AL 2026 Physics Support Seminar by Ministry of Education

[00:00:00]2026 support seminar paper by Ministry of Education. Question number one expression is given as a = a into e ^ - a t / a minus doesn't matter here in the dimension analysis the signs don't affect but we know the theory the quantities in the powers should not have dimensions. Therefore dimension of a to m should be equal to 1. Therefore dimension of a should be equal to m / t that is mtus one. The best answer is fifth one. Question number two circular scale equation is given by r = x + ny.

[00:00:52]X is the main scale reading preceding the zero of the circular scale.

[00:00:58]X is a main scale reading preceding the zero of the circular one scale that is 26.5° and n is a number of one scale divisions coinciding with the main scale. You can see there is seven here. Then we have to find out y is a least column of the scale. This count can be get from the equation one main scale division over total number of circular scale division.

[00:01:26]The main scale simulus division is 0.5° half degree. Number of uh circular scale division totally 30. Therefore 1 / 60° you know 1° 60 minutes. Therefore 1 60° 1 minute. We can substitute in the equation 266.5° + 7 into 1 minute that is equal to 266°.

[00:01:54]This is 30 minutes. Half degree means 30 minutes. There is uh 7 the half degree is 30 minutes. Here 7 minutes are there. Therefore totally 37 minutes are there. The best answer is third one.

[00:02:12]Question number three. Five displacement versus time graphs are given and they asking the acceleration.

[00:02:20]Therefore we know the theory the gradient of SD graphs gives velocity.

[00:02:23]First we will find the velocity. The straight line gradient is constant and acute angle gradient is plus. Here also straight line for gradient is constant but as it is obtuse angle it is minus.

[00:02:35]Here the gradient is high at the start and decreases towards zero.

[00:02:43]This is the third figure. Fourth one.

[00:02:46]Initially the gradient is high but in minus it's also moving towards zero.

[00:02:52]Here the initial gradient is zero increases in minus. These five are the v graphs. In these two graphs the acceleration is zero. The rate of change of velocity is zero. The gradient of BT graphs gives acceleration. There is no change of velocity. Therefore, accelerations are same. In the third figure and the fifth figure, accelerations are minus. The gradients are negative. Therefore, accelerations are minus. In fourth figure, acceleration is plus. Therefore, answer is fourth. Here the speed is decreasing but in minus. The speed decreases in minus. Therefore, it is acceleration. We can get from the gradient also. The angle is acute. Therefore, gradient is plus. Therefore, it is acceleration.

[00:03:41]The first two figures acceleration is zero. Third and fifth acceleration is minus that mean deceleration. In the fourth answer, it is accelerating.

[00:03:51]Therefore, question number three, the best answer is four. Question number four, there are three masses. Each of them is 2 kg.

[00:04:00]Therefore, the weight is 20 Newton. Here also 20 Newton totally 40 Newton in the right side. Let's try 20 Nton. Therefore acceleration is equal to net force over total mass. 40 - 20 net force is 20 Nton. Total mass is 6 kg. Therefore the acceleration is 10 by3.

[00:04:21]Then we have to find the tension.

[00:04:24]It is accelerating downward with 10 / 3.

[00:04:29]Tension is upward. If we can apply the equation f MA we can apply the equation f= ma downward to c 20 wait tension upward is equal to 2 into 10 / 3 then tension is equal to 20 - 20 / 3 is 40 / 3 that is 13.3.

[00:04:51]The best answer for question number four is 10.

[00:04:58]The first one will be the answer tension will be zero when at free fall. Second will be the answer if they asking acceleration and fifth one is greater than 20 that will happen when it is accelerating upward. Now it is accelerating downward.

[00:05:13]Therefore the tension should be less than 20. Question number five.

[00:05:21]The frequency of a transverse wave is given and the velocity also given. We can find the wavelength. Apply the equation V= F lambda. Lambda is equal V.

[00:05:31]That means 60 / 60 is 1 m.

[00:05:35]And we know the relationship between the wavelength and the phase difference. If it is one full wavelength, the phase difference is 2 pi rad.

[00:05:44]Therefore, for 1 m, this is 2 pi rad.

[00:05:48]They're asking between two points at 25 cm. That means 1x 4 m. 25 cm means 1x 4 m. Therefore we divide by 4 therefore 5 by 2 radian. Question number five answer number three is solve.

[00:06:08]Question number six.

[00:06:10]An object of 2 kilogram is kept on an inclined plane at 30°. The weight is 20 Newton. We can make it into two components. This is 20 sin 30.

[00:06:22]This is 20 cos 30.

[00:06:27]Therefore the weight component is 10 Newton. And what are the other forces?

[00:06:31]There will be a friction force opposing the motion that is mu into R. R is equal to for the equilibrium in the perpendicular direction of the plane. R is equal to mg cos theta that is 10 <unk>3. Therefore friction is equal to half into friction coicient of friction is half in reaction is 10 <unk>3 that is equal to 5 <unk>3 <unk>3 is given as 1.7 into 5 is equal to 8.5 newton therefore along the plane the weight component is 10 newton the friction is only giving 8.5 newton we to give extra 1.5 newton to balance it to keep in equilibrium we have to give extra 1.5 newton And therefore the best answer for question number six is first question number seven.

[00:07:26]A ray of light enters from vacuum.

[00:07:28]Vacuum means the refractive index is 1.0 and into the medium of N.

[00:07:35]There's a data angle of incidence is twice the angle of refraction. Let's mark angle of incidence I angle of refraction as R.

[00:07:45]The data is given I = 2R and at refraction we can do syn X 1 sin I = N sin R.

[00:08:04]Then we can use I = 2R sin 2R = N into sin R. Sin 2R can be written as 2 sin I cosine.

[00:08:18]Sin 2 R can be written as 2 sin R cos R into sin N sin R. The both sin R and sin R will be cut off cos R is equal to N /2.

[00:08:33]F_sub_R is equal to cos inverse N /2.

[00:08:37]There's a chance of KS by putting one that is angle of refraction. But the question is angle of incidence.

[00:08:45]Therefore, don't forget to multiply by two.

[00:08:49]I = 2 R. Therefore, 2 * cos inverse N / 2. For question number seven, answer number three is correct.

[00:09:02]Question number eight.

[00:09:04]Five equal forces are acting at equal angle symmetrically then the resultant is obviously zero. Otherwise you can draw vector polygon. If you draw all five forces in order they will complete the vector pendant. Therefore the resultant will be zero.

[00:09:22]Here they didn't mention angle. The angle will be 72°. That means 360 divided by number of forces five that will be 72.

[00:09:32]If there are four forces equal forces acting at 90° the resultant will be zero. If there are six equal forces acting at 60° with within each other result will be zero. Therefore here five angles acting here five forces acting symmetrically the resultant will be zero. Five here the five equal forces acting symmetrically therefore the resultant is zero. Question number nine.

[00:09:56]Total pressure is equal to sum of static pressure and the dynamic pressure. The total pressure is given as 8,000.

[00:10:04]Static pressure is given as 6,000.

[00:10:06]Therefore, dynamic pressure is equal to 2,000.

[00:10:12]We know the equation for dynamic pressure half row v² is given as 2,000.

[00:10:18]The density is 1.2 into 10 of 3.

[00:10:21]Therefore, half into 1.2 into 10 of 3.

[00:10:25]Velocity squared is equal to 2,00 3 and 3 will be cut off.

[00:10:31]If you cross multiply V² is equal to 4 1.2 from that V = 2 /<unk> 1.2.

[00:10:39]The question is flow rate. Flow rate is given by Q is equal AV crossel area is given as 2.5 into 10us 3. The velocity is 2 /<unk> 1 by 2. So multiply 5 into 10 - 3 /<unk>2.

[00:10:58]of 1 by 2 is roughly equal to 1.

[00:11:00]Therefore the most suitable answer is 5 into 10 - 3 second minus one. Answer is second.

[00:11:11]The actual answer will be 4.78 around.

[00:11:15]If you approximate the best answer is second bar and no need to worry in final papers the calculation will be easier not like this. Question number 10.

[00:11:26]Energy stored in the unit volume.

[00:11:28]The equation is equal to half into stress into strain.

[00:11:33]Here stress is not given by the J's modulus. We know the relationship between the stress and strain. Stress of strain is equal to J's modulus.

[00:11:41]Therefore, we can substitute stress is equal to J's modulus into strain. If you substitute, you will get the equation J's modulus into same square half. Junk modulus is given as 2 into 10 ^ 10. Strain is given percentage. Don't forget to make it a fraction. 0.06 over 100. Therefore, 0.06 squared over 100 squared.

[00:12:10]Therefore, 2 and 2 will be gone. 6 into 6 is 36. There will be only one answer in 36. First answer is correct. We can count the zeros also. uh 2 to 4 uh 6 8 10 of 10 minus t of 8 is equal to 10 of 2. Therefore answer is 3.6 into 3 jou mus 3. The answer is first question number 11.

[00:12:45]Container of depth D is half filled with a liquid of refractive index mu1 and the remaining is filled with mu2 and the depths are dx2 dx2. We know the equation refractive index is equal to real depth of apparent depth. Therefore we can subject append depth is equal to real depth over refractive index.

[00:13:06]Therefore, apparent depth 1 is equal to dx2 / mu1. Apparent depth 2 is equal to dx2 / mu2. The total apprent is equal to if you add both of them dx2 into 1 / mu1 + 1 / mu2. Answer is fourth one. Question number 12.

[00:13:38]A rim is rolling.

[00:13:44]Therefore, we asking the maximum height.

[00:13:48]Here at the maximum height, the velocity is zero. Therefore, all the kindinetic energies are zero. Therefore, the loss of kinetic energy is converted as gain of potential energy. The loss of kinetic energy is converted to gain of potential energy. Here it has two components of kinetic energy. First one is rotational kinetic energy. Other one is linear kinetic energy. Both are converted to gravitational potential energy.

[00:14:17]Rotational kinetic energy equation is half omega². Linear kinetic energy is half m² is equal to mgh.

[00:14:27]Moment of fatia of rim is m r² into omega² plus/ m v² is equal to m. In this question is independent. It will be cut off. As it is rolling without slipping we can just say equation v is equal to r omega. Therefore it is half v². Here also half v² is equal to gh. Therefore add them v ² is equal to gh. The question is h is equal to v² g. Question number 12. Answer number two is square.

[00:15:02]If it's a disk, it will be another half.

[00:15:06]Therefore, answer will be 3 v² [clears throat] over 4.

[00:15:12]Question number 13.

[00:15:17]The velocity of light in a medium of refragment x3 is given by n= c / v. Therefore, velocity is equal to c. C is velocity of light in vacuum.

[00:15:31]And is a refractive index. Therefore, velocity is given by 1 into 10^ 8 ms - 1. There's a time taken to travel through a distance of 4 mm.

[00:15:43]Therefore, equation of velocity is equal to distance over time. Then time is equal to distance over velocity.

[00:15:49]Distance travel is 4 mm.

[00:15:52]Velocity is 1 into 10^ 8 m/s. Then answer is 4 into 10^ of - 11 second. The best answer will be second.

[00:16:05]Fluid flowing through two capillary tubes in parallel connection. You know in series connection flow rate will be same. Pressure difference will be added.

[00:16:14]In the parall connection pressure difference will be same. The flow rate will be added. The total flow rate is equal to sum of flow rate through each pipe. And we know the equation for the flow rate. Q = pi 4 delta p / a e.

[00:16:34]Here everything is constants except radius. Here radius is R. The second tube radius is R /2. If we know the flow rate is proportional to R^ 4. The radius is half the flow rate will be 1 / 16 times. Therefore 8 into 1 / 16 is half cm cube second. From the first tube the flow rate is 8. From the second tube flow rate is half. If you add both of them the total flow rate is 8.5 cm second minus one. The first answer will be second answer will be correct.

[00:17:14]Right? In question number 15, the maximum heights are equal in three instances. Therefore, equation for maximum height is equal to v² / 2g.

[00:17:24]Maximum heights are equal. The gravity is equal for all of them. Therefore, vertical velocity should be equal to all of them.

[00:17:33]Answers 2 or 3 or five are possible.

[00:17:38]and time flight time is equal to 2 vertical velocity / g. Therefore g is same for all of them. Vertical velocity also same. Therefore flight time also same for all of them. If among 2 3 2 is eliminated 305 will be solved. And for the horizontal range we can apply c is equal to ut time is same for all of them. Therefore, if U is greater, then only the horizontal range will be greater. Therefore, J3 is greater than J2 is greater than J1. The best answer is third one.

[00:18:20]Question number 16.

[00:18:22]The mass ratios of the man in the vehicle is 1 is to 4.

[00:18:29]Therefore the distance traveled will be in the ratio of 4 is to 1 and the total distance is 5 m. Therefore we can get the distance traveled by the man is 4 m and the distance traveled by the vehicle is 1 m.

[00:18:46]Otherwise you can you can directly do this. Otherwise you can do say equation distance is equal to mass total distance over total mass mass of the if you want the mass if you want the distance tra by the man you have to put the mass of the trolley 200 the total distance is 5 m over total mass is 250 then if you divide the answer will be formed.

[00:19:20]You can do from the fundamental also.

[00:19:22]There's no net force acting on this system. Therefore, we can apply conservation of momentum also. Initial momentum is zero. Therefore, final momentum should be also zero. Let's take the distance traveled by the trolley in the opposite direction is x. Therefore, when the man moves a distance 5 m relative to trolley, trolley moves a distance of x backward. Therefore the distance traveled with the man related to the earth is 5 - x minus you're putting minus because trolley is moving in the opposite direction. The mass is 200 the distance is x to make it as velocity you can divide by time t both are simultaneous time they will be cut off. Then if you bring it left side 200 x is = 50 into 5 - x 50 and 200 cut of 4 4x = 5 - x 5x = 5 x = 1 you can do by this method also otherwise you can use you can find the center of gravity initially and finally they should be at same place from that method also you can find question number 16 answer number two will be correct Question number 17. It is given as equilateral prism. Therefore, every angle is 60.

[00:20:44]And it is mentioned that angle of incidence is 3x4 times of angular.

[00:20:50]Therefore, I = 3x4 into 60 45.

[00:20:55]The question is N. N is equal to sin I / sin R. I is known. We have to find R and the data is given. The angle of incident is equal to angle of emergence mean it is at minimum deviation. In the minimum deviation for equilateral oymmetrical prism the refracted r will be parall to the third phase. Therefore this is 60. This is also 60 equilateral triangle. The angle of refraction will be 30 because it is normal.

[00:21:31]Normal end service is 90° it is 60. Therefore remaining complimentary angle is 30. You can substitute n is equal to sin 45 / i is 45. R is 30. Sin 30. Sin 45 is 1 / <unk>2. Sin 30 is half. Then the fourth of 2 <unk>2 is equal to <unk>2.

[00:21:54]Or otherwise you can do from the direct equation n is equal to sin a + d minimum by 2 over sin a bit by here you can find the d minimum this a long way but you can do it also d minimum is equal to 2 i minus a i is 45 therefore 2 i is 90 angle prism is 60 minimum is 30 then if you substitute a 60 D is 30. It will be sin 90 by 4 2 is 44. Sin A by 2, sin 60 by 2 is sin 30.

[00:22:32]Therefore, here also the answer will be same. Question number 17, answer will be 40.

[00:22:40]Question number 18. Statement A is a definition is correct. In Asia is the reluctance of an object to change its state of motion is correct. Statement B.

[00:22:54]The force required to position object on a smooth horizontal surface is depends on the gravitational mass of the object is wrong. Because we are applying F is equal to MA. It's only depends on the inertial mass. It is independent on the gravitational weight or mass. It's independent on the gravity unless there's no friction.

[00:23:17]If there's a friction the gravity effects otherwise on a smooth plane it's only depends on the initial mass and independent on the weight or gravity. But in the C statement initial mass is measured by the spring balance is wrong. The spring balance measures the weight that means gravitational mass. The gravity effects in the moon and earth the weights are different. An object weighing 100 Newton on earth will measure only 70 Nton on moon. We already studied in order.

[00:23:55]If you measure the weight by a beam balance, it will be same whether you are measuring on earth or on moon. It is independent on the gravity. Therefore, a beam balance will measure the initial mass. A spring balance will measure the gravitational mass. Therefore the question B is also wrong. Question C.

[00:24:18]Therefore the statement B is also wrong.

[00:24:21]Statement C is also wrong. The correct answer is first one. Statement A is only correct.

[00:24:32]Question number 19. The hydrometers is given. Let's take the volume of the bulb is VB and the volume of the whole stem is VS. When it's mentioned as zero, it is floating in water. That means the relative density is one.

[00:24:49]It's showing 10. When it is floating in a liquid of density 1.5 when the hydrometer is floating, the weight of the hydrometer is equal to weight is always constant. Therefore will be also always equal. Upress is equal to V displace density of the liquid in G. G is always constant. Then we can always equate we can always equate the product of immersed volume and the densities of liquid.

[00:25:19]Initially VB when it is floating with a density of 1.5 and uh V B + V steed when it is floating in water. From that we can find the volume of stem.

[00:25:38]V ste is equal to half V.

[00:25:44]The question is the density at the middle in a liquid of density row bar it shows level five VB into 1.5 is equal to now VB plus half of the volume of stem the volume of stem is VB by 2 it's half of the volume of stem therefore VB by O into density of the liquid row bar. You can substitute VB. This is 3x2 5 by 4 into row bar.

[00:26:26]VB will be cut off. And if you cross multiply row bar is equal to 12 / 10 that is 1.20.

[00:26:34]Question number 19. Answer number three is correct. You can check by mind also in a hydrometer you know the densities are increasing downward but in a compressing manner it's a compressing scale nonlinear scale therefore 1.25 25 will be not in the middle that will be close to the 1.5. Therefore, you know the densities at the midpoint is less than 1.25. You can eliminate either 40 or five among the first three answers. The suitable one is 1.2.

[00:27:18]For question number 90, third answer is this is radius. This is also radius. The angle is theta.

[00:27:29]This is also theta because it's an isocles triangle. Therefore, distance traveled by the bead is r cos theta. R cos theta. Therefore, s is equal to 2 r cos theta.

[00:27:45]If acceleration also in this direction the vertically acceleration is gravity g in this direction g cos theta. We know the distance travel and we know the acceleration. The question is time no.

[00:27:59]Therefore we can apply s is equal to ut +/ a² s is 2 r cos theta u is c starting from rest half g cos theta into t² cos theta and cos theta will be cut off 2 and 2 will be 4 r / g = t² therefore t is equal to roo<unk> of 2 r g four will come out as 2. For question number 20, first answer is correct.