Solids & Fluids MIRROR 30 | JEE+NEET MCQS + Tricks- NEET 2026 Physics NST TNVK Class 11 Physics

Added: 2026-05-05

[00:00:02]Hello learners, welcome to meet strl. My name is Rangila Rajas. You can call me RR. Okay. I have seven plus years of teaching experience. I'm here to make physics simple, clear and actually interesting. Okay. So already physics and 2026 very few days and okayids and neat 30.

[00:00:52]Okay.

[00:00:53]2026 model.

[00:01:01]So that first question in the session Complete understands.

[00:01:39]Okay. First question, physics, chemistry, biology channel. Okay.

[00:02:14]Okay.

[00:02:16]Free crash course.

[00:02:18]Okay.

[00:02:30]Third option.

[00:02:44]Okay. Right. So, okay. Start. So, first question. First question. A small spherical ball of A small spherical ball of radius R falling through a viscous medium falling through a viscous medium of negligible density has the terminal velocity V. Okay.

[00:03:17]Another ball of same mass but of radius 2 r falling through the same viscous medium will have the terminal velocity.

[00:03:30]So jing 2024 question.

[00:03:46]Okay.

[00:03:56]Falling through a viscous medium. Okay.

[00:04:00]Fluids with negligible density. density velocity.

[00:04:21]Okay.

[00:04:25]Okay. Another ball with the same mass previous.

[00:04:30]Okay.

[00:04:47]Okay.

[00:04:50]Okay.

[00:05:00]Okay.

[00:05:03]Okay. the ball one.

[00:05:10]Okay. So, second ball radius.

[00:05:16]Okay. Right. So, in the case it is negligible. Okay.

[00:05:31]Same.

[00:05:32]Okay. Same medium.

[00:05:40]Okay.

[00:05:47]Okay.

[00:05:50]Verminal velocity.

[00:05:56]Okay. the Vdash second terminal velocity.

[00:06:03]Okay. Right.

[00:06:10]Understand first of all with the radius.

[00:06:31]Okay. With the radius velocity density.

[00:06:47]same timeinal velocity simple question. Okay. So terminal velocity formula velocity formula.

[00:07:14]Okay. Right.

[00:07:19]formula term velocity formula V is equal to 2 by 9 R² R² G / E okay R² G divided by EA okay so first velocity velocity Right. Okay.

[00:07:53]Same visual viscosity forminal.

[00:08:14]terminal velocity formula in the okay in the V in the R. So okay V is directly proportional to Rinal is directly proportional to RG is directly proportional to R density constant.

[00:08:48]Density con is directly proportant.

[00:09:27]second.

[00:09:29]So density mass volume.

[00:09:42]Okay.

[00:09:46]volume v = 2x 9 r² listen.

[00:10:09]So volume of sperical ball volume 4 okay 43 in terms of mass volume R² cancel Rat.

[00:10:46]Okay. G va 4xel.

[00:10:59]So velocity radius relation mass V is inversely proportional to R. V is inversely proportional to R. Density is directly proportional to R².

[00:11:18]Massinal velocity is inversely proportional to directly proportional to R. Directly proportional to R².

[00:11:35]It is inversely proportional to R. Okay.

[00:11:38]So conceptinal velocity rel is inversely proportional to R.

[00:11:57]Okay. V is inversely proportional to R.

[00:12:01]V is inversely proportional to R.

[00:12:05]Okay. Right. V is inversely proport proportional to radius.

[00:12:13]Okay. Then second radius.

[00:12:21]Okay.

[00:12:28]Right.

[00:12:32]Radius.

[00:12:34]Okay.

[00:12:39]Okay.

[00:12:44]So, velocity first half. So, V by 2 is the right answer.

[00:12:51]Okay. V by 2 V by option simple questionain because of okay question.

[00:13:40]Okay. Right. So density coninal velocity radius directly proportion.

[00:13:59]Okay.

[00:14:02]Second question. Beautiful question.

[00:14:04]Okay. J 2022 question. J 2022 question.

[00:14:13]So mostly calculation the mechanical properties of okay drops of radius 1 micrometer. A water drop of radius 1 micrometer falls in a situation where the effect of bion force is negligible radius 1 micrometer.

[00:14:46]Okay.

[00:14:55]So coefficient of viscosity of air 1.8^us Okay. Density is negligible as compared to the water that is 10 ^ 6. Okay.

[00:15:16]Terminal velocity of water drop.

[00:15:22]Terminal velocity V value.

[00:15:38]in the water radius 1 micrometer.

[00:15:53]Correct.

[00:15:57]The same right down the force is negligible.

[00:16:21]Okay. Yes.

[00:16:37]So the viscous force and downgal down.

[00:17:06]So net force net force.

[00:17:20]Okay.

[00:17:23]Rainceal.

[00:17:41]Okay. Right.

[00:17:51]First of all, viscous force formula 6 RV. Okay. Viscus 6 RV. Okay. So velocity.

[00:18:12]Okay.

[00:18:14]Velocity. Right. So normal terminal velocity of viscosity.

[00:18:31]Okay. Againity again density of waterminal velocity.

[00:18:44]Okay. mg is equal to 6 pi r. So velocity mg / 6 mg / 6.

[00:19:08]Okay.

[00:19:11]Okay. V is equal to mg / 6 pi e r. Okay.

[00:19:17]So vass density density density is equal to mass volume into density. Okay. So volume 43 volume 43 Okay.

[00:20:04]compared to the water density value.

[00:20:17]Okay.

[00:20:27]6 pi coefficient of viscosity. Coefficient of viscosity 1.8 into 10 ^ 5 10 ^us 5.8 into 10 ^ - 5. Okay. 1.8 into 10us value. So radius one micrometer volume.

[00:21:12]Okay.

[00:21:15]Then in the 3. Okay. the 4x3.

[00:21:27]Okay. Into 10 power. Okay. R² R² into 10 ^ 4 R² into 10 ^ into 10 radicus micro value 10 - okay into 10 ^ - 6 but the whole square r² so 10^ - 6 the whole square into 10 ^ 4 / 3 into 6 into 1.8 into 10 ^ - clear.

[00:22:41]Okay. Right.

[00:22:43]Right. Next step. Next.

[00:22:48]Okay.

[00:22:53]3 2 10^ 12. Okay. 10^ -2.5.

[00:23:17]Okay.

[00:23:20]numbers. Okay. So next 2 Okayid divided by 3 into whole number 3 into 3 into 1.8 10^ - 10^ 4 / 10 + right.

[00:23:52]Okay. So 10^ 12 okay right number 3 into 3 9 9 into 1.8 16.2 16.2 2 into 10us 3. Okay. 2 / 16 0.123 into 10 ^us 3 final answer. So in the final answer 0.12 Okay.

[00:25:01]1 2 3 1 2 3 into 10 ^ - 3 againus 3 final answer 1 2 3 into Okay 1 2 3 into 10 power option water drop radius 1 micrometer downg.

[00:25:33]force force force negligible means net force forceminal velocity velocity Okay= mg / 6al mass volume sorry volume.

[00:26:30]Okay. Then density value.

[00:26:36]Okay.

[00:26:55]1 micromeic 1 same value 1 2 3 okay 4 option but that's okay 1 2 3 into 6 Final answer.

[00:27:23]Okay. Right. So clear.

[00:27:27]Next question.

[00:27:30]Next question. Next question.

[00:27:34]Okay. Next question. J mean 2024 April.

[00:27:42]Okay. Right. Simple question.

[00:27:58]A big drop is formed by colling thousand small droplets of water. Okay. Big drop.

[00:28:06]Okay.

[00:28:09]Thousand small droplets.

[00:28:12]small thousand small droplet form clear. Okay. So in the mad condition the ratio of surface energy of thousand droplets to that of the energy of big drop is 10 by the value of x previous small Okay. Question 2024 questionual droplet.

[00:29:29]Okay.

[00:29:35]So,000 power 1 by3 * of the previous drops.

[00:29:40]Okay. So, r the big droplet new radius value 1,000 power 1x3 into r.

[00:29:49]So 1,000^ 1x3 is 10 times of the big drop 10face surface energy ratio surface energy ratio. So surface energy formula surface energy is equal to surface tension into Okay.

[00:30:43]Okay.

[00:30:47]Surface energy is equal to surface tension into surface tension into shape 4 r² 4 pi r².

[00:31:07]So surface energy surface surface r* 10² tension surface energy is equal to surface tension into 4<unk> into 10 square.

[00:31:52]Okay. Right.

[00:31:56]First one and divided by the second one.

[00:31:59]Okay.

[00:32:09]correct. Okay. So remaining value first equation second equation 100 right. So final answer 10 the ratio of surface energy and the thousand droplets with respect to big drop. Final answer 10. Answer by 10 value clear.

[00:32:53]Okay. Right.

[00:32:56]Next question. Okay.

[00:33:02]The excess pressure inside the soap bubble is twice the excess pressure inside the second soap bubble. The ratio between the volume of first and second bubble is okay.

[00:33:17]Okay.

[00:33:21]P1 second pressure.

[00:33:27]Okay. So the excess pressure inside the soap bubble is thrice the excess pressure inside the second so pressure * second so bubble excess pressure so P1 is equal to 3 into P2 3 * P2 P1 3 * P2 the first line right the ratio between volume of the first and second volume okay volume of the first bubble and second bubble excess pressure excess pressure formula 412.

[00:34:39]Okay. R1 is equal to 3 into 4 S by R2 right okay 4 S 4S cancel 4S cancel R1 by R2 okayirect directly proportional to rq right 4x3 r= 4x3 rirectly proportional volume directly proportional so r1 r2 the whole huh R1 by R2 the whole cube.

[00:35:42]So R1 by R2 value 1x3. So 1x3 the whole 1x3 the whole cube. Okay. So 1x 3 the whole cube 1 cube 1 cube is 1 3 cube 3 3 cube in the sense 3 into 3 into 3 correct 3 into 3 into 3 3 into 3 9 9 into 3 27 okay so 1 by 27 is the answer means 1 is to 27 is the answer okay so 1 is to 27 now in the option C look option C look clear simple Excess pressure volume is directly proportion.

[00:36:51]Next question. M okay a metal block of base area it is 0.2 m²ared is placed on a table a liquid film is thick film of thickness it is 0.25 25 mm is inserted between the block and the table. The block is pushed by the horizontal force of.1 Newton moves with constant speed. The viscosity of the liquid 5 into 10 - 3 pas okay viscosity. Yeah. Right. The speed of the speed of the block.

[00:37:37]So diagramal block in the metal block.

[00:37:57]Okay. Right.

[00:38:03]Okay.

[00:38:09]metal block 0.2 mal of thickness 0.2 m inserted between the block and table. So in the block. Okay.

[00:38:38]Okay. Right. So the block is pushed by the horizontal force. Socidity left side.1 Newton force.

[00:39:07]It's not a water. Okay. A kind of liquid. Okay.

[00:39:16]experience.

[00:39:29]Okay. Right.

[00:39:36]0.25 25 and the liquid viscosity and the block.

[00:39:54]This is the question.

[00:40:00]Okay. J mean 2023 session. J mean 2023 January.

[00:40:07]Okay. 2023 2023. Check. Okay.

[00:40:17]Okay.

[00:40:50]Okay.

[00:41:11]Okay.

[00:41:13]Almost constant.

[00:41:20]Okay.

[00:41:22]Okay.

[00:41:31]Almostant almost rest.

[00:41:48]Okay.

[00:41:56]Just movecity.

[00:42:18]Okay. Sorry. I'm sorry.

[00:42:22]Viscosity difference in velocity velocity so that okay change in velocity almost in distance between distance.

[00:43:06]Okay. Right.

[00:43:09]DV by DX return by DX return almost velocity change top to bottom velocity.

[00:43:40]Small change, very ministing in a constant speed.

[00:44:00]Okay. Block is moving at the constant speed. Block is moving.

[00:44:06]block is moving in uniform speed. Okay.

[00:44:12]With uniform speed.

[00:44:35]Okay. Right.

[00:44:39]So force.1 appce right? So since this block is moving with uniform speed the net force is zero force already block Speed force means.

[00:45:50]Okay. Right. So appi force.1.

[00:46:08]Okay.

[00:46:14]Viscosity means viscous.1.

[00:46:33]Okay. Right.

[00:46:39]Force value. Okay.

[00:46:43]Force is equal to E A DV by DX. This is a formula.

[00:46:49]Correct. E DV by DX.

[00:46:54]Coefficient coefficient of viscosity.

[00:46:57]Correct.

[00:47:01]Block.

[00:47:06]Coefficient of viscosity.

[00:47:08]coefficient of viscosity.

[00:47:13]Okay. DV by DX.

[00:47:30]thickness and the liquid velocity velocity almost by F is equal to EA DV by DX F.1 into the power minus 3. Okay. 5 into it is equal to equal to 5 into 10 ^ - 3 and area2 m² m². So 2 into dv means velocity.

[00:48:26]Okay. Dx.

[00:48:28]Okay. So, DX thickness almost thick. of the liquid thickness of the liquid.

[00:49:05]So dx is nothing but thickness.25.

[00:49:17]So 25 mmus 10us okay so.12 02 velocity correct right into 0.2 1 velocity is equal to 0.1 into 0.25 0.1 into 0.25 25 into the power 25us 3 m/s. Okay. So 25 into 10us 25 clear 203 question. Okay. Right. So max next question.

[00:50:28]Okay, next question. A surface of water.

[00:50:32]Okay, listen. A surface of water uh in a water tank of cross-section area 750 cm squared on the top of the house is h m above the tap level. The speed of the water coming out through the tap of cross-sectional area of 500 mm squared is 30 cm/s.

[00:50:59]At that instant the magnitude of dh by dt is x into 10 ^us 3 the value of x main 2023 February session question the surface of water tank first of all tank waterface surface of water surface level crosssectional area. Okay. The surface area 750 cm square 750 cm square.

[00:51:48]Okay. And the top of the house is H above the tap level in the water surface of water top of the distance.

[00:52:17]The surface of water in a water tank with a cross-sectional cross-section 750 cm square on the top of the house is H m above the tap above surface of water crosssection area 750 top water level surface of water 750 cm square Okay.

[00:52:51]The speed of the water coming out through the tap of cross-sectional area 500 mm² is 30 cm second.

[00:53:08]surface of water. Surface of water 30 cm.

[00:53:28]In the water open area cross-sectional value okayction Okay. Right. At that instant magnitude of DH by DT.

[00:54:08]Okay. Magnitude of DH by DT.

[00:54:13]Okay.

[00:54:15]by open velocity cmapsection A12 magnitude of DH by DT.

[00:54:58]What is DH by DT? DH by DT DH by DT. Now time height change height change. Okay. Right.

[00:55:14]Height is nothing but distance.

[00:55:23]Distance with respect to time.

[00:55:24]Disrespect to time.

[00:55:27]Disrespect with respect to time.

[00:55:36]Okay.

[00:55:40]Okay.

[00:55:46]Okay.

[00:55:50]Okay.

[00:56:01]Okay.

[00:56:03]Right.

[00:56:05]Clear.

[00:56:18]surface of waterc V2 simple question.

[00:56:34]Okay. Right. Simple question. So V1 dity.

[00:56:55]Okay. Equation of continuity formula A1 V1 = A2 V2. A1 V1 is equal to A2 V2.

[00:57:05]Okay. A1 and surface of water area 750 cm square.

[00:57:18]Next.

[00:57:22]Okay. A1 B1 A1 surface of water area that is 750 cmus square the power minus 4 into velocity DH by DT velocity by second 500 mm square value 500 - 6. So 500 into 10^ - 6 velocity 30 cm.3.

[00:58:17]Okay.3.

[00:58:28]Okay. 500 into 10 - 650 into 10us.

[00:58:37]Okay. Right. So next next okay us 0.3 V1 is equal to 500 into 0.3 that is 150.

[00:59:09]So 150 denominator 750us.

[00:59:18]So 0 15 by 75 15 by 75 by 75 by 75.

[00:59:33]So v_sub1=2 into 10us 3. Okay. So final answer already velocity.

[01:00:02]Okay. Simple question, right? Next question.

[01:00:11]Next question.

[01:00:16]Next question.

[01:00:19]The velocity of upper layer of the water in a river is 36 kilome.

[01:00:27]Okay.

[01:00:30]velocity 36 kilome.

[01:00:38]Okay. Right.

[01:00:41]So shearing stress between horizontal layer of water is 10 ^us layers horizontal layers layers already.

[01:01:09]Okay.

[01:01:13]36 kilome 36 2 shearing stress between the horizontal layer is 10 ^ -3 Newton m²ared.

[01:01:37]Shearing stress. Sharing stress. Okay.

[01:01:41]So, shearing stress. Shearing stress formula.

[01:01:45]Shearing stress.

[01:01:47]Shearing stress formula force by area.

[01:01:50]Correct. Shearing stress formula force by area force by area 10 ^us 3 m²us 3ce by area. Okay. Depth of the river is dash.

[01:02:11]So in the river of the river almost zero lower layer upper layer 36 river value.

[01:02:48]Okay. Coefficient of this particular water it is 10 ^ -2.

[01:02:55]Okay. Got it.

[01:03:03]Depth of the river. Depth of the river.

[01:03:07]Depth of the river value. Force by F= E DV by DX. Correct. F= E A DV by DX.

[01:03:26]Correct. Okay. So force by area dv by dx force by area sharing stress thenic of viscosity dv by dxenid flow Upperlay velocity lower dx dx dx. So depth.

[01:04:16]Okay.

[01:04:30]Sharing stress horizontal layers area forceal area dv by dx force area by dxce.

[01:05:15]Okay. Right.

[01:05:17]Next area.

[01:05:26]So 10^ - 2 into dv by dx change in velocity change in velocity.

[01:05:36]The lower change in velocity for the final velocity minus initial velocity. final.

[01:05:46]Okay.

[01:05:52]Okay.

[01:05:54]So, dvid dx correct. Okay. dx 10us 3usal dx. Okay.

[01:06:20]Simple.

[01:06:22]Okay. So 10 - 3 + 1 / dx.

[01:06:29]Okay. Right.

[01:06:32]Yes. So dx.

[01:06:38]So dx = 10 / 10 ^ -1. So 10us final value 100 m 100 m. Okay. So value 100 m22 simple question. Okay.

[01:07:08]Next question.

[01:07:10]Next question. Water flows in a large tank with a flat bottom. Okay. Water flows in a large tank with a flat bottom uh at the rate. Yeah. At the rate of 10 ^ -4 m cube/s.

[01:07:31]What is also sorry water is also leaking out the whole area of 1 cm squared at its bottom. If the height of the water tank remains steady then this height is J 2019 J 2025.

[01:07:55]So this is very important question start question. Okay right give me the answer.

[01:08:08]Okay. Question.

[01:08:11]Water flows in a large tank with a flat bottom. Okay.

[01:08:19]Flat bottom.

[01:08:33]-4 m.

[01:08:37]Water is also leaking out in the flat.

[01:08:46]Okay.

[01:09:01]Okay.

[01:09:04]If the height of the water in the tank remains steady height of the water in the bottom surface of water height remains steady. If the height of the water in the tank remains steady in the condition in the height amount of water. Okay.

[01:09:55]Fore direction.

[01:10:24]Okay.

[01:10:31]Height steady one meaning rate flow rate same. Okay.

[01:10:52]flow rate.

[01:10:55]Okay.

[01:11:03]10^us 10us 4 it is equal to and area and velocity. area and velocity.

[01:11:24]Right. Okay. Okay.

[01:11:31]Velocity refaccept okay so 10 ^ -4 then area into velocity. So then square 1 cm 10 - 4 cm square. So 1 into 10 - 4 cm - 2 cm² so - 4 okay 4 then velocity velocity roo<unk> of 2 gh velocity value roo<unk> of 2 gh<unk> of 2 gh okay right so next 10 ^us roo<unk> 2 G.

[01:12:52]Okay.

[01:12:56]2 gh it is equal to 1.

[01:13:00]Okay. Root of 2 GH it is 1.

[01:13:10]Okay.

[01:13:14]1 by 2. 1 by 2 G. 1x 2. Okay.

[01:13:22]Okay.

[01:13:24]1x 20, right? 1 by 20. Okay. 1x 20 1x 2.5 1x 20 5 cm 5 cm. Final answer. Okay. 5.1 cm 7 cm 4 cm 9 cm 5.1 Correct answer.

[01:14:04]Okay.

[01:14:06]Important question.

[01:14:28]Flow rate is same.

[01:14:31]Flow rate is same. Flow rate.

[01:14:56]Simple calculation.

[01:15:02]Okay.

[01:15:05]Next question.

[01:15:08]Okay. Next. J 19.

[01:15:14]Okay. J 2019.

[01:15:21]A fluid is flowing through a horizontal pipe is wearing a crosssection over a line diagram. Okay.

[01:15:42]Okay. Okay. So, a fluid is through a fluid is flowing through a horizontal pipe of varying cross-section.

[01:15:50]Horizontal pipe pipe horizontal pipe crosssection area.

[01:16:07]Okay.

[01:16:16]cross-sectional area with the velocction.

[01:16:54]Second second crosssection pressure cross-sectional area V small V okay small another point another pressure is P by2 pressure P by2 the speed.

[01:17:19]Okay.

[01:17:21]Capital V small.

[01:17:25]If the density of the fluid is row kilogram per meter cube and the flow is streamline then v is equal to okay in the density of the liquid flow Capital V small fluid is flowing through the horizontal pipe crosssection small pressure at another point means pressure.

[01:18:27]Okay. Density of the fluid.

[01:18:34]Streamline.

[01:18:45]Okay. Clear. Right. Question explanation. Right.

[01:18:52]Formula P +/ row V² is equal to constant formula P + half row V² is equal to constant cross-sectional area.

[01:19:19]Okay.

[01:19:21]Aal by 2 plus square.

[01:19:42]Okay.

[01:19:44]Capital V. So row capital V. Okay. So capital V² then it is equal to pressure small V plus half V² okay Pressure pressure.

[01:20:30]foreign.

[01:20:35]Okay.

[01:20:51]Okay. Right. Right.

[01:21:03]P2 Okay. The pressure minus P by 2 P minus P by 2 again P by 2. Okay. P minus P by 2 P by 2. Correct. Okay. Half row capital V² - small V² it is equal to P by 2.

[01:21:29]Similar terms cancel V square V square V square V square V square P by row P by capital V square V square Right side minus small V square right capital V square is equal to P by row capital V square= P by row small V square right V square so small V square plus small V square root small V² root of P by row + small V² square.

[01:22:34]Okay. Root of P by row + V² roo<unk> of P by row + V² root of P by row + V square. The option D. Option D value P and half of Pceptle.

[01:23:09]Okay. Right. Next question.

[01:23:14]Next question.

[01:23:17]Next level question. Okay.

[01:23:22]Okay.

[01:23:26]J 2023 question.

[01:23:30]Question. Water from a tap emerges vertically downward. Water from a tap emerges vertically downward. Okay. With an initial speed of 1 m/s.

[01:23:45]The cross-sectional area of the tap is 10 ^ -4 m²ared. Assume that pressure is constant throughout the stream of water and the flow also streamlined.

[01:23:58]Okay. The cross-sectional area of the stream 0.15 m below the tap would be cross-sectional area of the stream.

[01:24:13]Okay. Water from a tap emerges vertically down.

[01:24:21]Okay.

[01:24:24]Okay.

[01:24:35]flow stream. Okay.

[01:24:54]Initial speed 1 m.

[01:25:02]So crosssectional area of the tab 10us crosssection 10^ - 4 - 4ine flow. Okay.

[01:25:34]The crosssectional area of the stream and the streamsection area of cross-section area of crosssection.

[01:25:50]Okay. Right.

[01:26:01]Below the tap would be below the tap below the tap.

[01:26:16]Okay. Below the tapow in the crosssection below the tab and the value 0.15 m it is height 0.15 simple question. Okay, just question area foreign.

[01:27:11]Okay. So yes, first of all, right?

[01:27:37]Equation of continity formula. equation of continity formula A1 V1 = A2 V2 correct. Okay. Right.

[01:27:49]V12.

[01:27:56]Okay. A1 A2 we know this values. Okay. So velocity flow velocity previous row v².

[01:28:19]Okay. So half row v² plus row ghal= constant. So okay half row V² + G= Vacity V1 V1² plus row G H1 it is equal to velocity V2² + G H2 G H2. Okay. G H2 G H2.

[01:29:34]Okay. Row cancel 1x2 by cancel.

[01:29:40]Okay.

[01:29:48]Okay.

[01:29:50]So half V1² is equal to H.

[01:30:02]Okay. H1 value below the.5 tap Open water.

[01:30:47]Okay.

[01:30:59]value H2 H1 H1 H1 Okay.15 0.1 final The bottom point height Simple logic bottom ground table particular height 0.1 the final height 0us 0us 0.1 bottom h2 0.1 okay h1 value h1 value Okay. Half v1 square is equal to/ into v_sub_2² into h2 correct v_sub_2² into h2. So next step v1² is equal to 1 by2 into v2 square value 1 m/s h2 0.15. Okay.

[01:33:10]Okay. So V2 square is 1 square correct one square is one and height H2 0.15 height H2 0.1 V1²= 1 by 0.5 plus 015 0.15 0.15 0.15 m 0 0.15 mistake Grow Next 1 by 2 V1² + G H1X 2 V2² + 10 into 0.15 okay 10 into H2 correct in the term Okay. G value 10 into 0.15. 10 into 0.15. 10 into 0.15.

[01:34:48]Next 10 into 0.15 1.5.

[01:34:53]So half v1 square is equal to 0.15 + 1.5 0.1 + 1.5 right v1 square v1 square into 2 v1 square= v1 square v1 bottom velocity that is 2 m.

[01:35:33]Okay.

[01:35:37]Okay. So, a1 is equal to a1 is equal to a1 into v a2 value.

[01:35:51]Okay. 101 by 2 1 by 2 into 10^ -4 okay 1x 2 - 4 1x 2 is nothing but.5 so.5 into 10 - 4.5 into 10^ -4 okay.5 into 10us okay 5 into 10us Okay.

[01:36:25]Bottom area of cross-section 5 into 10 ^ - 5 m² 5 into 10 - 5 m square answer 5 into 10 ^ - 5 m² 5 into 10 - 5 m² okay half v half v square cancel.

[01:37:04]Okay. Right.

[01:37:17]Right.

[01:37:21]H2 H2 value 0.1 final value final H2O bottom height height 0.1 1.5.

[01:37:57]Okay. Right.

[01:38:00]Next question. Next question.

[01:38:05]A fully loaded Boeing aircraft has a mass of 5.4 into 10 ^ 10 ^ 5 kilog.

[01:38:13]Okay. Its total wing area is 500 m²ared.

[01:38:18]It is in level flight with a speed of uh,80 kilome per hour. If the density of the air is 1.2, Point to the fractional increase in the speed fractional increase in the speed of the air on the upper surface of the wing relative to the lower surface in the percent.

[01:38:43]J mean 2023 J mean 2023 question. First of all, Boeing aircraft.

[01:38:53]Okay. First of all, aircrafting.

[01:39:01]So, Boeing.

[01:39:09]Okay.

[01:39:13]lower top V2 V1.

[01:39:28]So in the flight velocity, 180 km original velocity,80 kilome lower velocity.

[01:39:55]Okay. Right. So density okay 1 kg okay fractional increase in the speed of the upper surface of the wing relative to the lower surface wings upper pressure lower pressure.

[01:40:32]So pressure down force. Correct.

[01:40:55]Right.

[01:41:06]Okay. Pressure is equal to force by area. Pressure is equal to pressure basic conceptal objecting.

[01:41:34]So upward pressure force pressure into area pressure into area. So pressure change right force is equal to change in pressure P2 - P1 into area P2us P1 into downward P= MG correct. P= MG. Pressure into area.

[01:42:54]So pressure and upper surface pressure fully loaded aircraft total wing.

[01:43:31]Okay. Right.

[01:43:45]It is relative to the lower surface percent.

[01:43:51]So normal upper velocity lowerface velocity normal common change in velocity surface pressure lowerface pressure force pressureal force area pressure.

[01:44:22]So always upward and downward. So Falgal mg is equal to pressure change up. So P2us P1 is equal into area.

[01:44:38]So pressure P2us P1 is equal to mg by A change in pressure in the formula.

[01:44:48]P1 P2US P1 is equal to mg by A. Okay.

[01:44:52]Right.

[01:44:54]P2US P1 is equal to MG 5.4 4 into 10 ^ 5. So 5.4 into 10 ^ 5 into okay gid by area. So wings area 500 m 500.

[01:45:24]Okay. Right. So 5.4 mean number 5.4 4 value 1.0 correct. So 1.08 8 into 10^ 6 - 2 6 - 2 is 4^ 4 pressure difference 1.08 8^ 4 just pressure difference.

[01:46:06]Okay. Pressure difference again= P= P row V²= constant correct. So down.

[01:46:37]Okay.

[01:46:40]Yes.

[01:46:45]V1 and half row velocity.

[01:46:51]Yes. V1 squared V1 square it is equal to second one P2 +/ row V2² okay again so P1 again Half V1 V1²us V2² right side V2 square V2 leftus V2 squareus V2² in V1 square V2 square formula A squareus B square formula A square B square for a square B square for A + B A minus B correct a square + B square A square B square for A + B A so half row A square B square A + B A B1 square B2 Now v_ub_1 + v_sub_2 and v_ub_1us v_sub_2 okay v1 - v_sub_2 clear it is equal to p2us p1 right okay velocity velocity average velocity Average velocity formula V average is equal to V1 + V_sub_2 / 2 correct v_sub_1 + v_sub_2 / 2 v1 + v2 / 2 right okay velocity average kilome kilome hour kilome.

[01:49:19]So,8 Okay. Right.

[01:49:30]16 300 m².

[01:49:38]So, average velocity,80 km 300 m/ second. Average velocity V1 + V2 / V1 + V2 simplif Right.

[01:50:06]V1 + V2 / 2.

[01:50:14]Okay.

[01:50:16]into row into V1 + V2. Okay, wait a minute. V1 + V_sub_2 / row 2 divided by 2. Okay into row into V_sub_1us V2 V_sub_1 + V_sub2id 2 into row into V_sub1 - V2. Yeah, right. It is equal to P2us P1. It is equal to P2 - V1. So v_sub1 + v_sub_2 divided by 2 nothing but v average correct. So v average into row into v_sub_1 minus v_sub_2 it is equal to p2us okay so 300 density value 1.2 kilog 1.2 V2 V2us V1. So V2US V1 P2US P1 change in velocity change in velocity 1.08 change in pressure 1.08 into power 4. So 1.08 1.08 into 10 clear okay 1.08 8 into 10 ^ 4 / 300 into 1.2 300 into 1.2.

[01:51:49]Okay.

[01:51:50]Right. So 1.08.

[01:51:58]30.

[01:52:00]So v_sub_2 minus v1 value 30. V_sub_2us v1 30. So just V2us V1 V2US V1 fractional increase fractional increase in the speed fractional increase in the speed v1us V2 correct okay fractional increase measurement fractional increase formula final change in velocity divided Original velocity into 100. Correct. Change in velocity divided by original velocity into 100.

[01:52:42]Final velocity minus initial velocity with respect to initial velocity into 100.

[01:52:47]Change in velocity with respect to original velocity into 100. Okay. So change in velocity 30. Original velocity average velocity 300 m.

[01:52:59]Okay.

[01:53:02]So 30% 10%.

[01:53:08]So fractional increase in the speed of the aircraft will answer right. Okay.

[01:53:35]fractional increase in the speed.

[01:53:38]Upper surface. Lower surface.

[01:53:47]Change pressure. Change in velocity.

[01:53:50]Change.

[01:53:58]Right.

[01:54:13]Okay. Right. So, next question.

[01:54:24]Figure below shows a liquid being pushed out of the tube by piston having the area of cross-section 2 cm squared. The area of cross-section at the outlet is 10 mm². If the piston is pushed at the speed of 4 cm/s, the speed of outgoing fluid in second.

[01:54:49]Okay.

[01:54:53]liquid. Okay. So, the liquid is being pushed out of the tube by the piston having the area 2 cm.

[01:55:02]Okay.

[01:55:05]Okay.

[01:55:12]2 cm square. The area of cross-section of the outlet is 10 m/s square. Right?

[01:55:23]So area of crosssection 2 cm square area of crosssection 2 cm square out area of crosssection 10 mm². Okay. If the piston is pushed at the speed of ceing okay Okay.

[01:56:06]Equation of continity. Okay.

[01:56:12]20 23.

[01:56:16]Okay.

[01:56:20]Okay.

[01:56:28]So again equation of continuity A1 V1 is equal to A2 V2 A1 V1 is equal to A2 V2 A1 value 2 cm² so 2 into cm² so 10 ^ - 4 into velocity 4 cm. So 4 into 10 ^ -2 it is equal to a2. Okay. Second outer port area 10 mm square. Okay. So 10 mm square V2 value.

[01:57:23]Right. Clear. And the right side - 6 v 10 - 10 into 10 - 6 1 - 6 power - 5 into v2 power -2 v2= 10 - 6 / 10 ^ - 5 correct 8 into 10 ^ - 10^ -1. Okay. 10 ^ -1. Okay. 8 into 10 - 1.

[01:58:16]Okay.

[01:58:21]Okay. 8 by 10.

[01:58:24]8 nothing but8 okay so 80 cm per second 80 cm per second okay okay next question yes next question a tube of length as shown in the figure Okay.

[01:58:54]Okay. The radius of crosssection at the point is area of crosssection. Area of crosssection radius of crosssection.

[01:59:05]Okay. For crosssection R1 value R1 value.1 is 2 cm. 2 cm. Okay. And 2 is 1 1 cm.

[01:59:18]Radius of crosssection 1 cm. If the velocity of the water entering at the point is 2 m/s velocity v1 that is 2 m v2.

[01:59:36]So again equation of 202 important question.

[01:59:44]Okay. So a1 v1 is equal to a2 v2 a1 v1 is equal to a2 v2 a area but area of crosssection radius of crosssection.

[01:59:55]Okay. Area of crosssection.

[01:59:58]Radius of crosssection. Radius of crosssection area is nothing but p<unk> r². Area is nothing but p<unk> r². So area r1² into v_sub_1 it is equal to p<unk> r2² into v_sub_2. Right?

[02:00:16]Yes.

[02:00:18]R1 square 2 square into V1 R2 V2 so V2 simple right so 2² is 4 4 into 2= V2 4 into 2 is 8 right V2 value 8 m/s 8 m/ Okay.

[02:00:53]Radius of crosssection 2 cm velocity part second.

[02:01:21]Radius first.

[02:01:41]Next question. An incompressible liquid is flowing through a horizontal pipe.

[02:01:46]Incompressible liquid incompressible liquid that is okay incompressible.

[02:01:58]Okay. Mostly fluid.

[02:02:08]Okay.

[02:02:13]Okay.

[02:02:24]The next perfect fluids are incompressible some cases.

[02:02:31]Okay. An incompressible liquid is flowing through a horizontal pipe.

[02:02:36]compressible.

[02:02:39]Okay.

[02:02:50]Okay.

[02:02:59]One branch of area.

[02:03:02]One branch of area 0.8 Okay.

[02:03:08]0.4.8.

[02:03:19]Okay.

[02:03:33]Okay. Fore velocity 6 m/sid.

[02:04:06]Okay. Right.

[02:04:17]Okay.

[02:04:30]AL simple basic concept equation of because area of A1 V1 port third port okay first portion A1 V1 second portion A2 A2 V2 third simple okay it is equal to B18 velocity 6 plus third port 3 * velocity.

[02:05:27]Okay.

[02:05:29]Right into correct. Okay. 8=8 into 68 into 4.8 4.8 + 3 V3. Correct. 4.8 8 + 3 V3 V3 48 - 4.8 is equal to 3 V3 correct 8 - 4.8 is equal to 3 V3 8 - 4.8 3.2 so 3.2 is equal to 3 V3 so V3 is equal to 3.2 / 3 3.2 / 3. Okay. So 1.1 1.120 clear okay right next to two cylindrical vessels of equal cross-sectional area. Okay. two cylindrical vessels of equal cross-sectional area 2 completely fluid related question. So again okay rent cylindrical vessels of equal cross-sectional area 2 m/ second contain water up to a height of 10 m and 6 m.

[02:07:15]Okay.

[02:07:24]Okay.

[02:07:32]6.

[02:07:35]If the vessels are connected at their bottom. If the vessels are connected at their bottom. So the vessels in the bottom the work by the force of gravity.

[02:07:56]By the force of gravity. Force of gravity.

[02:08:09]2025 question.

[02:08:13]Okay. So cylindid Work is equal to force into disce into disrece.

[02:09:16]Okay.

[02:09:29]Okay.

[02:09:30]If the vessels are connected at the bottom, the work done by the force.

[02:09:45]For example, the center port 6 height difference.

[02:10:14]Okay. Height difference for this It is 2 m height difference.

[02:10:30]Same level.

[02:10:43]Same level.

[02:10:46]height difference.

[02:10:50]Okay.

[02:11:00]Okay.

[02:11:04]Change in potential energy. Okay. Change in potential energy. So mass density in terms of density.

[02:11:30]So mass means density into volume density into volume.

[02:11:38]Density into volume.

[02:11:58]Okay.

[02:12:07]formula mg h because change in potential energy. So mgh m means mass mass direct value.

[02:12:19]So but density so mass into density sorry mass is equal to density into volume.

[02:12:34]Okay.

[02:12:41]Volume into density. Okay.

[02:12:46]Okay. Again volume area height again density of water.

[02:13:16]So 10^ 3 is th000 acceleration due to gravity is g that is 10. Okay again so 2 into 4 4 into 8 work done by force of gravity. Okay.

[02:13:39]Work done by force of gravity 8 into the^ 4. Option D. Simple question.

[02:13:47]Okay. Right. Next question.

[02:13:50]Yes.

[02:13:52]Next question. Next question.

[02:13:56]Yes. Next question.

[02:14:02]Next question.

[02:14:03]Mercury is filled.

[02:14:06]Right. Next question. Mercury is filled up. Yes. Mercury is filled in a tube of radius 2 cm up to a height of 30 cm. The force exerted by the mercury on the bottom of the tube.

[02:14:26]Okay.

[02:14:32]height 30 cmury radius.

[02:14:48]The force exerted by the mercury on bottom.

[02:15:02]value.

[02:15:08]Okay.

[02:15:13]Atmospheric pressure atmospheric pressure 22.

[02:15:26]Okay.

[02:15:31]pressure is equal to atmospheric pressure plus hog g. Formula formula.

[02:15:45]Okay. Pressure is equal to atmospheric pressure plus h.

[02:15:50]Okay. So pressure is equal to atmospheric pressure 10 ^ 5. So 10^ 5 plus height of the liquid 30 cm.

[02:16:02]Height of the liquid 30 cm 30 cm 30 cm.3.

[02:16:16]So 3 into density of mercury 1.36 into 10^ 4. So 1.36 into 10 ^ 4 into acceleration due to gravity is 10. So acceleration due to gravity is 10. Okay. So pressure is equal to 10 ^ 5 + 0.3 into 1.36 main number 0.3 into 1.36 value 0.4 48 10 power 10 power 4 + 1 5.

[02:17:02]So the power so 1 + 0.4 40.4.

[02:17:24]So 1.0 1 + 0.4 1.4 1 + 0.4 1.4. Okay. So pressure is equal to 1.4 into 10 ^ 5 pascal. Okay.

[02:17:38]Pressure 1.5 into 10^ 5 question. Okay.

[02:17:43]Force exerted by mercury on the bottom.

[02:17:47]Okay.

[02:17:49]Forceal intoceal pressure force area.

[02:18:04]Pressure into area. Correct. Pressure into area. So pressure into area pressure value 1.4 into 10 ^ 5 and area okay 1.4 4 into 10^ 22 by 7. Okay. R² R square R 2 cm.

[02:18:49]Okay. R squareus.

[02:18:53]Right. So next 14.4.

[02:19:09]So in the 2 in the 22 in the four 17.6 10^ 5 - 4 10^ 5 - 4 5 - 4 that is 1.

[02:19:30]So 17.6 into 17 17.6 6 176 new 176 new answer 176 new clear okay simple question right next questions Okay.

[02:20:07]Next question.

[02:20:16]The pressure acting on a submarine is 3 into 10 ^ 5 pascal at the certain depth.

[02:20:22]at the certain depth. If the depth is doubled, the percentage increase in the pressure acting on the submarine submarine would be J21.

[02:20:36]Okay, one minute guys.

[02:20:47]Pressure acting on the submarine value 3 into 10^ 5.

[02:20:52]Okay. At a certain depth. Okay. Okay. If the depth is double pressure acting on the submarine percent increase in the pressure the percent increase in the pressure.

[02:21:31]Okay.

[02:21:50]Pascal pressure is equal to 3 into 10 ^ 5 correct. Okay.

[02:22:01]Atmospheric pressure means dro okay it is equal to pressure value 3 into 10^ it is notic atmospheric pressure normal pressure meaning Okay.

[02:22:40]Acceleration due to gravity. Atmospheric pressure. Density. Acceleration due to gravity.

[02:22:50]Normal pressure on the submarine atmospheric pressure.

[02:23:03]Okay.

[02:23:09]Okay. Right.

[02:23:17]for Okay.

[02:23:44]Atmospheric pressure value 10 ^ 5 and d it is equal to 3 into 10 ^ 5 is equal to 3 into 10 ^ - 5 sorry 3 into 10 ^ 5 Right side g okay yeah d g is equal to 3 into 10 ^ 5 - 10 ^ 5. Okay. - 10^ 5 5 is nothing but 1 into 10 ^ 5. Correct? So 10^ 5. So 10^ 3 - 1 3 - 1 is nothing but 2 that is 2.

[02:24:59]Okay.

[02:25:03]Acceleration due to gravity. So depth value acceleration due to gravity.

[02:25:09]density value 2 into 10.

[02:25:12]Okay. 2 into 10 ^ 5mal pressure atmospheric pressure 10 balance.

[02:25:46]So value 2 value 2.

[02:25:57]Okay. Pressure pressure due to 2D. Okay. Normal dep* 2 into 2 into 10^ Okay. Normal depal 2 into 5 2 * of d 2 into 2 into 10 clear.

[02:26:40]Okay.

[02:26:56]pressure atmospheric pressure 2 into 10^ 5 into 2 into 10^ 5 4 Okay. Right.

[02:27:24]Okay. Pressure due to just the pressure due total pressure at okaytal sorry. Okay. Total pressure total pressure at Total P A plus in the 4 into 4 into 10^ 5.

[02:27:50]Okay.

[02:27:56]Again 1 5 into 10 ^ 5 total pressure at double times of depth.

[02:28:12]Okay. Total pressure at double times of depth. Total pressure at double times of depth.

[02:28:25]Okay.

[02:28:36]Percent increase in the pressure value.

[02:28:38]Percent increase percent change is equal to change by okay change by initial change into 100. Okay. Change final minus initial final value final value 5 into 10 ^ 5 initial value 3 into 10 ^ 5. So 5 into 10 ^ 5 - initial 3 into 10 ^ 5 / initial 3 into 10 ^ 5 okay into 100. So 5 into 10^ 5. So 10^ 5 into 100 2x 3 is 100 is nothing but 200 by 3. Okay. 200 by 3 is percentage change in pressure 200 by3.

[02:29:42]So 200 by 3 option clear 200 by option you got my point right. So the next last fluid question. Okay.

[02:30:08]Okay. Right. Next question. Four vessels A, B, C, D have different shape vessel shape different, the B, the C, the D. Okay. Which of the following is correct regarding the pressure at their bottom? Bottom pressure shape of the liquid. The shape of the container.

[02:30:58]Okay.

[02:31:03]pressure 2024.

[02:31:14]Okay.

[02:31:26]The left side.

[02:31:28]The left side. The right side.

[02:31:40]pressure.

[02:31:45]Left side bottom, right side.

[02:31:57]Okay.

[02:32:00]Main reason First conditioner number same condition satisfied.

[02:32:41]Acceleration Bure.

[02:33:02]Okay.

[02:33:08]pressure.

[02:33:16]Okay. Right.

[02:33:23]Last question.

[02:33:31]Okay.

[02:33:33]2021.

[02:33:37]Okay. Very very important question.

[02:33:43]Okay. Right.

[02:33:59]Okay.

[02:34:08]Ready. Okay.

[02:34:13]Okay.

[02:34:19]Okay.

[02:34:23]Yes.

[02:34:24]A cylindrical rod of length 1 m radius 4 m is just mounted vertically. Okay, just mounted vertically. It is subjected to the sheer force. Subjected to the sheer force of 10 ^ 5 Newton at the top.

[02:34:42]Okay. Considering considering infinitesimally small displacement in the upper edge, the angular displacement of the rod axis from its original position would be cylind vertically. It just subjected to the sheer force.

[02:35:26]Force in the direction subject. subjected to the sheer force that is 10 ^ considering small disperisal Force by G. Okay. Force by GA. Okay.

[02:36:37]So 10^ G power 10. So 10 power. Then area= r² r² 4 square so 4 square cm square 10 - 4 right square so r² into 10 - 16 into 10 ^ - 4 10 ^ 5 by 10 ^ 10 into 16 - 5. Okay. Right. So next 10^ 16 ^ 5 / 10 ^ 10us.

[02:37:41]So 10^ 16 power 10^ 6 final answer 1 / 16 into 10. So 1 / 160<unk> 1 divided by 160 pi angle value. Okay. Angle that is 1x 16 1x 16 pi.

[02:38:17]Clear.

[02:38:18]Okay. Right. So next question.

[02:38:26]Next question. Okay. So next question.

[02:38:31]Yes. A fractional the fractional compression it is delv by V of water depth 2.5 kilometer below the sea level below the sea level given the bulk modus of water is 2 into 10 ^ 9 density of water 10 ^ 3 acceleration due to gravity 10 J main 202 fractional compression value fractional compression of water del V by V Okay. Right. So dep Okay. Right.

[02:39:29]depth is equal to 2.5 kilome value 2.5 kilome below sea level modus is equal to change in pressure by change in volume by original volume.

[02:39:56]Change in volume compression.

[02:40:00]So change in volume change in pressure by change in pressure.

[02:40:10]Right? And the change in pressure.

[02:40:17]Change in pressure.

[02:40:21]H change in pressure value. H change in pressure. H change in pressure.

[02:40:35]D / bulk modulus. ded.

[02:40:41]fractional compression by equal to d by okay right change in v by original v is equal to d 2.5 kilome so 10 ^ 3 okay and value density of water and acceleration due to gravity okay Okay.

[02:41:12]Bulus value modus is 2 into 10 ^ 9. So 2 into 9 fraction change.

[02:41:25]Okay. Right. So main number 2.5 / 2.5id 2 is 1.25. Correct. 2.5id 2 is 1.25.

[02:41:38]1.25.

[02:41:48]3 10 3 that is 9 cancel okay right so I'm sorry guys 10^ 3.25 compress fractional compression it is 1.25 25.

[02:42:25]Okay. Right.

[02:42:28]Next question. Next question.

[02:42:39]A steel wire of length 2 m. Young's modus is 2 into 10^ 11 Newton per meter.

[02:42:49]Newton per me² by the force normal.

[02:42:53]steel wire change.

[02:43:07]If the poison ratio and the transverse strain for the wire is 0.2 in the case poison ratio value 0.2 into transverse strain value 10 ^ minus 3 transverse transverse strain value 10 ^us 3 transverse strain is nothing but lateral strain okay then the elastic potential energy of the energy density of the wire elastic potential energy density elastic potential energy density is equal to half y epilon square yilon Yes. Yes. Yes.

[02:43:57]Poison ratio.

[02:43:59]Poison ratio is equal to del r / l by.

[02:44:11]Okay. Del Rid L by L in the transverse strain is nothing but lateral strain. Transverse strain.

[02:44:22]Okay.

[02:44:28]So L by L is equal to R is so lateral strain divided by sigma P ratio. Okay. So by sigma sigma= 10 ^us 3 and sigma poison ratio value 10 ^us 3 / 10.2 2 into 10^ - 1 2us.

[02:45:02]Okay. So 1 by 2 1 by 2= 10us.

[02:45:11]So + 2. Okay. So 1 by 2.5.

[02:45:20]Okay. 2 - 3 - 22 right into okay value okay sigma value epilon value change.

[02:45:55]First of all, energy density elastic potential energy density.

[02:46:14]Okay. So the next u =/ epsilon y² okay y square epilon okay yepon square so 1 by 2 mod x modus value 2 into 10 ^ 11 correct 2 into 10 ^ 11 2 into 10 ^ 11 into 10 ^us 3 but the whole square.

[02:46:46]So next 2.

[02:46:50]So 10 ^ 11 5 square 25 10 10 ^ - 3 square 10 ^ - 6. So 25 main number 10 ^ 11 - 6.

[02:47:04]So 25 into 10 ^ 5.

[02:47:16]Okay. Right.

[02:47:20]J 2025 202.

[02:47:28]Okay. Right.

[02:47:32]Next question. Next question.

[02:47:40]Okay.

[02:47:51]question.

[02:47:53]Okay. Right. So, next question. A certain pressure P is applied to one liter. One liter of water and 2 L of water separately. certain pressure.

[02:48:17]Okay. Certain pressure one liter of water. Same amount of pressure 2 L of liquid.

[02:48:25]Water gets compressed by compressed to 0.1. Okay. 0 0.01 percentage. Whereas liquid gets compressed 0.03%.

[02:48:38]The ratio of bulk modus of water to that J 2023.

[02:49:04]Okay. Right. Compression change in volume by original volume is equal to change in pressure by bulk modus. Correct. Change in volume by original volume is equal to change in pressure by bulk modus.

[02:49:22]change volume.

[02:49:33]So change in volume by original volume it is inversely proportional to modus.

[02:49:40]It is inversely proportional to bul.

[02:49:48]Okay. V is inversely V dash is inversely proportional to bulk modus. Vash it is inversely proportional to bulus for water with respect to bulus for liquid. It is equal to inversely proportional inversely proportional water.

[02:50:29]So 0.03 water 0.01. So the answer 3 by 1 correct. Okay.

[02:50:44]One. Simple question.

[02:50:47]Okay. Right. In the next question. The elastic potential energy. The elastic potential energy stored in a steel wire of length 20 cm stretched through 2 cm is 80 jou. The crosssectional area of the wire J 2023 J 2023 elastic potential energy again elastic potial elastic potial energy = halfch Okayength change.

[02:51:41]Okay.

[02:51:43]The whole square del square. So u = 1 by 2 k value then l del² correct mod square. So value 1 2 Okay.

[02:52:16]And change in length.

[02:52:18]Change in value. 20 cm 2 cm 20 cm. 20 change in 2 cm. 2 cm 2 cm value.

[02:52:55]Okay. Right.

[02:53:00]Okay. So correct terms number 11 - 4 10^ 10 10^ 4 6 40 is equal to 10 ^ 6 into a okay 40 is equal to 10^ a is equal to 40 by 10 ^ 6 40 into 10 ^ wait a minute wait a minute okay 40 into 40 into 10 ^ - 6 area value 40 into 10 - 6 mm²m 40.

[02:54:18]Okay. Right.

[02:54:27]Okay.

[02:54:28]Next important question.

[02:54:31]The elastic behavior of the material for a linear stress and linear strain as shown in the figure.

[02:54:46]20 40ome straight.

[02:54:58]The energy density for a linear stream 5 into 10 ^us 4. Okay.

[02:55:06]Energy density for a linear strain value epsilon square correct half number.

[02:55:31]Okay. elastic strain.

[02:55:36]Okay.

[02:55:38]value.

[02:55:47]First of all, in the straight line.

[02:56:03]Okay.

[02:56:10]Reciprocal of reciprocal first.

[02:56:23]Okay.

[02:56:26]So the last stress Okay.

[02:56:38]Okay. dx dy correct. Okay. dy by dx formula slope formula dy by dx correct dyra stress by stress epilon epilon 4 into 10us 4 into 10 ^ - 10 / 80 4 into 10 - So 80 / 4 into 10 ^ - 10 / 4 into 10^ - okay right Sorry.

[02:57:39]10us 20.

[02:57:45]Okay. 20 2= 1x 2 yon².

[02:58:01]Now 1 2 and value 5 into 10 power - 4. So 5 into 10 - 4 square - 4² - 8. Okay. Next 2. So 25 into 11 - 8. So 25 into 3 joule per me cube joule per me cube. 25 into 3 25 into 3. Okay.

[02:58:38]So 25 into 10 ^ 3 is the right answer.

[02:58:41]25 into 10 ^ 3 is the right answer.

[02:58:45]Clear.

[02:58:47]Okay. Right.

[02:58:50]Next question. Next question. Two wires endg of materials of two wires y1 and y2 modus value y1 in the x modus value Y2.

[02:59:26]The combination behaves as a single wire combination combination combination value.

[02:59:44]Okay.

[02:59:51]1x 1 by k1 + 1 by k2 y 2 y1 y2 y1 + y2 simple. Okay. 2 y1 y2 y1 + y2 option.

[03:00:13]Okay. And this is the last question.

[03:00:16]This is the last question for solids.

[03:00:17]Match the following. Match list one and list two.

[03:00:23]Force that restores the elastic body of unit area to its original state.

[03:00:30]Force that restores an elastic body of unit area to its original state. Okay.

[03:00:36]Force per unit area. Force per unit area.

[03:00:51]Stress will be equal to force. Stress is equal to force.

[03:00:56]Okay. Force that restores the elastic body of area its option.

[03:01:07]option.

[03:01:18]Two equal and opposite forces are parallel to opposite faces.

[03:01:32]Right? equal opposite option second option.

[03:01:48]Second option option option C. Okay. Forces perpendicular everywhere to the surface per unit area and same everywhere.

[03:02:05]at the two equal forces opposite forces perpendicular to opposite forces that is basic right. So mechanical properties of fluids and solids.

[03:02:36]Okay. Right.

[03:02:51]Okay. Right. So, thank you so much. Take care. Bye.