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1lgyr5ir2 | physics | waves | superposition-and-reflection-of-waves | <p>A guitar string of length 90 cm vibrates with a fundamental frequency of 120 Hz. The length of the string producing a fundamental frequency of 180 Hz will be _________ cm.</p> | [] | null | 60 | <p>The fundamental frequency (also known as the first harmonic) of a vibrating string is given by the formula:</p>
<p>$f = \frac{v}{2L}$</p>
<p>where:</p>
<ul>
<li>(f) is the frequency,</li>
<li>(v) is the speed of the wave in the string, and</li>
<li>(L) is the length of the string.</li>
</ul>
<p>In this case, the speed of the wave in the string stays the same because it depends on the properties of the string and the tension in it, which we can assume to be constant.</p>
<p>We can write the equation for the fundamental frequency of the original string and the shorter string:</p>
<p>$f_1 = \frac{v}{2L_1}$<br/><br/>
$f_2 = \frac{v}{2L_2}$</p>
<p>where:</p>
<ul>
<li>$(f_1 = 120 \, \text{Hz})$ and $ (L_1 = 90 \, \text{cm})$ for the original string, and</li>
<li>$(f_2 = 180 \, \text{Hz})$ and $(L_2)$ is what we're trying to find for the shorter string.</li>
</ul>
<p>We can set up a ratio of these two equations:</p>
<p>$\frac{f_1}{f_2} = \frac{L_2}{L_1}$</p>
<p>Substituting in the given values, we get:</p>
<p>$\frac{120 \, \text{Hz}}{180 \, \text{Hz}} = \frac{L_2}{90 \, \text{cm}}$</p>
<p>Solving for ($L_2$) gives:</p>
<p>$L_2 = 90 \, \text{cm} \times \frac{120 \, \text{Hz}}{180 \, \text{Hz}} = 60 \, \text{cm}$</p>
<p>So, the length of the string producing a fundamental frequency of 180 Hz will be 60 cm.</p>
| integer | jee-main-2023-online-8th-april-evening-shift | 13,264 |
1lh02f07b | physics | waves | superposition-and-reflection-of-waves | <p>An organ pipe $$40 \mathrm{~cm}$$ long is open at both ends. The speed of sound in air is $$360 \mathrm{~ms}^{-1}$$. The frequency of the second harmonic is ___________ $$\mathrm{Hz}$$.</p> | [] | null | 900 | <p>An organ pipe that is open at both ends resonates at all harmonics, including the fundamental (first harmonic), second harmonic, third harmonic, etc.</p>
<p>The frequency $f$ of the $n$-th harmonic for a pipe open at both ends is given by:</p>
<p>$f_n = \frac{n v}{2L}$,</p>
<p>where:</p>
<ul>
<li>$n$ is the number of the harmonic,</li>
<li>$v$ is the speed of sound, and</li>
<li>$L$ is the length of the pipe.</li>
</ul>
<p>To find the frequency of the second harmonic ($n = 2$), we can substitute the given values into the formula:</p>
<p>$f_2 = \frac{2 \times 360}{2 \times 0.4} = 900 \, \text{Hz}$.</p>
<p>Therefore, the frequency of the second harmonic is $900 \, \text{Hz}$.</p>
| integer | jee-main-2023-online-8th-april-morning-shift | 13,265 |
lsblmve2 | physics | waves | superposition-and-reflection-of-waves | A tuning fork resonates with a sonometer wire of length $1 \mathrm{~m}$ stretched with a tension of $6 \mathrm{~N}$. When the tension in the wire is changed to $54 \mathrm{~N}$, the same tuning fork produces 12 beats per second with it. The frequency of the tuning fork is ________________ $\mathrm{Hz}$. | [] | null | 6 | <p>To solve this problem, we'll have to use the relationship between the frequency of a vibrating string and the tension applied to it. When a tuning fork resonates with a sonometer wire, their frequencies are equal.</p>
<p>The frequency of a vibrating string is given by the formula:</p>
<p>$f = \frac{1}{2L} \sqrt{\frac{T}{\mu}}$</p>
<p>Where:</p>
<ul>
<li>$ f $ is the frequency of the vibration</li><br>
<li>$ L $ is the length of the wire</li><br>
<li>$ T $ is the tension in the wire</li><br>
<li>$ \mu $ is the linear mass density of the wire</li>
</ul>
<p>The linear mass density ($ \mu $) of the wire remains constant.</p>
<p>Initially, when the string resonates with the tuning fork, the frequency of both is given by:</p>
<p>$f_1 = \frac{1}{2L} \sqrt{\frac{T_1}{\mu}}$</p>
<p>Here $ T_1 = 6 \mathrm{~N} $ and $ L = 1 \mathrm{~m} $, so we have:</p>
<p>$f_1 = \frac{1}{2 \cdot 1} \sqrt{\frac{6}{\mu}} = \frac{1}{2} \sqrt{\frac{6}{\mu}}$</p>
<p>Now, when the tension is changed to $ T_2 = 54 \mathrm{~N} $, the frequency of the wire changes to $ f_2 $ and it is given by:</p>
<p>$f_2 = \frac{1}{2L} \sqrt{\frac{T_2}{\mu}}$</p>
<p>Since $ L $ and $ \mu $ remain the same, substituting $ T_2 $:</p>
<p>$f_2 = \frac{1}{2 \cdot 1} \sqrt{\frac{54}{\mu}} = \frac{1}{2} \sqrt{\frac{54}{\mu}}$</p>
<p>Notice that $ 54 = 6 \times 9 $, therefore:</p>
<p>$f_2 = \frac{1}{2} \sqrt{\frac{6 \times 9}{\mu}} = \frac{1}{2} \sqrt{9} \sqrt{\frac{6}{\mu}} = \frac{3}{2} \sqrt{\frac{6}{\mu}} = 3 f_1$</p>
<p>When the tension was increased, the frequency became thrice the original frequency.</p>
<p>Since the second instance of the string produces 12 beats per second this means that the frequency of the tuning fork (and original string) and the new frequency (of the string with higher tension) differ by 12 Hz. If $ f_F $ is the frequency of the tuning fork, then:</p>
<p>Either $ f_2 = f_F + 12 $ Hz or $ f_2 = f_F - 12 $ Hz.</p>
<p>Since we have determined $ f_2 = 3 f_1 $ and $ f_1 = f_F $, we can state:</p>
<p>Either $ 3f_F = f_F + 12 $ or $ 3f_F = f_F - 12 $.</p>
<p>If $ 3f_F = f_F - 12 $, then :</p>
<p>$3f_F - f_F = -12$</p>
<p>$2f_F = -12$</p>
<p>This result is not possible since frequency cannot be negative.</p>
<p>So, we must consider the correct equation which is:</p>
<p>$f_2 = f_F + 12$</p>
<p>Now, substituting $ f_2 = 3f_F $:</p>
<p>$3f_F = f_F + 12$</p>
<p>$2f_F = 12$</p>
<p>$f_F = 6 \text{ Hz}$</p>
<p>Thus, the frequency of the tuning fork is 6 Hz.</p> | integer | jee-main-2024-online-1st-february-morning-shift | 13,266 |
lv5gst3k | physics | waves | superposition-and-reflection-of-waves | <p>A closed and an open organ pipe have same lengths. If the ratio of frequencies of their seventh overtones is $$\left(\frac{a-1}{a}\right)$$ then the value of $$a$$ is _________.</p> | [] | null | 16 | <p>$$\begin{aligned}
& f_o=\frac{v}{2 I} \quad \Rightarrow \quad f_{o_7}=8 \frac{v}{2 l} \\
& f_c=\frac{v}{4 I} \quad \Rightarrow \quad f_{c_7}=15 \frac{v}{4 I} \\
& \frac{f_{c_7}}{f_{o_7}}=15 \frac{v}{4 I} \frac{2 l}{8 v}=\frac{30}{32}=\frac{15}{16}
\end{aligned}$$</p> | integer | jee-main-2024-online-8th-april-morning-shift | 13,268 |
dciWqKRInJedFcSE | physics | work-power-and-energy | energy | If a body looses half of its velocity on penetrating $$3$$ $$cm$$ in a wooden block, then how much will it penetrate more before coming to rest? | [{"identifier": "A", "content": "$$1$$ $$cm$$ "}, {"identifier": "B", "content": "$$2$$ $$cm$$ "}, {"identifier": "C", "content": "$$3$$ $$cm$$ "}, {"identifier": "D", "content": "$$4$$ $$cm$$ "}] | ["A"] | null | We know the work energy theorem, $$W = \Delta K = FS$$
<br><br>For first penetration, by applying work energy theorem we get,
<br><br>$${1 \over 2}m{v^2} - {1 \over 2}m{\left( {{v \over 2}} \right)^2} = F \times 3\,\,...(i)$$
<br><br>For second penetration, by applying work energy theorem we get,
<br><br>$${1 \over 2}m{\left( {{v \over 2}} \right)^2} - 0 = F \times S\,...(ii)$$
<br><br>On dividing $$(ii)$$ by $$(i)$$
<br><br>$${{1/4} \over {3/4}} = S/3$$
<br><br>$$\therefore$$ $$S = 1\,cm$$ | mcq | aieee-2002 | 13,270 |
slG1cirAyiL5Igxz | physics | work-power-and-energy | energy | A wire suspended vertically from one of its ends is stretched by attaching a weight of $$200N$$ to the lower end. The weight stretches the wire by $$1$$ $$mm.$$ Then the elastic energy stored in the wire is | [{"identifier": "A", "content": "$$0.2$$ $$J$$ "}, {"identifier": "B", "content": "$$10$$ $$J$$ "}, {"identifier": "C", "content": "$$20$$ $$J$$ "}, {"identifier": "D", "content": "$$0.1$$ $$J$$ "}] | ["D"] | null | The elastic potential energy
<br><br>$$ = {1 \over 2} \times $$ Force $$ \times $$ extension
<br><br>$$= {1 \over 2} \times 200 \times 0.001 = 0.1\,J$$ | mcq | aieee-2003 | 13,271 |
mor1tIL395NgGm6o | physics | work-power-and-energy | energy | A particle moves in a straight line with retardation proportional to its displacement. Its loss of kinetic energy for any displacement $$x$$ is proportional to | [{"identifier": "A", "content": "$$x$$ "}, {"identifier": "B", "content": "$${e^x}$$ "}, {"identifier": "C", "content": "$${x^2}$$ "}, {"identifier": "D", "content": "$${\\log _e}x$$ "}] | ["C"] | null | Given that, retardation $$ \propto $$ displacement
<br><br>$$ \Rightarrow $$ $$a=-kx$$
<br><br>But we know $$a = v{{dv} \over {dx}}\,\,\,\,\,\,\,\,\,$$
<br><br>$$\therefore$$ $${{vdv} \over {dx}} = - kx $$
<br><br>$$\Rightarrow \int\limits_{{v_1}}^{{v_2}} v \,dv = - k\int\limits_0^x {xdx} $$
<br><br>$$\left( {v_2^2 - v_1^2} \right) = - k{{{x^2}} \over 2}$$
<br><br>$$ \Rightarrow {1 \over 2}m\left( {v_2^2 - v_1^2} \right) = {1 \over 2}mk\left( {{{ - x^2} \over 2}} \right)$$
<br><br>$$\therefore$$ Loss in kinetic energy is proportional to $${x^2}$$.
<br><br>$$\therefore$$ $$\Delta K \propto {x^2}$$ | mcq | aieee-2004 | 13,272 |
E06vjkbrljlbse8g | physics | work-power-and-energy | energy | A uniform chain of length $$2$$ $$m$$ is kept on a table such that a length of $$60$$ $$cm$$ hangs freely from the edge of the table. The total mass of the chain is $$4$$ $$kg.$$ What is the work done in pulling the entire chain on the table? | [{"identifier": "A", "content": "$$12$$ $$J$$ "}, {"identifier": "B", "content": "$$3.6$$ $$J$$ "}, {"identifier": "C", "content": "$$7.2$$ $$J$$ "}, {"identifier": "D", "content": "$$1200$$ $$J$$ "}] | ["B"] | null | Mass of hanging part $$(m') = {4 \over 2} \times \left( {0.6} \right)kg$$ = 1.2 kg
<br><br>Let at the surface $$PE=0$$
<br><br>Center of mass of hanging part $$=0.3$$ $$m$$ below the surface of the table
<br><br>$${U_i} = - m'gx = - 1.2 \times 10 \times 0.30$$ = - 3.6 J
<br><br>$$\Delta U = m'gx = 3.6 J = $$ Work done in putting the entire chain on the table. | mcq | aieee-2004 | 13,273 |
WXGbxhl2cLFb0XzO | physics | work-power-and-energy | energy | A particle is acted upon by a force of constant magnitude which is always perpendicular to the velocity of the particle, the motion of the particles takes place in a plane. It follows that | [{"identifier": "A", "content": "its kinetic energy is constant "}, {"identifier": "B", "content": "is acceleration is constant "}, {"identifier": "C", "content": "its velocity is constant "}, {"identifier": "D", "content": "it moves in a straight line "}] | ["A"] | null | Work done by such force is always zero when a force of constant magnitude always at right angle to the velocity of a particle when the motion of the particle takes place in a plane.
<br><br>$$\therefore$$ From work-energy theorem, $$ \Delta K = 0$$
<br><br>$$\therefore$$ $$K$$ remains constant. | mcq | aieee-2004 | 13,274 |
IFRAJQhvbCb0V715 | physics | work-power-and-energy | energy | A bullet fired into a fixed target loses half of its velocity after penetrating $$3$$ $$cm.$$ How much further it will penetrate before coming to rest assuming that it faces constant resistance to motion? | [{"identifier": "A", "content": "$$2.0$$ $$cm$$ "}, {"identifier": "B", "content": "$$3.0$$ $$cm$$ "}, {"identifier": "C", "content": "$$1.0$$ $$cm$$ "}, {"identifier": "D", "content": "$$1.5$$ $$cm$$"}] | ["C"] | null | Let $$K$$ be the initial kinetic energy and $$F$$ be the resistive force. Then according to work-energy theorem,
$$$W = \Delta K$$$
<br><br>i.e., $$3F = {1 \over 2}m{v^2} - {1 \over 2}m{\left( {{v \over 2}} \right)^2}...\left( 1 \right)$$
<br><br><b>Let the bullet will penetrate x cm more before coming to rest.</b>
<br><br>$$\therefore$$ $$Fx = {1 \over 2}m{\left( {{v \over 2}} \right)^2} - {1 \over 2}m{\left( 0 \right)^2}...\left( 2 \right)$$
<br><br>Dividing eq. $$(1)$$ and $$(2)$$ we get,
<br><br> $${x \over 3} = {1 \over 3}$$ or x = 1 cm | mcq | aieee-2005 | 13,275 |
k6wzlGYB11QeZqO3 | physics | work-power-and-energy | energy | The upper half of an inclined plane with inclination $$\phi $$ is perfectly smooth while the lower half is rough. A body starting from rest at the top will again come to rest at the bottom if the coefficient of friction for the lower half is given by | [{"identifier": "A", "content": "$$2\\,\\cos \\,\\,\\phi $$ "}, {"identifier": "B", "content": "$$2\\,sin\\,\\,\\phi $$ "}, {"identifier": "C", "content": "$$\\,\\tan \\,\\,\\phi $$ "}, {"identifier": "D", "content": "$$2\\,\\tan \\,\\,\\phi $$ "}] | ["D"] | null | Let the length of the inclined plane is = $$l$$. So only $${l \over 2}$$ part will have friction.
<br><br>According to work-energy theorem, $$W = \Delta k = 0$$
(Since initial and final speeds are zero)
<br><br>$$\therefore$$ Work done by friction + Work done by gravity $$=0$$
<br><br>i.e., $$ - \left( {\mu \,mg\,\cos \,\phi } \right){\ell \over 2} + mg\ell \,\sin \,\phi = 0$$
<br><br>or $${\mu \over 2}\cos \,\phi = \sin \phi $$
<br><br>or $$\mu = 2\,\tan \,\phi $$ | mcq | aieee-2005 | 13,276 |
8TUAunhq3xPQ0rm0 | physics | work-power-and-energy | energy | A mass of $$M$$ $$kg$$ is suspended by a weightless string. The horizontal force that is required to displace it until the string makes an angle of $${45^ \circ }$$ with the initial vertical direction is | [{"identifier": "A", "content": "$$Mg\\left( {\\sqrt 2 + 1} \\right)$$ "}, {"identifier": "B", "content": "$$Mg\\sqrt 2 $$ "}, {"identifier": "C", "content": "$${{Mg} \\over {\\sqrt 2 }}$$ "}, {"identifier": "D", "content": "$$Mg\\left( {\\sqrt 2 - 1} \\right)$$ "}] | ["D"] | null | <img class="question-image" src="https://imagex.cdn.examgoal.net/GJE1bpFt8OqeL1wYR/655wofzO7pEdvBG3YMIR9IsXvSstW/q5xzKVLPJ2RkcDAdixoSFo/image.svg" loading="lazy" alt="AIEEE 2006 Physics - Work Power & Energy Question 122 English Explanation">
From work energy theorem we can say,
<br><br>Work done by tension $$+$$ work done by force (applied) $$+$$ Work done by gravitational force $$=$$ change in kinetic energy
<br><br>Here Work done by tension is zero
<br><br>$$ \Rightarrow 0 + F \times AB - Mg \times AC = 0$$
<br>$$ \Rightarrow F = Mg\left( {{{AC} \over {AB}}} \right) = Mg\left[ {{{1 - {1 \over {\sqrt 2 }}} \over {{1 \over 2}}}} \right]$$
<br>[ as $$AB = \ell \sin {45^ \circ } = {\ell \over {\sqrt 2 }}$$
<br>and $$AC = OC - OA = \ell - \ell \,\cos \,{45^ \circ } = \ell \left( {1 - {1 \over {\sqrt 2 }}} \right)$$
<br>where $$\ell = $$ length of the string. ]
<br>$$ \Rightarrow F = Mg\left( {\sqrt 2 - 1} \right)$$ | mcq | aieee-2006 | 13,278 |
usA64o7LZ73SyIrB | physics | work-power-and-energy | energy | A ball of mass $$0.2$$ $$kg$$ is thrown vertically upwards by applying a force by hand. If the hand moves $$0.2$$ $$m$$ while applying the force and the ball goes upto $$2$$ $$m$$ height further, find the magnitude of the force. (consider $$g = 10\,m/{s^2}$$). | [{"identifier": "A", "content": "$$4N$$ "}, {"identifier": "B", "content": "$$16$$ $$N$$ "}, {"identifier": "C", "content": "$$20$$ $$N$$ "}, {"identifier": "D", "content": "$$22$$ $$N$$ "}] | ["D"] | null | According to energy conservation law,
<br><br>Work done by the hand and due to gravity = total change in the kinetic energy
<br><br>Initially the the ball is at rest and finally at top its velocity become zero so total change in kinetic energy $$\Delta K$$ = 0
<br><br>$${W_{hand}} + {W_{gravity}} = \Delta K$$
<br><br>[Here distance covered would be 0.2 meter for force by hand as force is applied while ball is in contact with hand.
<br>And gravity will still work while ball is in contact with hand so total distance due to gravity would be 2 + 0.2 = 2.2 meter.]
<br>$$ \Rightarrow F\left( {0.2} \right) - \left( {0.2} \right)\left( {10} \right)\left( {2.2} \right)$$ $$ = 0 \Rightarrow F = 22\,N$$
<br><br>$$\therefore$$ Option (D) is correct. | mcq | aieee-2006 | 13,279 |
UD5n3M6kkE7ZAY2I | physics | work-power-and-energy | energy | A particle of mass $$100g$$ is thrown vertically upwards with a speed of $$5$$ $$m/s$$. The work done by the force of gravity during the time the particle goes up is | [{"identifier": "A", "content": "$$-0.5J$$ "}, {"identifier": "B", "content": "$$-1.25J$$ "}, {"identifier": "C", "content": "$$1.25J$$ "}, {"identifier": "D", "content": "$$0.5J$$ "}] | ["B"] | null | Kinetic energy at point of throwing is converted into potential energy of the particle during rise.
<br><br>$$K.E = {1 \over 2}m{v^2} = {1 \over 2} \times 0.1 \times 25 = 1.25\,J$$
<br><br>$$W = - mgh = - \left( {{1 \over 2}m{v^2}} \right) = - 1.25\,J$$
<br><br>$$\left[ \, \right.$$ As we know, $$mgh = {1 \over 2}m{v^2}$$ by energy conservation $$\left. \, \right]$$
| mcq | aieee-2006 | 13,280 |
YUsFXpo3NaZrN29x | physics | work-power-and-energy | energy | The potential energy of a $$1$$ $$kg$$ particle free to move along the $$x$$-axis is given by $$V\left( x \right) = \left( {{{{x^4}} \over 4} - {{{x^2}} \over 2}} \right)J$$.
<p>The total mechanical energy of the particle is $$2J.$$ Then, the maximum speed (in $$m/s$$) is </p> | [{"identifier": "A", "content": "$${3 \\over {\\sqrt 2 }}$$ "}, {"identifier": "B", "content": "$${\\sqrt 2 }$$ "}, {"identifier": "C", "content": "$${1 \\over {\\sqrt 2 }}$$ "}, {"identifier": "D", "content": "$$2$$ "}] | ["A"] | null | Velocity is maximum when kinetic energy is maximum and when kinetic energy is maximum then potential energy should be minimum
<br><br> For minimum potential energy,
<br><br>$${{dV} \over {dx}} = 0 $$
<br><br>$$\Rightarrow {x^3} - x = 0 $$
<br><br>$$\Rightarrow x = \pm 1$$
<br><br>$$ \Rightarrow$$ Min. Potential energy (P.E.) =$$ {1 \over 4} - {1 \over 2} = - {1 \over 4}J$$
<br><br>$$K.E{._{\left( {\max .} \right)}} + P.E{._{\left( {\min .} \right)}} = 2\,$$ (Given)
<br><br>$$\therefore$$ $$K.E{._{\left( {\max .} \right)}} = 2 + {1 \over 4} = {9 \over 4}$$
<br><br>$$\therefore$$ $${1 \over 2}mv_{\max }^2$$ = $${9 \over 4}$$
<br><br>$$ \Rightarrow {1 \over 2} \times 1 \times {v^2}_{\max .} = {9 \over 4}$$
<br><br>$$ \Rightarrow {v_{\max }} = {3 \over {\sqrt 2 }}$$ m/s | mcq | aieee-2006 | 13,281 |
fikBwNmY84OD7I2h | physics | work-power-and-energy | energy | A particle is projected at $$60^\circ $$ to the horizontal with a kinetic energy K. The kinetic energy at the
highest point is | [{"identifier": "A", "content": "K/2"}, {"identifier": "B", "content": "K"}, {"identifier": "C", "content": "Zero"}, {"identifier": "D", "content": "K/4"}] | ["D"] | null | Let $$u$$ be the velocity with which the particle is thrown and $$m$$ be the mass of the particle. Then
<br><br>$$KE = {1 \over 2}m{u^2}.\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...\left( 1 \right)$$
<br><br>At the highest point the velocity is $$u$$ $$\cos \,{60^ \circ }$$ (only the horizontal component remains, the vertical component being zero at the top-most point).
<br><br>Therefore kinetic energy at the highest point,
<br><br>$${\left( {KE} \right)_H} = {1 \over 2}m{u^2}{\cos ^2}60^\circ $$
<br><br>$$\,\,\,\,\,\,\,\,\,\,\,\, = {K \over 4}$$ [ From eq $$(1)$$ ] | mcq | aieee-2007 | 13,282 |
VEKmSiNkuXyahSLy | physics | work-power-and-energy | energy | An athlete in the olympic games covers a distance of $$100$$ $$m$$ in $$10$$ $$s.$$ His kinetic energy can be estimated to be in the range | [{"identifier": "A", "content": "$$200J-500J$$ "}, {"identifier": "B", "content": "$$2 \\times {10^5}J - 3 \\times {10^5}J$$ "}, {"identifier": "C", "content": "$$20,000J - 50,000J$$ "}, {"identifier": "D", "content": "$$2,000J - 5,000J$$ "}] | ["D"] | null | The average speed of the athelete
<br><br>$$v = {{100} \over {10}} = 10m/s\,\,\,\,$$ $$\therefore$$ $$K.E. = {1 \over 2}m{v^2}$$
<br><br>If mass of athlete is $$40$$ $$kg$$ then, $$K.E.$$ $$ = {1 \over 2} \times 40 \times {\left( {10} \right)^2} = 2000J$$
<br><br>If mass of athlete is $$100$$ $$kg$$ then, $$K.E.$$ $$ = {1 \over 2} \times 100 \times {\left( {10} \right)^2} = 5000J$$
<br><br>His kinetic energy can be in the range = 2000 J to 5000 J. | mcq | aieee-2008 | 13,284 |
uWhjPEbP9mZUUT5B | physics | work-power-and-energy | energy | The potential energy function for the force between two atoms in a diatomic molecule is approximately given by $$U\left( x \right) = {a \over {{x^{12}}}} - {b \over {{x^6}}},$$ where $$a$$ and $$b$$ are constants and $$x$$ is the distance between the atoms. If the dissociation energy of the molecule is $$D = \left[ {U\left( {x = \infty } \right) - {U_{at\,\,equilibrium}}} \right],\,\,D$$ is | [{"identifier": "A", "content": "$${{{b^2}} \\over {2a}}$$ "}, {"identifier": "B", "content": "$${{{b^2}} \\over {12a}}$$ "}, {"identifier": "C", "content": "$${{{b^2}} \\over {4a}}$$"}, {"identifier": "D", "content": "$${{{b^2}} \\over {6a}}$$"}] | ["C"] | null | Given $$U\left( x \right) = {a \over {{x^{12}}}} - {b \over {{x^6}}}$$
<br><br>$${U\left( {x = \infty } \right)}$$ = 0
<br><br>We know $$F = - {{dU} \over {dx}} = - \left[ {{{12a} \over {{x^{13}}}} + {{6b} \over {{x^7}}}} \right]$$
<br><br>At equilibrium: $${{dU\left( x \right)} \over {dx}} = 0$$
<br><br>$$ \Rightarrow {{ - 12a} \over {{x^{13}}}} = {{ - 6b} \over {{x^7}}} $$
<br><br>$$\Rightarrow x = {\left( {{{2a} \over h}} \right)^{{1 \over 6}}}$$
<br><br>$$\therefore$$ $${U_{at\,\,equilibrium\,}} = {a \over {{{\left( {{{2a} \over b}} \right)}^2}}} - {b \over {\left( {{{2a} \over b}} \right)}}$$
<br><br>$$ = - {{{b^2}} \over {4a}}$$
<br><br>$$\therefore$$ $$D = 0 - \left( { - {{{b^2}} \over {4a}}} \right) = {{{b^2}} \over {4a}}$$ | mcq | aieee-2010 | 13,285 |
KY02Pg1xJZ6d8kDg | physics | work-power-and-energy | energy | This question has Statement $$1$$ and Statement $$2.$$ Of the four choices given after the Statements, choose the one that best describes the two Statements.
<br/><br/><p>If two springs $${S_1}$$ and $${S_2}$$ of force constants $${k_1}$$ and $${k_2}$$, respectively, are stretched by the same force, it is found that more work is done on spring $${S_1}$$ than on spring $${S_2}$$.
<br/><br/><b>STATEMENT 1:</b> If stretched by the same amount work done on $${S_1}$$, Work done on $${S_1}$$ is more than $${S_2}$$
<br/><b>STATEMENT 2:</b> $${k_1} < {k_2}$$</p> | [{"identifier": "A", "content": "Statement 1 is false, Statement 2 is true "}, {"identifier": "B", "content": "Statement 1 is true, Statement 2 is false"}, {"identifier": "C", "content": "Statement 1 is true, Statement 2 is true, Statement 2 is the correct explanation for Statement 1"}, {"identifier": "D", "content": "Statement 1 is true, Statement 2 is true, Statement 2 is not the correct explanation for Statement 1"}] | ["A"] | null | We know force (F) = kx
<br><br>$$W = {1 \over 2}k{x^2}$$
<br><br>$$W =$$ $${{{{\left( {kx} \right)}^2}} \over {2k}}$$ $$\,\,\,$$
<br><br>$$\therefore$$ $$W = {{{F^2}} \over {2k}}$$ [ as $$F=kx$$ ]
<br><br>When force is same then,
<br><br>$$W \propto {1 \over k}$$
<br><br>Given that, $${W_1} > {W_2}$$
<br><br>$$\therefore$$ $${k_1} < {k_2}$$
<br><br><b>Statement-2 is true.</b>
<br><br>For the same extension, x<sub>1</sub>
= x<sub>2</sub>
= x
<br><br>Work done on spring S<sub>1</sub> is W<sub>1</sub> = $${1 \over 2}{k_1}x_1^2 = {1 \over 2}{k_1}{x^2}$$
<br><br>Work done on spring S<sub>2</sub> is W<sub>2</sub> = $${1 \over 2}{k_2}x_2^2 = {1 \over 2}{k_2}{x^2}$$
<br><br>$$ \therefore $$ $${{{W_1}} \over {{W_2}}} = {{{k_1}} \over {{k_2}}}$$
<br><br>As $${k_1} < {k_2}$$ then $${W_1} < {W_2}$$
<br><br><b>So, Statement-1 is false.</b> | mcq | aieee-2012 | 13,286 |
u6Ea70S4r7bOa7MT | physics | work-power-and-energy | energy | A point particle of mass $$m,$$ moves long the uniformly rough track $$PQR$$ as shown in the figure. The coefficient of friction, between the particle and the rough track equals $$\mu .$$ The particle is released, from rest from the point $$P$$ and it comes to rest at point $$R.$$ The energies, lost by the ball, over the parts, $$PQ$$ and $$QR$$, of the track, are equal to each other , and no energy is lost when particle changes direction from $$PQ$$ to $$QR$$.
<p>The value of the coefficient of friction $$\mu $$ and the distance $$x$$ $$(=QR),$$ are, respectively close to: </p>
<img src="data:image/png;base64,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"/> | [{"identifier": "A", "content": "$$0.29$$ and $$3.5$$ $$m$$ "}, {"identifier": "B", "content": "$$0.29$$ and $$6.5$$ $$m$$ "}, {"identifier": "C", "content": "$$0.2$$ and $$6.5$$ $$m$$ "}, {"identifier": "D", "content": "$$0.2$$ and $$3.5$$ $$m$$ "}] | ["A"] | null | Using work energy theorem for the motion of the particle,
<br><br>Loss in $$P.E.=$$ Work done against friction from $$p \to Q$$
<br><br>$$ + $$ work done against friction from $$Q \to R$$
<br><br>$$mgh = \mu \left( {mg\cos \theta } \right)PQ + \mu mg\left( {QR} \right)$$
<br><br>$$h = \mu \,\cos \,\theta \times PQ + \mu \left( {QR} \right)$$
<br><br>$$2 = \mu \times {{\sqrt 3 } \over 2} \times {2 \over {\sin \,{{30}^ \circ }}} + \mu x$$
<br><br>$$2 = 2\sqrt 3 \mu + \mu x$$ $$\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,....(i)$$
<br><br>[ $$\sin \,\,{30^ \circ } = {2 \over {PQ}}$$ ]
<br><br>According to the question, work done $$P \to Q = $$ work done $$Q \to R$$
<br><br>$$ \Rightarrow $$$$\mu \left( {mg\cos \theta } \right)PQ + \mu mg\left( {QR} \right)$$
<br><br>$$\therefore$$ $$2\sqrt 3 \,\mu = \mu x$$
<br><br>$$\therefore$$ $$x \approx 3.5m$$
<br><br>From $$(i)$$
<br><br>$$2 = 2\sqrt 3 \mu + 2\sqrt 3 \mu = 4\sqrt 3 \mu $$
<br><br>$$\therefore$$ $$\mu = {2 \over {4\sqrt 3 }} = {1 \over {2 \times 1.732}} = 0.29$$ | mcq | jee-main-2016-offline | 13,288 |
qamYbZetJbvnaWDTNxEHv | physics | work-power-and-energy | energy | Velocity-time graph for a body of mass 10 kg is shown in figure. Work-done on
the body in first two seconds of the motion is :
<br/><br/><img src="data:image/png;base64,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"/> | [{"identifier": "A", "content": "12000 J"}, {"identifier": "B", "content": "$$-$$ 12000 J"}, {"identifier": "C", "content": "$$-$$ 4500 J"}, {"identifier": "D", "content": "$$-$$ 9300 J"}] | ["C"] | null | <img src="https://app-content.cdn.examgoal.net/fly/@width/image/1l82x41my/097d3d61-5618-4e49-8441-af0a000c1ad5/680d2aa0-34e2-11ed-b84c-a3c7c2456516/file-1l82x41mz.png?format=png" data-orsrc="https://app-content.cdn.examgoal.net/image/1l82x41my/097d3d61-5618-4e49-8441-af0a000c1ad5/680d2aa0-34e2-11ed-b84c-a3c7c2456516/file-1l82x41mz.png" loading="lazy" style="max-width: 100%; height: auto; display: block; margin: 0px auto; max-height: 40vh;" alt="JEE Main 2016 (Online) 10th April Morning Slot Physics - Work Power & Energy Question 92 English Explanation">
<br>Here u = 50 m/s , what t = 0
<br><br>$$\alpha $$ = $${{\Delta v} \over {\Delta t}}$$ = $${{50 - 0} \over {0 - 10}}$$ = $$-$$5 m/s<sup>2</sup>
<br><br>Speed of the body at t = 2 s
<br><br>v = u + at
<br><br>= 50 + ($$-$$ 5) $$ \times $$ 2
<br><br>= 40 m/s
<br><br>From work energy theorem,
<br><br>$$\Delta $$w = $${1 \over 2}m{v^2} - {1 \over 2}m{u^2}$$
<br><br>= $${1 \over 2}$$ m(v<sup>2</sup> $$-$$u<sup>2</sup>)
<br><br>= $${1 \over 2}$$ $$ \times $$ 10 $$ \times $$ (40<sup>2</sup> $$-$$ 50<sup>2</sup>)
<br><br>= 5 $$ \times $$ ($$-$$10)(90)
<br><br>= $$-$$ 4500 J | mcq | jee-main-2016-online-10th-april-morning-slot | 13,289 |
y1q1Cv23VTPqHYK4 | physics | work-power-and-energy | energy | A body of mass m = 10<sup>–2</sup> kg is moving in a medium and experiences a frictional force F = –kv<sup>2</sup>. Its initial speed is v<sub>0</sub> = 10 ms<sup>–1</sup>. If, after 10 s, its energy is $${1 \over 8}mv_0^2$$, the value of k will be: | [{"identifier": "A", "content": "10<sup>-1</sup> kg m<sup>-1</sup> s<sup>-1</sup>"}, {"identifier": "B", "content": "10<sup>-3</sup> kg m<sup>-1</sup>"}, {"identifier": "C", "content": "10<sup>-3</sup> kg s<sup>-1</sup>"}, {"identifier": "D", "content": "10<sup>-4</sup> kg m<sup>-1</sup>"}] | ["D"] | null | According to the question, final kinetic energy = $${1 \over 8}mv_0^2$$
<br><br>Let final speed of the body = V<sub>f</sub>
<br><br>So final kinetic energy = $${1 \over 2}mv_f^2$$
<br><br>According to question,
<br><br>$${1 \over 2}mv_f^2$$ = $${1 \over 8}mv_0^2$$
<br><br>$$ \Rightarrow {v_f} = {{{v_0}} \over 2}$$ = $${{10} \over 2}$$ = 5 m/s
<br><br>Given that, F = –kv<sup>2</sup>
<br><br>$$ \Rightarrow $$ $$m\left( {{{dv} \over {dt}}} \right)$$$$ = - k{v^2}$$
<br><br>$$ \Rightarrow {10^{ - 2}}\left( {{{dv} \over {dt}}} \right) = - k{v^2}$$
<br><br>$$ \Rightarrow \int\limits_{10}^5 {{{dv} \over {{v^2}}}} = - 100k\int\limits_0^{10} {dt} $$
<br><br>$$ \Rightarrow {1 \over 5} - {1 \over {10}} = 100k \times 10$$
<br><br>$$ \Rightarrow k = {10^{ - 4}}kg\,{m^{ - 1}}$$ | mcq | jee-main-2017-offline | 13,290 |
uJ5k3O0S7r7e3uvy | physics | work-power-and-energy | energy | A time dependent force F = 6t acts on a particle of mass 1 kg. If the particle starts from rest, the work done
by the force during the first 1 sec. will be: | [{"identifier": "A", "content": "18 J "}, {"identifier": "B", "content": "4.5 J "}, {"identifier": "C", "content": "22 J "}, {"identifier": "D", "content": "9 J"}] | ["B"] | null | Given that, F = 6t
<br><br>We know, F = ma = $$m{{dv} \over {dt}}$$
<br><br>$$\therefore$$ $$m{{dv} \over {dt}} = 6t$$
<br><br>$$ \Rightarrow $$ $$1.{{dv} \over {dt}} = 6t$$ [as m = 1]
<br><br>$$ \Rightarrow $$ $$\int\limits_0^v {dv} = \int {6t} dt$$
<br><br>$$ \Rightarrow $$ $$v = 6\left[ {{{{t^2}} \over 2}} \right]_0^1$$
<br><br>$$ \Rightarrow $$ $$v = {6 \over 2} = 3$$ m/s [ as given t = 1 sec ]
<br><br>Work done by the body during the first 1 form work-energy theorem,
<br><br>W = $$\Delta $$K.E = $${1 \over 2}m\left( {{V^2} - {v^2}} \right)$$
<br><br>= $${1 \over 2}.1.\left( {{3^2} - {0^2}} \right)$$ = 4.5 J
| mcq | jee-main-2017-offline | 13,291 |
C4qcuKUAna4e6RlkuWtgW | physics | work-power-and-energy | energy | An object is dropped from a height h from the ground. Every time it hits the ground it looses 50% of its kinetic energy. The total distance covered as t $$ \to $$ $$\infty $$ is : | [{"identifier": "A", "content": "3h"}, {"identifier": "B", "content": "$$\\infty $$"}, {"identifier": "C", "content": "$${5 \\over 3}$$h"}, {"identifier": "D", "content": "$${8 \\over 3}$$h"}] | ["A"] | null | Let,
<br><br>Kinetic energy (k) = $${1 \over 2}$$ m $$\upsilon $$<sup>2</sup> before it hit the ground.
<br><br>After hitting the ground kinetic energy
<br><br>(k') = $${1 \over 2}$$ m $$\upsilon $$$$_1^2$$
<br><br>$$\therefore\,\,\,$$According to the question,
<br><br> $${1 \over 2}$$ m$$\upsilon $$$$_1^2$$ = $${1 \over 2}$$ $$ \times $$ $${1 \over 2}$$ m$$\upsilon $$<sup>2</sup>
<br><br>$$ \Rightarrow $$$$\,\,\,$$ $$\upsilon $$<sub>1</sub> = $${v \over {\sqrt 2 }}$$
<br><br>After hitting the ground the object will bounce
<br><br>h' = $${{v_1^2} \over {2g}}$$ = $${{{v^2}} \over {4g}}$$ = $${h \over 2}$$ [ as h = $${{{v^2}} \over {2g}}$$ ]
<br><br>Total distance travelled from the time it first hits the ground to the next time it hits the ground is = $${h \over 2}$$ + $${h \over 2}$$ = h
<br><br>So, this will create a infinite geometric progression with the common ration $${1 \over 2}$$.
<br><br>$$\therefore\,\,\,$$ Total distance covered
<br><br>= h (distance travelled by the obhect when first dropped, before it hits the ground)
<br>+ (h + $${h \over 2}$$ + $${h \over 4}$$ + . . . . . . . .$$ \propto $$)
<br><br>= h + $${h \over {1 - {1 \over 2}}}$$
<br><br>= h + 2h
<br><br>= 3h | mcq | jee-main-2017-online-8th-april-morning-slot | 13,292 |
nQBrRdrD6Jc5VGZS | physics | work-power-and-energy | energy | A particle is moving in a circular path of radius $$a$$ under the action of an attractive potential $$U = - {k \over {2{r^2}}}$$ Its total energy is: | [{"identifier": "A", "content": "$$ - {3 \\over 2}{k \\over {{a^2}}}$$ "}, {"identifier": "B", "content": "Zero"}, {"identifier": "C", "content": "$$ - {k \\over {4{a^2}}}$$ "}, {"identifier": "D", "content": "$$ {k \\over {2{a^2}}}$$ "}] | ["B"] | null | We know, Total energy = Kinetic energy + Potential energy
<br><br>Potential energy given as $$U = - {k \over {2{r^2}}}$$
<br><br>We need to find Kinetic Energy.
<br><br>As Force acting on the particle (F) = $$ - {{dU} \over {dr}}$$
<br><br>$$ \Rightarrow F = - {d \over {dr}}\left( {{{ - k} \over {2{r^2}}}} \right)$$
<br><br>$$= {k \over 2} \times \left( { - 2} \right) \times {r^{ - 3}}$$
<br><br>$$ = - {k \over {{r^3}}}$$
<br><br>Because of this force particle is having circular motion so it will provide possible centripetal force.
<br><br>$$\left| F \right| = {{m{v^2}} \over r}$$
<br><br>$$ \Rightarrow {{m{v^2}} \over r} = {k \over {{r^3}}}$$
<br><br>$$ \Rightarrow $$ $$m{v^2} = {k \over {{r^2}}}$$
<br><br>We know kinetic energy of particle, K = $${1 \over 2}m{v^2}$$ = $${k \over {2{r^2}}}$$
<br><br>As Total energy = Kinetic energy + Potential energy
<br><br>So Total energy = $${k \over {2{r^2}}}$$ $$ - {k \over {2{r^2}}}$$ = 0 | mcq | jee-main-2018-offline | 13,293 |
XE1KidIvUZJDR05xGBii4 | physics | work-power-and-energy | energy | Two particles of the same mass m are moving in circular orbits because of force, given by $$F\left( r \right) = {{ - 16} \over r} - {r^3}$$
<br/><br/>The first particle is at a distance r = 1, and the second, at r = 4. The best estimate for the ratio of kinetic energies of the first and the second particle is closest to : | [{"identifier": "A", "content": "$$6 \\times {10^{ - 2}}$$"}, {"identifier": "B", "content": "$$3 \\times {10^{ - 3}}$$"}, {"identifier": "C", "content": "$${10^{ - 1}}$$"}, {"identifier": "D", "content": "$$6 \\times {10^{ 2}}$$"}] | ["A"] | null | In circular motion the force required
<br><br>$$\left| F \right| = {{m{v^2}} \over r}$$
<br><br>$$\therefore\,\,\,$$ $${{m{v^2}} \over r} = {{16} \over r} + {r^3}$$
<br><br>$$ \Rightarrow $$ mv<sup>2</sup> = 16 + r<sup>4</sup>
<br><br>$$\therefore\,\,\,$$ kinetic energy (K) = $${1 \over 2}$$ mv<sup>2</sup> = $${1 \over 2}$$ [ 16 + r<sup>4</sup>]
<br><br>$$\therefore\,\,\,$$ Kinetic energy of first particle (K<sub>1</sub>) = $${1 \over 2}$$ [16 + 1]
<br><br>Kinetic energy of second particle (K<sub>2</sub>) = $${1 \over 2}$$ [16 + 4<sup>4</sup>]
<br><br>$$\therefore\,\,\,\,$$ $${{{K_1}} \over {{K_2}}}$$ = $${{{{16 + 1} \over 2}} \over {{{16 + 256} \over 2}}}$$ = $${{17} \over {272}}$$
<br><br>$$ \Rightarrow $$ $$\,\,\,$$ $${{{K_1}} \over {{K_2}}} = 6 \times {10^{ - 2}}$$ | mcq | jee-main-2018-online-16th-april-morning-slot | 13,294 |
3uxZwzJ2S5Rygx7z1OOL6 | physics | work-power-and-energy | energy | A force acts on a 2 kg object so that its position is given as a function of time as x = 3t<sup>2</sup> + 5. What is the work done by this force in first 5 seconds ? | [{"identifier": "A", "content": "850 J"}, {"identifier": "B", "content": "950 J"}, {"identifier": "C", "content": "875 J"}, {"identifier": "D", "content": "900 J"}] | ["D"] | null | Displacement,
<br><br>x = 3t<sup>2</sup> + 5
<br><br>$$ \therefore $$ v = $${{dx} \over {dt}} = 6t$$
<br><br>At t = 0, velocity = 6 $$ \times $$ 0 = 0
<br><br>at t = 5, velocity = 5 $$ \times $$ 6 = 30 m/s
<br><br>we know from work energy theorem,
<br><br>Work (W) = change in kinetic energy ($$\Delta $$K)
<br><br>= $${1 \over 2}mv_F^2 - {1 \over 2}mv_i^2$$
<br><br>= $${1 \over 2}$$ $$ \times $$ 2 $$ \times $$ (30)<sup>2</sup> $$-$$ 0
<br><br>= 900 J | mcq | jee-main-2019-online-9th-january-evening-slot | 13,295 |
zUMzaA2w9ekwHAY5aXXxJ | physics | work-power-and-energy | energy | A particle moves in one dimension from rest
under the influence of a force that varies with
the distance travelled by the particle as shown
in the figure. The kinetic energy of the particle
after it has travelled 3m is :
<img src="data:image/png;base64,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"/> | [{"identifier": "A", "content": "6.5 J"}, {"identifier": "B", "content": "2.5 J"}, {"identifier": "C", "content": "5 J"}, {"identifier": "D", "content": "4 J"}] | ["A"] | null | According to work energy theorem.<br><br>
Work done by force on the particle = Change in KE<br><br>
Work done = Area under F-x graph = $$\int {F.dx} $$<br><br>
= $$2 \times 2 + {{(2 + 3) \times 1} \over 2}$$<br><br>
W = KE<sub>final</sub> – KE<sub>initial</sub> = 6.5<br><br>
KE<sub>initial</sub> = 0<br><br>
KE<sub>final</sub> = 6.5 J | mcq | jee-main-2019-online-8th-april-morning-slot | 13,296 |
moa8dmyYInAJFicbkaXoD | physics | work-power-and-energy | energy | A particle which is experiencing a force, given by $$\overrightarrow F = 3\widehat i - 12\widehat j,$$ undergoes a displacement of $$\overrightarrow d = 4\overrightarrow i $$ particle had a kinetic energy of 3 J at the beginning of the displacement, what is its kinetic energy at the end of the displacement ?
| [{"identifier": "A", "content": "9 J"}, {"identifier": "B", "content": "10 J"}, {"identifier": "C", "content": "12 J"}, {"identifier": "D", "content": "15 J"}] | ["D"] | null | Work done = $$\overrightarrow F \cdot \overrightarrow d $$
<br><br> $$=$$ 12 J
<br><br>work energy theorem
<br><br>w<sub>net</sub> $$=$$ $$\Delta $$K.E.
<br><br>12 $$=$$ K<sub>f</sub> $$-$$ 3
<br><br>K<sub>f</sub> = 15 J | mcq | jee-main-2019-online-10th-january-evening-slot | 13,297 |
9XJNuhT1F1TSc7O7wO7k9k2k5dusm7e | physics | work-power-and-energy | energy | A particle (m = 1 kg) slides down a frictionless track (AOC) starting from rest at a point A (height 2 m). After reaching C, the particle continues to move freely in air as a projectile. When it reaching its highest point P (height 1 m), the kinetic energy of the particle (in )) is :
<br/>(Figure drawn is schematic and not to scale; take g = 10 ms<sup>-2</sup>)
<img src="data:image/png;base64,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"/> | [] | null | 10 | KE = PE<sub>1</sub> – PE<sub>2</sub> = mgh<sub>1</sub> – mgh<sub>2</sub>
<br><br>= 1 × 10 × 2 – 1 × 10 × 1 = 10 J | integer | jee-main-2020-online-7th-january-morning-slot | 13,299 |
K2CFt4EwRFvx9qXghNjgy2xukf253soi | physics | work-power-and-energy | energy | A cricket ball of mass 0.15 kg is thrown
vertically up by a bowling machine so that it
rises to a maximum height of 20 m after leaving
the machine. If the part pushing the ball applies
a constant force F on the ball and moves
horizontally a distance of 0.2 m while launching
the ball, the value of F (in N) is (g = 10 ms<sup>–2</sup>)
____. | [] | null | 150 | Initial velocity, v = $$\sqrt {2gh} $$
<br><br>= $$\sqrt {2 \times 10 \times 20} $$
<br><br>= 20 m/s
<br><br>Now work done by the machine,
<br><br>W<sub>F</sub> = $$\Delta $$k
<br><br>$$ \Rightarrow $$ F.d = $$\Delta $$k
<br><br>$$ \Rightarrow $$ F = $${{\Delta k} \over d}$$
<br><br>= $${{{1 \over 2} \times 0.15 \times 400 - 0} \over {0.2}}$$
<br><br>= 150 N | integer | jee-main-2020-online-3rd-september-morning-slot | 13,301 |
TOQv34lCwymEwEZbG4jgy2xukf3w58ju | physics | work-power-and-energy | energy | A block starts moving up an inclined plane of inclination 30<sup>o</sup> with an initial velocity of v<sub>0</sub>
. It comes
back to its initial position with velocity $${{{v_0}} \over 2}$$. The value of the coefficient of kinetic friction between
the block and the inclined plane is close to $${I \over {1000}}$$. The nearest integer to I is____. | [] | null | 346 | <img src="https://res.cloudinary.com/dckxllbjy/image/upload/v1734264080/exam_images/h2xgqxpgbctx5ls8sdiw.webp" style="max-width: 100%;height: auto;display: block;margin: 0 auto;" loading="lazy" alt="JEE Main 2020 (Online) 3rd September Evening Slot Physics - Work Power & Energy Question 78 English Explanation">
<br>a = g sin 30 + $$\mu $$ g cos 30
<br><br>We know, v<sup>2</sup> = u<sup>2</sup> + 2as
<br><br>$$ \Rightarrow $$ 0 = $$v_0^2$$ - 2ad
<br><br>$$ \Rightarrow $$ $$v_0^2 = 2ad$$<br><br>$$d = {{v_0^2} \over {2a}}$$
<br><br>Total work done,
<br><br>$${W_f} = {k_f} - {k_i}$$<br><br>$$ \Rightarrow $$ $$ - 2\mu mg\,\cos 30{{v_0^2} \over {2a}} = {1 \over 2}m{{v_0^2} \over 4} - {1 \over 2}mv_0^2$$<br><br>$$ \Rightarrow $$ $${{ + \mu g\,\cos 30} \over a} = $$$${3 \over 8}$$<br><br>$$ \Rightarrow $$ $$8\mu g\,\cos 30 = 3g\,\sin 30 + 3\mu \,\cos 30$$<br><br>$$ \Rightarrow $$ $$5\mu g\,\cos 30 = 3g\,\sin 30$$<br><br>$$ \Rightarrow $$ $$\mu = {{3\tan 30} \over 5} = {{\sqrt 3 } \over 5}$$<br><br>$$ \Rightarrow $$ $${{\sqrt 3 } \over 5} = {I \over {1000}}$$<br><br>$$ \Rightarrow $$ $$I = 346$$ | integer | jee-main-2020-online-3rd-september-evening-slot | 13,302 |
3GgNSfzUmHCgcT4Ul4jgy2xukfrny787 | physics | work-power-and-energy | energy | If the potential energy between two molecules is given by
<br/>U = $$ - {A \over {{r^6}}} + {B \over {{r^{12}}}}$$,
<br/>then at equilibrium,
separation between molecules, and the potential energy are : | [{"identifier": "A", "content": "$${\\left( {{{2B} \\over A}} \\right)^{1/6}}$$, $$ - {{{A^2}} \\over {4B}}$$"}, {"identifier": "B", "content": "$${\\left( {{{2B} \\over A}} \\right)^{1/6}}, - {{{A^2}} \\over {2B}}$$"}, {"identifier": "C", "content": "$${\\left( {{B \\over A}} \\right)^{1/6}},0$$"}, {"identifier": "D", "content": "$${\\left( {{B \\over {2A}}} \\right)^{1/6}}, - {{{A^2}} \\over {2B}}$$"}] | ["A"] | null | U = $$ - {A \over {{r^6}}} + {B \over {{r^{12}}}}$$
<br><br>F = - $${{dU} \over {dr}}$$
<br><br>= – (A(–6r<sup>–7</sup>
)) + B(–12r<sup>–13</sup>)
<br><br>for equilibrium, F = 0
<br><br>$$ \therefore $$ 0 = $${{6A} \over {{r^7}}} - {{12B} \over {{r^{13}}}}$$
<br><br>$$ \Rightarrow $$ $${{6A} \over {12B}} = {1 \over {{r^6}}}$$
<br><br>$$ \Rightarrow $$ r = $${\left( {{{2B} \over A}} \right)^{{1 \over 6}}}$$
<br><br>$$ \therefore $$ U = $$ - {A \over {{{2B} \over A}}} + {B \over {{{\left( {{{2B} \over A}} \right)}^2}}}$$
<br><br>= $$ - {{{A^2}} \over {2B}} + {{{A^2}} \over {4B}}$$
<br><br>= $$ - {{{A^2}} \over {4B}}$$ | mcq | jee-main-2020-online-6th-september-morning-slot | 13,303 |
WQ1T9xWK8rZ5nipd1Z1klryxzzy | physics | work-power-and-energy | energy | The potential energy (U) of a diatomic molecule is a function dependent on r (interatomic distance) as <br/><br/>$$U = {\alpha \over {{r^{10}}}} - {\beta \over {{r^5}}} - 3$$<br/><br/>where, $$\alpha$$ and $$\beta$$ are positive constants. The equilibrium distance between two atoms will be $${\left( {{{2\alpha } \over \beta }} \right)^{{a \over b}}}$$, where a = ___________. | [] | null | 1 | $$F = - {{dU} \over {dr}}$$<br><br>$$F = - \left[ { - {{10\alpha } \over {{r^{11}}}} + {{5\beta } \over {{r^6}}}} \right]$$<br><br>for equilibrium, F = 0<br><br>$${{10\alpha } \over {{r^{11}}}} = {{5\beta } \over {{r^6}}}$$<br><br>$${{2\alpha } \over \beta } = {r^5}$$<br><br>$$r = {\left( {{{2\alpha } \over \beta }} \right)^{1/5}}$$<br><br>$$ \therefore $$ $$a = 1$$ | integer | jee-main-2021-online-25th-february-morning-slot | 13,304 |
q9NXhE2gBDHn1yOKdd1kmks0bk2 | physics | work-power-and-energy | energy | As shown in the figure, a particle of mass 10 kg is placed at a point A. When the particle is slightly displaced to its right, it starts moving and reaches the point B. The speed of the particle at B is x m/s. (Take g = 10 m/s<sup>2</sup>)<br/><br/>The value of 'x' to the nearest integer is __________.<br/><br/><img src="data:image/png;base64,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"/> | [] | null | 10 | By energy conservation<br><br>K<sub>i</sub> + U<sub>i</sub> = K<sub>f</sub> + U<sub>f</sub><br><br>0 + 10 $$\times$$ 10 $$\times$$ 10 = $${1 \over 2}$$ $$\times$$ 10 $$\times$$ v$$_B^2$$ + 10 $$\times$$ 10 $$\times$$ 5<br><br>1000 = 5v$$_B^2$$ + 500<br><br>v$$_B^2$$ = $${{500} \over 5}$$ = 100<br><br>V<sub>B</sub> = 10 m/s<br><br>x = 10 | integer | jee-main-2021-online-18th-march-morning-shift | 13,306 |
r87Aim6MWZGpyyQRLp1kmlwq7lp | physics | work-power-and-energy | energy | A ball of mass 4 kg, moving with a velocity of 10 ms<sup>$$-$$1</sup>, collides with a spring of length 8 m and force constant 100 Nm<sup>$$-$$1</sup>. The length of the compressed spring is x m. The value of x, to the nearest integer, is ____________. | [] | null | 6 | <picture><source media="(max-width: 1147px)" srcset="https://res.cloudinary.com/dckxllbjy/image/upload/v1734263809/exam_images/anxtagbtptmwcnrxmwf5.webp"><source media="(max-width: 320px)" srcset="https://res.cloudinary.com/dckxllbjy/image/upload/v1734266873/exam_images/bmzxmktdhf0lxbfdsnt5.webp"><source media="(max-width: 500px)" srcset="https://res.cloudinary.com/dckxllbjy/image/upload/v1734265602/exam_images/nhnexr8vh5sstchrgiwa.webp"><source media="(max-width: 680px)" srcset="https://res.cloudinary.com/dckxllbjy/image/upload/v1734267033/exam_images/vqzaiyvpenkuet0yiglw.webp"><img src="https://res.cloudinary.com/dckxllbjy/image/upload/v1734264355/exam_images/qpslldgazaaqe9gehs47.webp" style="max-width: 100%;height: auto;display: block;margin: 0 auto;" loading="lazy" alt="JEE Main 2021 (Online) 18th March Evening Shift Physics - Work Power & Energy Question 70 English Explanation"></picture>
<br>If spring compressed by x, <br><br>then work done by spring = 0 $$-$$ $${1 \over 2}$$ $$\times$$ 4 $$\times$$ 10<sup>2</sup><br><br>Applying work energy theorem, <br><br>$$-$$$${1 \over 2}$$ kx<sup>2</sup> = $$-$$$${1 \over 2}$$ $$\times$$ 4 $$\times$$ 10<sup>2</sup><br><br>$$ \Rightarrow $$ 100x<sup>2</sup> = 4 $$\times$$ 10<sup>2</sup><br><br>$$ \Rightarrow $$ x = 2<br><br>$$ \therefore $$ Final length of the spring = 8 $$-$$ 2 = 6 m | integer | jee-main-2021-online-18th-march-evening-shift | 13,307 |
1krppkijo | physics | work-power-and-energy | energy | In a spring gun having spring constant 100 N/m a small ball 'B' of mass 100 g is put in its barrel (as shown in figure) by compressing the spring through 0.05 m. There should be a box placed at a distance 'd' on the ground so that the ball falls in it. If the ball leaves the gun horizontally at a height of 2 m above the ground. The value of d is _________ m. (g = 10 m/s<sup>2</sup>).<br/><br/><img src="data:image/png;base64,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"/> | [] | null | 1 | Given, k = 100 N/m<br/><br/>m = 100 g = 0.1 kg<br/><br/>x = 0.05 m and H = 2 m<br/><br/>By energy conservation,<br/><br/>$${1 \over 2}k{x^2} = {1 \over 2}m{v^2} \Rightarrow v = x\sqrt {{k \over m}} $$<br/><br/>$$ = 0.05 \times \sqrt {{{100} \over {0.1}}} = 0.5\sqrt {10} $$ ms<sup>$$-$$1</sup> .... (i)<br/><br/>Time of flight of ball, $$t = \sqrt {{{2H} \over g}} $$<br/><br/>$$ \Rightarrow t = \sqrt {{{2 \times 2} \over {10}}} = {2 \over {\sqrt {10} }}$$s .... (ii)<br/><br/>$$\therefore$$ Range of ball, d = vt<br/><br/>$$ = 0.5\sqrt {10} \times \left( {{2 \over {\sqrt {10} }}} \right)$$ [From Eqs. (i) and (ii)]<br/><br/>$$ = 0.5 \times 2 = 1 m$$ | integer | jee-main-2021-online-20th-july-morning-shift | 13,308 |
1ks1arj1i | physics | work-power-and-energy | energy | A small block slides down from the top of hemisphere of radius R = 3 m as shown in the figure. The height 'h' at which the block will lose contact with the surface of the sphere is __________ m.<br/><br/>(Assume there is no friction between the block and the hemisphere)<br/><br/><img src="data:image/png;base64,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"/> | [] | null | 2 | <img src="https://res.cloudinary.com/dckxllbjy/image/upload/v1734264326/exam_images/xamklicnswtnjovr6fdj.webp" style="max-width: 100%;height: auto;display: block;margin: 0 auto;" loading="lazy" alt="JEE Main 2021 (Online) 27th July Evening Shift Physics - Work Power & Energy Question 65 English Explanation"><br>$$mg\cos \theta = {{m{v^2}} \over R}$$ .... (1)<br><br>$$\cos \theta = {h \over R}$$<br><br>Energy conservation<br><br>$$mg\{ R - h\} = {1 \over 2}m{v^2}$$ ..... (2)<br><br>from (1) & (2)
<br><br>$$\Rightarrow$$ $$mg\left\{ {{h \over R}} \right\} = {{2mg\{ R - h\} } \over R}$$<br><br>$$h = {{2R} \over 3}$$ = 2m | integer | jee-main-2021-online-27th-july-evening-shift | 13,310 |
1kth682v8 | physics | work-power-and-energy | energy | A block moving horizontally on a smooth surface with a speed of 40 ms<sup>$$-$$1</sup> splits into two equal parts. If one of the parts moves at 60 ms<sup>$$-$$1</sup> in the same direction, then the fractional change in the kinetic energy will be x : 4 where x = ___________. | [] | null | 1 | <img src="https://res.cloudinary.com/dckxllbjy/image/upload/v1734264773/exam_images/jrievdqbqsws3zewixut.webp" style="max-width: 100%;height: auto;display: block;margin: 0 auto;" loading="lazy" alt="JEE Main 2021 (Online) 31st August Morning Shift Physics - Work Power & Energy Question 60 English Explanation"><br>P<sub>i</sub> = P<sub>f</sub><br><br>m $$\times$$ 40 = $${m \over 2}$$ $$\times$$ v + $${m \over 2}$$ $$\times$$ 60<br><br>40 = $${v \over 2}$$ + 30<br><br>$$\Rightarrow$$ v = 20<br><br>(K. E.)<sub>I</sub> = $${1 \over 2}$$m $$\times$$ (40)<sup>2</sup> = 800 m<br><br>(K. E.)<sub>f</sub> = $${1 \over 2}$$$${m \over 2}$$ . (20)<sup>2</sup> + $${1 \over 2}$$ . $${m \over 2}$$ (60)<sup>2</sup> = 1000 m<br><br>| $$\Delta$$ K. E. | = | 1000m $$-$$ 800 m | = 200 m<br><br>$${{\Delta K.E.} \over {{{(K.E.)}_i}}} = {{200m} \over {800m}} = {1 \over 4} = {x \over 4}$$<br><br>x = 1 | integer | jee-main-2021-online-31st-august-morning-shift | 13,312 |
1ktjpowmd | physics | work-power-and-energy | energy | A block moving horizontally on a smooth surface with a speed of 40 m/s splits into two parts with masses in the ratio of 1 : 2. If the smaller part moves at 60 m/s in the same direction, then the fractional change in kinetic energy is :- | [{"identifier": "A", "content": "$${{1 \\over 3}}$$"}, {"identifier": "B", "content": "$${{2 \\over 3}}$$"}, {"identifier": "C", "content": "$${{1 \\over 8}}$$"}, {"identifier": "D", "content": "$${{1 \\over 4}}$$"}] | ["C"] | null | <picture><source media="(max-width: 320px)" srcset="https://res.cloudinary.com/dckxllbjy/image/upload/v1734263414/exam_images/ovdcd6ijejzouwkx2ser.webp"><source media="(max-width: 500px)" srcset="https://res.cloudinary.com/dckxllbjy/image/upload/v1734265347/exam_images/ec1jrorr3vpnfkassjet.webp"><source media="(max-width: 680px)" srcset="https://res.cloudinary.com/dckxllbjy/image/upload/v1734263456/exam_images/rtrj2wvg9fjwcslnpwod.webp"><source media="(max-width: 860px)" srcset="https://res.cloudinary.com/dckxllbjy/image/upload/v1734266105/exam_images/stupzrzb0ensztzg7ryj.webp"><source media="(max-width: 1040px)" srcset="https://res.cloudinary.com/dckxllbjy/image/upload/v1734265967/exam_images/yrbxrtciernwiqtkfezi.webp"><source media="(max-width: 1220px)" srcset="https://res.cloudinary.com/dckxllbjy/image/upload/v1734266053/exam_images/eavqpqenkjx8tqrby320.webp"><img src="https://res.cloudinary.com/dckxllbjy/image/upload/v1734263882/exam_images/jjqnjlmto57bmcwkprzs.webp" style="max-width: 100%;height: auto;display: block;margin: 0 auto;" loading="lazy" alt="JEE Main 2021 (Online) 31st August Evening Shift Physics - Work Power & Energy Question 57 English Explanation"></picture> <br>3MV<sub>0</sub> = 2MV<sub>2</sub> + MV<sub>1</sub><br><br>3V<sub>0</sub> = 2V<sub>2</sub> + V<sub>1</sub><br><br>120 = 2V<sub>2</sub> + 60 $$\Rightarrow$$ V<sub>2</sub> = 30 m/s<br><br>$${{\Delta K.E.} \over {K.E.}} = {{{1 \over 2}MV_1^2 + {1 \over 2}2MV_2^2 - {1 \over 2}3MV_0^2} \over {{1 \over 2}3MV_0^2}}$$<br><br>$$ = {{V_1^2 + 2V_2^2 - 3V_0^2} \over {3V_0^2}}$$<br><br>$$ = {{3600 + 1800 - 4800} \over {4800}} = {1 \over 8}$$ | mcq | jee-main-2021-online-31st-august-evening-shift | 13,313 |
1l5476l4l | physics | work-power-and-energy | energy | <p>A particle of mass 500 gm is moving in a straight line with velocity v = b x<sup>5/2</sup>. The work done by the net force during its displacement from x = 0 to x = 4 m is : (Take b = 0.25 m<sup>$$-$$3/2</sup> s<sup>$$-$$1</sup>).</p> | [{"identifier": "A", "content": "2 J"}, {"identifier": "B", "content": "4 J"}, {"identifier": "C", "content": "8 J"}, {"identifier": "D", "content": "16 J"}] | ["D"] | null | <p>$${W_{total}} = \Delta K$$</p>
<p>$$ = {1 \over 2}\left( {{1 \over 2}} \right)\left[ {{{\{ b{{(4)}^{5/2}}\} }^2} - 0} \right]$$</p>
<p>$$ = {{{b^2}} \over 4} \times {4^5}$$</p>
<p>$$ \Rightarrow {W_{total}} = 16\,J$$</p> | mcq | jee-main-2022-online-29th-june-morning-shift | 13,315 |
1l54vckob | physics | work-power-and-energy | energy | <p>In the given figure, the block of mass m is dropped from the point 'A'. The expression for kinetic energy of block when it reaches point 'B' is</p>
<p> <img src="data:image/png;base64,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"/></p> | [{"identifier": "A", "content": "$${1 \\over 2}mg\\,{y_0}^2$$"}, {"identifier": "B", "content": "$${1 \\over 2}mg\\,{y^2}$$"}, {"identifier": "C", "content": "$$mg(y - {y_0})$$"}, {"identifier": "D", "content": "$$mg{y_0}$$"}] | ["D"] | null | <p>Loss in potential energy = gain in kinetic energy</p>
<p>$$-$$ (mg(y $$-$$ y<sub>0</sub>) $$-$$ mgy) = KE $$-$$ 0</p>
<p>$$\Rightarrow$$ KE = mgy<sub>0</sub></p> | mcq | jee-main-2022-online-29th-june-evening-shift | 13,316 |
1l5al77dk | physics | work-power-and-energy | energy | <p>A uniform chain of 6 m length is placed on a table such that a part of its length is hanging over the edge of the table. The system is at rest. The co-efficient of static friction between the chain and the surface of the table is 0.5, the maximum length of the chain hanging from the table is ___________ m.</p> | [] | null | 2 | <p>$$(x)g\lambda = \mu (6 - x)\,g\lambda $$ where x is length of hanging part</p>
<p>$$ \Rightarrow x = 3 - 0.5x$$</p>
<p>$$ \Rightarrow x = 2$$ m</p> | integer | jee-main-2022-online-25th-june-morning-shift | 13,317 |
1l5al85mw | physics | work-power-and-energy | energy | <p>A 0.5 kg block moving at a speed of 12 ms<sup>$$-$$1</sup> compresses a spring through a distance 30 cm when its speed is halved. The spring constant of the spring will be _______________ Nm<sup>$$-$$1</sup>.</p> | [] | null | 600 | <p>$${1 \over 2}m\,{V^2} = {1 \over 2}k{x^2} + {1 \over 2}m{\left( {{v \over 2}} \right)^2}$$</p>
<p>$$ \Rightarrow {3 \over 8}m{v^2} = {1 \over 2}k{x^2}$$</p>
<p>$$ \Rightarrow k = {3 \over 4} \times {1 \over 2} \times {{144} \over 9} \times 100$$</p>
<p>$$ = 600$$</p>
<p>$$ \Rightarrow 600$$</p> | integer | jee-main-2022-online-25th-june-morning-shift | 13,318 |
1l5c4ot2s | physics | work-power-and-energy | energy | <p>A ball of mass 100 g is dropped from a height h = 10 cm on a platform fixed at the top of a vertical spring (as shown in figure). The ball stays on the platform and the platform is depressed by a distance $${h \over 2}$$. The spring constant is _____________ Nm<sup>$$-$$1</sup>.</p>
<p>(Use g = 10 ms<sup>$$-$$2</sup>)</p>
<p><img src="data:image/png;base64,UklGRt4LAABXRUJQVlA4INILAAAQugCdASrhAQADP4G82GY2LrknITD5GyAwCWlu4W8hG/Pz9TYwZiP+w3PfbXNtN7egeZP///rVIfDS0u6GtycndCsVzCKhrcnI2lafgmLGTa4rT8DIKnbNGe/sU1Uw2urbcR811YSoZNritPwTFjJtcNQV01D4wS0u3DZ/gmLGTa4rT8EvGSCx268HagnIpsx997Wu6FjP8ExYybXFafgmTB51Z2jKsNHcEExYybXFafgmLGV1+2IjYa3JvJixk2uK0/BMWMm1kPCT0u6GoCTa4rT8ExYybXFaSHOaulpd0LGf4Jixk2uK0/BMWF0fjZ/DS0uSLGTa4rT8ExYybXC2g1QACxZooJi1z4Jixk2uK0/BMWMmC2AAYFUlOB/7dxI6mpmrT9m0rT8ExYybXFafgl03gCjy6j1Z3t8M/CXK0/BMWMm1xWn4JiY1ZHu/sL285rkEHi1z4Jixk2uK0/BMWMmBcdr1ths/2bStPwTFjJtcVp+CXUsqh1UvXFajfXFafgmLGTa4rT8DfVmE/wTFrnwTFjJtcVp+CYsZMzJxYybZ7m61LyEKQqzCUS/fyYscZ22uKlzmN5f3Jtn9rg60nJDeM8/412QAs7sW494jG3CQ5D7myBjxnCClXkQOyvDwuHVY9IIaXRj3w7iKnbYrYbWpfDqjNFSTa22uFiTFhsqhBWXy382rig22hcPCP9kPRb48vAFRV8BguN+616Zi32+n2WCs7q2csOzYOHgKaOo+rBruS4XyVrg2eRyc8uA+ldr1hfqgww7risHfWdtrEJbl/poBswVVqglEDuQQ2qDfZ8kFcqOenTphfqRSkK3cjxF7Xk5bLwZDveRymw6vWUjm2wrdv6L7rEzhPXKKJVmOXc0X5qFnCGM+G6kkrnLOEMlpyen18xHhee61GVWOOXlmyhSKuMZzFhDs3SckjZ8xHfAsC6Q38quFb6Wk2jokxQtcG6acd1YZ6bET0LNMm1xWn4Jixk2uGlKEnZcjhy7yd2FmmTa4rT8ExYybXDgKJR3reVJFAKy5n1vhD92c9bt+F2hC8vxRyNl4a3BZtK0/BMWMm1xWnm/JCGSjtZQLiH6aLSmfAOuK1G+uK0/BMWMm1xWjTVbmwUq01P8Exa58ExYybXFafgmJ44y3ghGMBT3UW14vmVu2r3ZtQZ09X4VKhCTbP7GTa4rT8ExYybFBVY9G2tIrjyrHjzhvw7cNn+EuVp+CYsZNritPvJXMZ3itwAgeLGV4mLGTa4rT8ExYxBVBcFfjnFRBVyWbSv17UgL+LXXB/gn+afgmLGTa4rT8EsVesb6hdzSmjy8fxn8HlVoNrDbVQSbZ/YybXFafgmLGTZfplHzg1RbTP8JcrT8ExYybXFafdAKIA8LTmDU1Q0IYVzZQr7BNrnIZNritPwTFjJtcIDXo6Hii3caz50FVSUFuUjCoGtriH8JYVqN9cVp+CYsZNritIKc44voaAmEz8E/zT8ExYybXFafgeF4iK0wUa1XeSdtri65NritPwTFjJtcVANi4OWtyTVFv8uI3Sv7qrf8F68odABWWWt3nV+gn+afgmLGTa4rT8Dg5vvPzn6LaZ/hLlafgmLGTa4rT7v5z3wdMCbXFafs2lafgmLGTa4rT7hs7Pi8Q22m9tpuiDP8Exa58ExYybXFafgmLW7IM3RQ42a2f4J/mn4Jixk2uK0/BMmH/miK0/CXK0/BMWMm1xWn4Jicww7ritRvritPwTFjJtY4xoZPQh9dzxKQ0TU69nsnxLLOXEeTVs4GdTZSZ8/Id5ZNp5KNLk6uK0/BMWMm1tEz17rtLmF7CZNTbeiixpCYbVZCeSYhofI7MIuoKKubzABZETslWRB8ComkN5hIdu0KQAUhM/BMWMm1xWnlMHrlYkxYq9UALeOzBbIRJddTzR4NGwViFiuPcWCQZEHUgLFEAIJd8FlHjx591gAxGnaef06yoPFjJtcVp+CWkpxe9gGQNuL32w7meBs4iVQAA/vvPBGmVstZWMAYfrv76mKBfrXQbMvDw9S7/IYk5jKiECgp1dPgRMFbG2ZQ3bNfMhjeBwlQTmrwe+HMJe3sSzgOoArhi5CGf1qfvKq/GacVypSQ79jrwuCgAfLv2gxgGm2Xri+UAevVW/gGuyqr17A4BULRUgBVeyJosZDzyf7kbqaYU242CgAaXl6bOH+7EVA3krON6eEEr9qYCBgvrDG5EqEr7xi/yLASnyygQnwTSAQgAwAMBIAdfx88wGAD+LP/903yms9vH353h90coSve7G+RHhbY8JKlboi5+YbHpB/zLE88A9WKCQARsKcMDHsh/fKK97ryTJL7TKvYXO0NdyI7L1ZatYYNoJgNdqxzc521+Oc9hRJtis+OtQYHoHZll/quQ0Ed8ZSi1dPszA8FgDj9vRXnO1/uUZqEetd3gi8ywXvyag4ylVlP/kxdU1jDlJ7EuP8IgSKzZXKd+TV6S8rPCxnW9S6PhqeQHdE/FqnCp8Na5OjVCRCWNXP93T6YrSF4/kvIzYcRTi6ij3Yrprc/xF3XcP9FsO5YSxgQMMzdFy8O4dH9YGbA4jwZZd6Dg1N01Wr2KZT+cUMY9n0f+c6Rwr0V9VT0RezKGsA9MyOQp+7pMu2m/mzbih10EwuZGvbLfqNw+IiKkospUrS4S1UcOwjNFU3yHen5jrSrqPV4lDICefeMpKfpUOCuJT/C/o18NmAtJEx5qnXatrFg/OtD9mFD5SlcIHBdWhs8gnG8miz00fqOgf+ZmY2NNwHwxIBFYEU6MuOQaFmWhwxwEYexR4fJ3+CloQ+Qk67wCwlsudm4jUeoFAQom49eyxkT2XZ4kuQAAE2qAphaidwD5ACeSLyWHYWQhzHnZmXhEAezPg3596LiowjECzXrU05HhY1untXEL1SVgQVWb9OW7WtbNXQEvoJhu9I6UdWMts/kATHjOExdFmlU1dyxNwgdKI5Bb5AYH3jQkAWpiyboTEATMfbkklkcDjhHiseKE1zMJh2zL7C2G08MkWC3eZ/DKLEsBDwTHi8CxSP2az4e9XJmF+mY2d5m1dMVCoDT2EK1viaLjk0MBE09Ziq2+3W9zgnhPi4nUoeczo64b+Fdp4XI+2XGDQff+hQCV0UxQaodel7jkstLEHgiuRU8oUvoXMnRHuPfbCMzIehWi6hk/iSjxAUyWKOFVtmLvdcrHYHi1WvspEonlD4CmKRL/d4gXSWTh/2itS78QugR9ZXg9h+MLzVoUAQtIN7iV2v30ccZrzwBd1Cqf7CeWQg7mkMrzos6NAQdYloTVSaZh9865lt6Fu4PsxauogHXlTeoaYCrC1BJuRa+mGtYpUZPL1nbqiW5Xz7OHOt50fDao4GYEWp6cXVyxS6oTan+cCid88AxPtDGl0GIGS+sAal6z2DbM39b1iCxPHFw3F4l6xmd79FlsASG6v46g7rmKZYhxr++xE9W6rDz2yzRSpgTNZhNT8Es6mUzHp7tTMzwbRHdNkUFpxin4qkepVuAHJD94hA2EPuvBmE402mOLuJsAEpaoyPdMvnRZ+1oBp7tmtqZNmR2cDEKYgDebGx7UWASQJxYh2ZhIxyOiuoAABsTxAMz2lOaqhS36Xuzun4EPeYnuMBRTB2fEyRwXuIX4uRySZ0tO1zvxfwZgvyEuen4DEFcCHckDHMmMHobaiKIfEK785y0uVfDau7yQpHYO5YHWhc1DHuY3uxK+8bKDJdJFK9JrXR9tyDrv1w9AWkFz/XO+5CUNUANm/wEEZWW7cs/i/xL8VTVwn7fOf/HULmb2W05ZBImowi/ECFNhDY/6j0WhCBo4gOVLeZBbeenKGC52liLCADqliinoaFRpM/q1PfNSIgyAXr25scP754mWnk2BaEpzamfIONGA3NVt31q30phg3aN/CG1GUEXPemALHZ0Er98oEyM6lj5R7tZeMCGA+M+NjvquMeVuhtXTR3EzgsfIEHtzymuan/QgIn/4ZdztHfIi2jt+qkdA0maAkci6UgUcKMq3JMXmbHcUCqJvUaT/To0AAA=="/></p> | [] | null | 120 | <p>$$mg\left( {h + {h \over 2}} \right) = {1 \over 2}k{\left( {{h \over 2}} \right)^2}$$</p>
<p>$$ \Rightarrow 0.1 \times 10\times(0.15) = {1 \over 2}k{(0.05)^2}$$</p>
<p>$$ \Rightarrow k = 120$$ N/m</p> | integer | jee-main-2022-online-24th-june-morning-shift | 13,319 |
1l6dxxjk3 | physics | work-power-and-energy | energy | <p>A body of mass $$0.5 \mathrm{~kg}$$ travels on straight line path with velocity $$v=\left(3 x^{2}+4\right) \mathrm{m} / \mathrm{s}$$. The net workdone by the force during its displacement from $$x=0$$ to $$x=2 \mathrm{~m}$$ is :</p> | [{"identifier": "A", "content": "64 J"}, {"identifier": "B", "content": "60 J"}, {"identifier": "C", "content": "120 J"}, {"identifier": "D", "content": "128 J"}] | ["B"] | null | <p>$$v = 3{x^2} + 4$$</p>
<p>at $$x = 0$$, $${v_1} = 4$$ m/s</p>
<p>$$x = 2$$, $${v_2} = 16$$ m/s</p>
<p>$$\Rightarrow$$ Work done = $$\Delta$$ kinetic energy</p>
<p>$$ = {1 \over 2} \times m\left( {v_2^2 - v_1^2} \right)$$</p>
<p>$$ = {1 \over 4}(256 - 16)$$</p>
<p>$$ = 60$$ J</p> | mcq | jee-main-2022-online-25th-july-morning-shift | 13,320 |
1l6f57pk1 | physics | work-power-and-energy | energy | <p>A bag of sand of mass 9.8 kg is suspended by a rope. A bullet of 200 g travelling with speed 10 ms<sup>$$-$$1</sup> gets embedded in it, then loss of kinetic energy will be :</p> | [{"identifier": "A", "content": "4.9 J"}, {"identifier": "B", "content": "9.8 J"}, {"identifier": "C", "content": "14.7 J"}, {"identifier": "D", "content": "19.6 J"}] | ["B"] | null | <p>Loss in $$KE = {1 \over 2} \times {{{m_1}{m_2}} \over {{m_1} + {m_2}}} \times {v^2}$$</p>
<p>$$ = {1 \over 2} \times {{9.8 \times 0.2} \over {10}} \times {(10)^2}$$</p>
<p>$$= 9.8$$ J</p> | mcq | jee-main-2022-online-25th-july-evening-shift | 13,321 |
1l6gn1t3i | physics | work-power-and-energy | energy | <p>As per the given figure, two blocks each of mass $$250 \mathrm{~g}$$ are connected to a spring of spring constant $$2 \,\mathrm{Nm}^{-1}$$. If both are given velocity $$v$$ in opposite directions, then maximum elongation of the spring is :</p>
<p><img src="data:image/png;base64,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"/></p> | [{"identifier": "A", "content": "$$\\frac{v}{2 \\sqrt{2}}$$"}, {"identifier": "B", "content": "$$\\frac{v}{2}$$"}, {"identifier": "C", "content": "$$\\frac{v}{4}$$"}, {"identifier": "D", "content": "$$\n\\frac{v}{\\sqrt{2}}\n$$"}] | ["B"] | null | <p>$$\because$$ Loss in KE = Gain in spring energy</p>
<p>$$ \Rightarrow {1 \over 2}m{v^2} \times 2 = {1 \over 2}kx_m^2$$</p>
<p>$$ \Rightarrow 2 \times {1 \over 4} \times {v^2} = 2 \times x_m^2$$</p>
<p>$$ \Rightarrow {x_m} = \sqrt {{{{v^2}} \over 4}} = {v \over 2}$$</p> | mcq | jee-main-2022-online-26th-july-morning-shift | 13,322 |
1l6mbpdo0 | physics | work-power-and-energy | energy | <p>A block of mass '$$\mathrm{m}$$' (as shown in figure) moving with kinetic energy E compresses a spring through a distance $$25 \mathrm{~cm}$$ when, its speed is halved. The value of spring constant of used spring will be $$\mathrm{nE} \,\,\mathrm{Nm}^{-1}$$ for $$\mathrm{n}=$$ _________.</p>
<p><img src="data:image/png;base64,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"/></p> | [] | null | 24 | <p>$$\Delta KE = {W_{all}}$$</p>
<p>So $${E \over 4} - E = - {1 \over 2}K \times {(0.25)^2}$$</p>
<p>$${{3E} \over 4} = {1 \over 2}K \times {1 \over {16}}$$</p>
<p>$$ = K = 24E$$</p> | integer | jee-main-2022-online-28th-july-morning-shift | 13,323 |
1l6nqgokb | physics | work-power-and-energy | energy | <p>A bullet of mass $$200 \mathrm{~g}$$ having initial kinetic energy $$90 \mathrm{~J}$$ is shot inside a long swimming pool as shown in the figure. If it's kinetic energy reduces to $$40 \mathrm{~J}$$ within $$1 \mathrm{~s}$$, the minimum length of the pool, the bullet has to travel so that it completely comes to rest is</p>
<p><img src="data:image/png;base64,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"/></p> | [{"identifier": "A", "content": "45 m"}, {"identifier": "B", "content": "90 m"}, {"identifier": "C", "content": "125 m"}, {"identifier": "D", "content": "25 m"}] | ["A"] | null | <p>$${1 \over 2}m{x^2} = 90$$</p>
<p>$$ \Rightarrow {1 \over 2} \times 0.2 \times {x^2} = 90$$,</p>
<p>$${x^2} = 900$$</p>
<p>$$x = 30$$ m/s</p>
<p>$${1 \over 2}m{v^2} = 40 \Rightarrow v = {2 \over 3} \times 30 = 20$$ m/s</p>
<p>$$20 = 30 - a \times 1 \Rightarrow a = - 10$$ m/s<sup>2</sup></p>
<p>$$0 - {x^2} = 2as$$</p>
<p>$$s = {{{x^2}} \over { - 2a}} = {{30 \times 30} \over {2 \times 10}}$$</p>
<p>$$ = 45$$ m</p> | mcq | jee-main-2022-online-28th-july-evening-shift | 13,324 |
1l6p45uh5 | physics | work-power-and-energy | energy | <p>A ball is projected with kinetic energy E, at an angle of $$60^{\circ}$$ to the horizontal. The kinetic energy of this ball at the highest point of its flight will become :</p> | [{"identifier": "A", "content": "Zero"}, {"identifier": "B", "content": "$$\\frac{E}{2}$$"}, {"identifier": "C", "content": "$$\\frac{E}{4}$$"}, {"identifier": "D", "content": "E"}] | ["C"] | null | <p>$$K.E. = E = {1 \over 2}m{v^2}$$</p>
<p>at highest point</p>
<p>$$K.E' = {1 \over 2}m{v^2}{\cos ^2}\theta $$</p>
<p>$$ = {1 \over 2}m{v^2}\left( {{1 \over 4}} \right)$$</p>
<p>$$ = {E \over 4}$$</p> | mcq | jee-main-2022-online-29th-july-morning-shift | 13,325 |
1ldnyihho | physics | work-power-and-energy | energy | <p>A block is fastened to a horizontal spring. The block is pulled to a distance $$x=10 \mathrm{~cm}$$ from its equilibrium position (at $$x=0$$) on a frictionless surface from rest. The energy of the block at $$x=5$$ $$\mathrm{cm}$$ is $$0.25 \mathrm{~J}$$. The spring constant of the spring is ___________ $$\mathrm{Nm}^{-1}$$</p> | [] | null | 67 | <img src="https://app-content.cdn.examgoal.net/fly/@width/image/1leslapx4/9dbc9fcc-7365-43b5-b4d8-cbc16b99a228/e145b380-b9c9-11ed-9339-bd1a4f4808c2/file-1leslapx5.png?format=png" data-orsrc="https://app-content.cdn.examgoal.net/image/1leslapx4/9dbc9fcc-7365-43b5-b4d8-cbc16b99a228/e145b380-b9c9-11ed-9339-bd1a4f4808c2/file-1leslapx5.png" loading="lazy" style="max-width: 100%; height: auto; display: block; margin: 0px auto; max-height: 40vh;" alt="JEE Main 2023 (Online) 1st February Evening Shift Physics - Work Power & Energy Question 42 English Explanation 1">
<br>Spring energy at x = 10 cm,
<br><br>$$\mathrm{U}_{\mathrm{i}} =\frac{1}{2} \mathrm{kx}_0^2 $$
<br><br>Energy of the block at x = 10,
<br><br>$$\mathrm{~K}_{\mathrm{i}} =0$$
<br><br><img src="https://app-content.cdn.examgoal.net/fly/@width/image/1leslcgi9/aa4f5a9d-05d8-41ec-9010-c86a0d7f537d/119e6900-b9ca-11ed-ada3-a1b4b9e94835/file-1leslcgia.png?format=png" data-orsrc="https://app-content.cdn.examgoal.net/image/1leslcgi9/aa4f5a9d-05d8-41ec-9010-c86a0d7f537d/119e6900-b9ca-11ed-ada3-a1b4b9e94835/file-1leslcgia.png" loading="lazy" style="max-width: 100%; height: auto; display: block; margin: 0px auto; max-height: 40vh;" alt="JEE Main 2023 (Online) 1st February Evening Shift Physics - Work Power & Energy Question 42 English Explanation 2">
<br>Spring energy at x = 5 cm,
<br><br>$$\mathrm{U}_{\mathrm{f}}=\frac{1}{2} \mathrm{k}\left(\frac{\mathrm{x}_0}{2}\right)^2 $$
<br><br>Energy of the block at x = 5, (which is only kinetic energy, no potential energy of block presents as block is not moving in the vertical direction)
<br><br>$$
\mathrm{~K}_{\mathrm{f}}=0.25 \mathrm{~J} $$
<br><br>Applying energy conservation law,
<br><br>Initial energy of Spring + Initial energy of Block = Final energy of Spring + Final energy of Block
<br><br>$$
\frac{1}{2} \mathrm{kx}_0^2+0=\frac{1}{2} \mathrm{k} \frac{\mathrm{x}_0^2}{4}+0.25 $$
<br><br>$$
\frac{1}{2} \mathrm{kx}_0^2 \frac{3}{4}=\frac{1}{4} $$
<br><br>$$
\frac{1}{2} \mathrm{k} \frac{3}{100}=1 \Rightarrow \mathrm{k}=\frac{200}{3} \mathrm{~N} / \mathrm{m} $$
<br><br>$$
=67 \mathrm{~N} / \mathrm{m}
$$ | integer | jee-main-2023-online-1st-february-evening-shift | 13,326 |
1ldpmodsf | physics | work-power-and-energy | energy | <p>A lift of mass $$\mathrm{M}=500 \mathrm{~kg}$$ is descending with speed of $$2 \mathrm{~ms}^{-1}$$. Its supporting cable begins to slip thus allowing it to fall with a constant acceleration of $$2 \mathrm{~ms}^{-2}$$. The kinetic energy of the lift at the end of fall through to a distance of $$6 \mathrm{~m}$$ will be _____________ $$\mathrm{kJ}$$.</p> | [] | null | 7 | Given, $u=2 \mathrm{~m} / \mathrm{s}$
<br/><br/>$$
\begin{aligned}
& a=2 \mathrm{~m} / \mathrm{s}^{2} \\\\
& s=6 \mathrm{~m} \\\\
& v=? \\\\
& v^{2}=u^{2}+2 a s \\\\
& v^{2}=4+2 \times 2 \times 6 \\\\
& =28
\end{aligned}
$$
<br/><br/>So, $\mathrm{KE}=\frac{1}{2} m v^{2}=\frac{1}{2} \times 500 \times 28 \mathrm{~J}$
<br/><br/>$=7000 \mathrm{~J}$
<br/><br/>$=7 \mathrm{~kJ}$ | integer | jee-main-2023-online-31st-january-morning-shift | 13,327 |
1ldsojq5f | physics | work-power-and-energy | energy | <p>A stone is projected at angle $$30^{\circ}$$ to the horizontal. The ratio of kinetic energy of the stone at point of projection to its kinetic energy at the highest point of flight will be -</p> | [{"identifier": "A", "content": "1 : 4"}, {"identifier": "B", "content": "1 : 2"}, {"identifier": "C", "content": "4 : 3"}, {"identifier": "D", "content": "4 : 1"}] | ["C"] | null | <img src="https://app-content.cdn.examgoal.net/fly/@width/image/1lei2tyfx/a6b83f8a-8e8e-45f4-97f6-f032b2abb398/d44804d0-b401-11ed-bf7e-c52177c53cde/file-1lei2tyfy.png?format=png" data-orsrc="https://app-content.cdn.examgoal.net/image/1lei2tyfx/a6b83f8a-8e8e-45f4-97f6-f032b2abb398/d44804d0-b401-11ed-bf7e-c52177c53cde/file-1lei2tyfy.png" loading="lazy" style="max-width: 100%; height: auto; display: block; margin: 0px auto; max-height: 40vh;" alt="JEE Main 2023 (Online) 29th January Morning Shift Physics - Work Power & Energy Question 35 English Explanation"><br>$$
\mathrm{KE}_{\mathrm{in}}=\frac{1}{2} m v^{2}
$$<br><br>
$\mathrm{KE}_{\text {final }}=\frac{1}{2} m v^{2} \cos ^{2} 30^{\circ}=\frac{1}{2} m v^{2}\left(\frac{\sqrt{3}}{2}\right)^{2}$
<br><br>
$\frac{\mathrm{KE}_{\mathrm{in}}}{\mathrm{KE}_{\mathrm{f}}}=\frac{\frac{1}{2} m v^{2}}{\frac{1}{2} m v^{2}\left(\frac{3}{4}\right)}=\frac{4}{3}$ | mcq | jee-main-2023-online-29th-january-morning-shift | 13,328 |
1ldsqc59z | physics | work-power-and-energy | energy | <p>A 0.4 kg mass takes 8s to reach ground when dropped from a certain height 'P' above surface of earth. The loss of potential energy in the last second of fall is __________ J.</p>
<p>(Take g = 10 m/s$$^2$$)</p> | [] | null | 300 | Displacement is $8^{\text {th }}$ sec.
<br/><br/>
$\mathrm{S}_{8}=0+\frac{1}{2} \times 10 \times(2 \times 8-1)$
<br/><br/>
$\mathbf{S}_{8}=5 \times 15$
<br/><br/>
$\Delta \mathrm{U}=0.4 \times 10 \times 5 \times 15$
<br/><br/>
$\Delta \mathrm{U}=20 \times 15=300$ | integer | jee-main-2023-online-29th-january-morning-shift | 13,329 |
1lduicszj | physics | work-power-and-energy | energy | <p>An object of mass 'm' initially at rest on a smooth horizontal plane starts moving under the action of force F = 2N. In the process of its linear motion, the angle $$\theta$$ (as shown in figure) between the direction of force and horizontal varies as $$\theta=\mathrm{k}x$$, where k is a constant and $$x$$ is the distance covered by the object from its initial position. The expression of kinetic energy of the object will be $$E = {n \over k}\sin \theta $$. The value of n is ___________.</p>
<p><img src="data:image/png;base64,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"/></p> | [] | null | 2 | <p><img src="https://app-content.cdn.examgoal.net/fly/@width/image/1le5klj1b/dfe257bf-4dfb-476c-ae29-8b55b8f8c437/1dca32f0-ad21-11ed-8bc1-d3bd0941e5b5/file-1le5klj1c.png?format=png" data-orsrc="https://app-content.cdn.examgoal.net/image/1le5klj1b/dfe257bf-4dfb-476c-ae29-8b55b8f8c437/1dca32f0-ad21-11ed-8bc1-d3bd0941e5b5/file-1le5klj1c.png" loading="lazy" style="max-width: 100%; height: auto; display: block; margin: 0px auto; max-height: 40vh;" alt="JEE Main 2023 (Online) 25th January Morning Shift Physics - Work Power & Energy Question 34 English Explanation"></p>
$$
\text { Work done }=\Delta \mathrm{K} . \mathrm{E}
$$
<br><br>
$$
\begin{aligned}
& \therefore \int F \cdot d x=\frac{1}{2} m v^{2}=E \\\\
& \therefore E=\int_{0}^{x} 2 \cos (k x) d x \\\\
& E=\frac{2}{k}[\sin k x]_{0}^{x} \\\\
& =\frac{2}{k} \sin k x \\\\
& =\frac{2 \sin \theta}{k}
\end{aligned}
$$ | integer | jee-main-2023-online-25th-january-morning-shift | 13,330 |
1ldyf6ppy | physics | work-power-and-energy | energy | <p>A spherical body of mass 2 kg starting from rest acquires a kinetic energy of 10000 J at the end of $$\mathrm{5^{th}}$$ second. The force acted on the body is ________ N.</p> | [] | null | 40 | Let the force be $F$ so acceleration $a=\frac{F}{m}$
<br/><br/>
So displacement $S=\frac{1}{2} a t^{2}=\frac{F t^{2}}{2 m}$
<br/><br/>
So work done $W=F . S=\frac{F^{2} t^{2}}{2 m}$
<br/><br/>
From work energy Theorem
<br/><br/>
$\Delta K E=W$
<br/><br/>
$W=\frac{F^{2} t^{2}}{2 m}=10000$
<br/><br/>
$F=\sqrt{\frac{10000 \times 2 \times 2}{5^{2}}}$
<br/><br/>
$F=40 \mathrm{~N}$ | integer | jee-main-2023-online-24th-january-morning-shift | 13,331 |
1lgp17i3v | physics | work-power-and-energy | energy | <p>A car accelerates from rest to $$u \mathrm{~m} / \mathrm{s}$$. The energy spent in this process is E J. The energy required to accelerate the car from $$u \mathrm{~m} / \mathrm{s}$$ to $$2 \mathrm{u} \mathrm{m} / \mathrm{s}$$ is $$\mathrm{nE~J}$$. The value of $$\mathrm{n}$$ is ____________.</p> | [] | null | 3 | The kinetic energy of a moving object of mass $$m$$ and velocity $$v$$ is given by the formula:
<br/><br/>
$$K = \frac{1}{2}mv^2$$
<br/><br/>
The work done in accelerating an object from rest to velocity $$v$$ is equal to its change in kinetic energy. Therefore, the energy spent in accelerating the car from rest to $$u \mathrm{~m}/\mathrm{s}$$ is:
<br/><br/>
$$E = \frac{1}{2}mu^2$$
<br/><br/>
The energy required to accelerate the car from $$u \mathrm{~m}/\mathrm{s}$$ to $$2u \mathrm{~m}/\mathrm{s}$$ is:
<br/><br/>
$$\begin{aligned} nE &= \frac{1}{2}m(2u)^2 - \frac{1}{2}mu^2 \\\\ &= 2mu^2 - \frac{1}{2}mu^2 \\\\ &= \frac{3}{2}mu^2 \\\\
&= 3E \end{aligned}
$$
<br/><br/>$$ \therefore $$ n = 3
| integer | jee-main-2023-online-13th-april-evening-shift | 13,332 |
1lgq2uj1c | physics | work-power-and-energy | energy | <p>Two bodies are having kinetic energies in the ratio 16 : 9. If they have same linear momentum, the ratio of their masses respectively is :</p> | [{"identifier": "A", "content": "$$3: 4$$"}, {"identifier": "B", "content": "$$4: 3$$"}, {"identifier": "C", "content": "$$9: 16$$"}, {"identifier": "D", "content": "$$16: 9$$"}] | ["C"] | null | The kinetic energy of a body of mass $m$ and velocity $v$ is given by $K=\frac{1}{2}mv^2$. Since the bodies have the same linear momentum, we can write:
<br/><br/>
$$p=mv$$
<br/><br/>
where $p$ is the linear momentum of the bodies.
<br/><br/>
Let the masses of the two bodies be $m_1$ and $m_2$ and their kinetic energies be $K_1$ and $K_2$, respectively. Then, we have:
<br/><br/>
$$\frac{K_1}{K_2}=\frac{16}{9}$$
<br/><br/>
$$\frac{1}{2}m_1v_1^2\div\frac{1}{2}m_2v_2^2=\frac{16}{9}$$
<br/><br/>
Since $p=mv$, we have $v_1=\frac{p}{m_1}$ and $v_2=\frac{p}{m_2}$. Substituting these in the above equation, we get:
<br/><br/>
$$\frac{m_2}{m_1}=\frac{9}{16}$$
<br/><br/>
Therefore, the ratio of the masses of the two bodies is $\boxed{9:16}$. | mcq | jee-main-2023-online-13th-april-morning-shift | 13,333 |
1lgrh1w3m | physics | work-power-and-energy | energy | <p>Given below are two statements:</p>
<p>Statement I : A truck and a car moving with same kinetic energy are brought to rest by applying breaks which provide equal retarding forces. Both come to rest in equal distance.</p>
<p>Statement II : A car moving towards east takes a turn and moves towards north, the speed remains unchanged. The acceleration of the car is zero.</p>
<p>In the light of given statements, choose the most appropriate answer from the options given below</p> | [{"identifier": "A", "content": "Statement I is incorrect but Statement II is correct."}, {"identifier": "B", "content": "Statement $$\\mathrm{I}$$ is correct but Statement II is incorrect."}, {"identifier": "C", "content": "Both Statement I and Statement II are correct."}, {"identifier": "D", "content": "Both Statement I and Statement II are incorrect."}] | ["B"] | null | Statement I is correct: The kinetic energy of an object is given by $\frac{1}{2}mv^2$, where m is the mass of the object and v is its velocity. If a truck and a car are moving with the same kinetic energy and are brought to rest by applying brakes that provide equal retarding forces, both will come to rest in equal distances. This is because the distance required to stop an object depends on its initial kinetic energy and the force applied to bring it to rest. Since both the truck and car have the same initial kinetic energy and are subjected to the same retarding force, they will come to rest in the same distance.
<br/><br/>
Statement II is incorrect. When the car moves from east to north, even though its speed remains unchanged, its direction changes. Since velocity is a vector quantity that has both magnitude (speed) and direction, a change in direction implies a change in velocity. Acceleration is the rate of change of velocity, so when the velocity changes, there is acceleration. In this case, the car's acceleration is not zero as it turns from east to north. | mcq | jee-main-2023-online-12th-april-morning-shift | 13,334 |
1lgyr7u5w | physics | work-power-and-energy | energy | <p>A body of mass $$5 \mathrm{~kg}$$ is moving with a momentum of $$10 \mathrm{~kg} \mathrm{~ms}^{-1}$$. Now a force of $$2 \mathrm{~N}$$ acts on the body in the direction of its motion for $$5 \mathrm{~s}$$. The increase in the Kinetic energy of the body is ___________ $$\mathrm{J}$$.</p> | [] | null | 30 | <p>The increase in kinetic energy can be found using the work-energy theorem, which states that the work done on an object is equal to the change in its kinetic energy.</p>
<p>The work done by a force is given by the equation:</p>
<p>$ W = F \cdot d $</p>
<p>where ( F ) is the force and ( d ) is the distance over which the force is applied. </p>
<p>However, we don't have the distance in this problem. But we do know that the force is applied for a time of 5 seconds, and that the initial momentum of the body is 10 kg m/s. We can use these facts to find the work done.</p>
<p>First, we can use the equation for force, ( F = ma ), to find the acceleration of the body:</p>
<p>$a = \frac{F}{m} = \frac{2 \, \text{N}}{5 \, \text{kg}} = 0.4 \, \text{m/s}^2 $</p>
<p>Then, we can use the equation for distance in uniformly accelerated motion, ( $d = v_i t + \frac{1}{2} a t^2 $), where ( $v_i$ ) is the initial velocity of the body. We can find ( $v_i $) from the initial momentum and the mass of the body:</p>
<p>$ v_i = \frac{p}{m} = \frac{10 \, \text{kg m/s}}{5 \, \text{kg}} = 2 \, \text{m/s} $</p>
<p>Substituting ( $v_i$ ), ( a ), and ( t ) into the equation for ( d ) gives:</p>
<p>$ d = 2 \, \text{m/s} \cdot 5 \, \text{s} + \frac{1}{2} \cdot 0.4 \, \text{m/s}^2 \cdot (5 \, \text{s})^2 = 10 \, \text{m} + 5 \, \text{m} = 15 \, \text{m} $</p>
<p>Finally, we can substitute ( F ) and ( d ) into the equation for work to find the increase in kinetic energy:</p>
<p>$ \Delta KE = W = F \cdot d = 2 \, \text{N} \cdot 15 \, \text{m} = 30 \, \text{J} $</p>
<p>So, the increase in the kinetic energy of the body is 30 J.</p>
| integer | jee-main-2023-online-8th-april-evening-shift | 13,335 |
1lh265upm | physics | work-power-and-energy | energy | <p>A particle of mass $$10 \mathrm{~g}$$ moves in a straight line with retardation $$2 x$$, where $$x$$ is the displacement in SI units. Its loss of kinetic energy for above displacement is $$\left(\frac{10}{x}\right)^{-n}$$ J. The value of $$\mathrm{n}$$ will be __________</p> | [] | null | 2 | <p>The work done against the retarding force is indeed equal to the loss in kinetic energy. </p>
<p>The force acting on the particle due to retardation is given by $F = ma = -2mx$. </p>
<p>When we integrate this force over the displacement from $0$ to $x$, we get:</p>
<p>$$\Delta KE = W = \int F \cdot dx = \int (-2mx) \, dx = -mx^2$$</p>
<p>The negative sign indicates that this is a loss of kinetic energy. </p>
<p>The problem states that the loss in kinetic energy is also given by $\left(\frac{10}{x}\right)^{-n}$ J. Therefore, we have:</p>
<p>$$-mx^2 = \left(\frac{10}{x}\right)^{-n}$$</p>
<p>Because this is a loss of kinetic energy, we should consider the absolute value. Hence,</p>
<p>$$mx^2 = \left(\frac{10}{x}\right)^{-n}$$</p>
<p>Substituting the given mass $m = 10 \, \text{g} = 0.01 \, \text{kg}$, we get:</p>
<p>$$0.01x^2 = \left(\frac{10}{x}\right)^{-n}$$</p>
<p>This simplifies to:</p>
<p>$$x^2 = \left(\frac{10}{x}\right)^{-n}$$</p>
<p>Comparing the two sides, we can see that $n = 2$. </p> | integer | jee-main-2023-online-6th-april-morning-shift | 13,336 |
jaoe38c1lscpdeey | physics | work-power-and-energy | energy | <p>A bullet is fired into a fixed target looses one third of its velocity after travelling $$4 \mathrm{~cm}$$. It penetrates further $$\mathrm{D} \times 10^{-3} \mathrm{~m}$$ before coming to rest. The value of $$\mathrm{D}$$ is :</p> | [{"identifier": "A", "content": "23"}, {"identifier": "B", "content": "32"}, {"identifier": "C", "content": "42"}, {"identifier": "D", "content": "52"}] | ["B"] | null | <p>$$\begin{aligned}
& v^2-u^2=2 a S \\
& \left(\frac{2 u}{3}\right)^2=u^2+2(-a)\left(4 \times 10^{-2}\right) \\
& \frac{4 u^2}{9}=u^2-2 a\left(4 \times 10^{-2}\right) \\
& -\frac{5 u^2}{9}=-2 a\left(4 \times 10^{-2}\right) \ldots(1) \\
& 0=\left(\frac{2 u}{3}\right)^2+2(-a)(x) \\
& -\frac{4 u^2}{9}=-2 a x \ldots(2)
\end{aligned}$$</p>
<p>$$(1)/(2)$$</p>
<p>$$\begin{aligned}
& \frac{5}{4}=\frac{4 \times 10^{-2}}{\mathrm{x}} \\
& \mathrm{x}=\frac{16}{5} \times 10^{-2} \\
& \mathrm{x}=3 \cdot 2 \times 10^{-2} \mathrm{~m} \\
& \mathrm{x}=32 \times 10^{-3} \mathrm{~m}
\end{aligned}$$</p> | mcq | jee-main-2024-online-27th-january-evening-shift | 13,338 |
jaoe38c1lsf1whjx | physics | work-power-and-energy | energy | <p>The potential energy function (in $$J$$ ) of a particle in a region of space is given as $$U=\left(2 x^2+3 y^3+2 z\right)$$. Here $$x, y$$ and $$z$$ are in meter. The magnitude of $$x$$-component of force (in $$N$$ ) acting on the particle at point $$P(1,2,3) \mathrm{m}$$ is :</p> | [{"identifier": "A", "content": "4"}, {"identifier": "B", "content": "2"}, {"identifier": "C", "content": "8"}, {"identifier": "D", "content": "6"}] | ["A"] | null | <p>$$\begin{aligned}
& \text { Given } U=2 x^2+3 y^3+2 z \\
& F_x=-\frac{\partial U}{\partial x}=-4 x
\end{aligned}$$</p>
<p>At $$x=1$$ magnitude of $$F_x$$ is $$4 N$$</p> | mcq | jee-main-2024-online-29th-january-morning-shift | 13,339 |
jaoe38c1lsflp1g7 | physics | work-power-and-energy | energy | <p>A bob of mass '$$m$$' is suspended by a light string of length '$$L$$'. It is imparted a minimum horizontal velocity at the lowest point $$A$$ such that it just completes half circle reaching the top most position B. The ratio of kinetic energies $$\frac{(K . E)_A}{(K . E)_B}$$ is :</p>
<p><img src="data:image/png;base64,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"/></p> | [{"identifier": "A", "content": "5 : 1"}, {"identifier": "B", "content": "3 : 2"}, {"identifier": "C", "content": "1 : 5"}, {"identifier": "D", "content": "2 : 5"}] | ["A"] | null | <p>Apply energy conservation between A & B</p>
<p>$$\begin{aligned}
& \frac{1}{2} \mathrm{mV}_{\mathrm{L}}^2=\frac{1}{2} \mathrm{mV}_{\mathrm{H}}^2+\mathrm{mg}(2 \mathrm{~L}) \\
& \because \mathrm{V}_{\mathrm{L}}=\sqrt{5 \mathrm{gL}}
\end{aligned}$$</p>
<p>So, $$\mathrm{V}_{\mathrm{H}}=\sqrt{\mathrm{gL}}$$</p>
<p>$$\frac{(\mathrm{K} . \mathrm{E})_{\mathrm{A}}}{(\mathrm{K} . \mathrm{E})_{\mathrm{B}}}=\frac{\frac{1}{2} \mathrm{~m}(\sqrt{5 \mathrm{gL}})^2}{\frac{1}{2} \mathrm{~m}(\sqrt{\mathrm{gL}})^2}=\frac{5}{1}$$</p> | mcq | jee-main-2024-online-29th-january-evening-shift | 13,340 |
1lsgd50wc | physics | work-power-and-energy | energy | <p>A particle is placed at the point $$A$$ of a frictionless track $$A B C$$ as shown in figure. It is gently pushed towards right. The speed of the particle when it reaches the point B is :<br/><br/> (Take $$g=10 \mathrm{~m} / \mathrm{s}^2$$).</p>
<p><img src="data:image/png;base64,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"/></p> | [{"identifier": "A", "content": "$$2 \\sqrt{10} \\mathrm{~m} / \\mathrm{s}$$\n"}, {"identifier": "B", "content": "$$10 \\mathrm{~m} / \\mathrm{s}$$\n"}, {"identifier": "C", "content": "$$\\sqrt{10} \\mathrm{~m} / \\mathrm{s}$$\n"}, {"identifier": "D", "content": "$$20 \\mathrm{~m} / \\mathrm{s}$$"}] | ["C"] | null | <p>By COME</p>
<p>$$\begin{aligned}
& \mathrm{KE}_{\mathrm{A}}+\mathrm{U}_{\mathrm{A}}=\mathrm{KE}_{\mathrm{B}}+\mathrm{U}_{\mathrm{B}} \\
& 0+\mathrm{mg}(1)=\frac{1}{2} \mathrm{mv}^2+\mathrm{mg} \times 0.5 \\
& v=\sqrt{g}=\sqrt{10} \mathrm{~m} / \mathrm{s}
\end{aligned}$$</p> | mcq | jee-main-2024-online-30th-january-morning-shift | 13,341 |
lv0vxlra | physics | work-power-and-energy | energy | <p>If a rubber ball falls from a height $$h$$ and rebounds upto the height of $$h / 2$$. The percentage loss of total energy of the initial system as well as velocity ball before it strikes the ground, respectively, are :</p> | [{"identifier": "A", "content": "$$50 \\%, \\sqrt{2 \\mathrm{gh}}$$\n"}, {"identifier": "B", "content": "$$50 \\%, \\sqrt{\\mathrm{gh}}$$\n"}, {"identifier": "C", "content": "$$50 \\%, \\sqrt{\\frac{\\text { gh }}{2}}$$\n"}, {"identifier": "D", "content": "$$40 \\%, \\sqrt{2 \\mathrm{gh}}$$"}] | ["A"] | null | <p>To solve this problem, we need to analyze both the energy loss and the initial velocity of the rubber ball before it strikes the ground.</p>
<p>First, let's consider the energy loss. The energy involved here is gravitational potential energy. The initial potential energy of the ball when it is about to fall is given by $$U_i = mgh$$, where $$U_i$$ is the initial potential energy, $$m$$ is the mass of the ball, $$g$$ is the acceleration due to gravity, and $$h$$ is the initial height from which the ball falls. After the ball rebounds, it reaches a height of $$h/2$$. The potential energy at this new height is $$U_f = mg \cdot \frac{h}{2}$$.</p>
<p>The energy loss can be calculated as the difference between the initial and final potential energies, and to find the percentage energy loss, we divide this difference by the initial energy and multiply by 100:</p>
<p>$$\text{Energy loss percentage} = \frac{(U_i - U_f)}{U_i} \times 100$$</p>
<p>Substituting the values of $$U_i$$ and $$U_f$$ gives:</p>
<p>$$\text{Energy loss percentage} = \frac{(mgh - mg\frac{h}{2})}{mgh} \times 100$$</p>
<p>By simplifying, we find:</p>
<p>$$\text{Energy loss percentage} = \frac{mgh - \frac{1}{2} mgh}{mgh} \times 100 = \frac{1}{2} \times 100 = 50\%$$</p>
<p>This tells us that the energy loss percentage is indeed $$50\%$$.</p>
<p>Next, we'll find the velocity of the ball just before it strikes the ground. The velocity can be determined using the formula for the velocity of an object in free fall:</p>
<p>$$v = \sqrt{2gh}$$</p>
<p>Here, $$v$$ is the velocity of the ball just before impact, $$g$$ is the acceleration due to gravity, and $$h$$ is the height from which the ball falls. This formula shows that the initial velocity of the ball before it strikes the ground is $$\sqrt{2gh}$$, not taking into account air resistance and assuming it starts from rest.</p>
<p>Therefore, the correct answer is <strong>Option A</strong>: $$50\%$$, $$\sqrt{2gh}$$.</p> | mcq | jee-main-2024-online-4th-april-morning-shift | 13,342 |
lv2es437 | physics | work-power-and-energy | energy | <p>A body of $$m \mathrm{~kg}$$ slides from rest along the curve of vertical circle from point $$A$$ to $$B$$ in friction less path. The velocity of the body at $$B$$ is:</p>
<p><img src="data:image/png;base64,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"/></p>
<p>(given, $$R=14 \mathrm{~m}, g=10 \mathrm{~m} / \mathrm{s}^2$$ and $$\sqrt{2}=1.4$$)</p> | [{"identifier": "A", "content": "10.6 m/s"}, {"identifier": "B", "content": "19.8 m/s"}, {"identifier": "C", "content": "16.7 m/s"}, {"identifier": "D", "content": "21.9 m/s"}] | ["D"] | null | <p>From energy conservation $$\rightarrow$$</p>
<p>$$\begin{aligned}
& m g(R+R \sin 45)=\frac{1}{2} m v^2 \\
& \Rightarrow 10\left(1+\frac{1}{\sqrt{2}}\right) \times 14=\frac{1}{2} v^2 \\
& \Rightarrow 10\left(1+\frac{\sqrt{2}}{2}\right) \times 28=v^2 \\
& \Rightarrow 10(1+0.7) \times 28=v^2 \\
& \Rightarrow v=21.81
\end{aligned}$$</p> | mcq | jee-main-2024-online-4th-april-evening-shift | 13,343 |
lvb29egr | physics | work-power-and-energy | energy | <p>When kinetic energy of a body becomes 36 times of its original value, the percentage increase in the momentum of the body will be :</p> | [{"identifier": "A", "content": "60%"}, {"identifier": "B", "content": "500%"}, {"identifier": "C", "content": "6%"}, {"identifier": "D", "content": "600%"}] | ["B"] | null | <p>The relationship between kinetic energy (K.E) and momentum (p) of a body can be expressed through their respective definitions. Kinetic energy is given by $$K.E = \frac{1}{2} mv^2$$ where $m$ is the mass of the body and $v$ is its velocity. The momentum (p) of a body is given by $$p = mv$$. To express kinetic energy in terms of momentum, we can manipulate the expression for momentum as follows:</p>
<p>$$p = mv \implies v = \frac{p}{m}$$</p>
<p>Substituting $v$ in the kinetic energy formula, we get</p>
<p>$$K.E = \frac{1}{2} m\left(\frac{p}{m}\right)^2 = \frac{1}{2} \frac{p^2}{m}$$</p>
<p>Therefore, we see that kinetic energy is directly proportional to the square of the momentum $(K.E \propto p^2)$.</p>
<p>Now, given that the kinetic energy of a body becomes 36 times its original value, we can set up the proportionality as</p>
<p>$$\frac{K.E_{\text{final}}}{K.E_{\text{original}}} = 36$$</p>
<p>Since $K.E_{\text{final}} = 36 \times K.E_{\text{original}}$ and knowing $K.E \propto p^2$, we can express this relationship through the squares of the initial and final momentum:</p>
<p>$$\frac{p_{\text{final}}^2}{p_{\text{original}}^2} = 36$$</p>
<p>Taking the square root of both sides to find the ratio of final to initial momentum, we have</p>
<p>$$\frac{p_{\text{final}}}{p_{\text{original}}} = \sqrt{36} = 6$$</p>
<p>This indicates that the final momentum is 6 times the original momentum. To find the percentage increase in the momentum, we calculate the increase from the original to the final, subtracting the original momentum (which is considered 1 times itself):</p>
<p>$$\text{Percentage increase} = \left(\frac{p_{\text{final}} - p_{\text{original}}}{p_{\text{original}}}\right) \times 100\% = \left(\frac{6p - p}{p}\right) \times 100\%
= \left(6 - 1\right) \times 100\% = 5 \times 100\% = 500\%$$</p>
<p>Therefore, the correct answer is Option B: 500%.</p> | mcq | jee-main-2024-online-6th-april-evening-shift | 13,345 |
lvc586c4 | physics | work-power-and-energy | energy | <p>A bullet of mass $$50 \mathrm{~g}$$ is fired with a speed $$100 \mathrm{~m} / \mathrm{s}$$ on a plywood and emerges with $$40 \mathrm{~m} / \mathrm{s}$$. The percentage loss of kinetic energy is :</p> | [{"identifier": "A", "content": "$$44 \\%$$\n"}, {"identifier": "B", "content": "$$16 \\%$$\n"}, {"identifier": "C", "content": "$$84 \\%$$\n"}, {"identifier": "D", "content": "$$32 \\%$$"}] | ["C"] | null | <p>To find the percentage loss of kinetic energy of the bullet, we first calculate the initial kinetic energy before the bullet hits the plywood and the final kinetic energy after it emerges. The formula for kinetic energy (KE) is given by:</p>
<p>$$KE = \frac{1}{2} mv^2$$</p>
<p>where $m$ is the mass of the object and $v$ is its velocity.</p>
<p>Let's calculate the initial and final kinetic energies.</p>
<p><strong>Initial Kinetic Energy:</strong></p>
<p>$$KE_{\text{initial}} = \frac{1}{2} \times 50 \times (100)^2 = \frac{1}{2} \times 50 \times 10000 = 25 \times 10000 = 250000 \, \text{g.m}^2/\text{s}^2$$</p>
<p>Note: To keep units consistent, we used grams and meters per second. We can also convert the mass to kilograms (by dividing by 1000) which would result in the energy being calculated in Joules, but for the purpose of finding the percentage change, the form of units does not matter as long as they are consistent, since it will be a ratio.</p>
<p><strong>Final Kinetic Energy:</strong></p>
<p>$$KE_{\text{final}} = \frac{1}{2} \times 50 \times (40)^2 = \frac{1}{2} \times 50 \times 1600 = 25 \times 1600 = 40000 \, \text{g.m}^2/\text{s}^2$$</p>
<p>The loss of kinetic energy is then:</p>
<p>$$\Delta KE = KE_{\text{initial}} - KE_{\text{final}} = 250000 - 40000 = 210000 \, \text{g.m}^2/\text{s}^2$$</p>
<p>Finally, the percentage loss of kinetic energy can be calculated using the formula:</p>
<p>$$\text{Percentage loss of KE} = \left( \frac{\Delta KE}{KE_{\text{initial}}} \right) \times 100\%$$</p>
<p>$$\text{Percentage loss of KE} = \left( \frac{210000}{250000} \right) \times 100\% = 0.84 \times 100\% = 84\%$$</p>
<p>Thus, the percentage loss of kinetic energy is <strong>84%</strong>, which corresponds to <strong>Option C</strong>.</p> | mcq | jee-main-2024-online-6th-april-morning-shift | 13,346 |
snApUKYYNFM6rlsV | physics | work-power-and-energy | power | A body is moved along a straight line by a machine delivering a constant power. The distance moved by the body in time $$'t'$$ is proportional to | [{"identifier": "A", "content": "$${t^{3/4}}$$ "}, {"identifier": "B", "content": "$${t^{3/2}}$$"}, {"identifier": "C", "content": "$${t^{1/4}}$$"}, {"identifier": "D", "content": "$${t^{1/2}}$$"}] | ["B"] | null | We know that $$F \times v = $$ Power
<br><br>According to the question, power is constant.
<br><br>$$\therefore$$ $$F \times v = c\,\,\,\,$$ where $$c=$$ constant
<br><br>$$\therefore$$ $$m{{dv} \over {dt}} \times v = c$$ $$\,\,\,\,\left( \, \right.$$ $$\therefore$$ $$\left. {F = ma = {{mdv} \over {dt}}\,\,} \right)$$
<br><br>$$\therefore$$ $$m\int\limits_0^v {vdv = c\int\limits_0^t {dt} } \,\,\,\,\,\,\,\,\,$$ $$\therefore$$ $${1 \over 2}m{v^2} = ct$$
<br><br>$$\therefore$$ $$v = \sqrt {{{2c} \over m}} \times {t^{{\raise0.5ex\hbox{$\scriptstyle 1$}
\kern-0.1em/\kern-0.15em
\lower0.25ex\hbox{$\scriptstyle 2$}}}}$$
<br><br>$${{dx} \over {dt}} = \sqrt {{{2c} \over m}} \times {t^{{\raise0.5ex\hbox{$\scriptstyle 1$}
\kern-0.1em/\kern-0.15em
\lower0.25ex\hbox{$\scriptstyle 2$}}}}\,\,\,\,$$ where $$v = {{dx} \over {dt}}$$
<br><br>$$\therefore$$ $$\int\limits_0^x {dx = \sqrt {{{2c} \over m}} } \times \int\limits_0^t {{t^{{\raise0.5ex\hbox{$\scriptstyle 1$}
\kern-0.1em/\kern-0.15em
\lower0.25ex\hbox{$\scriptstyle 2$}}}}} dt$$
<br>$$x = \sqrt {{{2c} \over m}} \times {{2{t^{{\raise0.5ex\hbox{$\scriptstyle 3$}
\kern-0.1em/\kern-0.15em
\lower0.25ex\hbox{$\scriptstyle 2$}}}}} \over 3} \Rightarrow x \propto {t^{{\raise0.5ex\hbox{$\scriptstyle 3$}
\kern-0.1em/\kern-0.15em
\lower0.25ex\hbox{$\scriptstyle 2$}}}}$$ | mcq | aieee-2003 | 13,348 |
Y3CQEw8N0K3fp45s | physics | work-power-and-energy | power | A body of mass $$' m ',$$ acceleration uniformly from rest to $$'{v_1}'$$ in time $${T}$$. The instantaneous power delivered to the body as a function of time is given by | [{"identifier": "A", "content": "$${{m{v_1}{t^2}} \\over {{T}}}$$ "}, {"identifier": "B", "content": "$${{mv_1^2t} \\over {T^2}}$$ "}, {"identifier": "C", "content": "$${{m{v_1}t} \\over {{T}}}$$ "}, {"identifier": "D", "content": "$${{mv_1^2t} \\over {{T}}}$$ "}] | ["B"] | null | Assume acceleration of body be $$a$$
<br><br>$$\therefore$$ $${v_1} = 0 + a{T} \Rightarrow a = {{{v_1}} \over {{T}}}$$
<br><br>$$\therefore$$ $$v = at \Rightarrow v = {{{v_1}t} \over {{T}}}$$
<br><br>$${P_{inst}} = \overrightarrow F .\overrightarrow v = \left( {m\overrightarrow a } \right).\overrightarrow v $$
<br><br>$$= \left( {{{m{v_1}} \over {{T}}}} \right)\left( {{{{v_1}t} \over {{T}}}} \right)$$
<br><br>$$ = m{\left( {{{{v_1}} \over {{T}}}} \right)^2}t$$ | mcq | aieee-2004 | 13,349 |
Myw55b51DroWYrEJ | physics | work-power-and-energy | power | A body of mass $$m$$ is accelerated uniformly from rest to a speed $$v$$ in a time $$T.$$ The instantaneous power delivered to the body as a function of time is given by | [{"identifier": "A", "content": "$${{m{v^2}} \\over {{T^2}}}.{t^2}$$ "}, {"identifier": "B", "content": "$${{m{v^2}} \\over {{T^2}}}.t$$ "}, {"identifier": "C", "content": "$${1 \\over 2}{{m{v^2}} \\over {{T^2}}}.{t^2}$$ "}, {"identifier": "D", "content": "$${1 \\over 2}{{m{v^2}} \\over {{T^2}}}.t$$ "}] | ["B"] | null | $$u = 0;v = u + aT;v = aT$$
<br><br>Instantaneous power $$ = F \times v = m.\,a.\,at = m.{a^2}.t$$
<br><br>$$\therefore$$ Instantaneous power $$ = {{m{v^2}t} \over {{T^2}}}$$ | mcq | aieee-2005 | 13,350 |
NdfqSU4oZXegcvY0up9KF | physics | work-power-and-energy | power | A car of weight W is on an inclined road that rises by 100 m over a distance of 1 km
and applies a constant frictional force $${W \over 20}$$ on the car. While moving uphill on the road at a speed of 10 ms<sup>−1</sup>, the car needs power P. If it needs power $${p \over 2}$$ while moving downhill at speed v then value of $$\upsilon $$ is : | [{"identifier": "A", "content": "20 ms<sup>$$-$$1</sup>"}, {"identifier": "B", "content": "15 ms<sup>$$-$$1</sup>"}, {"identifier": "C", "content": "10 ms<sup>$$-$$1</sup>"}, {"identifier": "D", "content": "5 ms<sup>$$-$$1</sup>"}] | ["B"] | null | Here, tan$$\theta $$ = $${{100} \over {1000}} = {1 \over {10}}$$
<br><br>$$ \therefore $$ sin$$\theta $$ = $${1 \over {10}}$$ (as $$\theta $$ is very small),
<br><br>when car is moving uphill :
<br><br><img src="https://res.cloudinary.com/dckxllbjy/image/upload/v1734264844/exam_images/w4ba1kfvqtexl7o03gba.webp" style="max-width: 100%; height: auto;display: block;margin: 0 auto;" loading="lazy" alt="JEE Main 2016 (Online) 9th April Morning Slot Physics - Work Power & Energy Question 94 English Explanation 1">
<br><br>P = f $$ \times $$ u
<br><br>= (wsin$$\theta $$ + f) $$ \times $$ u
<br><br>= $$\left( {{w \over {10}} + {w \over {20}}} \right) \times 10$$
<br><br>P = $${{3w} \over {20}} \times 10$$ = $${{3w} \over 2}$$
<br><br>When car is moving down hill :
<br><br><img src="https://res.cloudinary.com/dckxllbjy/image/upload/v1734265860/exam_images/vcg0fso4anwqfxzos40i.webp" style="max-width: 100%; height: auto;display: block;margin: 0 auto;" loading="lazy" alt="JEE Main 2016 (Online) 9th April Morning Slot Physics - Work Power & Energy Question 94 English Explanation 2">
<br><br>$$ \therefore $$ $${P \over 2}$$ = (wsin$$\theta $$ $$-$$ f) $$ \times $$ v
<br><br>$$ \Rightarrow $$ $${{3w} \over 4}$$ = $$\left( {{w \over {10}} - {w \over {20}}} \right)$$ $$ \times $$ v
<br><br>$$ \Rightarrow $$ $${{w \over {20}} \times }$$ v = $${{3w} \over 4}$$
<br><br>$$ \Rightarrow $$ v = 15 m/s | mcq | jee-main-2016-online-9th-april-morning-slot | 13,351 |
ixRSOh2PSaBUBL6v437k9k2k5dgg4fj | physics | work-power-and-energy | power | A 60 HP electric motor lifts an elevator having a maximum total load capacity of 2000 kg. If the frictional force on the elevator is 4000 N, the speed of the elevator at full load is close to : <br/>(1 HP = 746 W, g = 10 ms<sup>-2</sup>) | [{"identifier": "A", "content": "1.5 ms<sup>-1</sup>"}, {"identifier": "B", "content": "1.7 ms<sup>-1</sup>"}, {"identifier": "C", "content": "2.0 ms<sup>-1</sup>"}, {"identifier": "D", "content": "1.9 ms<sup>-1</sup>"}] | ["D"] | null | <br><br>F = mg + f
<br><br>F = 20000 + 4000 = 24000 N
<br><br>We know, Power(P) = Fv
<br><br>$$ \Rightarrow $$ v = $${P \over F}$$ = $${{60 \times 746} \over {24000}}$$
<br><br>$$ \Rightarrow $$ v $$ \approx $$ 1.9 m/s | mcq | jee-main-2020-online-7th-january-morning-slot | 13,353 |
rgQmR27Xkry0uATPBi7k9k2k5f7e86q | physics | work-power-and-energy | power | An elevator in a building can carry a maximum of 10 persons, with the average mass of each
person being 68 kg, The mass of the elevator itself is 920 kg and it moves with a constant speed
of 3 m/s. The frictional force opposing the motion is 6000 N. If the elevator is moving up with its
full capacity, the power delivered by the motor to the elevator (g = 10 m/s<sup>2</sup>) must be at least : | [{"identifier": "A", "content": "48000 W"}, {"identifier": "B", "content": "62360 W"}, {"identifier": "C", "content": "56300 W"}, {"identifier": "D", "content": "66000 W"}] | ["D"] | null | Net force on motor will be
<br><br>F<sub>m</sub> = [920 + 68(10)]g + 6000
= 22000 N
<br><br>So, required power for motor
<br><br>P = F<sub>m</sub>.V = 22000$$ \times $$3 = 66000 W | mcq | jee-main-2020-online-7th-january-evening-slot | 13,354 |
tSTmRcKEdCVL7z6p0fjgy2xukf3rg4wa | physics | work-power-and-energy | power | A particle is moving unidirectionally on a horizontal plane under the action of a constant power
supplying energy source. The displacement (s) - time (t) graph that describes the motion of the
particle is (graphs are drawn schematically and are not to scale) : | [{"identifier": "A", "content": "<img src=\"https://res.cloudinary.com/dckxllbjy/image/upload/v1734263447/exam_images/op2mwxqpesnb73rzjfqm.webp\" style=\"max-width: 100%;height: auto;display: block;margin: 0 auto;\" loading=\"lazy\" alt=\"JEE Main 2020 (Online) 3rd September Evening Slot Physics - Work Power & Energy Question 79 English Option 1\">"}, {"identifier": "B", "content": "<img src=\"https://res.cloudinary.com/dckxllbjy/image/upload/v1734267060/exam_images/xwi4q8ovghwrolibbwn7.webp\" style=\"max-width: 100%;height: auto;display: block;margin: 0 auto;\" loading=\"lazy\" alt=\"JEE Main 2020 (Online) 3rd September Evening Slot Physics - Work Power & Energy Question 79 English Option 2\">"}, {"identifier": "C", "content": "<img src=\"https://res.cloudinary.com/dckxllbjy/image/upload/v1734265104/exam_images/ulmvcyyslj2kedelvuyx.webp\" style=\"max-width: 100%;height: auto;display: block;margin: 0 auto;\" loading=\"lazy\" alt=\"JEE Main 2020 (Online) 3rd September Evening Slot Physics - Work Power & Energy Question 79 English Option 3\">"}, {"identifier": "D", "content": "<img src=\"https://res.cloudinary.com/dckxllbjy/image/upload/v1734265474/exam_images/qq5ia0phrftagy9qumdo.webp\" style=\"max-width: 100%;height: auto;display: block;margin: 0 auto;\" loading=\"lazy\" alt=\"JEE Main 2020 (Online) 3rd September Evening Slot Physics - Work Power & Energy Question 79 English Option 4\">"}] | ["B"] | null | $$P = Fv$$<br><br>$$P = m{{dv} \over {dt}}v$$<br><br>$$ \Rightarrow $$ $$vdv = {P \over m}dt$$
<br><br>Integrating both sides, we get<br><br>$${V^2} = k't$$<br><br>$$ \Rightarrow $$ V = k"$$\sqrt t $$
<br><br>$$ \Rightarrow $$ $${{ds} \over {dt}}$$ = k"$$\sqrt t $$
<br><br>$$ \Rightarrow $$ $$\int {ds} = \int {k''\sqrt t } dt$$
<br><br>$$ \Rightarrow $$ s $$ \propto $$ t<sup>3/2</sup> | mcq | jee-main-2020-online-3rd-september-evening-slot | 13,355 |
XdbirJag4pQj3lfHZ8jgy2xukfotfvz8 | physics | work-power-and-energy | power | A body of mass 2 kg is driven by an engine
delivering a constant power of 1 J/s. The body
starts from rest and moves in a straight line.
After 9 seconds, the body has moved a
distance (in m) _______. | [] | null | 18 | <p>Let s be the required distance.</p>
<p><img src="https://app-content.cdn.examgoal.net/fly/@width/image/1l3831ie3/ce2eca9a-da42-4b77-9f5b-fc6f11cf3030/6eb182b0-d4bc-11ec-ad28-abe411cf4979/file-1l3831ie4.png?format=png" data-orsrc="https://app-content.cdn.examgoal.net/image/1l3831ie3/ce2eca9a-da42-4b77-9f5b-fc6f11cf3030/6eb182b0-d4bc-11ec-ad28-abe411cf4979/file-1l3831ie4.png" loading="lazy" style="max-width: 100%; height: auto; display: block; margin: 0px auto; max-height: 40vh;" alt="JEE Main 2020 (Online) 5th September Evening Slot Physics - Work Power & Energy Question 76 English Explanation"></p>
<p>From Work - Energy theorem,</p>
<p>Work = Change in kinetic energy</p>
<p>$$\Rightarrow$$ Power $$\times$$ Time = $$\Delta$$K</p>
<p>i.e., Pt = $$\Delta$$K $$\Rightarrow$$ Pt = $${1 \over 2}$$mv<sup>2</sup> ..... (i)</p>
<p>Given, P = 1 Js<sup>$$-$$1</sup>, t = 9 s, m = 2 kg</p>
<p>Substituting all the given values in eq. (i), we get</p>
<p>1 $$\times$$ 9 = $${1 \over 2}$$(2) v<sup>2</sup></p>
<p>v<sup>2</sup> = 9 $$\Rightarrow$$ v = 3 m/s (at t = 9 s)</p>
<p>As, Fv = P $$\Rightarrow$$ (ma)v = P [$$\because$$ F = ma]</p>
<p>$$ \Rightarrow m\left[ {{{dv} \over {dt}}} \right]v = P \Rightarrow m\left[ {{{ds} \over {dt}}{{dv} \over {ds}}} \right]v = P$$</p>
<p>$$ \Rightarrow m\left[ {v{{dv} \over {ds}}} \right]v = P$$</p>
<p>$$ \Rightarrow 2{v^2}dv = ds$$ {$$\because$$ P = 1 J/s and m = 2 kg}</p>
<p>Integrating both sides,</p>
<p>$$\int\limits_0^3 {2{v^2}dv = \int\limits_0^s {ds \Rightarrow {2 \over 3}[{v^3}]_0^3 = 8} } $$</p>
<p>$${2 \over 3}[27 - 0] = s \Rightarrow s = 18$$ m</p>
<p>Hence, after 9 s, the body has moved a distance of 18 m.</p> | integer | jee-main-2020-online-5th-september-evening-slot | 13,356 |
vr8ojlJgMPfgGFa8TV1kmlho355 | physics | work-power-and-energy | power | A constant power delivering machine has towed a box, which was initially at rest, along a horizontal straight line. The distance moved by the box in time 't' is proportional to :- | [{"identifier": "A", "content": "t<sup>2/3</sup>"}, {"identifier": "B", "content": "t<sup>3/2</sup>"}, {"identifier": "C", "content": "t"}, {"identifier": "D", "content": "t<sup>1/2</sup>"}] | ["B"] | null | $$P = F.v = mav$$<br><br>$$P = {{mvdv} \over {dt}}$$<br><br>$$\int\limits_0^t {Pdt} = m\int\limits_0^v {vdv} $$<br><br>$$Pt = {{m{v^2}} \over 2}$$<br><br>$$v = \sqrt {{{2Pt} \over m}} $$<br><br>$${{dx} \over {dt}} = \sqrt {{{2Pt} \over m}} $$<br><br>$$\int {dx} = \int {\sqrt {{{2Pt} \over m}} } dt$$<br><br>$$x \propto {t^{3/2}}$$ | mcq | jee-main-2021-online-18th-march-morning-shift | 13,357 |
1krqd8z4h | physics | work-power-and-energy | power | A body at rest is moved along a horizontal straight line by a machine delivering a constant power. The distance moved by the body in time 't' is proportional to : | [{"identifier": "A", "content": "$${t^{{3 \\over 2}}}$$"}, {"identifier": "B", "content": "$${t^{{1 \\over 2}}}$$"}, {"identifier": "C", "content": "$${t^{{1 \\over 4}}}$$"}, {"identifier": "D", "content": "$${t^{{3 \\over 4}}}$$"}] | ["A"] | null | P = constant<br><br>$${1 \over 2}$$mv<sup>2</sup> = Pt<br><br>$$\Rightarrow$$ v $$\propto$$ $$\sqrt t $$<br><br>$${{dx} \over {dt}} = C\sqrt t $$ [C = constant]<br><br>by integration.<br><br>$$x = C{{{t^{{1 \over 2} + 1}}} \over {{1 \over 2} + 1}}$$<br><br>$$x \propto {t^{3/2}}$$ | mcq | jee-main-2021-online-20th-july-evening-shift | 13,358 |
1ks1943x1 | physics | work-power-and-energy | power | An automobile of mass 'm' accelerates starting from origin and initially at rest, while the engine supplies constant power P. The position is given as a function of time by : | [{"identifier": "A", "content": "$${\\left( {{{9P} \\over {8m}}} \\right)^{{1 \\over 2}}}{t^{{3 \\over 2}}}$$"}, {"identifier": "B", "content": "$${\\left( {{{8P} \\over {9m}}} \\right)^{{1 \\over 2}}}{t^{{2 \\over 3}}}$$"}, {"identifier": "C", "content": "$${\\left( {{{9m} \\over {8P}}} \\right)^{{1 \\over 2}}}{t^{{3 \\over 2}}}$$"}, {"identifier": "D", "content": "$${\\left( {{{8P} \\over {9m}}} \\right)^{{1 \\over 2}}}{t^{{3 \\over 2}}}$$"}] | ["D"] | null | P = const.<br><br>$$P = Fv = {{m{v^2}dv} \over {dx}}$$<br><br>$$\int\limits_0^x {{P \over m}dx} = \int\limits_0^v {{v^2}dv} $$<br><br>$${{Px} \over m} = {{{v^3}} \over 3}$$<br><br>$${\left( {{{3Px} \over m}} \right)^{1/3}} = v = {{dx} \over {dt}}$$<br><br>$${\left( {{{3P} \over m}} \right)^{1/3}}\int\limits_0^t {dt} = \int\limits_0^x {{x^{ - 1/3}}} dx$$<br><br>$$ \Rightarrow x = {\left( {{{8P} \over {9m}}} \right)^{1/2}}{t^{3/2}}$$ | mcq | jee-main-2021-online-27th-july-evening-shift | 13,359 |
1l6jevuii | physics | work-power-and-energy | power | <p>Sand is being dropped from a stationary dropper at a rate of $$0.5 \,\mathrm{kgs}^{-1}$$ on a conveyor belt moving with a velocity of $$5 \mathrm{~ms}^{-1}$$. The power needed to keep the belt moving with the same velocity will be :</p> | [{"identifier": "A", "content": "1.25 W"}, {"identifier": "B", "content": "2.5 W"}, {"identifier": "C", "content": "6.25 W"}, {"identifier": "D", "content": "12.5 W"}] | ["D"] | null | <p>$${{dm} \over {dt}} = 0.5$$ kg/s</p>
<p>$$v = 5$$ m/s</p>
<p>$$F = {{vdm} \over {dt}} = 2.5$$ kg m/s<sup>2</sup></p>
<p>$$P = \overline F \,.\,\overline v = (2.5)(5)$$ W</p>
<p>$$ = 12.5$$ W</p> | mcq | jee-main-2022-online-27th-july-morning-shift | 13,360 |
ldqwacbo | physics | work-power-and-energy | power | A body of mass $2 \mathrm{~kg}$ is initially at rest. It starts moving unidirectionally under the influence of a source of constant power P. Its displacement in $4 \mathrm{~s}$ is $\frac{1}{3} \alpha^{2} \sqrt{P} m$. The value of $\alpha$ will be ______. | [] | null | 4 | <p>$$P = Fv$$</p>
<p>$$m{{vdv} \over {dt}} = P$$</p>
<p>$$m\int_0^v {vdv = \int_0^t {Pdt} } $$</p>
<p>$${{m{v^2}} \over 2} = Pt$$</p>
<p>$$v = \sqrt {{{2P} \over m}} {t^{1/2}}$$</p>
<p>$$\int_0^s {dx = \sqrt {{{2P} \over m}} \int_0^t {{t^{1/2}}dt} } $$</p>
<p>$$s = {2 \over 3}\sqrt {{{2P} \over m}} {t^{3/2}}$$</p>
<p>or $$s = {2 \over 3}\sqrt {{{2P} \over 2}} \times {4^{3/2}}$$</p>
<p>$$ = {{16} \over 3}\sqrt P ~m$$</p>
<p>So, $$\alpha = 4$$</p> | integer | jee-main-2023-online-30th-january-evening-shift | 13,361 |
1lgq2tf2b | physics | work-power-and-energy | power | <p>The ratio of powers of two motors is $$\frac{3 \sqrt{x}}{\sqrt{x}+1}$$, that are capable of raising $$300 \mathrm{~kg}$$ water in 5 minutes and $$50 \mathrm{~kg}$$ water in 2 minutes respectively from a well of $$100 \mathrm{~m}$$ deep. The value of $$x$$ will be</p> | [{"identifier": "A", "content": "16"}, {"identifier": "B", "content": "4"}, {"identifier": "C", "content": "2"}, {"identifier": "D", "content": "2.4"}] | ["A"] | null | Let us first find the power required to lift the water using each motor. Let $P_1$ be the power of the first motor, and $P_2$ be the power of the second motor.
<br/><br/>
The work done in lifting the water is given by $W = mgh$, where $m$ is the mass of water lifted, $g$ is the acceleration due to gravity, and $h$ is the height through which the water is lifted. In this case, $m = 300\mathrm{~kg}$ and $h = 100\mathrm{~m}$ for the first motor, and $m = 50\mathrm{~kg}$ and $h = 100\mathrm{~m}$ for the second motor.
<br/><br/>
The work done in lifting the water in 5 minutes by the first motor is:<br/><br/>
$$W_1 = mgh = (300\mathrm{~kg})(9.8\mathrm{~m/s^2})(100\mathrm{~m}) = 294000\mathrm{~J}$$
<br/><br/>
The power required to do this work in 5 minutes is:<br/><br/>
$$P_1 = \frac{W_1}{t_1} = \frac{294000\mathrm{~J}}{300\mathrm{~s}} = 980\mathrm{~W}$$
<br/><br/>
The work done in lifting the water in 2 minutes by the second motor is:<br/><br/>
$$W_2 = mgh = (50\mathrm{~kg})(9.8\mathrm{~m/s^2})(100\mathrm{~m}) = 49000\mathrm{~J}$$
<br/><br/>
The power required to do this work in 2 minutes is:<br/><br/>
$$P_2 = \frac{W_2}{t_2} = \frac{49000\mathrm{~J}}{120\mathrm{~s}} = 408.33\mathrm{~W}$$
<br/><br/>
The ratio of the powers of the two motors is:<br/><br/>
$$\frac{P_1}{P_2} = \frac{980\mathrm{~W}}{408.33\mathrm{~W}} \approx 2.4$$
<br/><br/>
We are given that this ratio is equal to:<br/><br/>
$$\frac{3 \sqrt{x}}{\sqrt{x}+1}$$
<br/><br/>
We can solve for $x$ as follows:<br/><br/>
$$\frac{3 \sqrt{x}}{\sqrt{x}+1} = 2.4$$<br/><br/>
$$3\sqrt{x} = 2.4(\sqrt{x}+1)$$<br/><br/>
$$3\sqrt{x} = 2.4\sqrt{x} + 2.4$$<br/><br/>
$$(3-2.4)\sqrt{x} = 2.4$$<br/><br/>
$$0.6\sqrt{x} = 2.4$$<br/><br/>
$$\sqrt{x} = 4$$<br/><br/>
$$x = 16$$<br/><br/>
Therefore, the value of $x$ is 16. | mcq | jee-main-2023-online-13th-april-morning-shift | 13,363 |
1lgsxpbkt | physics | work-power-and-energy | power | <p>A block of mass $$5 \mathrm{~kg}$$ starting from rest pulled up on a smooth incline plane making an angle of $$30^{\circ}$$ with horizontal with an affective acceleration of $$1 \mathrm{~ms}^{-2}$$. The power delivered by the pulling force at $$t=10 \mathrm{~s}$$ from the start is ___________ W.</p>
<p>[use $$\mathrm{g}=10 \mathrm{~ms}^{-2}$$ ]</p>
<p>(calculate the nearest integer value)</p> | [] | null | 300 | <p>To find the power delivered by the pulling force at t = 10 s, we first need to find the work done by the force. The work done is given by the product of force and displacement, and the power is the rate of work done.</p>
<p><b>Calculate the velocity (v) at t = 10 s:</b><br/><br/>
Since the block starts from rest and is pulled up with an effective acceleration of 1 m/s², we can use the equation of motion to find the velocity (v) at t = 10 s:
<br/><br/>
$$v = u + at$$
<br/><br/>
Here, u = 0 (initial velocity) and a = 1 m/s² (acceleration). Plugging in the values:
<br/><br/>
$$v_{10} = 0 + 1(10) = 10 \mathrm{~m/s}$$</p>
<ol>
<li><b>Calculate the net force acting on the block ($F_{net}$):</b><br/><br/>
The net force acting on the block along the incline plane is the difference between the pulling force (F) and the gravitational force component acting parallel to the incline (mgsinθ):</li>
</ol>
<p>$$F_\text{net} = F - mgsinθ$$</p>
<p>Since F_net = ma, we can write:</p>
<p>$$F = ma + mgsinθ$$</p>
<p>Plugging in the values (m = 5 kg, a = 1 m/s², g = 10 m/s², and θ = 30°):</p>
<p>$$F = 5(1) + 5(10)(\sin 30°) = 5 + 25 = 30 \mathrm{~N}$$</p>
<p><b>Calculate the power (P) at t = 10 s:</b><br/><br/>
The power (P) can be calculated as the product of force (F) and velocity (v):
<br/><br/>
$$P_{10} = Fv = 30(10) = 300 \mathrm{~W}$$
<br/><br/>
So, the power delivered by the pulling force at t = 10 s from the start is 300 W.</p> | integer | jee-main-2023-online-11th-april-evening-shift | 13,364 |
1lgvtejbv | physics | work-power-and-energy | power | <p>If the maximum load carried by an elevator is $$1400 \mathrm{~kg}$$ ( $$600 \mathrm{~kg}$$ - Passengers + 800 $$\mathrm{kg}$$ - elevator), which is moving up with a uniform speed of $$3 \mathrm{~m} \mathrm{~s}^{-1}$$ and the frictional force acting on it is $$2000 \mathrm{~N}$$, then the maximum power used by the motor is __________ $$\mathrm{kW}\left(\mathrm{g}=10 \mathrm{~m} / \mathrm{s}^{2}\right)$$</p> | [] | null | 48 | <p>First, let's find the total weight of the elevator and passengers:</p>
<p>Total weight = (mass of passengers + mass of elevator) × g<br/><br/>
Total weight = (600 kg + 800 kg) × 10 m/s²<br/><br/>
Total weight = 1400 kg × 10 m/s² = 14,000 N</p>
<p>Now, we need to calculate the total force acting on the elevator as it moves upwards. Since the elevator is moving at a constant speed, the net force acting on it is zero. Therefore, the tension in the cable must balance the total weight and frictional force:</p>
<p>Tension = Total weight + Frictional force
Tension = 14,000 N + 2,000 N = 16,000 N</p>
<p>The power used by the motor can be calculated using the formula:</p>
<p>Power = Force × Velocity</p>
<p>Here, the force is the tension in the cable, and the velocity is the speed of the elevator:</p>
<p>Power = 16,000 N × 3 m/s = 48,000 W</p>
<p>To convert the power to kilowatts, divide by 1,000:</p>
<p>Power = 48,000 W / 1,000 = 48 kW</p>
<p>So, the maximum power used by the motor is 48 kW.</p>
| integer | jee-main-2023-online-10th-april-evening-shift | 13,365 |
lv9s20qb | physics | work-power-and-energy | power | <p>A body is moving unidirectionally under the influence of a constant power source. Its displacement in time t is proportional to :</p> | [{"identifier": "A", "content": "t<sup>2/3</sup>"}, {"identifier": "B", "content": "t<sup>3/2</sup>"}, {"identifier": "C", "content": "t"}, {"identifier": "D", "content": "t<sup>2</sup>"}] | ["B"] | null | <p>When a body moves under the influence of a constant power, the relationship between displacement and time can be established through the concept of power. Power (P) is defined as the rate at which work is done, and it can also be expressed in terms of force (F) and velocity (v) as $ P = F \cdot v $.</p>
<p>For a constant power P and assuming the force acts in the direction of the velocity, we can analyze how displacement (s) changes with time (t). Since force can also be written as $ F = \frac{d(mv)}{dt} $ for a constant mass m, this simplifies to $ F = m \frac{dv}{dt} $, because mass doesn't change with time for most cases. Integrating force over a distance gives work (W), and power is the rate of doing work, thus we can connect these concepts.</p>
<p>The kinetic energy (K.E) of the body is given by $ K.E = \frac{1}{2}mv^2 $, and the work done by the force is equal to the change in kinetic energy. Considering power is constant, $ P = \frac{dW}{dt} = \frac{d(\frac{1}{2}mv^2)}{dt} $. Rearranging terms to focus on velocity and integrating with respect to time will give us a relation involving velocity and time.</p>
<p>For a constant mass system, and using $ P = F \cdot v = m \cdot a \cdot v = m \cdot \frac{dv}{dt} \cdot v $, and knowing that $ P = \text{constant} $, we rearrange to find the relationship between velocity and time.</p>
<p>Given $ P = m \cdot v \cdot \frac{dv}{dt} $, we rearrange to $ \frac{P}{m} dt = v dv $. Integrating both sides where the initial condition is when $ t = 0, v = 0 $, we get $ \frac{P}{m} t = \frac{1}{2} v^2 $, solving for $ v $ gives $ v \propto t^{1/2} $, so $ v = k \cdot t^{1/2} $ for some constant $ k $.</p>
<p>The displacement $ s $ is obtained by integrating the velocity with respect to time, $ s = \int v dt = \int k \cdot t^{1/2} dt = \frac{2}{3}k \cdot t^{3/2} $. Therefore, the displacement $ s $ is proportional to $ t^{3/2} $.</p>
<p>The correct answer is <strong>Option B</strong>, $ t^{3/2} $.</p> | mcq | jee-main-2024-online-5th-april-evening-shift | 13,367 |
LPEk2w5DjjNtUsdy | physics | work-power-and-energy | work | A spring of spring constant $$5 \times {10^3}\,N/m$$ is stretched initially by $$5$$ $$cm$$ from the unstretched position. Then the work required to stretch it further by another $$5$$ $$cm$$ is | [{"identifier": "A", "content": "$$12.50$$ $$N$$-$$m$$ "}, {"identifier": "B", "content": "$$18.75$$ $$N$$-$$m$$ "}, {"identifier": "C", "content": "$$25.00$$ $$N$$-$$m$$ "}, {"identifier": "D", "content": "$$625$$ $$N$$-$$m$$ "}] | ["B"] | null | Given $$k = 5 \times {10^3}N/m$$
<br><br>Work done when a spring stretched from x<sub>1</sub> cm to x<sub>2</sub> cm,
<br><br>$$W = {1 \over 2}k\left( {x_2^2 - x_1^2} \right) $$
<br><br>$$= {1 \over 2} \times 5 \times {10^3}\left[ {{{\left( {0.1} \right)}^2} - {{\left( {0.05} \right)}^2}} \right]$$
<br><br>$$ = {{5000} \over 2} \times 0.15 \times 0.05 = 18.75\,\,Nm$$ | mcq | aieee-2003 | 13,369 |
M83DLNrq3KVL75D3 | physics | work-power-and-energy | work | When a rubber-band is stretched by a distance $$x$$, it exerts restoring force of magnitude $$F = ax + b{x^2}$$ where $$a$$ and $$b$$ are constants. The work done in stretching the unstretched rubber-band by $$L$$ is : | [{"identifier": "A", "content": "$$a{L^2} + b{L^3}$$ "}, {"identifier": "B", "content": "$${1 \\over 2}\\left( {a{L^2} + b{L^3}} \\right)$$ "}, {"identifier": "C", "content": "$${{a{L^2}} \\over 2} + {{b{L^3}} \\over 3}$$ "}, {"identifier": "D", "content": "$${1 \\over 2}\\left( {{{a{L^2}} \\over 2} + {{b{L^3}} \\over 3}} \\right)$$ "}] | ["C"] | null | Given Restoring force, F = ax + bx<sup>2</sup>
<br><br>Work done in stretching the rubber-band by a distance $$dx$$ is
<br><br>$$\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,dW = F\,dx = \left( {ax + b{x^2}} \right)dx$$
<br><br>Intergrating both sides,
<br><br>$$W = \int\limits_0^L {axdx + \int\limits_0^L {b{x^2}dx}}$$
<br><br>= $$\left[ {a{{{x^2}} \over 2} + b{{{x^3}} \over 3}} \right]_0^L$$
<br><br>= $${{a{L^2}} \over 2} + {{b{L^3}} \over 3}$$ | mcq | jee-main-2014-offline | 13,371 |
Qyl0DPTtmSRlKUjMmxCnJ | physics | work-power-and-energy | work | A body of mass m starts moving from rest along x-axis so that its velocity varies as $$\upsilon = a\sqrt s $$ where a is a constant and s is the distance covered by the body. The total work done by all the forces acting on the body in the first t seconds after the start of the motion is : | [{"identifier": "A", "content": "$${1 \\over 8}\\,$$ m a<sup>4</sup> t<sup>2</sup>"}, {"identifier": "B", "content": "8 m a<sup>4</sup> t<sup>2</sup>"}, {"identifier": "C", "content": "4 m a<sup>4</sup> t<sup>2</sup>"}, {"identifier": "D", "content": "$${1 \\over 4}\\,$$ m a<sup>4</sup> t<sup>2</sup>"}] | ["A"] | null | Given,
<br><br>$$\upsilon $$ = a $$\sqrt s $$
<br><br>$$ \Rightarrow $$$$\,\,\,$$ $${{ds} \over {dt}} = a\sqrt s $$
<br><br>$$ \Rightarrow $$$$\,\,\,$$ $$\int\limits_0^t {{{ds} \over {\sqrt s }}} = \int\limits_0^z {a\,dt} $$
<br><br>$$ \Rightarrow $$$$\,\,\,$$ 2$$\sqrt s $$ = at
<br><br>$$ \Rightarrow $$$$\,\,\,$$ s = $${{{a^2}{t^2}} \over 4}$$
<br><br>= $${1 \over 2}.{{{a^2}} \over 2}.{t^2}$$
<br><br>$$\therefore\,\,\,\,$$ acceleration = $${{{a^2}} \over 2}$$
<br><br>$$\therefore\,\,\,$$ Force (F) = m $$ \times $$ $${{{a^2}} \over 2}$$
<br><br>$$\therefore\,\,\,\,$$ Work done = F. S
<br><br>= $${{m{a^2}} \over 2} \times {{{a^2}{t^2}} \over 4}$$
<br><br>= $${{m{a^4}{t^2}} \over 8}$$ | mcq | jee-main-2018-online-16th-april-morning-slot | 13,372 |
MJ21TJXZmf9efjhfBcTrN | physics | work-power-and-energy | work | A block of mass m is kept on a platform which starts from rest with constant acceleration g/2 upward, as shown in figure. Work done by normal reaction on block in time is -
<br/><br/><img src="data:image/png;base64,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"/> | [{"identifier": "A", "content": "$${{m{g^2}{t^2}} \\over 8}$$"}, {"identifier": "B", "content": "$${{3m{g^2}{t^2}} \\over 8}$$"}, {"identifier": "C", "content": "$$-$$ $${{m{g^2}{t^2}} \\over 8}$$"}, {"identifier": "D", "content": "0"}] | ["B"] | null | N $$-$$ mg = $${{mg} \over 2}$$ $$ \Rightarrow $$ N = $${{3mg} \over 2}$$
<br><br>The distance travelled by the system in time t is
<br><br>S = ut + $${1 \over 2}a{t^2} = 0 + {1 \over 2}\left( {{g \over 2}} \right){t^2} = {1 \over 2}{g \over 2}{t^2}$$
<br><br>Now, work done
<br><br>W = N.S = $$\left( {{3 \over 2}mg} \right)\left( {{1 \over 2}{g \over 2}{t^2}} \right)$$
<br><br>$$ \Rightarrow $$ W = $${{3m{g^2}{t^2}} \over 8}$$ | mcq | jee-main-2019-online-10th-january-morning-slot | 13,373 |
dTeBzFNTsV1tfh1F4RZ9l | physics | work-power-and-energy | work | A uniform cable of mass 'M' and length 'L' is
placed on a horizontal surface such that its (1/n)<sup>th</sup>
part is hanging below the edge of the
surface. To lift the hanging part of the cable upto
the surface, the work done should be : | [{"identifier": "A", "content": "$${{2MgL} \\over {{n^2}}}$$"}, {"identifier": "B", "content": "nMgL"}, {"identifier": "C", "content": "$${{MgL} \\over {2{n^2}}}$$"}, {"identifier": "D", "content": "$${{MgL} \\over {{n^2}}}$$"}] | ["C"] | null | To solve this problem, we need to determine the work done to lift the hanging part of the cable up to the surface. The work done lifting a small element of the cable will be the weight of the element times the distance it has to be lifted.
<br/><br/>Let's take a small section of the cable at a depth $x$ below the surface. This section has a length of $dx$, so its mass is $(M/L)dx$ where $(M/L)$ is the linear mass density of the cable.
<br/><br/>The work $dW$ done to lift this small section up to the surface is the weight of the section times the distance it has to be lifted :
<br/><br/>$dW = (M/L)gdx \times x$.
<br/><br/>Integrating this expression from 0 to L/n (the length of the hanging part of the cable) gives the total work done :
<br/><br/>$$W = \int\limits_{0}^{L/n} (M/L)gxdx$$
<br/><br/>$$= (Mg/L) \int\limits_{0}^{L/n} xdx$$
<br/><br/>$$= (Mg/L) \times [x^2/2]_{0}^{L/n}$$
<br/><br/>$$= (Mg/L) \times [L^2/(2n^2)]$$
<br/><br/>$$= MgL/(2n^2)$$
<br/><br/>So the work done to lift the hanging part of the cable up to the surface is $MgL/(2n^2)$.
<br/><br/>Therefore, the correct answer is Option C :
<br/><br/>$$\frac{MgL}{2n^2}$$ | mcq | jee-main-2019-online-9th-april-morning-slot | 13,374 |
s5U4YwqfEnzUatOPjP7k9k2k5i6usth | physics | work-power-and-energy | work | Consider a force $$\overrightarrow F = - x\widehat i + y\widehat j$$
. The work done
by this force in moving a particle from point
A(1, 0) to B(0, 1) along the line segment is :
(all quantities are in SI units)
<img src="data:image/png;base64,UklGRmQMAABXRUJQVlA4IFgMAAAQVACdASpYARcBPm02mEikIyKhIrH5kIANiWlu4XShG/OT8Yfx7td/un5ZdeN6a9mfVM/cP4B1Of8l263un8a/uv/O+UH7f/cfBngBepP8R/Jv6hwVgAPp7/Zf5P+2n9d+I70v/Bf0z+U+xX1e/0X6AfIF+nX+o/k39Z9K//FfwDyvPJPYN/Kv+7/nf9H+I39Z/2v8T/m/7ve4D5r/7X8d/uvuhf7P+Rf1n9mvly/QD35/q5+gHw/fq5+gAvPjf8fVF3gsjcPzrS/E33wqrFCphnihUve/b4IrQy3lMPh5Fvv0cknAfbvk2FlVYpYCyFBLH5ABAsALhq5YVzWAGPC8wGoEbJl3EFgM3LKqxQqYZrEQjGyp6o2954ybgBZYPs1BYUkBzRcCP6S091YGbeWw7evCxMMCxsUKmGeKFR6GMursziP0r5KdjgqAcnd8mwsqrFA+F7DZNtl6RdQqFISShUwzxKo9kiqcjBmwCyqcVIPJKFTDN0Ywc4//1Q0Lc5la0HklCphnigfrOGXglmA0IFplf1u/JsLKqxQqQfMW+oKMf5pzrzW9SDyShUwzdGQeMPfh9fxkulxJQqYZ4oURumwrf+oCD8MWS3VedxUg8koH6zuKizxcV86DCcWSrmQZAVIEwWT0TDtgOjzMB0zlYvSFY0srjPmpfgE4v9SRVXxqLCen4gYYqXS4F6pVheRs7YXBng5veMapyDNH0lb42Y9Bj5jxSpchpDx04i5QeR31wjxZKMEzkjiPCEaHNg5YFB4h4BONpwoSLBxbXHyM3CA7i6Q8goH2nbVlU32zCc7uLIYFnS5Iy1VEU8T/Mi9iFda2vkNGbPwB4WVVihUvhLvXHX2eAvnreRqf0s4kqhAMO5jpts45Dg6T3klCphnihUxRG0HkkgAA/v+d6ZQDdkarCgQ/lMW7YZlOAuvXWOxxkcuv3I4F2sTy/XXEPuld5I2Xv8DDynuFWyt8PFtOVmQLt+msMDzuZZxcfH/tpMDfu88y/IzoDfQY4QXQqZ1n+wP7nv/gdeX+uOMto2p/1T1p6CS/rGhlVgEXwC3LfQe1xuoMfoQAys9FkI9Zo+DiEQ/mBAWXTe7ag7fgGzN3MnD0rrAnsfLoYh5ZzmIZB1OUmEfzQ3f1vEwOqw2wnhbN1rvxgaJd6H5jIr+kcP8HhTkNoaWdLGbv0ibGyzCjoXpY1NQKLNXkKGNUik+DD7EZWSpsacDkpQDRSsw3Qi18kYHmRONGkj3DkzJfBnj9eTCRYFJe/fI/4XBw3EyBRg7CL/BS/VScJ6umpMh/okG0omXFWStjhZBTFbQ+khRsOJiorbeQSggsnJ2n+GCnMK5Efp+EQot9cifK+I9Egni2JiY0jKL93dVW36kO/ctATyViqxg1b7J3Cz7v0omYijIWWb32p182gfzE5mgHoozy3GvK3kFkscP6/1W+qvA7aVr032fNsrPhQ2D5NkyHwrJ9hbXrhWHExHVFC19vjcdHEeMtrUzPVrN1PG16c8bfoWMzFrHTMhKTGDLNHqBb3dh1YoRPuMaTi6RGOlEBoDJkKLMkDaCQjxNa/gshWLFR+A4UVvW0kkqlN00KMQxaN8rBHBhmVeV9gJ6cmM7+3coEHIeVlvJBnTp7O/NFQY/h4MNycFQP9r4H/F/0Qg1qUPY618JnoWCsEXwzg3rD6R+VPXv98rvWnL5wiHqmJFIaWIpmbElBXyQVcjDXcyVpHxTChuqdyFd8mWrDmwx62zYJwCLAWm7wkpMeSvVKEJpIOkZxhkWQFSVgQwAo1mvn7FFic+LJq1krqJpqXT6MB0mA5BdA1UncsAzgKQqynwFhpKhIvXv/iNtPrQ/FKZR69R+j6tNhRtVNc4a2MrgdFrWM0aTa1uXwNQE+XFdaCg2D78T9/Fe0tLB34bVIONIRVQf+j4q9P7GxqoyECqyxkoRNrkaz8Cwv2b7SJcXxyoX2MUbEqT0MicF+jhgTijlkuUPzCDrE48q1Djb/Wj6V2UiOUJ7pAGybQGPXgQFAL6x/luznoS2biTeROPxfoO0rGfd8/RPykKGv9EO9ptUTHgW5AcG04h2AhYWMC/02kZNJgE7lt9YCDzTt2N/PgtPBOfLK9iZBzE5Rm+MYRGO/mXbpRWDsWkotHy1ggOu7SITEDAR8/Y54NdXTu1bZWgkNMelnMv52PVI9LGDBGrYrMs0GDkXQeQV7qifCkNt/AsEMG2tJRws+HeaoPjQVlINtjVns5CnwfXmXGseRdCLB2g2LJGWpXtZqk+Ri6sWqpEe6AwMo/YEWJQt68hB8oBt+Q9wplnvzg0f6P0SXleGo3Y0h51+pEuANmiIeqwRVQY4usQwch+vGNNKNdpeFKGreV6bN+qDxagpKs4f5fg+/CKE5gkbUK4/tsrKSZ9xho4XvB470TQFGgn9cMrgRS/7prfTVDqyzDzQaXJFjxVNyjnodVLzenAXb1UFmli8QCw7pi4EusyeuPTU2cKKOLZ8XEVOVfQ1XUhR5pLTG1x6JjRi3aRP4pzQvraCqznadR2/AN5OdzImh92+BYhn0HbOTg5Pf8LB3f+aZ3ohRTbpr6TENWJL+q624N8+r8IfiKKgVbJTQMkHCL+IySPgXzE1va+Cu7OmxBRlRxkeaWAx/jfmD/2NM/5k/pRzFdjPylNiLLqnfRaZvFKB78a3IXpkWl9XMqxAuXvrYpMTrAl6oQ7+/evGC/fARIzjzSSDotXleZJo1VA8HDp4fVd5RROl5X9ngtMyTG399UjRNHpMHysgHw+yQUg0j1H2sHa8K0JkZMQT1MZojWXzdEjmT/8S2ZllBt6EDhK9kjvNWUcqknrurba7U/IGqk/hWuoNqnV4Bxy9Pv5HYZERLUuk13oIXUgQzNtP+WgCwiOxkNtwFnJ8CB4n7rzJwCTHoKyy+/H7eHeqCf/iMiMo7iibQkQeP+RIQeQ26E0CzQob0yEz2CKxyoPNK0asDpigMtleh7RtYWTwTFCv1t+WDDpvsz6VeSFAWAG4VrrsVAYzg3OUYyMZulVb0EEQtQ+9oYUxC0z71g9JGcXlVz7Kt5rml0ee7u+sJl9pH8xneaYNSQdcqxs/GF99ZthLCU/Xj4jQ23MTt6/dIkjhYjp1qv+ds0EM/GkSo7hdj12ioivHQHJo2iKySQew8GmtzvjsoIdB2qbvZhKKPUSLoWSQ0Is7n4/7HmV5zgzaEYzX3HO3D9pPzPrz/kcEkLOyq1fjRvLU2Jpuy5gfA4Otc3QLYnXt+LNgOLosJbbUUmg6HDpRtUUFefg4aN4LPiUfXkMFWR4wGSTyoQbaEaUG0Dn5tzpoaMSmLa9Y9ZBExSnLky2zyixaahnEUaPPZv6zkAAs5XpG4o5446TIFXqCx6urIFx5pzJsnBfF3n5l50XNfvhHlZhs08a0aOGWkv61P/rfJqbXoYX2iyjBz8IKcZFHrZE9kXrq5cfOj87Sr8L8HIEaQp52eKM3fmpwCddNeA8t1K2YaxBMCquX2f1NnnJ8qGf5NL/T3lH5c+H8UoNA5xz4g+l4faj825UXK04vYTXxiz6RLP8ykBWCim0OR4YEkaUO4VwSygEX/iEaeq+D0quecAhivmn2VVjlSkclH3Wax9bR3aq+N674QeYl4lW+IvsFY3W8X4rG8hZWk2Hr2g5LzGasl7LTQmLvNP+J2gJqr8G3d+1A+1X1nf9TjnU6oSkcXkyEIWE9rHY2h/k3juCmB8JDPENFOled3irnk/5dmLP43unbItCbt75O56LaQolKdZ65P3NYOAjaf9HxymZ148EC6ZRHVWLYFTSfKQCDcQejS3HW3wUbV745qxMBOfrJQSKuH58gpN08QScftmAMSPCnGH26pmtMYgozjpHYEYnItFYHFzDCBeXDeyFMPOFxlAlORPJiFrMebyZeCbXeBoCQVWprxOAGgDU83aiF00oHY83F2ushRDEG7znLSsTL/tNU8k/6ASLB/SQROL7vIxEVD6wDwEnCOLvFiD4fIASb91I3UG2NcOVjxPDOKFiadurI864b06g9/M62ZJ528bN1qF3YqDm3S5j/5djWgQhgC4KVLv+n7/XtuV0A0lCVc1t3RTBafd8f68wnnr/yOf0fBnxn87SNBWkLI6RlA4EIH4BMkzUIPRJDAoDlliUFdzkrx3UUihYFZdxUI6Yf8zStUGGF+OdljHbdnCaO4z3PxgO9WqLpGeItVwAAA"/>
| [{"identifier": "A", "content": "2"}, {"identifier": "B", "content": "$${1 \\over 2}$$"}, {"identifier": "C", "content": "1"}, {"identifier": "D", "content": "$${3 \\over 2}$$"}] | ["C"] | null | W = $$\int {\overrightarrow F } .d\overrightarrow r $$
<br><br>= $$\int {\left( { - x\widehat i + y\widehat j} \right)} .\left( {dx\widehat i + dy\widehat j} \right)$$
<br><br>= $$ - \int\limits_1^0 {xdx} + \int\limits_0^1 {ydy} $$
<br><br>= $${1 \over 2} + {1 \over 2}$$ = 1 J | mcq | jee-main-2020-online-9th-january-morning-slot | 13,375 |
ZjHmqNwNiJbLX81qPwjgy2xukfahetcu | physics | work-power-and-energy | work | A person pushes a box on a rough horizontal plateform surface. He applies a force of 200 N over a
distance of 15 m. Thereafter, he gets progressively tired and his applied force reduces linearly with
distance to 100 N. The total distance through which the box has been moved is 30 m. What is the
work done by the person during the total movement of the box? | [{"identifier": "A", "content": "5690 J"}, {"identifier": "B", "content": "5250 J"}, {"identifier": "C", "content": "2780 J"}, {"identifier": "D", "content": "3280 J"}] | ["B"] | null | <img src="https://res.cloudinary.com/dckxllbjy/image/upload/v1734263269/exam_images/uppfh1pvufarfecpjfb7.webp" style="max-width: 100%;height: auto;display: block;margin: 0 auto;" loading="lazy" alt="JEE Main 2020 (Online) 4th September Evening Slot Physics - Work Power & Energy Question 77 English Explanation">
<br><br>Work done = area of ABCEO
<br><br>= area of trapezium ABCD + area of rectangle ODCE
<br><br>= $${1 \over 2}$$ $$ \times $$ 45 $$ \times $$ 30 + 100 $$ \times $$ 30 = 5250J | mcq | jee-main-2020-online-4th-september-evening-slot | 13,376 |
1krsu3bwl | physics | work-power-and-energy | work | A porter lifts a heavy suitcase of mass 80 kg and at the destination lowers it down by a distance of 80 cm with a constant velocity. Calculate the work done by the porter in lowering the suitcase.<br/><br/>(take g = 9.8 ms<sup>$$-$$2</sup>) | [{"identifier": "A", "content": "+627.2 J"}, {"identifier": "B", "content": "$$-$$62720.0 J"}, {"identifier": "C", "content": "$$-$$627.2 J"}, {"identifier": "D", "content": "784.0 J"}] | ["C"] | null | $$W = - N \times \Delta x$$<br><br>$$ = - 80 \times 9.8 \times {{80} \over {100}}$$<br><br>$$ = - 627.2$$ J | mcq | jee-main-2021-online-22th-july-evening-shift | 13,377 |
1krwcnrm2 | physics | work-power-and-energy | work | A force of F = (5y + 20)$$\widehat j$$ N acts on a particle. The work done by this force when the particle is moved from y = 0 m to y = 10 m is ___________ J. | [] | null | 450 | F = (5y + 20)$$\widehat j$$<br><br>$$W = \int {Fdy = \int\limits_0^{10} {(5y + 20)dy} } $$<br><br>$$ = \left( {{{5{y^2}} \over 2} + 20y} \right)_0^{10}$$<br><br>$$ = {5 \over 2} \times 100 + 20 \times 10$$<br><br>$$ = 250 + 200 = 450$$ J | integer | jee-main-2021-online-25th-july-evening-shift | 13,378 |
1ktmn70rw | physics | work-power-and-energy | work | A body of mass 'm' dropped from a height 'h' reaches the ground with a speed of 0.8$$\sqrt {gh} $$. The value of workdone by the air-friction is : | [{"identifier": "A", "content": "$$-$$0.68 mgh"}, {"identifier": "B", "content": "mgh"}, {"identifier": "C", "content": "1.64 mgh"}, {"identifier": "D", "content": "0.64 mgh"}] | ["A"] | null | Given, the mass of the body = m<br/><br/>The height from which the body dropped = h<br/><br/>The speed of the body when reached the ground, $${v_f} = 0.8\sqrt {gh} $$<br/><br/>Initial velocity of the body, v = 0 m/s<br/><br/>Using the work-energy theorem,<br/><br/>Work done by gravity + Work done by air-friction = Final kinetic energy $$-$$ Initial kinetic energy.<br/><br/>$${W_{mg}} + {W_{air - friction}} = {1 \over 2}mv_f^2 - {1 \over 2}mv_i^2$$<br/><br/>Here, work done by gravity = mgh<br/><br/>$$ \Rightarrow mgh + {W_{air - friction}} = {1 \over 2}m{(0.8\sqrt {gh} )^2} - {1 \over 2}m{(0)^2}$$<br/><br/>$$ \Rightarrow {W_{air - friction}} = {{0.64mgh} \over 2} - mgh$$<br/><br/>$$ \Rightarrow 0.32mgh - mgh = - 0.68mgh$$<br/><br/>The value of the work done by the air friction is $$-$$ 0.68 mgh. | mcq | jee-main-2021-online-1st-september-evening-shift | 13,380 |
1l58hfe08 | physics | work-power-and-energy | work | <p>Arrange the four graphs in descending order of total work done; where W<sub>1</sub>, W<sub>2</sub>, W<sub>3</sub> and W<sub>4</sub> are the work done corresponding to figure a, b, c and d respectively.</p>
<p><img src="data:image/png;base64,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"/>
<img src="data:image/png;base64,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"/></p>
<p><img src="data:image/png;base64,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"/></p>
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<p>$$\Rightarrow$$ W<sub>c</sub> > W<sub>b</sub> > W<sub>a</sub> > W<sub>d</sub></p>
<p>$$\Rightarrow$$ W<sub>3</sub> > W<sub>2</sub> > W<sub>1</sub> > W<sub>4</sub></p> | mcq | jee-main-2022-online-26th-june-evening-shift | 13,381 |
1l5c3r2k8 | physics | work-power-and-energy | work | <p>A particle experiences a variable force $$\overrightarrow F = \left( {4x\widehat i + 3{y^2}\widehat j} \right)$$ in a horizontal x-y plane. Assume distance in meters and force is newton. If the particle moves from point (1, 2) to point (2, 3) in the x-y plane, then Kinetic Energy changes by :</p> | [{"identifier": "A", "content": "50.0 J"}, {"identifier": "B", "content": "12.5 J"}, {"identifier": "C", "content": "25.0 J"}, {"identifier": "D", "content": "0 J"}] | ["C"] | null | <p>$$W = \int {\overrightarrow F \,.\,d\overrightarrow r } $$</p>
<p>$$ = \int\limits_1^2 {4xdx + \int\limits_2^3 {3{y^2}dy} } $$</p>
<p>$$ = [2{x^2}]_1^2 + [{y^3}]_2^3$$</p>
<p>$$ = 2 \times 3 + (27 - 8)$$</p>
<p>$$ = 25$$ J</p> | mcq | jee-main-2022-online-24th-june-morning-shift | 13,382 |
1ldnzricg | physics | work-power-and-energy | work | <p>A force $$\mathrm{F}=\left(5+3 y^{2}\right)$$ acts on a particle in the $$y$$-direction, where $$\mathrm{F}$$ is in newton and $$y$$ is in meter. The work done by the force during a displacement from $$y=2 \mathrm{~m}$$ to $$y=5 \mathrm{~m}$$ is ___________ J.</p> | [] | null | 132 | $\begin{aligned} & W=\int F d y=\int_2^5\left(5+3 y^2\right) d y \\\\ & =\left.\left(5 y+y^3\right)\right|_2 ^5 \\\\ & =(15+125-8) \mathrm{J} \\\\ & =132 \mathrm{~J}\end{aligned}$ | integer | jee-main-2023-online-1st-february-evening-shift | 13,383 |
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