ $\; {\text{I}}_2=\;{ {\text{m}}_2 {\text{R}}^2}/{4}+ {\text{m}}_2 {\text{R}}^2$ $\; {\text{I}}_2=\;\frac{5}{4} {\text{m}}_2 {\text{R}}^2$ $\; {\text{I}}_2=5 {\text{R}}^2$ $\;\;{ {\text{I}}_2}/{ {\text{I}}_1}=\;\frac{5}{7}$ $\; {\text{x}}=5$

Q13JEE Main 2023NAT

Moment of inertia of a dise of mass $M$ and radius ' ${\text{R}}$ ' about any of its diameter is $\;{ {\text{MR}}^2}/{4}$ . The moment of inertia of this disc about an axis normal to the disc and passing through a point on its edge will be, $\;{ {\text{x}}}/{2} {\text{MR}}^2$ . The value of ${\text{x}}$ is ______. [1-Feb-2023 Shift 2]

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📖 Explanation

$\; {\text{I}}= {\text{I}}_{ {\text{cm}}}+ {\text{Md}}^2$ $\;=\;{ {\text{MR}}^2}/{2}+ {\text{MR}}^2$ $\;=\;\frac{3}{2} {\text{MR}}^2$ $\; {\text{x}}=3$

Q14JEE Main 2023NAT

A solid sphere is rolling on a horizontal plane without slipping. If the ratio of angular momentum about axis of rotation of the sphere to the total energy of moving sphere is $\pi : 22$ the, the value of its angular speed will be ${\text{rad}} / {\ s}$ . [13-Apr-2023 shift 1]

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📖 Explanation

Angular momentum, $\; {\text{L}}=I \omega$ $\; {\text{L}}=\;\frac{2}{5} {\text{MR}}^2 \omega$ $\;\;\text{ Energy }\;=\;\frac{1}{2} {\text{MV}}^{2+}\;\frac{1}{2} {\text{I}} \omega ^2$ $\; {\text{E}}=\;\frac{1}{2} {\text{M}}(\omega {\text{R}})^{2+}\;\frac{1}{2} (\;\frac{2}{5} {\text{MR}}^2) \omega ^2$ $\;=\;\frac{7}{10} {\text{M}} \omega ^2 {\text{R}}^2$ $\;\;{ {\text{L}}}/{ {\text{E}}}=\;\frac{4}{7 \omega }=\;\frac{\pi }{22}$ $\;\omega =\;\frac{88}{7 \pi }=\;\frac{88}{7 \times \;\frac{22}{7}}=4 {\text{rad}} / {\ s}$

Q15JEE Main 2023NAT

A uniform disc of mass $0.5 {\text{kg}}$ and radius ${\text{r}}$ is projected with velocity $18 {\text{m}} / {\ s}$ at ${\text{t}}=0 {\ s}$ on a rough horizontal surface. It starts off with a purely sliding motion at ${\text{t}}=0 {\ s}$ . After $2 {\ s}$ it acquires a purely rolling motion (see figure). The total kinetic energy of the disc after $2 {\ s}$ will be _______ J. (given, coefficient of friction is $0.3$ and ${\text{g}}=10 {\text{m}} / {\ s}^2)$ [30-Jan-2023 Shift 2]

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📖 Explanation

$\; {\text{a}}=-\mu _{ {\text{k}}} {\text{g}}=-3$ $\; {\text{V}}=18-3 \times 2$ $\; {\text{V}}=12 {\text{m}} / {\ s}$ $\; {\text{KE}}=\;\frac{1}{2} {\ mv}^{2+}\;\frac{1}{2} \;{ {\text{mr}}^2}/{2} \;{ {\text{v}}^2}/{ {\text{r}}^2}$ $\; {\text{KE}}=\;\frac{3}{4} {\ mv}^2$ $\; {\text{KE}}=3 \times 18=54 {\text{J}}$

Q16JEE Main 2023NAT

A solid sphere of mass $1 {\text{kg}}$ rolls without slipping on a plane surface. Its kinetic energy is $7 \times 10^{-3} {\text{J}}$ . The speed of the centre of mass of the sphere is _______ ${\text{cm}} {\ s}^{-1}$ . [31-Jan-2023 Shift 1]

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📖 Explanation

$\;\;\frac{1}{2} {\ mv}^{2+}\;\frac{1}{2} {\text{I}}^2=7 \times 10^{-3}$ $\;\;\frac{1}{2} {\ mv}^{2+}\;\frac{1}{2} (\;\frac{2}{5} {\text{MR}}^2) (\;{ {\text{V}}}/{ {\text{R}}}) ^2=7 \times 10^{-3}$ $\;\;\frac{1}{2} {\text{MV}}^2 [1+\;\frac{2}{5}] =7 \times 10^{-3}$ $\;\;\frac{1}{2}(1) ( {\text{V}}^2) (\;\frac{7}{5}) =7 \times 10^{-3}$ $\; {\text{V}}^2=10^{-2}$ $\; {\text{V}}=10^{-1}=0.1 {\text{m}} / {\ s}=10 {\text{cm}} / {\ s}$ $\;\;\text{ Ans: }\; 10$

Q17JEE Main 2023NAT

A thin uniform rod of length $2 {\text{m}}$ . cross sectional area ' ${\text{A}}$ ' and density ' ${\text{d}}$ ' is rotated about an axis passing through the centre and perpendicular to its length with angular velocity $\omega$ . If value of $\omega$ in terms of its rotational kinetic energy $E$ is $\sqrt{\;\frac{\alpha E}{A d}}$ then the value of $\alpha$ is ________. [30-Jan-2023 Shift 1]

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📖 Explanation

$\;( {\text{KE}})_{\;\text{Rotational }\;}=\;\frac{1}{2} {\text{I}} \omega ^2= {\text{E}}$ $\; {\text{E}}=\;\frac{1}{2} \;{ {\text{m}} ℓ^2}/{12} \omega ^2$ $\; {\text{E}}=\;\frac{1}{2} \;{ {\text{dA}} ℓ^3}/{12} \omega ^2$ $\; {\text{E}}=\;{ {\text{dA}}(2)^3}/{24} \omega ^2$ $\;\sqrt{\;{3 {\text{E}}}/{ {\text{dA}}}}=\omega$ $\;\alpha =3 \;\text{ Ans. }\;$

Q18JEE Main 2023NAT

A force of $- {\text{P}} { {\text{k}}^{\^}}$ acts on the origin of the coordinate system. The torque about the point $(2,-3)$ is ${\text{P}}(\hat{a} \hat{i}+ {\text{bj}})$ , The ratio of $\;\frac{a}{b}$ is $\;\frac{x}{2}$ . The value of $x$ is _______. [10-Apr-2023 shift 2]

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📖 Explanation

$\vec{\tau }={ {\text{r}}}^{\to } \times { {\text{F}}}^{\to }$ $\;\vec{\tau }=\;\text{ head }\;- \;\text{ tail }\;$ $\;=(0-2) \hat{i}+(0(-3)) \hat{j}$ $\;=-2 \hat{i}+3 \hat{j}$ $\;\tau =(-2 \hat{i}+3 \hat{j}) \times (-p \hat{k})$ $\;=-2 \hat{p}-3 p \hat{i}$ $\;=-p(3 p \hat{i}+2 \hat{j}) \;\frac{a}{b}=\;\frac{3}{2}=\;\frac{x}{2} \;\; x=3$

Q19JEE Main 2023NAT

For a rolling spherical shell, the ratio of rotational kinetic energy and total kinetic energy is $\;\frac{x}{5}$ . The value of $x$ is ______. [12-Apr-2023 shift 1]

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📖 Explanation

$\; {\text{KE}}_{ {\text{T}}} \;\frac{1}{2} {\text{mV}}^2$ $\; {\text{KE}}_{ {\text{R}}}=\;\frac{1}{2} (\;\frac{2}{5} {\text{mR}}^2) \omega ^2=\;\frac{1}{3} {\text{mR}}^2 \cdot \;{ {\text{V}}^2}/{ {\text{R}}^2}=\;\frac{1}{3} {\text{mV}}^2$ $\; {\text{KE}}_{ {\text{T}}}=\;\frac{1}{2} {\text{mV}}^{2+}\;{ {\text{mV}}^2}/{3}$ $\; {\text{KE}}_{ {\text{T}}}=\;\frac{5}{6} {\text{mV}}^2$ $\;\;{ {\text{K}}_{ {\text{R}}}}/{ {\text{K}}_{ {\text{T}}}}=\;{1 / 3 {\text{mV}}^2}/{5 / 6 {\text{mV}}^2}=\;\frac{2}{5}$ $\; {\text{x}}=2$

Q20JEE Main 2023NAT

A particle of mass $100 {\text{g}}$ is projected at time ${\text{t}}=0$ with a speed $20 {\text{ms}}^{-1}$ at an angle $45^{\circ}$ to the horizontal as given in the figure. The magnitude of the angular momentum of the particle about the starting point at time $t=2 s$ is found to be $\sqrt{ {\text{K}}} {\text{kg}} {\text{m}}^2 / {\ s}$ . The value of ${\text{K}}$ is ________. $(.$ Take $g=10 {\text{ms}}^{-2})$ [29-Jan-2023 Shift 2]

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📖 Explanation

$\;\;\text{ Use }\; \Delta {\text{L}}=∫_{0}^{ {\text{t}}} \tau {\text{dt}}$ $\; {\text{L}}_0=∫_{0^2 {\text{mg}} ( {\text{v}}}^{ {\text{x}}} {\text{t}}) {\text{dt}}$ $\;=m g v_x \;\frac{t^2}{2}=(0.1)(10)(10 {\sqrt{2}}) \;\frac{2^2}{2}$ $\;=20 {\sqrt{2}}$ $\;={\sqrt{800}} {\text{kgm}}^2 / {\ s}$ $\;$

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