NEET 2026 PHYSICS TOP MCQs Quick Solving tricks||Score 140+||NEET 9

Added: 2026-05-09

[00:00:03]Hello dear students welcome to my channel already so need question papers max very important concepts maximum questions so you question.

[00:00:33]So very important concepts top.

[00:00:41]So definite scoring 140 plus video. So definite question [clears throat] first mechanical properties of solids.

[00:01:02]The maximum elongation of a steel wire of 1 m length length 1 m the [clears throat] uh if the elastic limit of steel and its ends modus elastic limit so that is stress okay and ang modus y is given the formula is y = stress divided by strain is delta L / So we want delta L. Okay. So stress value elastic limit value is given that is 8 into 10^ 8 divided by Y. Y value is 2 into 10^ 11. L value is 1. So 2 1 2 4. So this is 4 into 10^ - 11 + 800. So -3.

[00:02:03]So -300 this is 4 mm.

[00:02:07]4 mm. The right answer is fourth option.

[00:02:10]Fourth option is the right answer.

[00:02:14]Next question number two. A horizontal force F is applied to uniform solid sphere at rest so that its line of action passes through the midpoint of vertical radius O. O beam center of the sphere. The acceleration of the uppermost point acceleration actually lead to accelerations.

[00:02:41]So acceleration tangential acceleration acceleration due to center of mass. So tangential acceleration acceleration due to center of mass that is f by m because f is equal to m into a cm. So actually acceleration at point a f by m + 80. So this is more than f by m. So second option greater than f by m. So that is the right answer.

[00:03:17]Greater than f by m. Second option is the right answer.

[00:03:21]So next [clears throat] question number three. The terminal velocity of a ball of radius 2 mm falling through an oil tank of viscosity.

[00:03:35]So this is coefficient of viscosity 1 kg per meter/s and v is given 6 cm/s. So viscous drag. So stock law. So the formula f is equal to 6<unk>i or v. So f is = 6 pi ea is 1 and r is 2 into 10^ - 3. So v is 6 into 10^ - 2 10^ - 2. Okay. So 6 into 6 36. So 36 9 into 4<unk> into 2 10 - 3 - 2 - 5. So 9 into 4 pi is 12.5 then 25 by 2 to 10^ - 5 22 cancel 25 into 9 is 225 225 into 10^ - 5 2.25 25 10^ 3 10 - 5 5 10 10 2 10 - 5 10 - 3 2.25 10 2 10 2 into 10 - 10 - 3 So the right answer is not here second option is new 2.26 into 6 into 10^ - 3 Newton.

[00:05:06]Next question number four. The displacement of the point P on the wheel as a wheel completes of revolution while rolling on the ground.

[00:05:19]The horizontal surface initial location is a final location. We want displacement strike line.

[00:05:28]Okay.

[00:05:30]So horizontally off rotation complete.

[00:05:34]So to distance cover that is very important. So horizontally.

[00:05:40]So to distance that off revolution complete rotation.

[00:05:47]So off rotation and so P A this is 2 R² + 4 R² R square is common take it as outside displacement Pdash is equal to roo<unk>² R² square + 4 r square. R square is common. Take it as outside. The root becomes r² +<unk> square + 4<unk>² + 4.

[00:06:31]You noting the right answer is fourth option r into roo<unk> square + 4.

[00:06:39]So next question number five. the maximum electric field of a plain electromagnetic wave traveling through vacuum. So E not is given the maximum magnetic field B formula is QVB is equal to Q. So QQ cancels Q cancel and C B is equal to E not. So this is the formula. So we want b is equal to a by c. So b is equal to e not is [clears throat] 3 into 10^2 and this is 3 into 10^ 8. So 33 cancels 10^ - 8 + 10 - 6 tesla. The right answer is second option.

[00:07:29]Second option is the right answer. Next question number six. The capacitors are initially uncharged. The three points A, B, C are maintained at potentials VA, VB, VC is given. The potential at X. Now calculate. So conservation of charge Q A plus QB plus Q C and the charge uh across the first capacitor second capacitor third capacitor is equal to zero. The charge on the first capacitor Q is equal to CV and I do C A into potential difference that is Vx minus VA plus QB charge on this capacitor CB this is CA and this is C C charge on B CB into VX minus VB plus charge on C is cc into vx minus vc is equal to z. So we want vx.

[00:08:45]So remaining term now shift shift CA minus CA VA minus CB VB minus C VC shift plus side CA VA plus CB VB plus C VC divided by CA + CB + CC CA into VB Uh so 1 micro into 4 1 into 4 4 plus CB into VB. So 3 into 1 3 + 6 into VC and 2 into VC is 1 2 / 1 + 3 4 + 2 4 4 + 2 is 6 5 + 4 9 by 6 3 2's are 3 are 3x 2 so this is 1.5 volt 1.5 volt but the right answer is the second option. Second option is the right answer.

[00:09:56]Next question number seven. During certain atmospheric processes, a pocket of air near the earth surface rises upwards very rapidly into upper regions of atmosphere.

[00:10:15]Actually heat temperature anywhere.

[00:10:20]So [clears throat] down the right answer is second option because adabatic cooling level one question. So next question number eight. A power plate capacitor is charged by connecting it to battery through a resistor. If I is a current in the circuit then the gap between the plates. So actually any conduction current inver DQ by DT but electric field between the plates E is equal to sigma / epsilon. Sigma is Q by A. So Q is equal to EA. EA is into epsilon. So Q place substitute. So epsilon is constant as epil and conduction current displacement current both are equal magnitude same even direction is also same. So the right answer is a displacement current of magnitude equal to I flows in the same direction as I displacement current conduction current value magnitude both are same. The right answer is the first option.

[00:11:41]Next uh question number question number nine. The wheel of a motor accelerates uniformly from rest in the first second it rotates through the angle 2 radian.

[00:11:58]What will be the angle through which it rotates in the next second and the second second? The formula theta is equal to omega t + of alpha t² omega t = to z because starting from rest.

[00:12:17]So theta_1 of ala 1 square theta_2. So of ala 2² is 4. So change in angular disment theta_2us theta 1. So of alpha 4 - 1 is 3. So of alpha value is given 2 radians.

[00:12:39]So this is 2 into 3 of alpha is 2 radian. So 2 into 3 is 6. 6 radian per.

[00:12:50]So the right answer is third option.

[00:12:56]Next question number 10. A spring lengthens by L and a block of mass M is suspended from it. This block is suspended from the same spring and the system is allowed to oscillate vertically in space where the gravity 1 nth of its original value the time period of small oscillation. So time period of oscillations of mass attached to a spring 2 pi roo<unk> m equivalent or l by g but time period does not depends on acceleration due to gravity. It depends only on mass and spring constant. So acceleration due to gravity new time period 2 pi roo<unk> m.

[00:13:57]So m okay the right answer is first option.

[00:14:07]The next question number 11.

[00:14:09][clears throat] An ideal spring obeys a hook's law.

[00:14:14]So f= minus kx mass is given and spring stretches x value is given the value of force constant. So the formula f is equal to magnitude kx. So we want k f by xx f is mg mg is 0.5 1 by2 g is 10 and x is 0.75 3x 4 10 - 1 okay [clears throat] so here mass is 0.5 x value 0.075 075 0.75 into 10 -1 3x 4 into 10 -1 mg divided by mg / x okay 1 2 shift 20 divided by 3 into 10 so 3 um 66 6.6 [clears throat] into 10 66 Newton per meter. The right answer is the fourth option.

[00:15:27]Fourth option is the right answer.

[00:15:30]Next question number.

[00:15:33]Well, a ball thrown vertically upwards is observed to move upward with speed v1 at time t1. and a speed v_sub_2 downward at time t2 ball upward.

[00:15:52]So velocity so v1 negative down the velocity positive acceleration the average acceleration so that is equal to v_sub_2 that is positive minus of minus v_sub1 because moving in opposite direction. So this is change in time average acceleration change in velocity divided by change in time V_sub_2 + V_sub_1 divided by TS2 minus T1. The right answer is second option.

[00:16:27]Second option is the right answer.

[00:16:30]Next question number 13. Two identical capacitors each of capacitor and C are connected in series and are charged uh by means of ideal battery or CMF.

[00:16:42]they are disconnected and reconnected in parl and connected to the same battery.

[00:16:47]During this reconnection, the positive terminals of the capacitors are connected to positive terminal of the battery and the negative terminals are similarly connected together. Then the work done by the battery during the first connection W1 during second connection. First connection W1 is equal to [clears throat] formula V= W by Q W= VQ. So V is nothing but E. So Q uh in the series combination uh and charge on first one charge on second one series combination charge on first capacitor charge on second capacitor charge on the whole combination same CS into V and E CS effective capacity if the capacitors are connected in series effective capacity 1x CS is equal to 1x C plus 1x C 2x CS is equal to C by 2 >> [clears throat] >> C by okay E into N C by 2. So which means C E² by 2. In the second case two capacitors disconnect reconnect.

[00:18:01]So now two capacitors are connected in par. Already they get charge C by2 initial charge C by2 C by2 C by2 plus E by the capacitors are connected in par charge add Q1 plus Q2 already by 2 D by 2 + C by 2 C. So battery connected CP 2 C 2 C into E. So delta Q S2 charge already C 2 C.

[00:18:55]So 2 C minus C delta Q. So this is charge required.

[00:19:03]2 C minus C C. So W2 is equal to E into delta Q.

[00:19:12]So E into delta Q is C and C into E C².

[00:19:19]Okay.

[00:19:21]W= C² is W2.

[00:19:27]Okay. C² / 2. W2X2. So W2 is equal to 2 * W1.

[00:19:34]The right answer is second option.

[00:19:37]Second option is the right answer.

[00:19:40]Next question number 14. The ratio of molar specific heat capacity at constant pressure to that at constant volume that is gamma. Gamma is equal to CP by CV. CP minus CV is equal to R. So in terms of degrees of freedom formula gamma is equal to 1 + 2x f gamma = 1 + 2x f Actually CV is equal to 3x2 FX2 FX2 FX by 2 RV = FX2 R. So now simplify CP is equal to R + CV. CV is FX by 2 R.

[00:20:53]So simplify R is actually common 1 + FX2. So R of 2 + F / 2. So gamma is CP by CV. So CP is u of 2 + 1 + fx 2. Okay.

[00:21:27]Divided by FX by 2 R. So R R cancel so 2 cancel 2 + F / F 2 by F + 1 1 + 2 F. So actually it does not depends on temperature equation equation observe two equations degrees of freedom dep by CV ratio depend so it does not depends on temperature gamma is proportional to t power zero it is independent of temperature right answer is the first option first option is the right answer.

[00:22:18]The next uh question number 15. A particle of mass m is kept at height 3 from the surface of earth 4.

[00:22:34]R is a radius of earth. M is a mass of earth. The minimum speed with which it should be projected so that it does not return. And e infinity is equal to ei.

[00:22:45]So E infinity is zero. Ei kinetic energy of M v² and potential minus G mm divided by radius is 4R. So G M / 4R is equal to M V². So Mm cancels here.

[00:23:05]goes on. So v² it is equal to g m / 2 r and v =<unk> gm / 2 r gm / 2 r power 1 by 2 gm / 2 r power 1x2 the right answer is the first option first option is the right answer.

[00:23:34]The next question number 16.

[00:23:38]At what distance on the axis?

[00:23:42]At what distance on the axis from the center of a circular current carrying coil of radius ordered as a magnetic field becomes 1x 27th of the magnetic field at the center. So BA is equal to 1x 27* magnetic field at center. BA is mu / 4<unk>i 2 2 pi ni r² / r² + x² power 3x2 so 1 by 27 is 3 okay magnetic field at center m by 4 pi 2 pi n i r² / x^ 3x2 x so then mu by 4<unk> 2 pi ni r² term cancel.

[00:24:33]So cross multiplication 3q r c 3q x cq is equal to r² + x² power 3x2. So now apply the cube root on both sides cube root on both sides. 3x = r² + x² power cube and cube root cancels 1x2 next squaring squaring on both sides squaring so 9 x² = I'm sorry uh so this is actually x=0 x=0 R.

[00:25:28]Okay. So this is not x r 3 r² r² + x² 9 r².

[00:25:41]This one is 9 r².

[00:25:45]This is 9 r² = r² + x². So x² = 9 - 8 8 4 into 2= 4 into 2 r² x =<unk> 4 is 2 2<unk>2 r the right answer is third option magnetic field x become z r² + 0 r² power 3x2 rq so that option is right answer Next question number 17. A block weighing 9.8 Newton is placed on a table exerts upward normal force that is 10 Newton and the weight 9.8 Newton. Uh the following statements which of the following statements is correct actually downward 9.8 8 upward 1000. So the block is moving upwards with acceleration. So this is 10 newton and this is 9.8 weight 9.8.

[00:27:05]Okay. The block exerts force of 10 newton on the table. Yes, that is true.

[00:27:11]Okay. Block exerts a force 19.8. Wrong.

[00:27:15]9.8.

[00:27:16]Wrong. The block has upward isolation.

[00:27:19]True. A and D statements A and D are correct.

[00:27:26]A and D only. Third option is the right answer. Next. Uh an iron ball sorry iron bar of length L has magnetic moment M. It is bent at the middle at its length. such that the two arms makes an angle 60° with each other. The magnetic moment of this new magnet like this.

[00:28:00]Okay.

[00:28:04]Lusmus QM. What is this plus QM? So this angle is 60° exact midpoint.

[00:28:21]So magnetic length the distance between two pole strengths magnetic this is equal triangle. Okay.

[00:28:39]So original So the original magnetic moment M is equal to QM into L. So the new magnetic moment Mdash is equal to QM into L by2.

[00:29:05]Okay. And this is M by 2. The right answer is the first option. First option is the right answer.

[00:29:19]Okay.

[00:29:22]Next question number 19.

[00:29:26]Okay. The two flaws shown in the figure have no ambient atmosphere. that is uh they exist in a chamber in which the atmosphere has been removed both flaws containing mercury and the same liquid to a height hight same liquid same and pressure same P1 is equal to P2 the volume of mercury in the second class is three times the volume of mercury in the first class the area at the bottom of the second fl is twice that of the first flask. Pressure P1 is the uh pressure at the bottom of the first flask. It is the pressure at the bottom of the second flask. Which equation holds good? Actually pressure inside the liquid does not depends on the shape of the container and P is equal to formula row GH. So your row GH complete constant P1 is exactly equal to P2 because H same density is same G same and P1 is equal to P2 the right answer second option second option is the right answer next question number 20. The planet has mass equal to planet has mass equal to 16 * mass of the earth and the radius of the planet is 4 * radius of earth. So escape velocity is earth is 2 gm by r.

[00:31:01]Planet is <unk>2 g mm becomes 16 m is 16 m divided by 4 r. So for 1 4 remaining that is Vicate planet divided by escapity of planet to that of earth is 2 by 1 2 is to 1. First option first option is the right answer.

[00:31:32]Next question number 21 for the given uh LCR series circuit.

[00:31:40]So XL is given X is given R is given.

[00:31:42]The value of power factor the formula cos= r / z where xl is more than x that is equal to roo<unk> r² + xl - xl - x² so z is equal to<unk> r² means 64 into 10² + x - x 130 - 80 is 50 [clears throat] uh 25 into 10^2 10^2 common take it as outside 10 64 + 25. So that is 89. So roo<unk> 89.

[00:32:22]So r value is 80. 80 divided by 10 roo<unk> 89. So 0 0 cancels here. So 8 divided by roo<unk> 89. The right answer is second option.

[00:32:35]Second option is the right answer.

[00:32:40]Next question number 22. For the uniform T-shaped structure with mass 3M, the moment of inertia about an axis normal to the plane passing through O. So here three rods connect. So system moment of inertia I= I1 + I2 + I3. So this is net moment of inertia at this point O. So I1 I2 I3 all are same that is 3 I. The moment of inertia of roden ais passing through just one end one.

[00:33:16]So ais moment of inertia is = ml² by3.

[00:33:23]So 3 into m l² by3 33 cancel and this is m l² m l². The right answer is the second option. Second option is the right answer. Uh next question number 23.

[00:33:42]Which of the following relation does not give the equation of an adabatic process? Actually adabatic means P V power gamma is equal to constant. But uh ideal gas equation PV is equal to mu RT.

[00:33:58]So P place RT divided by V. So v power gamma is equal to constant. Okay. So mu is constant and t v power gamma -1 is equal to constant. T v power gamma conant write t v power gus m / p. So um p into v mu rt divided by p power gamma is equal to constant. So p power 1 - gamma shift p power minus gamma p power 1 p power 1 - gamma mu power gamma r is constant t power gamma is equal to conant this is very important question so [clears throat] true okay false powers 1us so this This is not correct. The right option is fourth one. Fourth option is the right answer. Next question number before the following figure shows the arrangement of bar magnets in different configuration. Each magnet has magnetic dipole.

[00:35:28]Which configuration have highest net magnetic dipole moment?

[00:35:34]First case magnetic moments are at right angles and roo<unk> m² + m² + 2 m² cos 90 roo<unk> m² + m² 2 * m² so this is <unk>2 m second case magnetic moment or in opposite direction mus cancel so resultant magnetic moment roo<unk> m² + m² m² + 2 m² cos 30 is <unk>3 by 2 and m² is common as outside. So <unk>2 + 1.73 2 + 1.73 MR equ=<unk> m² + m² 2 2 m² + m² 2 2 m² cos 60 is 1x2 so 22 cancel roo<unk>3 <unk>3 m 1.73 m 3 this is maximum magnetic moment the cost of value cos of theta theta cos theta value.

[00:36:55]The next question number from T5 pi. For a sound wave traveling through the air in positive x direction the variation of pressure with position x the amplitude of the wave amplitude of the wave. So this is progressive wave equation for s= a into sin of omega t minus kx. So bulk modulus is given here. Bulk modulus quoted.

[00:37:31]So bulk modulus is 5 into 10^ 5 5 into 10^ 5.

[00:37:38]This is 5 into 10^ 5. So I think your formula you know modus is p / actually delta v by v. So uh this can also be written as dx divided by d s by dx d s by dx. But maximum maximum knowitude maximum.

[00:38:13]So the formula different with respect to x dx.

[00:38:19]So a into second of kx - omega t kx - omega t. So this is a k sin of kx - omega t. And this is the maximum value of ds by dx. This is the maximum value. So p maximum is equal to k a k a into b. So we want p maximum.

[00:39:07]This is P maximum maximum pressure and this is the wavelength lambda 0.2 2 into 10^ -1. Okay. So a is equal to P maximum divided by KB. P maximum is 5<unk> into 10^ 2 divided by K is 2<unk>i by lambda.

[00:39:27]Lambda is 2 into 10 - 1. So b 5 into 10^ 5 cancel 5 cancel 5 cancel 2 2 cancel okay a = 10 power shift + 1 5 6 6 shift - 6 + - 4 second option is the right Amplitude of the wind.

[00:40:07]Next question number 26.

[00:40:11]What is the direction of electric field at point? So test charge due to this semi circular path.

[00:40:21]Semic E negative off that is E minus E plus E minus both are moving so resultant E plus plus E minus the resultant positive X-axis first option is the right Question number 27. Consider the circuit shown and figure the current and ampere entering at the node A in one situation. The current is allowed to leave from node B. First situation B current.

[00:41:12]So but uh not from outgoing wire see outgoing.

[00:41:20]So first situation first situation but in the second situation allowed to leave from C but not from B that is second situation I1 I2.

[00:41:46]So in situation one situation leaving 10 15 10 15 both are connected here 25 same resistance the current equal that 10 is divided equally so that is 5 I1 Five I to 5. The right answer is the first option.

[00:42:21]First option is the right answer.

[00:42:27]Next question number 28. In a charge free region electric field lines can be taken to be discontinuous curves. Charge free electric charge no charge continuous but discontinuous that is not correct. Wrong statement. Concept of field was first introduced by F that is true statement when true is a false incorrect first one second one correct second option second option is the right answer number 29 imagine that there exists a planet whose mass and radius are both of that of earth.

[00:43:16]Uh the acceleration due to gravity on the planet will be the formula is g = gm by r².

[00:43:30]So mass of given radius of r² by 4 to 1 to 2 shift 2 gm by r² is g 2 g. Third option next note. A current of 2.5 a flows through a coil.

[00:43:49]Inductance is given magnetic flux with coil. The formula 5 is equal to l. Just a simple question. L is five and I is 2.55 by 2 25 by2 so this is 12.5 the right answer is second option second option is the right answer question number 31 given below are two statements law of electromagnetic induction is not consistent with law of conservation of energy that is not correct actually first So whenever the magnet first linked with the coord changes emf induces produces current and in mechanical power electric power it is based on conservation of energy but not consistent wrong lenses law is the consequence of what is the consequence of lenses law of conservation of energy that is a basis of lenses law this is true a true or false first option a True or false?

[00:45:00]Sorry, sorry, sorry.

[00:45:02]A is false or is true. A is false.

[00:45:08]A is false or is true. Second option is the right option.

[00:45:18]One false. Second one is Yeah. So next uh question number 32. A boy throws uh two balls A B from window of a tall multi-storyried uh building.

[00:45:34]Ball A is dropped from west and ball B is given an initial horizontal speed 30 m/s. The position of the ball is at a level of 18 m above the ground and balls are thrown simultaneously. Ignore air resistance. Take G is equal to 10. Ball A reaches the ground. Formula S is equal to UT + first situation S= U plus off².

[00:45:58]So dropped means UT first situation ball a dropped and hit UT goes to zero. So S is height that is 80. A becomes G that is 10 T² 0 cancel a 2 16 is equal to T².

[00:46:14]So T is equal to 4 second. The right answer is second option. 4 second is the right answer.

[00:46:22]Next [clears throat] question number 33.

[00:46:23]Choose a correct circuit which can achieve the bridge balance. Bridge balance.

[00:46:30]So P by Q is equal to R S P Q and R. What is this S diode P and junction? So this is P region. This is N region. P region is connected to negative. N is connected to positive. Right? So reverse bias.

[00:46:51]So wrong answer. So connected to positive and region is connected to negative but resistance is short and effective resistance.

[00:47:04]So this is also 15. So it is also not correct. So the P region is connected to negative and the reverse bias current.

[00:47:13]Wrong answer. The P region is connected to positive forward bias connection.

[00:47:19]So 10 ohm 10 + 5 15 10 by 10 is equal to 15 by 15 balance.

[00:47:28]Right answer fourth option is the right answer.

[00:47:33]Fourth option balance forward bias reverse bias identify.

[00:47:38]Next question number 34. The following four wires are made of the same material.

[00:47:46]Which of these will have the largest extension when same tension is applied?

[00:47:53]So the formula y is equal to f by a delta l by l. So f is tension. So delta l is nothing but tl divided by<unk> r² pi d² by 4. So delta L is proportional to L by D². In the first case delta L value is 100 divided by 1 square 100 / 1 square in the second case delta L is equal to 200 divided by 2² 4 is 50.

[00:48:34]So next case delta L is 300 by 900 93 something the last case delta hill is = 50 / 1 by 2² 1 by 4 200 maximum extension is maximum four case largest extension produced in fourth fourth case.

[00:49:14]Next question number 35. A vehicle travels off the distance with speed v and the remaining distance with speed 2 v. It's average speed direct formula that 2 into v_sub1 v2 divided by v_sub1 + v_sub2 because same distance. So the formula 2 into v_sub1 is v. V_sub2 is 2 v. So v + 2 v is 3 v. So v cancel this is 4 v by 3.

[00:49:43]Fourth option is the answer.

[00:49:47]The next question number is six. A piece of alloy of mass 250 g specific heat capacity 0.1 into that of water is placed in a furnace and then put into calorie meter containing to 40 g of water at 20°C. The water equivalent of calorie meter is 10 g. The final temperature of the mixture is 50°C.

[00:50:16]uh the temperature of the furnace is so heat lost is equal to heat gain formula.

[00:50:25]So alloy alloy 50= mc into delta t. So heat loss alloy. So mass of alloy mass of alloy 250 250 g into specific heat of alloy 0.1 times specific heat of water.

[00:50:51]So into delta t so temperature so tus 50 t is maximum. So 240 + 10 water equivalent of the calorie meter 10 g 240 + 10 250 into specific heat of water into temperature difference 50 - 20 the initial temperature is the final temperature 50 - 20 so 250 CW cancel so T - 50 is equal to 50 - 20 is 30 shift 30 into 10 - 1 300 300 + 50 350 350°C this is 350°C the right answer is second option Second option is the right answer.

[00:52:10]Question number 37. The coefficient of cubical expansion of water negative from 0°C to 4°C because from 0 to 4° the water volume gamma is delta v / v into delta t. So this is positive with a positive delta V final volume minus initial volume. So initial volume is more than the final and then minimum minus maximum negative delta V delta V VF minus VI. So this is negative because VF is less compared to VI. So the right answer is first option 4° C.

[00:53:09]Next question number 38.

[00:53:12]A spring uh gun of spring constant 90 newton per cm comes to 12 cm by ball of mass 16 g. The trigger is pulled at the velocity of the ball. So of m v² = k x² elastic potential energy kindinetic energy. So we want velocity alone. So v² is equal to k.

[00:53:40]So k value is put 90 newton per t and 10 -2 x² 12 into 10^ -2 12 into 10 -2 divided by m is 16 into 10 - 3 kg. 4 are 4 3 4 3 4 1 3 3 are 3 into 3 9 9 into 9 9² 9² 10 uh 10 into 10^ - 10^ - 1 shift + 3 - 1 2 9 into 10^ 2 so v² is equal to 9² into 9 square into 10 square. So v is equal to 9 into 10 90 square root roo<unk> 9 square is 9<unk> 10 square is 10 9 into 10 10 9 into 10 90 m/ second.

[00:54:47]The right answer is fourth fourth option.

[00:54:54]Question number 39.

[00:54:56]A non ideal coil, a capacitor and an AC source of RMS voltage VRMS is given here.

[00:55:06]VRMS put V RMS 24 ving frequency maximum current is given maximum current 6 amp. So current maximum LCR series that should be in resonance X= XC and V RMS is equal to I RMS into R. So current becomes maximum because impedance minimum that is R. So R is equal to V RMS is 24 IRMS maximum current is 6 1 4 the resistance is 4 ohm. This is first part. Now the car is connected to DC battery of EMF internal resistance DC current. The formula is equal to R + E / R E is given 12. R is 4 small R is 2. So 12 by this is DC 4 + 2 is 6 1 are 2. This is two and the right answer is third option.

[00:56:09]Option is the right answer.

[00:56:13]Question number 14.

[00:56:15]A circular metallic wire is connected at the opposite ends of diameter to the battery. The magnetic field at the center of the circular wire B1 B1 actually B1 first segment magnetic field inverse. Second segment magnetic field is outwards. Resultant magnetic field zero. magnet only inverse second case. So here B2 is not equal to zero.

[00:56:49][clears throat] On the other hand the terminals of the battery connected to the ends of the wire the wire is shaped into full circle. Imagine if we at the center is B2. This is B2. B2 is not equal to Z. B1 is equal to zero. B1 0 B2 is not equal to 0. Right answer is fourth option. P1 is zero. B2 is not equal to zero. Next question number 41. Mass is attached to thin wire and while in a vertical circle the wire is most likely to break when the tension becomes maximum. Tension is maximum when the mass is comes at the lowest point. Lowest point.

[00:57:31]Okay. So mg maxion net force t minus mg actually= mv² by r. So maximum tension is mg + mv² by r. So the mass is at the lowest point only breakout chances because tension becomes maximum at the lowest point.

[00:58:05]Okay. So next question number 42.

[00:58:09]Let the wave number of a sound wave be nut.

[00:58:15]Okay. The wavelength of wave be lambda.

[00:58:19]The dimension of the product. So n is 2 pi by lambda. The lambda lambda lambda cancel 2 pi is dimensionless. Fourth option is the right answer.

[00:58:31]So next question number 43.

[00:58:34]A 1 kg object strikes a wall.

[00:58:40]1 kg object.

[00:58:43]It will strike say wall bounce 60° with the wall.

[00:58:56]Okay.

[00:58:58]Two components mg sin 30. Okay.

[00:59:09]So this is 30. This one is 30. So what is this? Mg cos 30 and this one is mg sin 30. Mg sin 30 mg sin 30 same direction same value. So change in momentum along y direction zero. So force is change in momentum along x direction mg cos 30 plus mg cos 30 minus of minus plus 2 mg cos 30 / delta t. So f is equal to 2 into mass 1 kg g 10 cos 30 is <unk>3 by 2 divided by delta t 0.1 - 1 so 2 2 cancel 2 m into g 2 mg 2 into 1 into gan cos 30 cos 30 is <unk>3 by 2 okay [clears throat] sorry mg so this is actually velocity momentum 2 mv this is 2 mv Actually change in momentum divided by So this is divided into two parts. This one is also divided into parts.

[01:01:23]So MV initial final 60 angle made by uh you know direction of velocity in the wall this is 60 this one is 30 mv cos 30 mv cos 30 mv sin 30. So this is uh this is angle 30 because angle of incidence angle of reflection same angle 30. So MV cos 30 MV sin 30. So change in momentum along Y direction.

[01:02:12]So F is equal to change in momentum along X direction 2 * MV cos 30 divided by delta T. So f = m is 1 v is 1 cos is <unk>3 by 2 / delta t is 10^ -1 2 * 2 cancel tan <unk>3 f = tan <unk>3 the right answer is third option third option is the right answer next question number 44. A mixture of two or more ideal gases does not follow the ideal gas equation does not follow the first law of thermodynamics as same RMS speed for each gas has same average translational kindinetic energy of each type of gas molecule. This is right answer because average kinetic energy of each molecule 3x2 KBT depends only on temperature does not depends on nature of gas. Different gases through more number of gases through average kinetic energy of each type of gas molecule that will remain same. Fourth option is the right answer.

[01:03:33]Next question number 45. During simple harmonic motion of a body the energy at the extreme position.

[01:03:41]So during simple harmonic motion at extreme position the body has only potential kinetic is zero purely potential. Fourth option is the right option one question.

[01:03:55]So okay thank you everyone complete watch.

[01:04:02]So all the very best videos like money.