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387eAeopEkCP9blfALjgy2xukfaw5rij | physics | rotational-motion | moment-of-inertia | For a uniform rectangular sheet shown in the figure, the ratio of moments of inertia about the axes
perpendicular to the sheet and passing through O (the centre of mass) and O' (corner point) is :
<img src="data:image/png;base64,UklGRmoIAABXRUJQVlA4IF4IAADQQQCdASoUAeEAPm0ulUkkIiGhIZFK4IANiWlu4W8hG/OT8W/xj8b/B7+xdEB439s... | [{"identifier": "A", "content": "$${1 \\over 2}$$"}, {"identifier": "B", "content": "$${1 \\over 4}$$"}, {"identifier": "C", "content": "$${1 \\over 8}$$"}, {"identifier": "D", "content": "$${2 \\over 3}$$"}] | ["B"] | null | $${I_O} = {M \over {12}}\left( {{a^2} + {b^2}} \right)$$ = $${M \over {12}}\left[ {{{80}^2} + {{60}^2}} \right]$$
<br><br>I<sub>O'</sub> = I<sub>O</sub> + Md<sup>2</sup> {parallel axis theorem}
<br><br>= $${M \over {12}}\left[ {{{80}^2} + {{60}^2}} \right]$$ + M[50]<sup>2</sup>
<br><br>$$ \therefore $$ $${{{I_O}} \over... | mcq | jee-main-2020-online-4th-september-evening-slot | 12,547 |
qgdOpGBjkYAFKHd8zUjgy2xukfaj3opk | physics | rotational-motion | moment-of-inertia | Consider two uniform discs of the same thickness and different radii R<sub>1</sub>
= R and<br/> R<sub>2</sub>
= $$\alpha $$R made of
the same material. If the ratio of their moments of inertia I<sub>1</sub>
and I<sub>2</sub>
, respectively, about their axes
is I<sub>1</sub>
: I<sub>2</sub>
= 1 : 16 then the value ... | [{"identifier": "A", "content": "$$\\sqrt 2 $$"}, {"identifier": "B", "content": "2"}, {"identifier": "C", "content": "$$2\\sqrt 2 $$"}, {"identifier": "D", "content": "4"}] | ["B"] | null | Moment of inertia of disc, $$I = {{M{R^2}} \over 2} = {{\left[ {p\left( {\pi {R^2}} \right)t} \right]{R^2}} \over 2}$$<br><br>$$I = K{R^4}$$<br><br>$${{{I_1}} \over {{I_2}}} = {\left( {{{{R_1}} \over {{R_2}}}} \right)^4}$$<br><br>$${1 \over {16}} = {\left( {{R \over {\alpha R}}} \right)^4} \Rightarrow \alpha = {\left(... | mcq | jee-main-2020-online-4th-september-evening-slot | 12,548 |
2p0E2TIZzrOYy5YhX67k9k2k5djt7ib | physics | rotational-motion | moment-of-inertia | The radius of gyration of a uniform rod of length $$l$$, about an axis passing through a
point $${l \over 4}$$ away from the centre of the rod,
and perpendicular to it, is : | [{"identifier": "A", "content": "$${1 \\over 8}l$$"}, {"identifier": "B", "content": "$${1 \\over 4}l$$"}, {"identifier": "C", "content": "$$\\sqrt {{7 \\over {48}}} l$$"}, {"identifier": "D", "content": "$$\\sqrt {{3 \\over 8}} l$$"}] | ["C"] | null | <img src="https://res.cloudinary.com/dckxllbjy/image/upload/v1734264632/exam_images/du89ilosutgelad9wfu6.webp" style="max-width: 100%;height: auto;display: block;margin: 0 auto;" loading="lazy" alt="JEE Main 2020 (Online) 7th January Morning Slot Physics - Rotational Motion Question 132 English Explanation">
<br><br>I ... | mcq | jee-main-2020-online-7th-january-morning-slot | 12,552 |
mmd27YeQWLKUu4xQ2u7k9k2k5f5eapa | physics | rotational-motion | moment-of-inertia | Mass per unit area of a circular disc of radius $$a$$ depends on the distance r from its centre as $$\sigma \left( r \right)$$ = A + Br
. The moment of inertia of the disc about the axis, perpendicular to the plane and
assing through its centre is: | [{"identifier": "A", "content": "$$2\\pi {a^4}\\left( {{A \\over 4} + {{aB} \\over 5}} \\right)$$"}, {"identifier": "B", "content": "$$\\pi {a^4}\\left( {{A \\over 4} + {{aB} \\over 5}} \\right)$$"}, {"identifier": "C", "content": "$$2\\pi {a^4}\\left( {{{aA} \\over 4} + {B \\over 5}} \\right)$$"}, {"identifier": "D", ... | ["A"] | null | <img src="https://res.cloudinary.com/dckxllbjy/image/upload/v1734264879/exam_images/szal3sqxcs1kkhxpytk7.webp" style="max-width: 100%;height: auto;display: block;margin: 0 auto;" loading="lazy" alt="JEE Main 2020 (Online) 7th January Evening Slot Physics - Rotational Motion Question 130 English Explanation">
<br><br>dI... | mcq | jee-main-2020-online-7th-january-evening-slot | 12,553 |
A2QvSLq161HSt2HT6J7k9k2k5inxn5k | physics | rotational-motion | moment-of-inertia | One end of a straight uniform 1m long bar is
pivoted on horizontal table. It is released from
rest when it makes an angle 30º from the
horizontal (see figure). Its angular speed when
it hits the table is given as $$\sqrt n $$ s<sup>-1</sup>, where n is
an integer. The value of n is _________.
<img src="data:image/png;b... | [] | null | 15 | By energy conservation
<br><br>U<sub>i</sub> + K<sub>i</sub> = U<sub>f</sub> + K<sub>f</sub>
<br><br>$$ \Rightarrow $$ $$mg{l \over 2}\sin 30^\circ $$ + 0 = 0 + $${1 \over 2}I{\omega ^2}$$
<br><br>$$ \Rightarrow $$ $$mg{1 \over 2} \times {1 \over 2}$$ = $${1 \over 2} \times {{m{{\left( 1 \right)}^2}} \over 3}{\omega ^2... | integer | jee-main-2020-online-9th-january-morning-slot | 12,554 |
A1UXoiLznn3BR7xzfR7k9k2k5in4qlp | physics | rotational-motion | moment-of-inertia | <img src="data:image/png;base64,UklGRtQbAABXRUJQVlA4IMgbAABwlgCdASqbATYBPm0wlUikIqIhI1Dq4IANiWlu/HyYhsV+gGipTdBM+KP55+SfgL/b/6H/g/+P/Yf+L7P/jvzD90/nv7P/1j/te3l/U/xv+n9aZ/kfzP+m9+Z8x/Ev5x/1fzJ+/H6T/kP4b/M/K/4+f2P819gv1V/gP4t/N/+D/dvTt/uf49/IvASsB/x/6r7Bfbz/D/w3+bf+D+9/Kj6h/qv5H/KPZn6zf6T+AfxP4C/zP/T/xr+Z+lF/yf4B/FPO15Ae... | [{"identifier": "A", "content": "$${{13} \\over {23}}$$"}, {"identifier": "B", "content": "$${{23} \\over {13}}$$"}, {"identifier": "C", "content": "$${{15} \\over {13}}$$"}, {"identifier": "D", "content": "$${{13} \\over {15}}$$"}] | ["A"] | null | AB = d
<br><br>$$ \therefore $$ OA = $${{d \over {\sqrt 3 }}}$$
<br><br>From parallel axis theorem
<br><br>I<sub>0</sub> = $$3 \times \left[ {{2 \over 5}M{{\left( {{d \over 2}} \right)}^2} + M{{\left( {{d \over {\sqrt 3 }}} \right)}^2}} \right]$$ = $${{13} \over {10}}M{d^2}$$
<br><br>I<sub>A</sub> = I<sub>0</sub> + 3M$... | mcq | jee-main-2020-online-9th-january-morning-slot | 12,555 |
D69V8gdvhXmgTfrdPW1klrifuhg | physics | rotational-motion | moment-of-inertia | Moment of inertia (M. I.) of four bodies, having same mass and radius, are reported as; <br/><br/>I<sub>1</sub> = M.I. of thin circular ring about its diameter,<br/><br/>I<sub>2</sub> = M.I. of circular disc about an axis perpendicular to disc and going through the centre,<br/><br/>I<sub>3</sub> = M.I. of solid cylinde... | [{"identifier": "A", "content": "I<sub>1</sub> = I<sub>2</sub> = I<sub>3</sub> > I<sub>4</sub>"}, {"identifier": "B", "content": "I<sub>1</sub> + I<sub>3</sub> < I<sub>2</sub> + I<sub>4</sub>"}, {"identifier": "C", "content": "I<sub>1</sub> = I<sub>2</sub> = I<sub>3</sub> < I<sub>4</sub>"}, {"identifier": "D",... | ["A"] | null | Let M and R be the mass and radius of four bodies. Then, as per
question, their moment of inertia are
<br/><br/>I<sub>1</sub> = $${{M{R^2}} \over 2}$$,
<br/><br/>I<sub>2</sub> = $${{M{R^2}} \over 2}$$,
<br/><br/>I<sub>3</sub> = $${{M{R^2}} \over 2}$$,
<br/><br/>I<sub>4</sub> = $${2 \over 5}M{R^2}$$
<br/><br/>$$ \theref... | mcq | jee-main-2021-online-24th-february-morning-slot | 12,556 |
umi1KK6mmaGVfwcv5z1klrp546p | physics | rotational-motion | moment-of-inertia | A uniform thin bar of mass 6 kg and length 2.4 meter is bent to make an equilateral hexagon. The moment of inertia about an axis passing through the centre of mass and perpendicular to the plane of hexagon is _______ $$\times$$ 10<sup>$$-$$1</sup> kg m<sup>2</sup>. | [] | null | 8 | <img src="https://res.cloudinary.com/dckxllbjy/image/upload/v1734266418/exam_images/hvjpvmaaww2y50u9kavq.webp" style="max-width: 100%;height: auto;display: block;margin: 0 auto;" loading="lazy" alt="JEE Main 2021 (Online) 24th February Evening Shift Physics - Rotational Motion Question 103 English Explanation">
<br>MOI... | integer | jee-main-2021-online-24th-february-evening-slot | 12,557 |
GggYuVILuLH4LnAfDQ1kmlvtyoe | physics | rotational-motion | moment-of-inertia | Consider a uniform wire of mass M and length L. It is bent into a semicircle. Its moment of inertia about a line perpendicular to the plane of the wire passing through the center is : | [{"identifier": "A", "content": "$${1 \\over 4}{{M{L^2}} \\over {{\\pi ^2}}}$$"}, {"identifier": "B", "content": "$${1 \\over 2}{{M{L^2}} \\over {{\\pi ^2}}}$$"}, {"identifier": "C", "content": "$${2 \\over 5}{{M{L^2}} \\over {{\\pi ^2}}}$$"}, {"identifier": "D", "content": "$${{M{L^2}} \\over {{\\pi ^2}}}$$"}] | ["D"] | null | <picture><source media="(max-width: 1577px)" srcset="https://res.cloudinary.com/dckxllbjy/image/upload/v1734264308/exam_images/pqxjbrdq84tqsbtksawk.webp"><source media="(max-width: 320px)" srcset="https://res.cloudinary.com/dckxllbjy/image/upload/v1734267293/exam_images/uweciyztvzukvpi1emuf.webp"><source media="(max-wi... | mcq | jee-main-2021-online-18th-march-evening-shift | 12,559 |
1krudvskt | physics | rotational-motion | moment-of-inertia | Given below are two statements : one is labelled as Assertion A and the other is labelled as Reason R. <br/><br/>Assertion A : Moment of inertia of a circular disc of mass 'M' and radius 'R' about X, Y axes (passing through its plane) and Z-axis which is perpendicular to its plane were found to be I<sub>x</sub>, I<sub>... | [{"identifier": "A", "content": "Both A and R are correct but R is NOT the correct explanation of A."}, {"identifier": "B", "content": "A is not correct but R is correct."}, {"identifier": "C", "content": "A is correct but R is not correct."}, {"identifier": "D", "content": "Both A and R are correct and R is the correc... | ["B"] | null | I<sub>z</sub> = I<sub>x</sub> + I<sub>y</sub> (using perpendicular axis theorem) & I = mk<sup>2</sup> (K : radius of gyration)<br><br>so, mK<sub>z</sub><sup>2</sup> = mK<sub>x</sub><sup>2</sup> + mK<sub>y</sub><sup>2</sup><br><br>K<sub>z</sub><sup>2</sup> = K<sub>x</sub><sup>2</sup> + K<sub>y</sub><sup>2</sup><br><... | mcq | jee-main-2021-online-25th-july-morning-shift | 12,560 |
1krytnal1 | physics | rotational-motion | moment-of-inertia | <table class="tg">
<thead>
<tr>
<th class="tg-amwm">List-I</th>
<th class="tg-amwm">List-II</th>
</tr>
</thead>
<tbody>
<tr>
<td class="tg-0lax">(a) MI of the rod (length L, Mass M, about an axis $$ \bot $$ to the rod passing through the midpoint)</td>
<td class="tg-0lax">(i) $$8M{L^2}/3$$</td>
</tr>
<tr>
<td class="tg... | [{"identifier": "A", "content": "(a)-(ii), (b)-(iii), (c)-(i), (d)-(iv)"}, {"identifier": "B", "content": "(a)-(ii), (b)-(i), (c)-(iii), (d)-(iv)"}, {"identifier": "C", "content": "(a)-(iii), (b)-(iv), (c)-(ii), (d)-(i)"}, {"identifier": "D", "content": "(a)-(iii), (b)-(iv), (c)-(i), (d)-(ii)"}] | ["C"] | null | <img src="https://res.cloudinary.com/dckxllbjy/image/upload/v1734264015/exam_images/zvv21dd7erzbvxbsqcfi.webp" style="max-width: 100%;height: auto;display: block;margin: 0 auto;" loading="lazy" alt="JEE Main 2021 (Online) 27th July Morning Shift Physics - Rotational Motion Question 78 English Explanation"> | mcq | jee-main-2021-online-27th-july-morning-shift | 12,561 |
1ks1a0rxv | physics | rotational-motion | moment-of-inertia | In the given figure, two wheels P and Q are connected by a belt B. The radius of P is three times as that of Q. In case of same rotational kinetic energy, the ratio of rotational inertias $$\left( {{{{I_1}} \over {{I_2}}}} \right)$$ will be x : 1. The value of x will be _____________.<br/><img src="data:image/png;base6... | [] | null | 9 | <img src="https://app-content.cdn.examgoal.net/fly/@width/image/1kx3uqjs0/1b12c7f4-f665-4985-a48c-3f368fd0e868/85f5aa00-5b9f-11ec-a5b4-139d631bd0b4/file-1kx3uqjs1.png?format=png" data-orsrc="https://app-content.cdn.examgoal.net/image/1kx3uqjs0/1b12c7f4-f665-4985-a48c-3f368fd0e868/85f5aa00-5b9f-11ec-a5b4-139d631bd0b4/fi... | integer | jee-main-2021-online-27th-july-evening-shift | 12,563 |
1ktah602o | physics | rotational-motion | moment-of-inertia | Consider a badminton racket with length scales as shown in the figure.<br/><br/><img src="data:image/png;base64,UklGRgIZAABXRUJQVlA4IPYYAAAQfwCdASrTASkBPm00lkkkIqIhIdEKwIANiWlu9mj+meq5LwHck/mewW1M3/I/gH9V8GbJigPwZYV/ir+XfiT+I30Z7zvqv9j/Yn+0+RH59+2/2P9l/Wm/s/4B1RH+T/Ou8D95/jv86/2/9q9cP7n/Nv6j/wP7n6e+4b9Q/oH4T/Iv6Z/tP8O/... | [] | null | 52 | <img src="https://res.cloudinary.com/dckxllbjy/image/upload/v1734263819/exam_images/rq6w2gycfungnzzsezey.webp" style="max-width: 100%;height: auto;display: block;margin: 0 auto;" loading="lazy" alt="JEE Main 2021 (Online) 26th August Morning Shift Physics - Rotational Motion Question 75 English Explanation"><br>$$I = \... | integer | jee-main-2021-online-26th-august-morning-shift | 12,564 |
1ktbo6v5p | physics | rotational-motion | moment-of-inertia | The solid cylinder of length 80 cm and mass M has a radius of 20 cm. Calculate the density of the material used if the moment of inertia of the cylinder about an axis CD parallel to AB as shown in figure is 2.7 kg m<sup>2</sup>.<br/><br/><img src="data:image/png;base64,UklGRi4LAABXRUJQVlA4ICILAAAQUgCdASpNAR8BPm00mEekIy... | [{"identifier": "A", "content": "14.9 kg/m<sup>3</sup>"}, {"identifier": "B", "content": "7.5 $$\\times$$ 10<sup>1</sup> kg/m<sup>3</sup>"}, {"identifier": "C", "content": "7.5 $$\\times$$ 10<sup>2</sup> kg/m<sup>3</sup>"}, {"identifier": "D", "content": "1.49 $$\\times$$ 10<sup>2</sup> kg/m<sup>3</sup>"}] | ["D"] | null | Parallel axis theorem<br><br>I = I<sub>CM</sub> + Md<sup>2</sup><br><br>$$I = {{M{r^2}} \over 2} + M{\left( {{L \over 2}} \right)^2}$$<br><br>$$2.7 = M{{{{(0.2)}^2}} \over 2} + M{\left( {{{0.8} \over 2}} \right)^2}$$<br><br>$$2.7 = M\left[ {{2 \over {100}} + {{16} \over {100}}} \right]$$<br><br>M = 15 kg<br><br>$$ \Rig... | mcq | jee-main-2021-online-26th-august-evening-shift | 12,565 |
1kte0gtvo | physics | rotational-motion | moment-of-inertia | Moment of inertia of a square plate of side l about the axis passing through one of the corner and perpendicular to the plane of square plate is given by : | [{"identifier": "A", "content": "$${{M{l^2}} \\over 6}$$"}, {"identifier": "B", "content": "$${M{l^2}}$$"}, {"identifier": "C", "content": "$${{M{l^2}} \\over {12}}$$"}, {"identifier": "D", "content": "$${2 \\over 3}M{l^2}$$"}] | ["D"] | null | According to perpendicular Axis theorem.<br><br><img src="https://res.cloudinary.com/dckxllbjy/image/upload/v1734266871/exam_images/f28emgcmgnceaprxbrmt.webp" style="max-width: 100%;height: auto;display: block;margin: 0 auto;" loading="lazy" alt="JEE Main 2021 (Online) 27th August Morning Shift Physics - Rotational Mot... | mcq | jee-main-2021-online-27th-august-morning-shift | 12,566 |
1ktjn2zdu | physics | rotational-motion | moment-of-inertia | A system consists of two identical spheres each of mass 1.5 kg and radius 50 cm at the end of light rod. The distance between the centres of the two spheres is 5 m. What will be the moment of inertia of the system about an axis perpendicular to the rod passing through its midpoint? | [{"identifier": "A", "content": "18.75 kgm<sup>2</sup>"}, {"identifier": "B", "content": "1.905 $$\\times$$ 10<sup>5</sup> kgm<sup>2</sup>"}, {"identifier": "C", "content": "19.05 kgm<sup>2</sup>"}, {"identifier": "D", "content": "1.875 $$\\times$$ 10<sup>5</sup> kgm<sup>2</sup>"}] | ["C"] | null | <picture><source media="(max-width: 320px)" srcset="https://res.cloudinary.com/dckxllbjy/image/upload/v1734265338/exam_images/zahks7szvg4zvvmtytuo.webp"><source media="(max-width: 500px)" srcset="https://res.cloudinary.com/dckxllbjy/image/upload/v1734264325/exam_images/kotlhj90p5w6zydzn5ni.webp"><img src="https://res.c... | mcq | jee-main-2021-online-31st-august-evening-shift | 12,567 |
1l54vvmmr | physics | rotational-motion | moment-of-inertia | <p>The moment of inertia of a uniform thin rod about a perpendicular axis passing through one end is I<sub>1</sub>. The same rod is bent into a ring and its moment of inertia about a diameter is I<sub>2</sub>. If $${{{I_1}} \over {{I_2}}}$$ is $${{x{\pi ^2}} \over 3}$$, then the value of x will be ____________.</p> | [] | null | 8 | <p>$${I_1} = {{M{L^2}} \over 3}$$ ..... (1)</p>
<p>For ring : $${I_2} = {{M{R^2}} \over 2}$$</p>
<p>and $$2\pi R = L$$</p>
<p>$$ \Rightarrow {I_2} = {M \over 2}\left( {{{{L^2}} \over {4{\pi ^2}}}} \right)$$ ...... (2)</p>
<p>$$ \Rightarrow {{{I_1}} \over {{I_2}}} = {{8{\pi ^2}} \over 3}$$</p>
<p>$$ \Rightarrow x = 8$$<... | integer | jee-main-2022-online-29th-june-evening-shift | 12,568 |
1l568o323 | physics | rotational-motion | moment-of-inertia | <p>Match List-I with List-II</p>
<p><style type="text/css">
.tg {border-collapse:collapse;border-spacing:0;}
.tg td{border-color:black;border-style:solid;border-width:1px;font-family:Arial, sans-serif;font-size:14px;
overflow:hidden;padding:10px 5px;word-break:normal;}
.tg th{border-color:black;border-style:solid;bo... | [{"identifier": "A", "content": "A - II, B - I, C - IV, D - III"}, {"identifier": "B", "content": "A - I, B - II, C - IV, D - III"}, {"identifier": "C", "content": "A - II, B - I, C - III, D - IV"}, {"identifier": "D", "content": "A - I, B - II, C - III, D - IV"}] | ["A"] | null | <p>(A) Moment of inertia of solid sphere of radius R about a tangent $$ = {2 \over 5}M{R^2} + M{R^2} = {7 \over 5}M{R^2}$$</p>
<p>$$\Rightarrow$$ A $$-$$ (II)</p>
<p>(B) Moment of inertia of hollow sphere of radius R about a tangent $$ = {2 \over 3}M{R^2} + M{R^2} = {5 \over 3}M{R^2}$$</p>
<p>$$\Rightarrow$$ B $$-$$ (I... | mcq | jee-main-2022-online-28th-june-morning-shift | 12,569 |
1l59pv08s | physics | rotational-motion | moment-of-inertia | <p>Moment of Inertia (M.I.) of four bodies having same mass 'M' and radius '2R' are as follows:</p>
<p>I<sub>1</sub> = M.I. of solid sphere about its diameter</p>
<p>I<sub>2</sub> = M.I. of solid cylinder about its axis</p>
<p>I<sub>3</sub> = M.I. of solid circular disc about its diameter</p>
<p>I<sub>4</sub> = M.I. of... | [] | null | 5 | <p>$$2\left( {{1 \over 2} + {1 \over 4}} \right) \times M{(2R)^2} + {1 \over 2}M{(2R)^2} = x{2 \over 5}M{(2R)^2}$$</p>
<p>$$ \Rightarrow 1 + {1 \over 2} + {1 \over 2} = x \times {2 \over 5}$$</p>
<p>$$ \Rightarrow x = 5$$</p> | integer | jee-main-2022-online-25th-june-evening-shift | 12,570 |
1l6i21mj2 | physics | rotational-motion | moment-of-inertia | <p>The radius of gyration of a cylindrical rod about an axis of rotation perpendicular to its length and passing through the center will be ___________ $$\mathrm{m}$$.</p>
<p>Given, the length of the rod is $$10 \sqrt{3} \mathrm{~m}$$.</p> | [] | null | 5 | <p>$$l = {{M{L^2}} \over {12}} = M{K^2}$$</p>
<p>$$K = {L \over {\sqrt {12} }} = {{10\sqrt 3 } \over {\sqrt {12} }} = 5\,m$$</p> | integer | jee-main-2022-online-26th-july-evening-shift | 12,572 |
1l6kolj1m | physics | rotational-motion | moment-of-inertia | <p>A solid cylinder length is suspended symmetrically through two massless strings, as shown in the figure. The distance from the initial rest position, the cylinder should be unbinding the strings to achieve a speed of $$4 \mathrm{~ms}^{-1}$$, is ____________ cm. (take g = $$10 \mathrm{~ms}^{-2}$$)</p>
<p><img src="da... | [] | null | 120 | In case of rotational motion of a rigid body like this, the kinetic energy is not just due to the linear motion, but also due to its rotation.
<br/><br/>For a solid cylinder, the moment of inertia I is given by $\frac{1}{2} m r^2$. The kinetic energy due to rotation is given by $\frac{1}{2} I \omega^2$. But we also k... | integer | jee-main-2022-online-27th-july-evening-shift | 12,573 |
1l6mbrtzb | physics | rotational-motion | moment-of-inertia | <p>Four identical discs each of mass '$$\mathrm{M}$$' and diameter '$$\mathrm{a}$$' are arranged in a small plane as shown in figure. If the moment of inertia of the system about $$\mathrm{OO}^{\prime}$$ is $$\frac{x}{4} \,\mathrm{Ma}^{2}$$. Then, the value of $$x$$ will be ____________.</p>
<p><img src="data:image/png... | [] | null | 3 | <p>$$I = 2 \times \left( {{{M{{\left( {{a \over 2}} \right)}^2}} \over 4}} \right) + 2 \times \left( {{{M{{\left( {{a \over 2}} \right)}^2}} \over 4} + M{{\left( {{a \over 2}} \right)}^2}} \right)$$</p>
<p>$$ = {{M{a^2}} \over 8} + {{5M{a^2}} \over 8}$$</p>
<p>$$ = {{6M{a^2}} \over 8} = {3 \over 4}M{a^2}$$</p> | integer | jee-main-2022-online-28th-july-morning-shift | 12,574 |
1ldnz6bcx | physics | rotational-motion | moment-of-inertia | <p>Moment of inertia of a disc of mass '$$M$$' and radius '$$R$$' about any of its diameter is $$\frac{M R^{2}}{4}$$. The moment of inertia of this disc about an axis normal to the disc and passing through a point on its edge will be, $$\frac{x}{2}$$ MR$$^{2}$$. The value of $$x$$ is ___________.</p> | [] | null | 3 | <img src="https://app-content.cdn.examgoal.net/fly/@width/image/1le49tel4/c421ea45-fada-4167-b61d-a3ff44aa75a8/2bc14270-ac6a-11ed-a805-097bbaf48d41/file-1le49tel5.png?format=png" data-orsrc="https://app-content.cdn.examgoal.net/image/1le49tel4/c421ea45-fada-4167-b61d-a3ff44aa75a8/2bc14270-ac6a-11ed-a805-097bbaf48d41/fi... | integer | jee-main-2023-online-1st-february-evening-shift | 12,575 |
ldo7la9t | physics | rotational-motion | moment-of-inertia | Two discs of same mass and different radii are made of different materials such that their
thicknesses are $1 \mathrm{~cm}$ and $0.5 \mathrm{~cm}$ respectively. The densities of materials are in the ratio $3: 5$. The
moment of inertia of these discs respectively about their diameters will be in the ratio of $\frac{x}... | [] | null | 5 | $m=\rho \pi R^{2} t$
<br/><br/>$$
\begin{aligned}
& \text { so } R^{2}=\frac{m}{\rho \pi t} \\\\
& I=\frac{m R^{2}}{4}=\frac{m^{2}}{4 \rho \pi t}
\end{aligned}
$$
<br/><br/>So $\frac{I_{1}}{I_{2}}=\frac{\rho_{2} t_{2}}{\rho_{1} t_{1}}=\frac{5}{3} \times \frac{0.5}{1}=\frac{5}{6}$
<br/><br/>So $x=5$ | integer | jee-main-2023-online-31st-january-evening-shift | 12,576 |
1ldtzxyz7 | physics | rotational-motion | moment-of-inertia | <p>If a solid sphere of mass 5 kg and a disc of mass 4 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 ___________.</p> | [] | null | 5 | <p>Solid Sphere :</p>
<p><img src="https://app-content.cdn.examgoal.net/fly/@width/image/1ledlxt6u/8e60848a-1705-4865-bc97-be2834e72041/a1aada60-b18c-11ed-a682-13f364283dca/file-1ledlxt6v.png?format=png" data-orsrc="https://app-content.cdn.examgoal.net/image/1ledlxt6u/8e60848a-1705-4865-bc97-be2834e72041/a1aada60-b18c-... | integer | jee-main-2023-online-25th-january-evening-shift | 12,578 |
1ldui9ys1 | physics | rotational-motion | moment-of-inertia | <p>$$\mathrm{I_{CM}}$$ is the moment of inertia of a circular disc about an axis (CM) passing through its center and perpendicular to the plane of disc. $$\mathrm{I_{AB}}$$ is it's moment of inertia about an axis AB perpendicular to plane and parallel to axis CM at a distance $$\frac{2}{3}$$R from center. Where R is th... | [] | null | 17 | <p><img src="https://app-content.cdn.examgoal.net/fly/@width/image/1le5l1b1b/df35c6af-e7ed-4854-99e3-26d3046438ef/d48e6af0-ad22-11ed-8bc1-d3bd0941e5b5/file-1le5l1b1c.png?format=png" data-orsrc="https://app-content.cdn.examgoal.net/image/1le5l1b1b/df35c6af-e7ed-4854-99e3-26d3046438ef/d48e6af0-ad22-11ed-8bc1-d3bd0941e5b5... | integer | jee-main-2023-online-25th-january-morning-shift | 12,579 |
1ldws2rrh | physics | rotational-motion | moment-of-inertia | <p>A uniform solid cylinder with radius R and length L has moment of inertia I$$_1$$, about the axis of the cylinder. A concentric solid cylinder of radius $$R'=\frac{R}{2}$$ and length $$L'=\frac{L}{2}$$ is carved out of the original cylinder. If I$$_2$$ is the moment of inertia of the carved out portion of the cylind... | [] | null | 32 | $I_{1}=\frac{\left(\rho \pi R^{2} L\right) R^{2}}{2} \quad$ ( $\rho$ : density of cylinder)
<br/><br/>
$$
\begin{aligned}
& I_{2}=\frac{\left[\rho \pi\left(\frac{R}{2}\right)^{2} \frac{L}{2}\right]\left(\frac{R}{2}\right)^{2}}{2} \\\\
& \therefore \quad \frac{I_{1}}{I_{2}}=\frac{32}{1}
\end{aligned}
$$ | integer | jee-main-2023-online-24th-january-evening-shift | 12,580 |
1ldyezter | physics | rotational-motion | moment-of-inertia | <p>Solid sphere A is rotating about an axis PQ. If the radius of the sphere is 5 cm then its radius of gyration about PQ will be $$\sqrt x$$ cm. The value of $$x$$ is ________.</p>
<p><img src="data:image/png;base64,UklGRkoLAABXRUJQVlA4ID4LAADw2gCdASoAA70CP4HA22Y2MD+nIRIo8/AwCWlu4W8FEmNwvx6i/2/q7usOOG+nbzNeBytCn3M///q9... | [] | null | 110 | $$
\begin{aligned}
& \mathrm{I}_{\mathrm{cm}}=\frac{2}{5} \mathrm{MR}^2 \\\\
& \mathrm{I}_{\mathrm{PQ}}=\mathrm{I}_{\mathrm{cm}}+\mathrm{md}^2 \\\\
& \mathrm{I}_{\mathrm{PQ}}=\frac{2}{5} \mathrm{mR}^2+\mathrm{m}(10 \mathrm{~cm})^2
\end{aligned}
$$<br/><br/>
For radius of gyration $\mathrm{I}_{\mathrm{PQ}}=\mathrm{mk}^2... | integer | jee-main-2023-online-24th-january-morning-shift | 12,581 |
lgnz71sz | physics | rotational-motion | moment-of-inertia | A solid sphere and a solid cylinder of same mass and radius are rolling on a horizontal surface without slipping. The ratio of their radius of gyrations respectively $\left(k_{\text {sph }}: k_{\text {cyl }}\right)$ is $2: \sqrt{x}$. The value of $x$ is ____________ . | [] | null | 5 | Let the mass of both the solid sphere and the solid cylinder be $M$, and let their common radius be $R$. The moment of inertia $I$ of a solid sphere and a solid cylinder are given by:
<br/><br/>
$I_{sph} = \frac{2}{5}MR^2$
<br/><br/>
$I_{cyl} = \frac{1}{2}MR^2$
<br/><br/>
The radius of gyration $k$ is related to the mo... | integer | jee-main-2023-online-15th-april-morning-shift | 12,582 |
1lh25v73h | physics | rotational-motion | moment-of-inertia | <p>Two identical solid spheres each of mass $$2 \mathrm{~kg}$$ and radii $$10 \mathrm{~cm}$$ are fixed at the ends of a light rod. The separation between the centres of the spheres is $$40 \mathrm{~cm}$$. The moment of inertia of the system about an axis perpendicular to the rod passing through its middle point is ____... | [] | null | 176 | <p>The problem requires calculating the moment of inertia of the system consisting of two identical solid spheres fixed at the ends of a light rod. We need to find the moment of inertia about an axis perpendicular to the rod and passing through its midpoint.</p>
<p>First, let’s identify the moment of inertia of each s... | integer | jee-main-2023-online-6th-april-morning-shift | 12,584 |
1lh31p7nb | physics | rotational-motion | moment-of-inertia | <p>A ring and a solid sphere rotating about an axis passing through their centers have same radii of gyration. The axis of rotation is perpendicular to plane of ring. The ratio of radius of ring to that of sphere is $$\sqrt{\frac{2}{x}}$$. The value of $$x$$ is ___________.</p> | [] | null | 5 | <p>Given that the radii of gyration for the ring and the solid sphere are equal, we have:</p>
<p>$$
K_1 = K_2
$$</p>
<p>For the ring, the moment of inertia is:</p>
<p>$$
I_{ring} = mR_1^2 = mK_1^2
$$</p>
<p>Thus, the radius of gyration for the ring is:</p>
<p>$$
K_1 = R_1
$$</p>
<p>For the solid sphere, the moment of i... | integer | jee-main-2023-online-6th-april-evening-shift | 12,585 |
jaoe38c1lsc4dwm8 | physics | rotational-motion | moment-of-inertia | <p>Four particles each of mass $$1 \mathrm{~kg}$$ are placed at four corners of a square of side $$2 \mathrm{~m}$$. Moment of inertia of system about an axis perpendicular to its plane and passing through one of its vertex is _____ $$\mathrm{kgm}^2$$.</p>
<p><img src="data:image/png;base64,UklGRrAIAABXRUJQVlA4IKQIAACQw... | [] | null | 16 | <p><img src="https://app-content.cdn.examgoal.net/fly/@width/image/6y3zli1lscua8ln/5603b55a-4f97-4d82-ac3f-d1f9b8def47e/82f86db0-c64b-11ee-9d8b-f1be86a1b2f3/file-6y3zli1lscua8lo.png?format=png" data-orsrc="https://app-content.cdn.examgoal.net/image/6y3zli1lscua8ln/5603b55a-4f97-4d82-ac3f-d1f9b8def47e/82f86db0-c64b-11ee... | integer | jee-main-2024-online-27th-january-morning-shift | 12,586 |
jaoe38c1lsd8itby | physics | rotational-motion | moment-of-inertia | <p>Two identical spheres each of mass $$2 \mathrm{~kg}$$ and radius $$50 \mathrm{~cm}$$ are fixed at the ends of a light rod so that the separation between the centers is $$150 \mathrm{~cm}$$. Then, moment of inertia of the system about an axis perpendicular to the rod and passing through its middle point is $$\frac{x}... | [] | null | 53 | <p><img src="https://app-content.cdn.examgoal.net/fly/@width/image/6y3zli1lsiiik6n/2a5f80e0-fd01-45e1-a1b6-3286c2c79074/30bcf3f0-c96a-11ee-b416-eff853096672/file-6y3zli1lsiiik6o.png?format=png" data-orsrc="https://app-content.cdn.examgoal.net/image/6y3zli1lsiiik6n/2a5f80e0-fd01-45e1-a1b6-3286c2c79074/30bcf3f0-c96a-11ee... | integer | jee-main-2024-online-31st-january-evening-shift | 12,587 |
lv7v4rjv | physics | rotational-motion | moment-of-inertia | <p>Ratio of radius of gyration of a hollow sphere to that of a solid cylinder of equal mass, for moment of Inertia about their diameter axis $$A B$$ as shown in figure is $$\sqrt{8 / x}$$. The value of $$x$$ is :</p>
<p><img src="data:image/png;base64,UklGRrQOAABXRUJQVlA4IKgOAADQ0ACdASoAAwMCP4HA3GS2MTunI1O5E3AwCWlu4WxD... | [{"identifier": "A", "content": "34"}, {"identifier": "B", "content": "51"}, {"identifier": "C", "content": "67"}, {"identifier": "D", "content": "17"}] | ["C"] | null | <p>For hollow sphere</p>
<p>$$\begin{aligned}
& I_1=\frac{2}{3} m R^2=\left(\sqrt{\frac{2}{3}} R\right)^2 m \\
& \therefore \quad K_1=\sqrt{\frac{2}{3}} R
\end{aligned}$$</p>
<p>For solid cylinder</p>
<p>$$\begin{aligned}
& I_2=\frac{1}{4} m R^2+\frac{m}{3}(4 R)^2 \\
& =\left(\sqrt{\frac{67}{12}} R\right)^2 m \\
& \the... | mcq | jee-main-2024-online-5th-april-morning-shift | 12,589 |
lvb29eoa | physics | rotational-motion | moment-of-inertia | <p>Three balls of masses $$2 \mathrm{~kg}, 4 \mathrm{~kg}$$ and $$6 \mathrm{~kg}$$ respectively are arranged at centre of the edges of an equilateral triangle of side $$2 \mathrm{~m}$$. The moment of inertia of the system about an axis through the centroid and perpendicular to the plane of triangle, will be ________ $$... | [] | null | 4 | <img src="https://app-content.cdn.examgoal.net/fly/@width/image/jaoe38c1lycativ4/73b62d51-7a39-440e-a119-b71e415f033d/b0f77d70-3cc8-11ef-b1db-75e49a1f78b6/file-jaoe38c1lycativb.png?format=png" data-orsrc="https://app-content.cdn.examgoal.net/image/jaoe38c1lycativ4/73b62d51-7a39-440e-a119-b71e415f033d/b0f77d70-3cc8-11ef... | integer | jee-main-2024-online-6th-april-evening-shift | 12,590 |
CIUadGcAS8eenG8f | physics | rotational-motion | torque | Let $$\overrightarrow F $$ be the force acting on a particle having position vector $$\overrightarrow r ,$$ and $$\overrightarrow \tau $$ be the torque of this force about the origin. Then | [{"identifier": "A", "content": "$$\\overrightarrow {r.} \\overrightarrow \\tau = 0\\,\\,$$ and $$\\overrightarrow {F.} \\overrightarrow \\tau \\ne 0\\,\\,$$ "}, {"identifier": "B", "content": "$$\\overrightarrow {r.} \\vec \\tau \\ne 0{\\mkern 1mu} {\\mkern 1mu} $$ and $$\\overrightarrow {F.} \\overrightarrow \\tau ... | ["D"] | null | <img class="question-image" src="https://imagex.cdn.examgoal.net/uqESmXEz4kNJy0XZA/6jwHiTCEx0WFEcOivn07bvBwR6o1q/x3HiteVMKop7uhSoUkb54o/image.svg" loading="lazy" alt="AIEEE 2003 Physics - Rotational Motion Question 204 English Explanation">
<br>As we know $$\overrightarrow \tau = \overrightarrow r \times \overright... | mcq | aieee-2003 | 12,591 |
tD2xnNx3ZNaZCq8b | physics | rotational-motion | torque | A force of $$ - F\widehat k$$ acts on $$O,$$ the origin of the coordinate system. The torque about the point $$(1, -1)$$ is
<img src="data:image/png;base64,UklGRqgGAABXRUJQVlA4IJwGAABwQwCdASrbAUoBPm02mUmkIqKhILYomIANiWlu4Xcx/V1qGT+pP8n7av7t0ans0+Y/Ufx33C/jv91/E/8Ofa3af/w38Z/YDg0AAfSz+u/xD2amgHqgfpXyRcbX6geo5/k/xX+5/r... | [{"identifier": "A", "content": "$$F\\left( {\\widehat i - \\widehat j} \\right)$$ "}, {"identifier": "B", "content": "$$ - F\\left( {\\widehat i + \\widehat j} \\right)$$ "}, {"identifier": "C", "content": "$$F\\left( {\\widehat i + \\widehat j} \\right)$$ "}, {"identifier": "D", "content": "$$ - F\\left( {\\widehat i... | ["C"] | null | We know, Torque $$\overrightarrow \tau = \overrightarrow r \times \overrightarrow F $$
<br><br>$$= \left( {\widehat i - \widehat j} \right) \times \left( { - F\widehat k} \right) $$
<br><br>$$= F\left( {\widehat i + \widehat j} \right)$$ | mcq | aieee-2006 | 12,592 |
xmtiN91wgRQTentL | physics | rotational-motion | torque | A pulley of radius $$2$$ $$m$$ is rotated about its axis by a force $$F = \left( {20t - 5{t^2}} \right)$$ newton (where $$t$$ is measured in seconds) applied tangentially. If the moment of inertia of the pulley about its axis of rotation is $$10kg$$-$${m^2}$$ the number of rotation made by the pulley before its direct... | [{"identifier": "A", "content": "more than $$3$$ but less than $$6$$ "}, {"identifier": "B", "content": "more than $$6$$ but less than $$9$$ "}, {"identifier": "C", "content": "more than $$9$$ "}, {"identifier": "D", "content": "less than $$3$$ "}] | ["A"] | null | Given $$F = 20t - 5{t^2}$$, R = 2 m and $$I$$ = 10 kg m<sup>2</sup>
<br><br>Torque applied on pulley $$\tau = FR$$
<br><br>$$\therefore$$ $$\alpha = {{FR} \over I}$$ [ as $$\tau = I\alpha $$ ]
<br><br>$$ \Rightarrow $$ $$\alpha = {{\left( {20t - 5{t^2}} \right) \times 2} \over {10}}$$
<br><br>$$ \Rightarrow $$ $$\... | mcq | aieee-2011 | 12,593 |
SEush98CufvOvXvQ | physics | rotational-motion | torque | A roller is made by joining together two cones at their vertices $$0$$. It is kept on two rails $$AB$$ and $$CD$$, which are placed asymmetrically (see figure), with its axis perpendicular to $$CD$$ and its center $$O$$ at the center of line joining $$AB$$ and $$CD$$ (see figure). It is given a light push so that it st... | [{"identifier": "A", "content": "go straight "}, {"identifier": "B", "content": "turn left and right alternately "}, {"identifier": "C", "content": "turn left "}, {"identifier": "D", "content": "turn right "}] | ["C"] | null | <img src="https://app-content.cdn.examgoal.net/fly/@width/image/1l7zddqs7/b98de344-c455-4d3c-8e96-ec16052b158d/e53a4580-32ee-11ed-8cf6-c1445513adbd/file-1l7zddqs8.png?format=png" data-orsrc="https://app-content.cdn.examgoal.net/image/1l7zddqs7/b98de344-c455-4d3c-8e96-ec16052b158d/e53a4580-32ee-11ed-8cf6-c1445513adbd/fi... | mcq | jee-main-2016-offline | 12,595 |
MJRqack07GUqVV3WcQzZQ | physics | rotational-motion | torque | In a physical balance working on the principle of moments, when 5 mg weight is placed on the left pan, the beam becomes horizontal. Both the empty pans of the balance are of equal mass. Which of the following statements is <b>correct</b> ?
| [{"identifier": "A", "content": "Left arm is longer than the right arm"}, {"identifier": "B", "content": "Both the arms are of same length"}, {"identifier": "C", "content": "Left arm is shorter than the right arm"}, {"identifier": "D", "content": "Every object that is weighed using this balance appears lighter than its... | ["C"] | null | From principle of moment we know, the anticlockwise moment is equal to clockwise moment when a system is stable or balance.
<br><br>$$\therefore\,\,\,$$ load $$ \times $$ load arm = effect $$ \times $$ effect arm
<br><br>When 5 mg weight is placed on the left pan, load arm shift to left side, hence left arm become sho... | mcq | jee-main-2017-online-8th-april-morning-slot | 12,597 |
shugQNB8jJKFVrgp4ot79 | physics | rotational-motion | torque | A uniform disc of radius R and mass M is free to rotate only about its axis. A string is wrapped over its rim and a body of mass m is tied to the free end of the string as shown in the figure. The body is released from rest. Then the acceleration of the body is :
<br/><br/><img src="data:image/png;base64,UklGRnYPAABXRU... | [{"identifier": "A", "content": "$${{2\\,\\,mg} \\over {2\\,m + M}}$$ "}, {"identifier": "B", "content": "$${{2\\,\\,Mg} \\over {2\\,m + M}}$$ "}, {"identifier": "C", "content": "$${{2\\,\\,mg} \\over {2\\,M + m}}$$ "}, {"identifier": "D", "content": "$${{2\\,\\,Mg} \\over {2\\,M + M}}$$ "}] | ["A"] | null | <img src="https://app-content.cdn.examgoal.net/fly/@width/image/1l83nur35/58fa9866-026c-430e-84e7-3a339d20849e/fc1b4110-354a-11ed-8c5c-f3f0850e6b94/file-1l83nur36.png?format=png" data-orsrc="https://app-content.cdn.examgoal.net/image/1l83nur35/58fa9866-026c-430e-84e7-3a339d20849e/fc1b4110-354a-11ed-8c5c-f3f0850e6b94/fi... | mcq | jee-main-2017-online-8th-april-morning-slot | 12,598 |
neeivCZGZMIeBaGwBIyu8 | physics | rotational-motion | torque | A force of $$40$$ $$N$$ acts on a point $$B$$ at the end of an $$L$$-shaped object, as shown in the figure. The angle $$\theta $$ that will produce maximum moment of the force about point $$A$$ is given by :
<br/><br/><img src="data:image/png;base64,UklGRgQOAABXRUJQVlA4IPgNAADQ1gCdASoAA20CP4HA3GU2MS2nIRNJUsAwCWlu8p8d/... | [{"identifier": "A", "content": "$$\\tan \\theta = {1 \\over 2}$$"}, {"identifier": "B", "content": "$$\\tan \\theta = 2$$ "}, {"identifier": "C", "content": "$$\\tan \\theta = 4$$"}, {"identifier": "D", "content": "$$\\tan \\theta = {1 \\over 4}$$ "}] | ["A"] | null | We can produce maximum moment of force, when line of action of force is perpendicular to line AB.
<br><br><img src="https://app-content.cdn.examgoal.net/fly/@width/image/1l7u6zsvh/1e42bedd-6565-423f-8087-53d1dc6f2cef/ff7e09d0-3015-11ed-ab58-4f43f7f8c851/file-1l7u6zsvi.png?format=png" data-orsrc="https://app-content.cd... | mcq | jee-main-2018-online-15th-april-morning-slot | 12,599 |
hxxseQetp6B4gyc4ZPhis | physics | rotational-motion | torque | <img src="data:image/png;base64,UklGRmAJAABXRUJQVlA4IFQJAACwngCdASoAA9ABP4HA2GY2L7+nILPZW/AwCWlu4XHEcmNwvV6X9H13NyT4C7k5mdv/0Ri/0feQ76rvjModDWxtYHo5SbbZnCuiUoSiQIzfVfKJ8NHzaucaJb31c40P5iwelpX01IGTX83QJDD+htuAjZmEz2LhylNiurGEINpp6PP5MCMmQczPc+0qbb31c40S3vq5xolvfVziw21gyX4Z6IrQGH9PDSrSShjVpK42Uu6GvWEQE5i5LjSwl04TCA5UfNq5... | [{"identifier": "A", "content": "$$m,x$$ "}, {"identifier": "B", "content": "$$m,{1 \\over x}$$ "}, {"identifier": "C", "content": "$$m,{1 \\over {{x^2}}}$$ "}, {"identifier": "D", "content": "$$m,{x^2}$$ "}] | ["B"] | null | <img src="https://app-content.cdn.examgoal.net/fly/@width/image/1l7u7n8ko/20f81bdc-a422-4358-bbec-14511e74da77/8b3afa80-3018-11ed-ab58-4f43f7f8c851/file-1l7u7n8kp.png?format=png" data-orsrc="https://app-content.cdn.examgoal.net/image/1l7u7n8ko/20f81bdc-a422-4358-bbec-14511e74da77/8b3afa80-3018-11ed-ab58-4f43f7f8c851/fi... | mcq | jee-main-2018-online-15th-april-morning-slot | 12,600 |
P2ZmuhDyMqDIgR4NS83rsa0w2w9jx3e9jgg | physics | rotational-motion | torque | A uniform rod of length $$\ell $$ is being rotated in a horizontal plane with a constant angular speed about an axis
passing through one of its ends. If the tension generated in the rod due to rotation is T(x) at a distance x from
the axis, then which of the following graphs depicts it most closely? | [{"identifier": "A", "content": "<img src=\"https://res.cloudinary.com/dckxllbjy/image/upload/v1734267387/exam_images/fyfz8tprrfpylccpzcyb.webp\" style=\"max-width: 100%; height: auto;display: block;margin: 0 auto;\" loading=\"lazy\" alt=\"JEE Main 2019 (Online) 12th April Morning Slot Physics - Rotational Motion Ques... | ["B"] | null | $$T = \int\limits_{x = x}^{x = \ell } {dm{\omega ^2}} x = \int\limits_{x = x}^{x = \ell } {{m \over \ell }} dx{\omega ^2}x\,T$$<br><br>
$$ = {{m{\omega ^2}} \over {2\ell }}\left( {{\ell ^2} - {x^2}} \right)$$<br><br>$$ \therefore T = {{m{\omega ^2}} \over {2\ell }}\left( {{\ell ^2} - {x^2}} \right)$$ | mcq | jee-main-2019-online-12th-april-morning-slot | 12,603 |
KNqMILrsamP2z6oBIv3rsa0w2w9jwzhlt2a | physics | rotational-motion | torque | A metal coin of mass 5 g and radius 1 cm is fixed to a thin stick AB of negligible mass as shown in the
figure. The system is initially at rest. The constant torque, that will make the system rotate about AB at 25
rotations per second in 5s, is close to :
<img src="data:image/png;base64,UklGRugOAABXRUJQVlA4INwOAACwbACd... | [{"identifier": "A", "content": "7.9 \u00d7 10<sup>\u20136</sup> Nm"}, {"identifier": "B", "content": "4.0 \u00d7 10<sup>\u20136</sup> Nm"}, {"identifier": "C", "content": "2.0 \u00d7 10<sup>\u20135</sup> Nm"}, {"identifier": "D", "content": "1.6 \u00d7 10<sup>\u20135</sup> Nm"}] | ["C"] | null | m = 5 × 10<sup>–3</sup> kg, r = 10<sup>–2</sup> m<br>
$$\omega $$ = 25 × 2$$\pi $$ rad/5<br>
= 50 $$\omega $$ rad/sec<br><br>
$$\omega = {\tau \over I}t$$<br>
$$\tau = {{I\omega } \over t} = {{5m{r^2}} \over 4} \times {\omega \over t}$$<br><br>
$$ = {{5 \times 5 \times {{10}^{ - 3}} \times {{10}^{ - 4}} \times 50\p... | mcq | jee-main-2019-online-10th-april-evening-slot | 12,604 |
lIeeCoZEUWeWBWBfNL18hoxe66ijvzncp98 | physics | rotational-motion | torque | A particle of mass m is moving along a
trajectory given by<br/>
x = x<sub>0</sub> + a cos$$\omega $$<sub>1</sub>t<br/>
y = y<sub>0</sub> + b sin$$\omega $$<sub>2</sub>t<br/>
The torque, acting on the particle about the
origin, at t = 0 is : | [{"identifier": "A", "content": "Zero"}, {"identifier": "B", "content": "+my<sub>0</sub>a $$\\omega _1^2$$$$\\widehat k$$"}, {"identifier": "C", "content": "$$ - m\\left( {{x_0}b\\omega _2^2 - {y_0}a\\omega _1^2} \\right)\\widehat k$$"}, {"identifier": "D", "content": "m (\u2013x<sub>0</sub>b + y<sub>0</sub>a) $$\\omeg... | ["B"] | null | $$\overrightarrow F = m\overrightarrow a = m\left[ { - a\omega _1^2\cos \omega ,t\widehat i - b\omega _2^2\sin {\omega _2}t\widehat j} \right]$$<br><br>
$${\overrightarrow f _{t = 0}} = - ma\omega _1^2\widehat i$$<br><br>
$${\overrightarrow r _{t = 0}} = \left( {{X_0} + a} \right)\widehat i + y\widehat j$$<br><br>
$... | mcq | jee-main-2019-online-10th-april-morning-slot | 12,605 |
7IgcOr0hYSrQ4geKwMzue | physics | rotational-motion | torque | The magnitude of torque on a particle of mass 1 kg is 2.5 Nm about the origin. If the force acting on it is 1 N, and the distance of the particle from the origin is 5m, the angle between the force and the position vector is (in radians) : | [{"identifier": "A", "content": "$${\\pi \\over 8}$$"}, {"identifier": "B", "content": "$${\\pi \\over 6}$$"}, {"identifier": "C", "content": "$${\\pi \\over 4}$$"}, {"identifier": "D", "content": "$${\\pi \\over 3}$$"}] | ["B"] | null | 2.5 = 1 $$ \times $$ 5 sin $$\theta $$
<br><br>sin$$\theta $$ = 0.5 = $${1 \over 2}$$
<br><br>$$\theta $$ = $${\pi \over 6}$$ | mcq | jee-main-2019-online-11th-january-evening-slot | 12,607 |
J4UGaA7Sbv6AC8krkWEm0 | physics | rotational-motion | torque | A slab is subjected to two forces $$\overrightarrow {{F_1}} $$ and $$\overrightarrow {{F_2}} $$ of same magnitude F as shown in the figure. Force $$\overrightarrow {{F_2}} $$ is in XY-plane while force $$\overrightarrow {{F_1}} $$ acts along z = axis at the point $$\left( {2\overrightarrow i + 3\overrightarrow j } \r... | [{"identifier": "A", "content": "$$\\left( {3\\widehat i - 2\\widehat j - 3\\widehat k} \\right)F$$"}, {"identifier": "B", "content": "$$\\left( {3\\widehat i + 2\\widehat j - 3\\widehat k} \\right)F$$"}, {"identifier": "C", "content": "$$\\left( {3\\widehat i + 2\\widehat j + 3\\widehat k} \\right)F$$"}, {"identifier"... | ["D"] | null | <p>To determine the moment of the forces about point O, we need to calculate the moments of each force and then combine them vectorially.</p>
<p>First, consider force $$\overrightarrow{F_1}$$. This force acts along the z-axis at the point $$(2\overrightarrow{i} + 3\overrightarrow{j})$$.</p>
<p>The position vector of ... | mcq | jee-main-2019-online-11th-january-morning-slot | 12,608 |
vMaqrY8yal6coKj5AlsEU | physics | rotational-motion | torque | A rigid massless rod of length 3l has two masses attached at each end as shown in the figure. The rod is pivoted at point P on the horizontal axis (see figure). When released from initial horizontal position, its
instantaneous angular acceleration will be -
<br/><br/><img src="data:image/png;base64,UklGRhwIAABXRUJQVlA... | [{"identifier": "A", "content": "$${g \\over {13l}}$$"}, {"identifier": "B", "content": "$${g \\over {2l}}$$"}, {"identifier": "C", "content": "$${g \\over {3l}}$$"}, {"identifier": "D", "content": "$${7g \\over {3l}}$$"}] | ["A"] | null | <img src="https://res.cloudinary.com/dckxllbjy/image/upload/v1734264995/exam_images/frgrlsfo90f2gzmijvd7.webp" style="max-width: 100%; height: auto;display: block;margin: 0 auto;" loading="lazy" alt="JEE Main 2019 (Online) 10th January Evening Slot Physics - Rotational Motion Question 160 English Explanation">
<br>App... | mcq | jee-main-2019-online-10th-january-evening-slot | 12,609 |
99ytypmVzB3HdMTD16VFF | physics | rotational-motion | torque | To mop-clean a floor, a cleaning machine presses a circular mop of radius R vertically down with a total force F and rotates it with a constant angular speed about its axis. If the force F is distributed uniformly over the mop and if coefficient of friction between the mop and the floor is $$\mu $$, the torque, applied... | [{"identifier": "A", "content": "$$\\mu $$FR/2"}, {"identifier": "B", "content": "$$\\mu $$FR/3"}, {"identifier": "C", "content": "$$\\mu $$FR/6"}, {"identifier": "D", "content": "$${2 \\over 3}$$$$\\mu $$FR"}] | ["D"] | null | <img src="https://res.cloudinary.com/dckxllbjy/image/upload/v1734263633/exam_images/dh3kvipt5kpzdkepgc9e.webp" style="max-width: 100%; height: auto;display: block;margin: 0 auto;" loading="lazy" alt="JEE Main 2019 (Online) 10th January Morning Slot Physics - Rotational Motion Question 162 English Explanation">
<br><br... | mcq | jee-main-2019-online-10th-january-morning-slot | 12,610 |
60q1AvE0F7mWohQowPjgy2xukfhqolwj | physics | rotational-motion | torque | A force $$\overrightarrow F = \left( {\widehat i + 2\widehat j + 3\widehat k} \right)$$ N acts at a point <br/>$$\left( {4\widehat i + 3\widehat j - \widehat k} \right)$$ m. Then the magnitude of torque
<br/>about the point $$\left( {\widehat i + 2\widehat j + \widehat k} \right)$$ m will be $$\sqrt x $$ N m.
<br/>The... | [] | null | 195 | $$\overrightarrow \tau = \overrightarrow r \times F = (3\widehat i + \widehat j - 2\widehat k) \times (\widehat i + 2\widehat j + 3\widehat k)$$<br><br>$$ = \left| {\matrix{
i & j & k \cr
3 & 1 & { - 2} \cr
1 & 2 & 3 \cr
} } \right|$$<br><br>$$ = \widehat i(3 + 4) - \widehat ... | integer | jee-main-2020-online-5th-september-morning-slot | 12,611 |
hdFSrrJ2kxtBiY45N57k9k2k5iodr83 | physics | rotational-motion | torque | A body of mass m = 10 kg is attached to one
end of a wire of length 0.3 m. The maximum
angular speed (in rad s<sup>–1</sup>) with which it can be
rotated about its other end in space station is :
<br/>(Breaking stress of wire = 4.8 × 10<sup>7</sup> Nm<sup>–2</sup> and
<br/>area of cross-section of the wire = 10<sup>–2<... | [] | null | 4 | T = m$${\omega ^2}l$$
<br><br>Breaking stress = $$\sigma $$ = $${{m{\omega ^2}l} \over A}$$
<br><br>$$ \Rightarrow $$ $${\omega ^2}$$ = $${{4.8 \times {{10}^7} \times \left( {{{10}^{ - 2}} \times {{10}^{ - 4}}} \right)} \over {10 \times 0.3}}$$ = 16
<br><br>$$ \Rightarrow $$ $$\omega $$ = 4 | integer | jee-main-2020-online-9th-january-morning-slot | 12,612 |
NMunNWfWOtf3IRdAgWjgy2xukf3u81sh | physics | rotational-motion | torque | A uniform rod of length ‘$$l$$’ is pivoted at one of its ends on a vertical shaft of negligible radius.
When the shaft rotates at angular speed $$\omega $$ the rod makes an angle $$\theta $$ with it (see figure). To find $$\theta $$
equate the rate of change of angular momentum (direction going into the paper) $${{m{l^... | [{"identifier": "A", "content": "$$\\cos \\theta = {{2g} \\over {3l{\\omega ^2}}}$$"}, {"identifier": "B", "content": "$$\\cos \\theta = {{3g} \\over {2l{\\omega ^2}}}$$"}, {"identifier": "C", "content": "$$\\cos \\theta = {g \\over {2l{\\omega ^2}}}$$"}, {"identifier": "D", "content": "$$\\cos \\theta = {g \\over ... | ["B"] | null | <picture><source media="(max-width: 320px)" srcset="https://res.cloudinary.com/dckxllbjy/image/upload/v1734266302/exam_images/on9bbrmzaj0ym4tteuzb.webp"><source media="(max-width: 500px)" srcset="https://res.cloudinary.com/dckxllbjy/image/upload/v1734267304/exam_images/hbunfcbhvhbvfdwlleum.webp"><source media="(max-wid... | mcq | jee-main-2020-online-3rd-september-evening-slot | 12,613 |
JrssBZ51hSXvZY2mV2jgy2xukev12neo | physics | rotational-motion | torque | A uniform cylinder of mass M and radius R is to
be pulled over a step of height a (a < R) by
applying a force F at its centre ‘O’
perpendicular to the plane through the axes of
the cylinder on the edge of the step (see
figure). The minimum value of F required is :
<img src="data:image/png;base64,UklGRkwKAABXRUJQVlA4... | [{"identifier": "A", "content": "$$Mg\\sqrt {1 - {{\\left( {{{R - a} \\over R}} \\right)}^2}} $$"}, {"identifier": "B", "content": "$$Mg\\sqrt {1 - {{{a^2}} \\over {{R^2}}}} $$"}, {"identifier": "C", "content": "$$Mg{a \\over R}$$"}, {"identifier": "D", "content": "$$Mg\\sqrt {{{\\left( {{R \\over {R - a}}} \\right)}^2... | ["A"] | null | <img src="https://res.cloudinary.com/dckxllbjy/image/upload/v1734266267/exam_images/ghmikxyaiuvfqgqgdixp.webp" style="max-width: 100%;height: auto;display: block;margin: 0 auto;" loading="lazy" alt="JEE Main 2020 (Online) 2nd September Morning Slot Physics - Rotational Motion Question 121 English Explanation">
<br><br>... | mcq | jee-main-2020-online-2nd-september-morning-slot | 12,614 |
lDFg2L8arkAJ0UK1bR7k9k2k5fdi1on | physics | rotational-motion | torque | <img src="data:image/png;base64,UklGRqIHAABXRUJQVlA4IJYHAACwOQCdASoMAc4APm00lkikIqIhIVObOIANiWlu/FsgfKm6Bj8V/1z8UPAf+19Er7E9m8oD9s/kf7EfLH9L/Sjx74AXpj++fkr5xv0r/pvf+S5eoF2W/1n8S/qf+38oD+c/pH7YcyH/sP4x/PPeT/a+D74R+gHq6/yn8o/JT4Pf7/+Nf2b9o/b79Gf87+S/173Qf9h/HP6z/yv8L8r/rz/V32RP1SI8fG252uZrv/jYoklNlyfJKq5fc8w5Z4MXpauWprFb... | [] | null | 50 | <img src="https://res.cloudinary.com/dckxllbjy/image/upload/v1734263448/exam_images/t1fetsj6sl0wy8kael49.webp" style="max-width: 100%;height: auto;display: block;margin: 0 auto;" loading="lazy" alt="JEE Main 2020 (Online) 7th January Evening Slot Physics - Rotational Motion Question 129 English Explanation">
<br>Torque... | integer | jee-main-2020-online-7th-january-evening-slot | 12,615 |
wXxeYhCfgk1xEq5wzGjgy2xukev0ufcj | physics | rotational-motion | torque | <img src="data:image/png;base64,UklGRj4IAABXRUJQVlA4IDIIAAAQPwCdASqFAaQAPm02mUgkIyKhJHKJaIANiWlu8p8dxWvIhz8Z/0ftY/xX9I6i/z/4uewD/AfwDqcL5/8F/If5D6Gf4D+Ufy/9aeE9/h/5R43/8t/GvAlAJ9U/87/Jv53+sfoCfzP8m/sPKh/or/S/Rd/Rj+AejX3f+mfrQ/4f9A/hfxIftf/I/hn839hv0P/0P4V/WvlM/Wn/gfxz+T+8cNJwgJUbgu2OEBKjcFuf3IitjC2JAJftXN58vguplQxBl+yF... | [{"identifier": "A", "content": "0.5 mg"}, {"identifier": "B", "content": "2 mg"}, {"identifier": "C", "content": "0.75 mg"}, {"identifier": "D", "content": "1 mg"}] | ["D"] | null | <picture><source media="(max-width: 320px)" srcset="https://res.cloudinary.com/dckxllbjy/image/upload/v1734264582/exam_images/ibp7adjgagnjb1qiz11j.webp"><source media="(max-width: 500px)" srcset="https://res.cloudinary.com/dckxllbjy/image/upload/v1734265448/exam_images/ykoawjwufoeijtj1xeuh.webp"><source media="(max-wid... | mcq | jee-main-2020-online-2nd-september-morning-slot | 12,616 |
HBw8KBk3vF4Bcjh6p11kmhpfulc | physics | rotational-motion | torque | Consider a 20 kg uniform circular disk of radius 0.2 m. It is pin supported at its center and is at rest initially. The disk is acted upon by a constant force F = 20 N through a massless string wrapped around is periphery as shown in the figure.<br/><br/><img src="data:image/png;base64,UklGRpoGAABXRUJQVlA4II4GAADwLQCdA... | [] | null | 20 | $$\alpha = {\tau \over I} = {{F.R.} \over {m{R^2}/2}} = {{2F} \over {mR}}$$<br><br>$$\alpha = {{2 \times 200} \over {20 \times (0.2)}} = 10$$ rad/s<sup>2</sup><br><br>$${\omega ^2} = {\omega _0}^2 + 2\alpha \Delta \theta $$<br><br>$${(50)^2} = {0^2} + 2(10)\Delta \theta $$<br><br>$$ \Rightarrow \Delta \theta = {{25... | integer | jee-main-2021-online-16th-march-morning-shift | 12,617 |
ZpW3Yp2GclYzmoxzkc1kmhpt3jc | physics | rotational-motion | torque | Consider a frame that is made up of two thin massless rods AB and AC as shown in the figure. A vertical force $$\overrightarrow P $$ of magnitude 100 N is applied at point A of the frame.<br/><br/><img src="data:image/png;base64,UklGRvAPAABXRUJQVlA4IOQPAABQ3ACdASqiAQADP4G+1mW2L6wnIjMpisAwCWlu6dBYIuGsLlojaCds+Jzhmq7ecxT... | [] | null | 82 | <img src="https://app-content.cdn.examgoal.net/fly/@width/image/1kx3uzit8/65e3b73a-d0b4-4c00-a994-abf09c954f6c/7f7f0ad0-5ba0-11ec-a5b4-139d631bd0b4/file-1kx3uzit9.png?format=png" data-orsrc="https://app-content.cdn.examgoal.net/image/1kx3uzit8/65e3b73a-d0b4-4c00-a994-abf09c954f6c/7f7f0ad0-5ba0-11ec-a5b4-139d631bd0b4/fi... | integer | jee-main-2021-online-16th-march-morning-shift | 12,618 |
EBY84tAudUadnrGYPm1kmiq73xb | physics | rotational-motion | torque | A force $$\overrightarrow F $$ = $${4\widehat i + 3\widehat j + 4\widehat k}$$ is applied on an intersection point of x = 2 plane and x-axis. The magnitude of torque of this force about a point (2, 3, 4) is ___________. (Round off to the Nearest Integer) | [] | null | 20 | $$\overrightarrow \tau = \overrightarrow r \times \overrightarrow F $$<br><br>$$ = \left[ {(2 - 2)\widehat i + (0 - 3)\widehat j + (0 - 4)\widehat k} \right] \times (4\widehat i + 3\widehat j + 4\widehat k)$$<br><br>$$ = ( - 3\widehat j - 4\widehat k) \times (4\widehat i + 3\widehat j + 4\widehat k)$$<br><br>$$ = -... | integer | jee-main-2021-online-16th-march-evening-shift | 12,619 |
wQEYxGZIs3cHnLxTc61kmlw4y83 | physics | rotational-motion | torque | A solid cylinder of mass m is wrapped with an inextensible light string and, is placed on a rough inclined plane as shown in the figure. The frictional force acting between the cylinder and the inclined plane is :<br/><br/><img src="data:image/png;base64,UklGRmgNAABXRUJQVlA4IFwNAACQSQCdASq/AFYBPm00lkkkIqIhIbO6mIANiWlu/... | [{"identifier": "A", "content": "0"}, {"identifier": "B", "content": "5 mg"}, {"identifier": "C", "content": "$${7 \\over 2}$$ mg"}, {"identifier": "D", "content": "$${{mg} \\over 5}$$"}] | ["D"] | null | <img src="https://res.cloudinary.com/dckxllbjy/image/upload/v1734266976/exam_images/pggpmhfzefans2xoagv3.webp" style="max-width: 100%;height: auto;display: block;margin: 0 auto;" loading="lazy" alt="JEE Main 2021 (Online) 18th March Evening Shift Physics - Rotational Motion Question 89 English Explanation">
<br>For tra... | mcq | jee-main-2021-online-18th-march-evening-shift | 12,621 |
1krwcrwzr | physics | rotational-motion | torque | A solid disc of radius 20 cm and mass 10 kg is rotating with an angular velocity of 600 rpm, about an axis normal to its circular plane and passing through its centre of mass. The retarding torque required to bring the disc at rest in 10 s is ____________ $$\pi$$ $$\times$$ 10<sup>$$-$$1</sup> Nm. | [] | null | 4 | $$\tau = {{\Delta L} \over {\Delta t}} = {{I({\omega _f} - {\omega _i})} \over {\Delta t}}$$<br><br>$$\tau = {{{{m{R^2}} \over 2} \times [0 - \omega ]} \over {\Delta t}}$$<br><br>$$ = {{10 \times {{(20 \times {{10}^{ - 2}})}^2}} \over 2} \times {{600 \times \pi } \over {30 \times 10}}$$<br><br>$$ = 0.4\pi = 4\pi \t... | integer | jee-main-2021-online-25th-july-evening-shift | 12,622 |
1l55jt87j | physics | rotational-motion | torque | <p>A $$\sqrt {34} $$ m long ladder weighing 10 kg leans on a frictionless wall. Its feet rest on the floor 3 m away from the wall as shown in the figure. If E<sub>f</sub> and F<sub>w</sub> are the reaction forces of the floor and the wall, then ratio of $${F_w}/{F_f}$$ will be :</p>
<p>(Use g = 10 m/s<sup>2</sup>.)</p>... | [{"identifier": "A", "content": "$${6 \\over {\\sqrt {110} }}$$"}, {"identifier": "B", "content": "$${3 \\over {\\sqrt {113} }}$$"}, {"identifier": "C", "content": "$${3 \\over {\\sqrt {109} }}$$"}, {"identifier": "D", "content": "$${2 \\over {\\sqrt {109} }}$$"}] | ["C"] | null | <p><img src="https://app-content.cdn.examgoal.net/fly/@width/image/1l5hwefzs/f8b080d4-5c58-4a1b-90b2-07807d04ebb3/bbb612d0-01ba-11ed-85a8-43d162d2b7e8/file-1l5hwefzx.png?format=png" data-orsrc="https://app-content.cdn.examgoal.net/image/1l5hwefzs/f8b080d4-5c58-4a1b-90b2-07807d04ebb3/bbb612d0-01ba-11ed-85a8-43d162d2b7e8... | mcq | jee-main-2022-online-28th-june-evening-shift | 12,623 |
1l55mk131 | physics | rotational-motion | torque | <p>A uniform disc with mass M = 4 kg and radius R = 10 cm is mounted on a fixed horizontal axle as shown in figure. A block with mass m = 2 kg hangs from a massless cord that is wrapped around the rim of the disc. During the fall of the block, the cord does not slip and there is no friction at the axle. The tension in ... | [] | null | 10 | <p><img src="https://app-content.cdn.examgoal.net/fly/@width/image/1l5hwn9ui/d78045da-e6b2-4941-b5f9-af239ee9374c/b13e0fa0-01bb-11ed-85a8-43d162d2b7e8/file-1l5hwn9uj.png?format=png" data-orsrc="https://app-content.cdn.examgoal.net/image/1l5hwn9ui/d78045da-e6b2-4941-b5f9-af239ee9374c/b13e0fa0-01bb-11ed-85a8-43d162d2b7e8... | integer | jee-main-2022-online-28th-june-evening-shift | 12,624 |
1l5c4u6zp | physics | rotational-motion | torque | <p>A metre scale is balanced on a knife edge at its centre. When two coins, each of mass 10 g are put one on the top of the other at the 10.0 cm mark the scale is found to be balanced at 40.0 cm mark. The mass of the metre scale is found to be x $$\times$$ 10<sup>$$-$$2</sup> kg. The value of x is ___________.</p> | [] | null | 6 | <img src="https://app-content.cdn.examgoal.net/fly/@width/image/6y3zli1lizupxtg/59ea8ced-e71a-4808-97df-913b2b75673d/14bbdc30-0cf9-11ee-8ead-7d21004ddad7/file-6y3zli1lizupxth.png?format=png" data-orsrc="https://app-content.cdn.examgoal.net/image/6y3zli1lizupxtg/59ea8ced-e71a-4808-97df-913b2b75673d/14bbdc30-0cf9-11ee-8e... | integer | jee-main-2022-online-24th-june-morning-shift | 12,626 |
1l6jk2emh | physics | rotational-motion | torque | <p>A pulley of radius $$1.5 \mathrm{~m}$$ is rotated about its axis by a force $$F=\left(12 \mathrm{t}-3 \mathrm{t}^{2}\right) N$$ applied tangentially (while t is measured in seconds). If moment of inertia of the pulley about its axis of rotation is $$4.5 \mathrm{~kg} \mathrm{~m}^{2}$$, the number of rotations made by... | [] | null | 18 | <p><img src="https://app-content.cdn.examgoal.net/fly/@width/image/1l6xtmozl/c04d538a-7804-40e1-afb0-fa443317884f/a782f710-1e48-11ed-9c61-4529b721806b/file-1l6xtmozm.png?format=png" data-orsrc="https://app-content.cdn.examgoal.net/image/1l6xtmozl/c04d538a-7804-40e1-afb0-fa443317884f/a782f710-1e48-11ed-9c61-4529b721806b... | integer | jee-main-2022-online-27th-july-morning-shift | 12,627 |
1lduhy6i3 | physics | rotational-motion | torque | <p>An object of mass 8 kg is hanging from one end of a uniform rod CD of mass 2 kg and length 1 m pivoted at its end C on a vertical wall as shown in figure. It is supported by a cable AB such that the system is in equilibrium. The tension in the cable is (Take g = 10 m/s$$^2$$)</p>
<p><img src="data:image/png;base64,U... | [{"identifier": "A", "content": "90 N"}, {"identifier": "B", "content": "240 N"}, {"identifier": "C", "content": "30 N"}, {"identifier": "D", "content": "300 N"}] | ["D"] | null | <p><img src="https://app-content.cdn.examgoal.net/fly/@width/image/1le5jn288/986708ca-be49-48e6-8efe-557d557409a2/5f4b9880-ad1d-11ed-8bc1-d3bd0941e5b5/file-1le5jn289.png?format=png" data-orsrc="https://app-content.cdn.examgoal.net/image/1le5jn288/986708ca-be49-48e6-8efe-557d557409a2/5f4b9880-ad1d-11ed-8bc1-d3bd0941e5b5... | mcq | jee-main-2023-online-25th-january-morning-shift | 12,629 |
1lgp0utju | physics | rotational-motion | torque | <p>A light rope is wound around a hollow cylinder of mass 5 kg and radius 70 cm. The rope is pulled with a force of 52.5 N. The angular acceleration of the cylinder will be _________ rad s$$^{-2}$$.</p> | [] | null | 15 | In this problem, the net force on the cylinder is the tension $$T$$ in the rope, which is equal to the force applied to the rope:
<br/><br/>
$$F=T=52.5~\mathrm{N}$$
<br/><br/>
The force causes the cylinder to accelerate with an angular acceleration $$\alpha$$, which is related to its linear acceleration $$a$$ and the r... | integer | jee-main-2023-online-13th-april-evening-shift | 12,630 |
1lgvs5aid | physics | rotational-motion | torque | <p>A force of $$-\mathrm{P} \hat{\mathrm{k}}$$ acts on the origin of the coordinate system. The torque about the point $$(2,-3)$$ is $$\mathrm{P}(a \hat{i}+b \hat{j})$$, The ratio of $$\frac{a}{b}$$ is $$\frac{x}{2}$$. The value of $$x$$ is -</p> | [] | null | 3 | <p>Let the point where the force acts be A, the origin of the coordinate system (0, 0, 0), and let the point about which the torque is calculated be B (2, -3, 0). The force vector is given by $$\vec{F} = -P\hat{k}$$.</p>
<p>To find the torque, we first find the position vector of point A with respect to point B:</p>
<p... | integer | jee-main-2023-online-10th-april-evening-shift | 12,631 |
jaoe38c1lscp2nwn | physics | rotational-motion | torque | <p>A heavy iron bar of weight $$12 \mathrm{~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}$$ with the horizontal, the weight experienced by the man is :</p> | [{"identifier": "A", "content": "$$3 \\mathrm{~kg}$$\n"}, {"identifier": "B", "content": "$$6 \\mathrm{~kg}$$\n"}, {"identifier": "C", "content": "$$6 \\sqrt{3} \\mathrm{~kg}$$\n"}, {"identifier": "D", "content": "$$12 \\mathrm{~kg}$$"}] | ["B"] | null | <p><img src="https://app-content.cdn.examgoal.net/fly/@width/image/6y3zli1lu533t2l/40d6f847-fc62-4927-98e2-437a284c3fa2/2bd802c0-e9a0-11ee-b9cd-e33b518f616d/file-6y3zli1lu533t2m.png?format=png" data-orsrc="https://app-content.cdn.examgoal.net/image/6y3zli1lu533t2l/40d6f847-fc62-4927-98e2-437a284c3fa2/2bd802c0-e9a0-11ee... | mcq | jee-main-2024-online-27th-january-evening-shift | 12,632 |
luz2uzm2 | physics | rotational-motion | torque | <p>A string is wrapped around the rim of a wheel of moment of inertia $$0.40 \mathrm{~kgm}^2$$ and radius $$10 \mathrm{~cm}$$. The wheel is free to rotate about its axis. Initially the wheel is at rest. The string is now pulled by a force of $$40 \mathrm{~N}$$. The angular velocity of the wheel after $$10 \mathrm{~s}$$... | [] | null | 100 | <p>To find the angular velocity ($\omega$) of the wheel after $10$ seconds, we first need to understand the relationship between the force applied through the string, the torque produced by this force, and how this torque affects the wheel's angular acceleration ($\alpha$).
<p>The torque ($\tau$) produced by the force... | integer | jee-main-2024-online-9th-april-morning-shift | 12,633 |
B7GUvm9882JuXuow | physics | simple-harmonic-motion | forced,-damped-oscillations-and-resonance | If a simple pendulum has significant amplitude (up to a factor of $$1/e$$ of original ) only in the period between $$t = 0s\,\,to\,\,t = \tau \,s,$$ then $$\tau \,$$ may be called the average life of the pendulum When the spherical bob of the pendulum suffers a retardation (due to viscous drag) proportional to its velo... | [{"identifier": "A", "content": "$${{0.693} \\over b}$$ "}, {"identifier": "B", "content": "$$b$$ "}, {"identifier": "C", "content": "$${1 \\over b}$$ "}, {"identifier": "D", "content": "$${2 \\over b}$$ "}] | ["D"] | null | The equation of motion for the pendulum, suffering retardation
<br><br>$$I\alpha = - mg\left( {\ell \sin \theta } \right) - mbv\left( \ell \right)$$ where $$I = m{\ell ^2}$$
<br><br>and $$\alpha = {d^2}\theta /d{t^2}$$
<br><br>$$\therefore$$ $${{{d^2}\theta } \over {d{t^2}}} = - {g \over \ell }\tan \theta + {{bv... | mcq | aieee-2012 | 12,635 |
VgQ1HInwgYLBv8GP | physics | simple-harmonic-motion | forced,-damped-oscillations-and-resonance | The amplitude of a damped oscillator decreases to $$0.9$$ times its original magnitude in $$5s$$. In another $$10s$$ it will decrease to $$\alpha $$ times its original magnitude, where $$\alpha $$ equals | [{"identifier": "A", "content": "$$0.7$$ "}, {"identifier": "B", "content": "$$0.81$$ "}, {"identifier": "C", "content": "$$0.729$$ "}, {"identifier": "D", "content": "$$0.6$$ "}] | ["C"] | null | as $$\,\,A = {A_0}{e^{{{bt} \over {2m}}}}$$ (where, $${A_0} = $$ maximum amplitude)
<br><br>According to the questions, after $$5$$ second,
<br><br>$$0.9{A_0} = {A_0}{e^{ - {{b\left( 5 \right)} \over {2m}}}}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...\left( i \right)$$
<br><br>After $$10$$ more second,
<br><br>$$A = {A_0}{e^... | mcq | jee-main-2013-offline | 12,636 |
ceMdTgOe0w5MurlETAo8s | physics | simple-harmonic-motion | forced,-damped-oscillations-and-resonance | A block of mass 0.1 kg is connected to an elastic spring of spring constant 640 Nm<sup>−1</sup> and oscillates in a damping medium of damping constant 10<sup>−2</sup> kg s<sup>−1</sup> . The system dissipates its energy gradually. The time taken for its mechanical energy of vibration to drop to half of its initial va... | [{"identifier": "A", "content": "2 s"}, {"identifier": "B", "content": "3.5 s"}, {"identifier": "C", "content": "5 s"}, {"identifier": "D", "content": "7 s"}] | ["D"] | null | <p>To determine the time taken for the mechanical energy of the damped oscillator to drop to half its initial value, we'll use the principles of damped harmonic motion.</p>
<h3><strong>Given:</strong></h3>
<p><p><strong>Mass of the block ($ m $)</strong>: 0.1 kg</p></p>
<p><p><strong>Spring constant ($ k $)</strong>: ... | mcq | jee-main-2017-online-9th-april-morning-slot | 12,637 |
xMJeEMmsuWemlT9ZvztNR | physics | simple-harmonic-motion | forced,-damped-oscillations-and-resonance | A closed organ pipe has a fundamental frequency of 1.5 kHz. The number of overtones that can be distinctly heard by a person with this organ pipe will be (Assume that the highest frequency a person can hear is 20,000 Hz) | [{"identifier": "A", "content": "4"}, {"identifier": "B", "content": "7"}, {"identifier": "C", "content": "6"}, {"identifier": "D", "content": "5"}] | ["B"] | null | For closed organ pipe, resonate frequency is odd multiple of fundamental frequency.
<br><br>$$ \therefore $$ (2n + 1) f<sub>0</sub> $$ \le $$ 20,000
<br><br>(f<sub>0</sub> is fundamental frequency = 1.5 KHz)
<br><br>$$ \therefore $$ n = 6
<br><br>$$ \therefore $$ Total number of overton... | mcq | jee-main-2019-online-10th-january-evening-slot | 12,638 |
TwVP2ZUg0hlMZaoFyMbOb | physics | simple-harmonic-motion | forced,-damped-oscillations-and-resonance | A simple pendulum of length 1 m is oscillating with an angular frequency 10 rad/s. The support of the pendulum starts oscillating up and down with a small angular frequency of 1 rad/s and an amplitude of 10<sup>–2</sup> m. The relative change in the angular frequency of the pendulum is best given by : | [{"identifier": "A", "content": "1 rad/s"}, {"identifier": "B", "content": "10<sup>$$-$$3</sup> rad/s"}, {"identifier": "C", "content": "10<sup>$$-$$1</sup> rad/s"}, {"identifier": "D", "content": "10<sup>$$-$$5</sup> rad/s"}] | ["B"] | null | Angular frequency of pendulum
<br><br>$$\omega $$ = $$\sqrt {{{{g_{eff}}} \over \ell }} $$
<br><br>$$ \therefore $$ $${{\Delta \omega } \over \omega }$$ = $${1 \over 2}$$ $${{\Delta {g_{eff}}} \over {{g_{eff}}}}$$
<br><br>$$\Delta $$$$\omega $$ = $${1 \over 2}$$ $${{\Delta g} \over g} \times \omega $$
<br><... | mcq | jee-main-2019-online-11th-january-evening-slot | 12,639 |
DvgRFow7lD0kFK2GSn18hoxe66ijvzt49s0 | physics | simple-harmonic-motion | forced,-damped-oscillations-and-resonance | The displacement of a damped harmonic
oscillator is given by<br/>
x(t ) = e<sup>–0.1t </sup>cos (10$$\pi $$t + f).<br/> Here t is in seconds.
The time taken for its amplitude of vibration to
drop to half of its initial value is close to : | [{"identifier": "A", "content": "27 s"}, {"identifier": "B", "content": "13 s"}, {"identifier": "C", "content": "7 s"}, {"identifier": "D", "content": "4 s"}] | ["C"] | null | Amplitude at (t = 0) A<sub>0</sub> = e<sup>–0.1× 0</sup> = 1<br><br>
$$ \therefore $$ $$at\,t = t$$ if $$A = {{{A_0}} \over 2}$$<br><br>
$$ \Rightarrow $$ $${1 \over 2} = {e^{ - 0.1t}}$$<br><br>
t = 10 ln 2 = 7 s | mcq | jee-main-2019-online-10th-april-morning-slot | 12,641 |
U23891NbmcviFxD9 | physics | simple-harmonic-motion | simple-harmonic-motion | If a spring has time period $$T,$$ and is cut into $$n$$ equal parts, then the time period of each part will be | [{"identifier": "A", "content": "$$T\\sqrt n $$ "}, {"identifier": "B", "content": "$$T/\\sqrt n $$ "}, {"identifier": "C", "content": "$$nT$$ "}, {"identifier": "D", "content": "$$T$$ "}] | ["B"] | null | Let the spring constant of the original spring be $$k.$$
<br><br>Then its time period $$T = 2\pi \sqrt {{m \over k}} $$ where $$m$$ is the mass of oscillating body.
<br>When the spring is cut into $$n$$ equal parts, the spring constant of one part becomes $$nk.$$ Therefore the new time period,
<br><br>$$T' = 2\pi \sqrt... | mcq | aieee-2002 | 12,644 |
4nLFC4HQ4tyheKRV | physics | simple-harmonic-motion | simple-harmonic-motion | In a simple harmonic oscillator, at the mean position | [{"identifier": "A", "content": "kinetic energy is minimum, potential energy is maximum "}, {"identifier": "B", "content": "both kinetic and potential energies are maximum "}, {"identifier": "C", "content": "kinetic energy is maximum, potential energy is minimum "}, {"identifier": "D", "content": "both kinetic and pote... | ["C"] | null | $$K.E = {1 \over 2}k\left( {{A^2} - {x^2}} \right);\,\,\,U = {1 \over 2}k{x^2}$$
<br><br/><b>At the mean position</b> $$x=0$$
<br/><br>$$\therefore$$ $$K.E. = {1 \over 2}k{A^2} = $$ Maximum and $$U=0$$ | mcq | aieee-2002 | 12,645 |
8prRwQnTYiNrCcsA | physics | simple-harmonic-motion | simple-harmonic-motion | The displacement of particle varies according to the relation
<br/>$$x=4$$$$\left( {\cos \,\pi t + \sin \,\pi t} \right).$$ The amplitude of the particle is | [{"identifier": "A", "content": "$$-4$$ "}, {"identifier": "B", "content": "$$4$$ "}, {"identifier": "C", "content": "$$4\\sqrt 2 $$ "}, {"identifier": "D", "content": "$$8$$ "}] | ["C"] | null | $$x = 4\left( {\cos \pi t + \sin \pi t} \right)$$
<br><br>$$ = \sqrt 2 \times 4\left( {{{\sin \pi t} \over {\sqrt 2 }} + {{\cos \pi t} \over {\sqrt 2 }}} \right)$$
<br><br>$$x = 4\sqrt 2 \sin \left( {\pi t + {{45}^ \circ }} \right)$$ | mcq | aieee-2003 | 12,646 |
nj1E14ufjeMxKVxU | physics | simple-harmonic-motion | simple-harmonic-motion | A body executes simple harmonic motion. The potential energy $$(P.E),$$ the kinetic energy $$(K.E)$$ and total energy $$(T.E)$$ are measured as a function of displacement $$x.$$ Which of the following statements is true ? | [{"identifier": "A", "content": "$$K.E$$ is maximum when $$x=0$$ "}, {"identifier": "B", "content": "$$T.E$$ is zero when $$x=0$$ "}, {"identifier": "C", "content": "$$K.E$$ is maximum when $$x$$ is maximum "}, {"identifier": "D", "content": "$$P.E$$ is maximum when $$x=0$$ "}] | ["A"] | null | $$K.E. = {1 \over 2}m{\omega ^2}\left( {{a^2} - {x^2}} \right)$$
<br><br>When $$x=0,$$ $$K.E$$ is maximum and is equal to $${1 \over 2}m{\omega ^2}{a^2}.$$ | mcq | aieee-2003 | 12,647 |
06Xs52fuhsUOl4iW | physics | simple-harmonic-motion | simple-harmonic-motion | The total energy of particle, executing simple harmonic motion is | [{"identifier": "A", "content": "independent of $$x$$ "}, {"identifier": "B", "content": "$$ \\propto \\,{x^2}$$ "}, {"identifier": "C", "content": "$$ \\propto \\,x$$ "}, {"identifier": "D", "content": "$$ \\propto \\,{x^{1/2}}$$ "}] | ["A"] | null | At any instant the total energy is
<br><br>$${1 \over 2}k{A^2} = \,\,$$ constant, where $$A=$$ amplitude
<br><br>hence total energy is independent of $$x.$$ | mcq | aieee-2004 | 12,648 |
UyF3fOCwppXJBgLo | physics | simple-harmonic-motion | simple-harmonic-motion | The function $${\sin ^2}\left( {\omega t} \right)$$ represents | [{"identifier": "A", "content": "a periodic, but not $$SHM$$ with a period $${\\pi \\over \\omega }$$ "}, {"identifier": "B", "content": "a periodic, but not $$SHM$$ with a period $${{2\\pi } \\over \\omega }$$ "}, {"identifier": "C", "content": "a $$SHM$$ with a period $${\\pi \\over \\omega }$$ "}, {"identifier": ... | ["A"] | null | y = sin<sup>2</sup>$$\omega $$t
<br><br>= $${{1 - \cos 2\omega t} \over 2}$$
<br><br>$$ = {1 \over 2} - {1 \over 2}\cos \,2\omega t$$
<br><br>$$ \therefore $$ Angular speed = 2$$\omega $$
<br><br>$$ \therefore $$ Period (T) = $${{2\pi } \over {angular\,speed}}$$ = $${{2\pi } \over {2\omega }}$$ = $${\pi \over \omega... | mcq | aieee-2005 | 12,649 |
SbAnufkqtDJhxdk1 | physics | simple-harmonic-motion | simple-harmonic-motion | Two simple harmonic motions are represented by the equations $${y_1} = 0.1\,\sin \left( {100\pi t + {\pi \over 3}} \right)$$ and $${y_2} = 0.1\,\cos \,\pi t.$$ The phase difference of the velocity of particle $$1$$ with respect to the velocity of particle $$2$$ is | [{"identifier": "A", "content": "$${\\pi \\over 3}$$ "}, {"identifier": "B", "content": "$${{ - \\pi } \\over 6}$$ "}, {"identifier": "C", "content": "$${\\pi \\over 6}$$ "}, {"identifier": "D", "content": "$${{ - \\pi } \\over 3}$$ "}] | ["B"] | null | $${v_1} = {{d{y_1}} \over {dt}} = 0.1 \times 100\pi \cos \left( {100\pi t + {\pi \over 3}} \right)$$
<br><br>$${v_2} = {{d{y_2}} \over {dt}} = - 0.1\pi sin\pi t = 0.1\pi cos\left( {\pi t + {\pi \over 2}} \right)$$
<br><br>$$\therefore$$ Phase diff. $$ = {\phi _1} - {\phi _2} = {\pi \over 3} - {\pi \over 2} = {{2\p... | mcq | aieee-2005 | 12,650 |
6uyRBI1wBMOJBn9Y | physics | simple-harmonic-motion | simple-harmonic-motion | If a simple harmonic motion is represented by $${{{d^2}x} \over {d{t^2}}} + \alpha x = 0.$$ its time period is | [{"identifier": "A", "content": "$${{2\\pi } \\over {\\sqrt \\alpha }}$$ "}, {"identifier": "B", "content": "$${{2\\pi } \\over \\alpha }$$ "}, {"identifier": "C", "content": "$$2\\pi \\sqrt \\alpha $$ "}, {"identifier": "D", "content": "$$2\\pi \\alpha $$ "}] | ["A"] | null | $${{{d^2}x} \over {d{t^2}}} = - \alpha x = - {\omega ^2}x$$
<br><br>$$ \Rightarrow \omega = \sqrt \alpha $$ $$\,\,\,\,$$ or $$\,\,\,\,$$ $$T = {{2\pi } \over \omega } = {{2\pi } \over {\sqrt \alpha }}$$ | mcq | aieee-2005 | 12,651 |
CtaC83vmguYf3d6j | physics | simple-harmonic-motion | simple-harmonic-motion | A coin is placed on a horizontal platform which undergoes vertical simple harmonic motoin of angular frequency $$\omega .$$ The amplitude of oscillation is gradually increased. The coin will leave contact with the platform for the first time | [{"identifier": "A", "content": "at the mean position of the platform "}, {"identifier": "B", "content": "for an amplitude of $${g \\over {{\\omega ^2}}}$$ "}, {"identifier": "C", "content": "For an amplitude of $${{{g^2}} \\over {{\\omega ^2}}}$$ "}, {"identifier": "D", "content": "at the height position of the platfo... | ["B"] | null | Coin $$A$$ is moving in simple harmonic motion in vertical direction. Now we are assuming coin will leave contact with the platform when platform is at a distance of $$x$$ from the mean position which is also called amplitude.
<img class="question-image" src="https://imagex.cdn.examgoal.net/Zb4ajf7bKuWUrM5hu/qptedzBVPp... | mcq | aieee-2006 | 12,652 |
J8uRzrAhyPfJjsch | physics | simple-harmonic-motion | simple-harmonic-motion | The maximum velocity of a particle, executing simple harmonic motion with an amplitude $$7$$ $$mm,$$ is $$4.4$$ $$m/s.$$ The period of oscillation is | [{"identifier": "A", "content": "$$0.01$$ $$s$$ "}, {"identifier": "B", "content": "$$10$$ $$s$$ "}, {"identifier": "C", "content": "$$0.1$$ $$s$$ "}, {"identifier": "D", "content": "$$100$$ $$s$$ "}] | ["A"] | null | Maximum velocity,
<br><br>$${v_{\max }} = a\omega ,\,\,\,\,\,{v_{\max }} = a \times {{2\pi } \over T}$$
<br><br>$$ \Rightarrow T = {{2\pi a} \over {{v_{\max }}}} = {{2 \times 3.14 \times 7 \times {{10}^{ - 3}}} \over {4.4}} \approx 0.01\,s$$ | mcq | aieee-2006 | 12,653 |
ebtUyonKOJW3vNQj | physics | simple-harmonic-motion | simple-harmonic-motion | Starting from the origin a body oscillates simple harmonically with a period of $$2$$ $$s.$$ After what time will its kinetic energy be $$75\% $$ of the total energy? | [{"identifier": "A", "content": "$${1 \\over 6}s$$ "}, {"identifier": "B", "content": "$${1 \\over 4}s$$"}, {"identifier": "C", "content": "$${1 \\over 3}s$$"}, {"identifier": "D", "content": "$${1 \\over 12}s$$"}] | ["A"] | null | $$K.E.\,$$ of a body undergoing $$SHM$$ is given by,
<br><br>$$K.E. = {1 \over 2}m{a^2}{\omega ^2}{\cos ^2}\,\omega t,$$
<br><br>$$T.E. = {1 \over 2}m{a^2}{\omega ^2}$$
<br><br>Given $$K.E.=0.75T.E.$$
<br><br>$$ \Rightarrow 0.75 = {\cos ^2}\omega t \Rightarrow \omega t = {\pi \over 6}$$
<br><br>$$ \Rightarrow t = {\pi... | mcq | aieee-2006 | 12,654 |
pByDQlRirgmfmwsq | physics | simple-harmonic-motion | simple-harmonic-motion | A particle of mass $$m$$ executes simple harmonic motion with amplitude a and frequency $$v.$$ The average kinetic energy during its motion from the position of equilibrium to the end is | [{"identifier": "A", "content": "$$2{\\pi ^2}\\,m{a^2}{v^2}$$ "}, {"identifier": "B", "content": "$${\\pi ^2}\\,m{a^2}{v^2}$$ "}, {"identifier": "C", "content": "$${1 \\over 4}\\,m{a^2}{v^2}$$ "}, {"identifier": "D", "content": "$$4{\\pi ^2}m{a^2}{v^2}$$ "}] | ["B"] | null | <b>KEY CONCEPT :</b> The instantaneous kinetic energy of a particle executing $$S.H.M.$$ is given by
<br><br>$$K = {1 \over 2}m{a^2}{\omega ^2}{\sin ^2}\omega t$$
<br><br>$$\therefore$$ average $$K.E. = < K > = < {1 \over 2}m{\omega ^2}{a^2}{\sin ^2}\omega t > $$
<br><br>$$ = {1 \over 2}m\omega {}^2{a^2}... | mcq | aieee-2007 | 12,655 |
F3Vl3EIFeVzopM9K | physics | simple-harmonic-motion | simple-harmonic-motion | Two springs, of force constant $${k_1}$$ and $${k_2}$$ are connected to a mass $$m$$ as shown. The frequency of oscillation of the mass is $$f.$$ If both $${k_1}$$ and $${k_2}$$ are made four times their original values, the frequency of oscillation becomes
<img src="data:image/png;base64,UklGRrYTAABXRUJQVlA4IKoTAAAQmQ... | [{"identifier": "A", "content": "$$2f$$ "}, {"identifier": "B", "content": "$$f/2$$ "}, {"identifier": "C", "content": "$$f/4$$"}, {"identifier": "D", "content": "$$4f$$ "}] | ["A"] | null | The two springs are in parallel.
<br><br>$$f = {1 \over {2\pi }}\sqrt {{{{K_1} + {K_2}} \over m}} \,\,\,\,\,\,\,\,\,\,\,\,\,\,...\left( i \right)$$
<br><br>$$f' = {1 \over {2\pi }}\sqrt {{{4{K_1} + 4{K_2}} \over m}} $$
<br><br>$$ = {1 \over {2\pi }}\sqrt {{{4\left( {{K_1} + 4{K_2}} \right)} \over m}} $$
<br><br>$$ = 2\... | mcq | aieee-2007 | 12,656 |
LzZpukGgNRPtGq8G | physics | simple-harmonic-motion | simple-harmonic-motion | A point mass oscillates along the $$x$$-axis according to the law $$x = {x_0}\,\cos \left( {\omega t - \pi /4} \right).$$ If the acceleration of the particle is written as $$a = A\,\cos \left( {\omega t + \delta } \right),$$ then | [{"identifier": "A", "content": "$$A = {x_0}{\\omega ^2},\\,\\,\\delta = 3\\pi /4$$ "}, {"identifier": "B", "content": "$$A = {x_0},\\,\\,\\delta = - \\pi /4$$ "}, {"identifier": "C", "content": "$$A = {x_0}{\\omega ^2},\\,\\,\\delta = \\pi /4$$ "}, {"identifier": "D", "content": "$$A = {x_0}{\\omega ^2},\\,\\,\\de... | ["A"] | null | Here,
<br><br>$$x = {x_0}\cos \left( {\omega t - \pi /4} \right)$$
<br><br>$$\therefore$$ Velocity, $$v = {{dx} \over {dt}} = - {x_0}\omega \sin \left( {\omega t - {\pi \over 4}} \right)$$
<br><br>Acceleration,
<br><br>$$a = {{dv} \over {dt}} = - {x_0}{\omega ^2}\cos \left( {\omega t - {\pi \over 4}} \right)$$
<br>... | mcq | aieee-2007 | 12,657 |
62HlH0BFHZTEnXmT | physics | simple-harmonic-motion | simple-harmonic-motion | If $$x,$$ $$v$$ and $$a$$ denote the displacement, the velocity and the acceleration of a particle executing simple harmonic motion of time period $$T,$$ then, which of the following does not change with time? | [{"identifier": "A", "content": "$$aT/x$$ "}, {"identifier": "B", "content": "$$aT + 2\\pi v$$ "}, {"identifier": "C", "content": "$$aT/v$$ "}, {"identifier": "D", "content": "$${a^2}{T^2} + 4{\\pi ^2}{v^2}$$ "}] | ["A"] | null | For an $$SHM,$$ the acceleration $$a = - {\omega ^2}x$$ where $${\omega ^2}$$ is a constant. Therefore $${a \over x}$$ is a constant. The time period $$T$$ is also constant. Therefore $${{aT} \over x}$$ is a constant. | mcq | aieee-2009 | 12,659 |
nc8I9HTKAlvVNXMY | physics | simple-harmonic-motion | simple-harmonic-motion | A mass $$M,$$ attached to a horizontal spring, executes $$S.H.M.$$ with amplitude $${A_1}.$$ When the mass $$M$$ passes through its mean position then a smaller mass $$m$$ is placed over it and both of them move together with amplitude $${A_2}.$$ The ratio of $$\left( {{{{A_1}} \over {{A_2}}}} \right)$$ is : | [{"identifier": "A", "content": "$${{M + m} \\over M}$$ "}, {"identifier": "B", "content": "$${\\left( {{M \\over {M + m}}} \\right)^{{1 \\over 2}}}$$ "}, {"identifier": "C", "content": "$${\\left( {{{M + m} \\over M}} \\right)^{{1 \\over 2}}}$$ "}, {"identifier": "D", "content": "$${M \\over {M + m}}$$ "}] | ["C"] | null | The net force becomes zero at the mean point. Therefore, linear momentum must be conserved.
<br><br>$$\therefore$$ $$M{v_1} = \left( {M + m} \right){v_2}$$
<br><br>$$M{A_1}\sqrt {{k \over M}} = \left( {M + m} \right){A_2}\sqrt {{k \over {m + M}}} $$
<br><br>$$\therefore$$ $$\left( {V = A\sqrt {{k \over M}} } \right)$$... | mcq | aieee-2011 | 12,660 |
YgHkasDCKsJtSxFH | physics | simple-harmonic-motion | simple-harmonic-motion | Two particles are executing simple harmonic motion of the same amplitude $$A$$ and frequency $$\omega $$ along the $$x$$-axis. Their mean position is separated by distance $${X_0}\left( {{X_0} > A} \right)$$. If the maximum separation between them is $$\left( {{X_0} + A} \right),$$ the phase difference between their... | [{"identifier": "A", "content": "$${\\pi \\over 3}$$ "}, {"identifier": "B", "content": "$${\\pi \\over 4}$$ "}, {"identifier": "C", "content": "$${\\pi \\over 6}$$ "}, {"identifier": "D", "content": "$${\\pi \\over 2}$$"}] | ["A"] | null | <p>We know that, equation for SHM along x-axis is given by $$x = A\sin (\omega t + \phi )$$</p>
<p>Let mean position for 1st particle is at x = 0</p>
<p>So, the SHM equation for 1st particle,</p>
<p>$${x_1} = A\sin (\omega t + {\phi _1})$$</p>
<p>Now, as the separation between mean positions of both the particle is x$_... | mcq | aieee-2011 | 12,661 |
cFCwAnr5uP5w4PPM | physics | simple-harmonic-motion | simple-harmonic-motion | A particle moves with simple harmonic motion in a straight line. In first $$\tau s,$$ after starting from rest it travels a distance $$a,$$ and in next $$\tau s$$ it travels $$2a,$$ in same direction, then: | [{"identifier": "A", "content": "amplitude of motion is $$3a$$ "}, {"identifier": "B", "content": "time period of oscillations is $$8\\tau $$ "}, {"identifier": "C", "content": "amplitude of motion is $$4a$$ "}, {"identifier": "D", "content": "time period of oscillations is $$6\\tau $$ "}] | ["D"] | null | In simple harmonic motion, starting from rest,
<br><br>At $$t=0,$$ $$x=A$$
<br><br>$$x = A\cos \omega t\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...\left( i \right)$$
<br><br>When $$t = \tau ,\,\,x = A - a$$
<br><br>When $$t = 2\,\tau ,\,x = A - 3a$$
<br><br>From equation $$(i)$$
<br><br>$$A - a = A\cos \omega \,\tau \,\,\,\,... | mcq | jee-main-2014-offline | 12,662 |
jejAjmwC6yYZiJYj | physics | simple-harmonic-motion | simple-harmonic-motion | A particle performs simple harmonic motion with amplitude $$A.$$ Its speed is trebled at the instant that it is at a distance $${{2A} \over 3}$$ from equilibrium position. The new amplitude of the motion is: | [{"identifier": "A", "content": "$$A\\sqrt 3 $$ "}, {"identifier": "B", "content": "$${{7A} \\over 3}$$ "}, {"identifier": "C", "content": "$${A \\over 3}\\sqrt {41} $$ "}, {"identifier": "D", "content": "$$3A$$ "}] | ["B"] | null | We know that $$V = \omega \sqrt {{A^2} - {x^2}} $$
<br><br>Initially $$v = \omega \sqrt {{A^2} - {{\left( {{{2A} \over 3}} \right)}^2}} $$
<br><br>Finally $$3v = \omega \sqrt {A_{new}^2 - {{\left( {{{2A} \over 3}} \right)}^2}} $$
<br><br>where $${A_{new}}$$ = final amplitude (Given at $$x = {{2A} \over 3},$$ velocity t... | mcq | jee-main-2016-offline | 12,663 |
4aY9Zg26IPzekAJj8MjXe | physics | simple-harmonic-motion | simple-harmonic-motion | In an engine the piston undergoes vertical simple harmonic motion with amplitude 7 cm. A washer rests on top of the piston and moves with it. The motor speed is slowly increased. The frequency of the piston at which the washer no longer stays in contact with the piston, is close to :
| [{"identifier": "A", "content": "0.1 Hz "}, {"identifier": "B", "content": "1.2 Hz "}, {"identifier": "C", "content": "0.7 Hz "}, {"identifier": "D", "content": "1.9 Hz "}] | ["D"] | null | Here,
<br><br>Amplitude, A = 7 cm = 0.07 m
<br><br>When washer is no longer stays in contact with the piston, then the normal force on the washer is = 0
<br><br>$$ \therefore $$ Maximum acceleration of the washer,
<br><br>a<sub>max</sub> = $$\omega $$<sup>2</sup>A = g
<br><br>$$ \Rightarrow $$ &n... | mcq | jee-main-2016-online-10th-april-morning-slot | 12,664 |
q0JavxzwOukBQqLiO4Uvl | physics | simple-harmonic-motion | simple-harmonic-motion | Two particles are performing simple harmonic motion in a straight line about
the same equilibrium point. The amplitude and time period for both particles are same and equal to A and <i>I</i>, respectively. At time t = 0 one particle has
displacement A while the other one has displacement $${{ - A} \over 2}$$ and they ... | [{"identifier": "A", "content": "$${T \\over 6}$$ "}, {"identifier": "B", "content": "$${5T \\over 6}$$"}, {"identifier": "C", "content": "$${T \\over 3}$$"}, {"identifier": "D", "content": "$${T \\over 4}$$"}] | ["A"] | null | <img src="https://res.cloudinary.com/dckxllbjy/image/upload/v1734266323/exam_images/iapjkhe26uiroktehmls.webp" style="max-width: 100%; height: auto;display: block;margin: 0 auto;" loading="lazy" alt="JEE Main 2016 (Online) 9th April Morning Slot Physics - Simple Harmonic Motion Question 114 English Explanation">
<br>A... | mcq | jee-main-2016-online-9th-april-morning-slot | 12,665 |
J0ztRZyOpDhCkCo5Py6tj | physics | simple-harmonic-motion | simple-harmonic-motion | A 1 kg block attached to a spring vibrates with a frequency of 1 Hz on a frictionless horizontal table. Two springs identical to the original spring are attached in parallel to an 8 kg block placed on the same table. So, the frequency of vibration of the 8 kg block is :
| [{"identifier": "A", "content": "$${1 \\over 4}Hz$$ "}, {"identifier": "B", "content": "$${1 \\over {2\\sqrt 2 }}Hz$$ "}, {"identifier": "C", "content": "$${1 \\over 2}Hz$$"}, {"identifier": "D", "content": "$$2$$ $$Hz$$"}] | ["C"] | null | <b>For 1 kg block : </b>
<br><br><img src="https://res.cloudinary.com/dckxllbjy/image/upload/v1734266032/exam_images/pzkzms5ndlv5ss6m8saj.webp" style="max-width: 100%; height: auto;display: block;margin: 0 auto;" loading="lazy" alt="JEE Main 2017 (Online) 8th April Morning Slot Physics - Simple Harmonic Motion Questi... | mcq | jee-main-2017-online-8th-april-morning-slot | 12,666 |
eAYBP42rGLjuQR2SS2H9g | physics | simple-harmonic-motion | simple-harmonic-motion | The ratio of maximum acceleration to maximum velocity in a simple harmonic motion is 10 s<sup>−1</sup> . At, t = 0 the displacement is 5 m. What is the maximum acceleration ? The initial phase is $${\pi \over 4}$$. | [{"identifier": "A", "content": "500 m/s<sup>2</sup> "}, {"identifier": "B", "content": "500 $$\\sqrt 2 m/$$ s<sup>2</sup> "}, {"identifier": "C", "content": "750 m/s<sup>2</sup> "}, {"identifier": "D", "content": "750 $$\\sqrt 2 $$m / s<sup>2</sup> "}] | ["B"] | null | Mximum velocity, V<sub>max</sub> = a$$\omega $$
<br><br>Maximum acceleration, A<sub>max</sub> = a$$\omega $$<sup>2</sup>
<br><br>Given that,
<br><br>$${{a{\omega ^2}} \over {a\omega }}$$ = 10
<br><br>$$ \Rightarrow $$$$\,\,\,$$ $$\omega $$ = 10 s<sup>$$-$$1</sup>
<br><br>Displacement, x = a sin ($$\omega $$t + $${\pi... | mcq | jee-main-2017-online-8th-april-morning-slot | 12,667 |
pzQYl7SkQ5ctjpNR | physics | simple-harmonic-motion | simple-harmonic-motion | A particle is executing simple harmonic motion with a time period T. At time t = 0, it is at its position of
equilibrium. The kinetic energy – time graph of the particle will look like: | [{"identifier": "A", "content": "<img src=\"https://app-content.cdn.examgoal.net/fly/@width/image/1l7ymg2w5/783ab36a-a9ac-4816-9cf9-96cbedfba369/8fb66d50-3285-11ed-8893-19b23ee4c66d/file-1l7ymg2w6.png?format=png\" data-orsrc=\"https://app-content.cdn.examgoal.net/image/1l7ymg2w5/783ab36a-a9ac-4816-9cf9-96cbedfba369/8fb... | ["A"] | null | For a particle executing SHM
<br><br>At mean position; t = 0, $$\omega $$t = 0, y = 0, V = V<sub>max</sub> = a$$\omega $$
<br><br>$$ \therefore $$ K.E = K.E<sub>max</sub> = $${1 \over 2}m{\omega ^2}{a^2}$$
<br><br>At extreme position : t =
$${T \over 4}$$
, $$\omega $$t =
$${\pi \over 2}$$
, y = A, V = V<sub>min</sub>... | mcq | jee-main-2017-offline | 12,668 |
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