| {"id": "483870", "subject": "Physics", "difficulty_level": "hard", "image": ["images/question_images/1750458_question.jpg"], "conversations": [{"from": "human", "value": "<image>\nQuestion: In the figure shown a plate a plate of mass $$60\\ \\mathrm{gm} $$ is at rest and in equilibrium, A particle of mass $$ m=30\\ \\mathrm{gm} $$ is released from height $$ \\dfrac{4.5 \\mathrm{mg}}{\\mathrm{k}} $$ from the plate.The particle sticks to the plate. Neglecting the duration of collision find time from the collision of the particle and the plate to the moment when the spring has maximum compression. Spring has force constant $$1\\ N/m$$. Calculate value of time in the form $$ {\\pi}/x $$ and find the value of $$ x . $$"}, {"from": "gpt", "value": "The correct answer is Unknown.\n\nVelocity of the particle just before collision.\n$$u=\\sqrt{2g\\times \\dfrac{4.5\\ mg}{K}}\\Rightarrow u=3g\\sqrt{\\dfrac{m}{K}}$$\n\nNow it collides with the plate.\n\nNow just after collision velocity $$(V)$$ of the system of plate + particle.\n\n$$mu=3mV\\Rightarrow V=\\dfrac{u}{3}=g\\sqrt{\\dfrac{m}{K}}$$\n\nNow the system performs $$SHM$$ with time period \n\n$$T=2\\pi\\sqrt{3m/K}\\left(\\omega =\\sqrt{\\dfrac{K}{3m}}\\right)$$ and mean position as $$mg/K$$ distance below the point of collision.\n\nLet the equation of motion is \n\n$$y=A\\sin (\\omega t+\\phi)$$\n\n$$v=\\dfrac{dy}{dt}=A\\omega \\cos (\\omega t+\\phi)$$\n\nAt $$t=0, y=\\dfrac{mg}{K}$$ and $$v=g\\sqrt{\\dfrac{m}{K}}$$\nfrom Eqs (i) and (ii) $$\\dfrac{mg}{K}=A\\sin \\phi$$\n\n$$A=\\dfrac{2mg}{K}$$\nfrom Eqs (iii) and (iv) $$\\Rightarrow \\phi=\\dfrac{5\\pi}{6}$$\n\n$$y=\\dfrac{2mg}{K}\\sin \\left(\\sqrt{\\dfrac{K}{3m}}t+\\dfrac{5\\pi}{6}\\right)$$\n\nHence equation of $$SHM$$ should be\n\n$$y=-A=-\\dfrac{2mg}{K}=\\dfrac{2mg}{K}\\sin \\left(\\sqrt{\\dfrac{K}{3m}}t+\\dfrac{5\\pi}{6}\\right)$$\nThe plate will be at rest again when\n\n$$y=-A=-\\dfrac{2mg}{K}=\\dfrac{2mg}{K}\\sin \\left(\\sqrt{\\dfrac{K}{3m}}t+\\dfrac{5\\pi}{6}\\right)$$\n\n$$\\Rightarrow \\sin \\left(\\sqrt{\\dfrac{K}{3m}}t+\\dfrac{5\\pi}{6}\\right)=-1=\\sin \\dfrac{3\\pi}{2}$$\n\n$$\\Rightarrow \\sqrt{\\dfrac{K}{3m}}t+\\dfrac{5\\pi}{6}=\\dfrac{3\\pi}{2}\\Rightarrow t=\\dfrac{2\\pi}{3}\\sqrt{\\dfrac{3m}{K}}$$\n\nUsing value $$t=\\pi/5\\ s$$."}]} |
| {"id": "1827867", "subject": "Physics", "difficulty_level": "hard", "image": ["images/question_images/1804530_question.jpg"], "conversations": [{"from": "human", "value": "<image>\nQuestion: Fig 3.13 shows a nut cracker.\nIn the diagram, mark the position of the fulcrum F and the line of action of load L and effort E."}, {"from": "gpt", "value": "The correct answer is Unknown."}], "solution_image_path": "images/solution_images/1804530_solution.jpg"} |
| {"id": "524103", "subject": "Maths", "difficulty_level": "easy", "image": ["images/question_images/524103_question.jpg"], "conversations": [{"from": "human", "value": "<image>\nQuestion: the angle of reflection of the ray from mirror PQ"}, {"from": "gpt", "value": "The correct answer is Unknown.\n\nAngle of incidence is angle of ray with normal to the mirror PQ.\nAngle of incidence $$\\angle LGA=90-30=60^0$$\nFor plane mirror angle of incidence is equal to angle of reflection.\nSo angle of reflection $$\\angle AGO=60^0$$ "}], "solution_image_path": "images/solution_images/524103_solution.jpg"} |
| {"id": "1874", "subject": "Physics", "difficulty_level": "medium", "image": ["images/question_images/40221_question.jpg"], "conversations": [{"from": "human", "value": "<image>\nQuestion: Shown in the figure is network of capacitors and resistors. The potentials of some of the nodes are given, Find the\n(a) potential of $$P$$\n(b) energy stored in the capacitor connected between R and S\n(c) energy stored in the capacitor connected between P and Q"}, {"from": "gpt", "value": "The correct answer is Unknown.\n\nAt steady state, capacitor offers infinite resistance to DC, therefore, no current passes through $$1\\mu F$$ capacitor and current passes through $$1\\Omega$$ resistor across PQ. Since three identical resistors of resistance $$\\cfrac { 1 }{ 3 } \\Omega $$ each are connected in series. This series combination is connected in parallel with another resistor of $$1\\Omega$$ across PQ\nThe effective resistance between P and Q will be $$\\cfrac { 1 }{ 2 } \\Omega $$\nApplying KCL at P we obtain\n$${ i }_{ 1 }+{ i }_{ 2 }+{ i }_{ 3 }=0\\quad \\Rightarrow \\cfrac { { V }_{ p }-10 }{ 1 } +\\cfrac { { V }_{ p }-3 }{ 0.3 } +\\cfrac { { V }_{ p }-(-20) }{ 1 } =0\\quad \\Rightarrow { V }_{ p }\\left[ 1+\\cfrac { 1 }{ 0.3 } +1 \\right] =10+10-20$$\n(b) $${ V }_{ p }=0;{ i }_{ 1 }=10amp;{ i }_{ 2 }=-10amp,{ i }_{ 3 }=20amp$$\nThe potential difference across PQ\n$${ V }_{ PQ }={ R }_{ PQ }.{ i }_{ 3 }\\Rightarrow { V }_{ PQ }=(1/2)20=10V$$\nSince $${ V }_{ p }=0,{ V }_{ Q }=0-10V$$\nie., $${ V }_{ PQ }=+10V$$\nThe potential difference between R and S $$={V}_{RS}$$\nSince $$\\quad { i }_{ PR }={ I }_{ PQ }/2={ i }_{ 3 }/2=\\cfrac { 20 }{ 10 } =10;{ V }_{ S }=0;{ V }_{ P }=0$$\n$${ V }_{ R }={ V }_{ P }-\\left( { I }_{ PR } \\right) \\left( { R }_{ PR } \\right) =-\\left[ 10\\times 1/3 \\right] =-\\cfrac { 10 }{ 3 } V$$\n(b) the energy stored in the capacitor $$(0.4\\mu F)$$\n$$=\\cfrac { 1 }{ 6 } \\left( 0.4\\times { 10 }^{ -6 } \\right) { \\left( \\cfrac { 10 }{ 3 } \\right) }^{ 2 }=2.2\\times { 10 }^{ -6 }J\\quad $$\n(c) the Energy stored in the capacitor $$\\left( 1\\mu F \\right) $$\n$$=\\cfrac { 1 }{ 2 } \\left( 1\\times { 10 }^{ -6 } \\right) { 10 }^{ 2 }=50\\mu J$$"}], "solution_image_path": "images/solution_images/40221_solution.jpg"} |
| {"id": "181273", "subject": "Biology", "difficulty_level": "hard", "image": ["images/question_images/41582_question.jpg"], "conversations": [{"from": "human", "value": "<image>\nQuestion: Three identical conducting large plates are fixed at separation of $$d$$ from each other as shown. The area of each plate is $$A$$. Plate $$1$$ is given charge $$+Q$$ while plate $$2\\;and\\;3$$ are neutral and are connected to each other through coil of inductances $$L$$ and switch $$S$$. If resistance of all connected wires is neglected the maximum current flow, through coil after closing switch is $$(C=\\varepsilon_0\\,A/d)$$ (neglect fringe effect)\nOptions:\nA. $$\\displaystyle\\frac{Q_0}{\\sqrt{LC}}$$\nB. $$\\displaystyle\\frac{Q_0}{\\sqrt{2LC}}$$\nC. $$\\displaystyle\\frac{2Q_0}{\\sqrt{LC}}$$\nD. $$\\displaystyle\\frac{Q_0}{2\\sqrt{LC}}$$"}, {"from": "gpt", "value": "The correct answer is D.\n\nThe charge distribution on the plates surfaces will be as shown in figure. Thus the charge on the capacitor in between plates 2 and 3 is Q/2. \nHere, energy stored in the capacitor (between 2 and 3 plate) $$=$$ energy stored in inductor.\nor $$\\dfrac{(Q/2)^2}{2C}=\\dfrac{1}{2} LI_0^2 \\Rightarrow I_0=\\dfrac{Q}{2\\sqrt{LC}}$$"}], "solution_image_path": "images/solution_images/41582_solution.jpg"} |
| {"id": "1780612", "subject": "Chemistry", "difficulty_level": "medium", "image": ["images/question_images/1780612_question.jpg"], "conversations": [{"from": "human", "value": "<image>\nQuestion: Write the IUPAC name of the given compound:"}, {"from": "gpt", "value": "The correct answer is Unknown.\n\nAmides are named by changing the -oic acid ending of the corresponding carboxylic acid to -amide. If alkyl groups are attached to the nitrogen, they are named as N-alkyl substituents.\nHence the name is N-Phenyl ethanamide."}], "solution_image_path": "images/solution_images/1780612_solution.jpg"} |
| {"id": "1124808", "subject": "Maths", "difficulty_level": "medium", "image": ["images/question_images/1786597_question.jpg"], "conversations": [{"from": "human", "value": "<image>\nQuestion: Draw a line segment of length $$ 7.6 cm $$ and divides it in the ratio $$ 5 : 8 $$ Measure the two parts. "}, {"from": "gpt", "value": "The correct answer is Unknown.\n\nSteps of construction :\nStep I : Draw a line segment $$ AB = 7.6 cm $$\nStep II : Draw any ray $$ AX $$ making an acute angle $$ \\angle BAX $$ with AB .\nStep III : On ray $$ AX $$ , starting from $$ A $$ mark $$ 5 + 8 = 13 $$ equal ares.\n$$ AA_1,A_1,A_2,A_2A_3,A_3A_4 ......A_{11}A_{12} \\quad and \\quad A_{12}A_{13} .$$\nStep IV : join $$ A_{13} zB $$\nStep V : From $$ A_5$$ Draw $$ A_5 P || A_{13} B $$ meeting $$ B $$ at $$ P $$.\nthus $$ P $$ divides $$ AB $$ in the ratio $$ 5 : 8 $$ on measuring the two parts . We find $$ AP = 2.9 cm $$ and $$ PB = 4.7 (approx) $$\njustification : \nIn $$ \\Delta ABA_{13} , PA || BA_{13} $$\n$$ \\therefore $$ By basic proportionality theorem \n$$ \\frac {AP}{PB} = \\frac {AA_5}{A_5A_{13} } = \\frac {5}{8} $$\n$$ \\Rightarrow \\frac {AP}{PB} = \\frac {5}{8} $$\n$$ \\therefore AP : PB = 5 : 8 $$"}]} |
| {"id": "1370301", "subject": "Biology", "difficulty_level": "medium", "image": ["images/question_images/332961_question.jpg"], "conversations": [{"from": "human", "value": "<image>\nQuestion: The IUPAC name of the following compound is :\nOptions:\nA. 3 - Ethyl - 2 - methylbut - 3 - ene\nB. 2 - Ethyl - 3 - methyl but - 1 - ene\nC. 2 - Methyl - 3 - ethylbut - 3 - ene\nD. 3 - Methyl - 2 - ethylbut - 1 - ene"}, {"from": "gpt", "value": "The correct answer is B.\n\nB. $$2-Ethyl-3-methyl but-1-ene$$ \nDouble bond gets priority than alkyl groups in the carbon chain."}]} |
| {"id": "1714229", "subject": "Maths", "difficulty_level": "medium", "image": ["images/question_images/1834135_question.jpg"], "conversations": [{"from": "human", "value": "<image>\nQuestion: Calculate AB."}, {"from": "gpt", "value": "The correct answer is Unknown.\n\nIn $$\\triangle AMO$$, \n$$cos\\, 30^0 = AO/MO$$ \n$$\\sqrt{3}/2 = AO/6$$ \n$$AO = 5.20 m$$ \nIn$$ \\triangle BNO$$, \n$$sin\\, 47^0 = OB/NO$$ \n$$0.73 = OB/5$$\n$$OB = 3.65 m$$ \nSo, $$AB = OA + OB$$ \n$$AB = 5.20 + 3.65$$ \n$$AB = 8.85 m$$ "}]} |
| {"id": "536580", "subject": "Biology", "difficulty_level": "medium", "image": ["images/question_images/490865_question.jpg"], "conversations": [{"from": "human", "value": "<image>\nQuestion: Which element is required by the process designated as (III)?\nOptions:\nA. Chlorine\nB. Nitrogen\nC. Oxygen\nD. Potassium"}, {"from": "gpt", "value": "The correct answer is C.\n\nElectron transport chain (III) includes downhill flow of electrons from NADH and $$FADH_2$$ to final electron acceptor (Oxygen) through a chain of membrane-bound carriers to facilitate the uphill transport of protons across a proton-impermeable membrane. Since inner mitochondrial membrane is impermeable to protons; protons can reenter the matrix only through $$F_o$$ particles of ATP synthase. The proton-motive force that drives protons back into the matrix provides the energy for ATP synthesis. Correct answer is C."}]} |
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