Doctors medical academy khammam is live

Added: 2026-04-29

[00:00:01]And I think okay Is it water?

[00:00:56]>> Okay.

[00:00:59]Part test. Last one.

[00:01:06]First one.

[00:01:09]A ray is incant at an angle of incidence I.

[00:01:14]So first I uh and emerges normally from the opposite face.

[00:01:27]Angle of incidence zero, angle of refraction Z is 90.

[00:01:32]The angle made by the refractor with the normal is known as angle of refraction 90= I for small I= Second develop lens makers formula the convex of focal length 25 cm and made of glass of pre is 1.5 is immersing water change in focal length of the lens is change in focal length absolute and plus first focal length 25 cm.

[00:02:35]So first air 1 by 25 is equal to new class that is 1.5 into something something right next same thing just the same thing cancel so h by 25 = 1.5 5 - 1.5 and 1 by 2 3x 1.5 3x 3 3x 2x 4x 3 9 by 8 - 1 1 by 8 x= 100 so 75 change Third test formula a point source of light is kept below the surface of water. The radius of the circular path the formula for r if you remember h by roo<unk> of mu² - 1.

[00:03:44]So here radius h means roo<unk> 7 m roo<unk> of mu² 7 u mu 4x 3 and 16 by 9 - 1 that is 7 by 9<unk> 7 roo<unk> 7 cancel root 9 and 3 3 cm 3 3 m fourth term all zero for the red diagon shown in the general false statement is the formula for D that is angle of deviation is given by D= I1 + I2US A where A= R1 so second one correct first one correct so third fourth is equal to sin I1 by sin I1. Similarly sin I21= I2 by sin sin I1 by sin R1. That is third option is correct. So wrong is fourth.

[00:05:08]Angulation for sin of a + 2 and 60 d 60 + 30 by 2 90 by 2 sin 45 by sin a by 2. A mean 60 by sin 1x<unk>2 by 1x2 2x <unk>2<unk>2 area of light passes from glass glass out of 1.62 is 28. So density where it moves away from the angle of incense which the angle of refraction is twice the angle of incidence angle of the angle of inance. Okay.

[00:06:10]1= sin 2 I by sin a sin 2 sin I cos by sin Sin A sin cancel 8 = cos I = cos inverse 4x 5 or sin inverse 3x 5 8 by 10 8 by 10 cos I is equal to adjent by hypotenuse adjent by hypotal inverse 3x Seven level zero astronomical telescope normal adjustment. You have to understand even nothing is mentioned tube.

[00:07:08]So f_sub = 5 f_sub by f_s = 20 clear 100 by 5 20 100 + 5 f 100 sol from this f from this f by 20 f what is f from this 5 - f= sub then you get 100 100 is the Five.

[00:07:43]Eighth one.

[00:08:00]And so I'm going to light one previous question 180 - 2 index is a by 2 So mu is equal to mu a= sin of a + 2 by sin by the formula and cos by sin.

[00:08:40]So cancelus is nothing but cosal of a + 2 180 by 2= 2 180 - 2 a convex mirror.

[00:09:10]has focal length f that mean of course a real object is placed at a distance when the object is at focus image will be at infinity that is not true for convex m or concave lens that is only true for concave mirror and convex lens not here u = we know the formula 1x f= 1 v 1x u - 1x v = 1x f + 1x that is 2x therefore u= fos within the purpose 10th one level zero In red light travels faster than others that equal velocity of light 3 into 10 level. So 1x font lo formula 1x f = 1x f_sub_1 + 1x f_sub_2 1x f2= 1x f_sub_1 in same plus 1x f_sub_2us 1x 10 so 1x 10 - 1x 10 f= infinity it behav motion velocity of sound for a particular sound. The path difference is equivalent to phal 1.6 6 the formula is delta 5= 2 by lambda into delta x that is that is called as lambda which is given by d by 2 by 1.65 65 into 40 cm 5 cancel 2 8 5 and 50 cm N is equal V by lambda V is given as 330 m/s by lambda 50 by Okay.

[00:11:48]660.

[00:12:07]Y= A.

[00:12:11]Maximum particle velocity particles are in SHM SH velocity mean position that J omega and Y omega 2 pi is equal * veloc by 2 14th a wave equation which gives the displacement along the y direction is given by okayave traveling along.

[00:12:55]So first formula 2<unk> by lambda = 2 lambda =<unk> frequency omega frequency 60 = 2 pi n = 30 by pi correct so that one is correct 15th level zero.

[00:13:28]Sound waves do not exhibit the property reflection, refraction, depraction, polarization, polarization.

[00:13:35]Sound waves can be draed easily.

[00:13:53]sound waves.

[00:13:59]So polarization it explains only transverse extra 16 level zero frequency of first two of closed first and fundamental first and third 1 3 3 V by 4 LC is equal to that of third fundamental first second third 1 2 3 4 open so 4 V by 2 L the ratio of L by C cancel.

[00:14:33]So the answer is L by LC = 8 by 3 8 to 3 17.

[00:14:45]The equation traveling the medium is given by K. The velocity we can omega means 600 by K.

[00:15:08]So three four 18th level zero this definition a line of surface on which the disturbance has the same phase at all points is called as wave front equipotential surface is equif known as wave and light ray is always perpendicular to wave just like electric lines are perpendicular to equip potential surface.

[00:15:39]So this also v= x and k speed of pro same y= omega by k 3 m. So 300 20.

[00:16:03]So density if at the same temperature and pressure the densities of two diatomic gases the ratio velocity of sound the formula for velocity of sound is root of gamma rt by matic constant constant Let me use V= P by D because answers are given. So now gamma P is same. So V is inversal to roo<unk> V1 by V= root of D2 by D1 21.

[00:16:51]The equation for the vibration fits at both ends in third harmonic that is given by a stationary wave y= 2 a cos a sin omega third harmonic third harmonic three loops 1 2 3 lambda by 2 lambda by 2 lambda is nothing but 3 lambda by 2 leng of the lambda 6 lambda is equal 0.6 lambda = 2<unk>i by 6 by 10 20 by 6 3x 2 into 20 by 6 5 pi 5 5 into 3 15.7 20 to level zero. It is observed to have resonant frequency at 420 and 315.

[00:18:00]They are successive frequenc. The difference between any two successive frequencies equal to fundamental lowest frequency that is also true for strings and also open pipe. But whereas in the case of closed pipe that is equal to twice the fundamental. So lowest frequency means difference 85 20.

[00:18:28]Now it is capacitors.

[00:18:47]When an additional charge of 2 kum is given to a capaci energy store increase the original charge the formula for energy is q ² by 2 c² so therefore u1 by u2 is =1 by q2² first increase 21% 121 q1 H+ square root 10 by 11 = x by x + 2. So 10 x + 20 = 11 x s x = 20 24 level 2 I think you'll get 3x 22 or 22 by 3 c 23 This is C 2 C 3 C in force connected to the rest of charges are C2 and C4 C2 and C4 series capac 2 C + 1x 3 C LCM 6 C 6 + 3 + 2 11 6 C by 11 so I'm putting 6 C by 11 already 4 C connected to V= so 6 C by 11 4 C 4 C by 11 6 C by 11.

[00:21:05]So 6 C by 11 V / 4 C cancel 2 2's are 3 3 is to 20 25 level formula C square= half into C1 into 1 square So 26 formula.

[00:21:39]Okay. Electrical between the plates sigma by epsilon KN. So, a key q potential difference v two capacitors with capacitor C1 and C2 are charged to potential V_sub1 and V_sub_2. When they are connected in parallel, the ratio of their energies charges are here. Q is equal to C adjacent V Q is proportional to C. Q1 by Q is equal to C1 by C2. First option 28 option 28 So anyway it is found that by together the potential and one of one can be made 0 each one can be made zero and the common potential is zero each. So we formula C1 V1 C1 into 120 plus C2 V2.

[00:23:08]So into 200 by C1 + C2 for zero. So therefore C 120 12 C1 is equal to 20 C2 3 4 4 3 4 3 C1 is equal to 52 C1 C2 capacitor never so 29 9 the charger four microfarad capacitors.

[00:24:07]Okay.

[00:24:21]connected 8 vol 4 vol the charge Q= C 4 16 30th level Z a PPC charged a potential difference B After disconnecting the charging battery and Q is constant. When battery is disconnected, Q is constant. Remaining connected V is constant.

[00:25:04]The distance between the plates capacity is increased. C= A. As D is increases, C decreases.

[00:25:11]As a result, potential difference between the plates. So V Q is equal to CV RAM. Q constant V is inversely proportional to C. You see like paragali.

[00:25:21]So increases that one horizont and m2 moving velocities opposite direction. m_sub_1 v_sub_1 m_sub_2 v_sub_2 elastible when they are of equal masses but in fact that also should be given v1 v2 opposite so they exchange their velocities they interchange their velocities means Elastic important elastic elastics.

[00:26:36]Another way to two identical particles M going moving in opposite direction with the velocities V 2 V. Okay.

[00:27:15]equation value of = 1 m 2 for mus 1 + M into U1 + M2 2 M2 by M1 + M2 into U 33. A ball hits the floor ands= x2 remains same during the collision. Yes, correct. There is no loss in kindin energy. kind doesn't depend upon directional correct the moment of the ball just after the collision is same as just no because direction change moment same the total moment of the ball and earth is conserved total moment of the ball and earth is consider Okay, third option is correct.

[00:29:07]they regain it original size and shape.

[00:29:10]So after collision, before not during during 34 ball A of mass 50 g moving with the velocity 10 m/s collides with other ball be of mass 10 g opposite direction 15. Okay.

[00:29:40]Determine the final speed of the B2 is 2. Again, V2 = formula M1 into another number.

[00:30:13]Unknown plus m_sub_2 - m1 e by m1 + m_sub_2 into m1 and 50 into 1 + e 1 + 2 x 5 7 by 5 by m1 + m2 60 into u1 10 - 15 m2 - m1 m2 10 - 15 into 50 into 2x 5 5 * 20 by m1 + m2 60 cancel so that is 0 0 cancel 5 70 by 6 - 10 1 by 6 and 15 by 6 85 by 6 85 by 6 equation opposite E= V2 - V1 by U1 + U 35 35 is important.

[00:31:35]A ball falls from a height of 20 m and rebounds to a height of 5 m.

[00:31:42]Time of contact 0.02 secondation.

[00:31:46]Aelation during impact acceleration equal to change in velocity by time final velocity minus initial velocity velocity<unk>2 gx <unk>2 into 10 into 20 that is 202 that is 10. So therefore the formula for a bar 10 jcus u bar - 20 jc and 20 j by 0.0 30 by 0.0500 1500% opposite change that it's direct x1 level 37 m velocity B enters a hanging pad of mass capital M with rest.

[00:33:02]It is raising the velocity of the sphere.

[00:33:17]M into V + = M + M into V <unk>2 GH last M + M by M Stop.

[00:33:58]That's it. actually integration.

[00:34:01]But the mass per unit of a non varies as 2x3 1.5x2 and only option is first option same problem formula is integral of x into dm by integral of TM lambda uh dx length of that dx by 1 into lambda lambda into dx by lambda into dx X 39 all this formula HC = masses M1 X1 M into 1 + M2 S2 2 M3 S3 3 M1 + M2 M3 3 M 2 to gamma 2 40 two bars thrown simal m1 into same due to gravity Gap plus M2 G1.

[00:36:16]So G1 cancel the answer.

[00:36:23]Therefore G is equal to G. First one doesn't depend on that is gravity 41.

[00:36:42]Two objects of mass 10 kg and 20 kg are respected to end of mass resid of length 10 mg 10 m. the distance of CF of the system from 10.

[00:37:02]So second chance 10 - x 10 into x = 20 into 10 - x what you do 2% of mass 55 k 60 respect are opposite side the length of the body is 3 m the weight is 100 kg the 55 kg walks up to the 65 gas and sits with him. If the boat in SH water the central mass of the system sits by zero because that is internal force internal force never change the motion of zero. No 43 direct level zero. RC bar is equal to M1 M1 into R1 bar 1 into ICAP + 2 Jcap + KCAP plus M2 into R2 bar - 9 ICAP - 6 Jcap + 3 KCAP by M1 + M2 4. So I - 9 - 8 by 4 - 2 opt 2 + - 6 - 4 by 4 - 9 k + 3 4k by 4 k 44 Easy versus the P Q R the distance of system from Pord 0 PQ 0 in H P R H= M1 X1= M2 H2 M into PQ plus M into P R by M1 + M2 + M3 Publish 45 a thick strike wire of length 5 m. So down the length m at its midpoint and then bent in the form of a circle.

[00:40:01]The shift in the center of mass. First r 2 r= 1 by 2.5.

[00:40:23]Okay. So it's the last part for this academic year and tomorrow we'll be having last but one grand test.

[00:40:40]Previously test Easy.

[00:41:05]Okay.