33 Years NEET-AIPMT Chapterwise Solutions - Physics 2020
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33 Years NEET-AIPMT Chapterwise Solutions - Physics 2020

All previous year questions , and these are downloaded from the sources in the internet. I don't own these resources and these are copyrighted by MTG. All previous year questions , and these are downloaded from the sources in the internet. I don't own these resources and these are copyrighted by MTG.

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Telegram @unacademyplusdiscounts82 NEET-AIPMT Chapterwise Topicwise Solutions Physicsd = npr + r(1 – p) = (np + 1 – p)rd = {1 + (n – 1)p} r3. (d) : Let the speed of the ejection of the liquidthrough the holes be v. Then according to the equation ofcontinuity, pR 2 V = npr 2 v or2 2πRVVRv = =2 2nπrnr4. (b) : According to Torricelli’s theorem,Velocity, v= 2gh= 2× 10 × 2 = 632 . m/s.From equation of continuity,Volume of liquid flowing per second, Q = Av= 2 × 10 –6 × 6.32 = 12.6 × 10 –6 m 3 /s5. (a)6. (a) : According to equation of continuity,Av = constantTherefore, velocity is maximum at the narrowest part andminimum at the widest part of the pipe.According to Bernoulli’s theorem for a horizontal pipe,P+ 2ρ v = constant2Hence, when a fluid flow across a horizontal pipe ofvariable area of cross-section its velocity is maximumand pressure is minimum at the narrowest part and viceversa.7.2r( ρ−σ)g(d) : Terminal velocity, v =9ηRatio of terminal velocity of spherical metal balls,2 2v()( 2 21 9 1 8 ρ − 01 . ρ )=v 22 2()( 2 29 2 ρ − 01 . ρ )⇒v179 . ρ279= =v2409 (. ρ2) 368. (c) : The viscous drag force, F = 6phrv;where v = terminal velocity\ The rate of production of heat = power= force × terminal velocity⇒ Power = 6phrv · v = 6phrv 2∵22r( ρ−σ ) gTerminal velocity v =;9η⎡ 4 24r( ρ−σ)⎤Now,power = 6 ⎢2πηrg ⎥⎣⎢281η⎦⎥or Power ∝ r 59. (c) : Force of surface tension balances the weight ofwater in capillary tube.F s = T cos q (2pr) = mgm ∝ rHence, m ′ r′m = ⇒ ′ 2r=m r 5 g r⇒ m′ = 10 g10. (d) : The pressure at a point Z 0 below the surface ofwater, P Z0 = P 0 + r g Z 0Also, pressure inside a soap bubble, P = P 0 + 4T RAs per question, P Z0 = P\ P 0 + 4T R = P 0 + r gZ 0−24T4 25 10Z 0 =Rρ g= × . ×= 1 × 10−3 3–2 m = 1 cm1× 10 × 10 × 1011. (b) : Work done = Surface tension of film × Changein area of the filmor, W = T × DAHere, A 1 = 4 cm × 2 cm = 8 cm 2 , A 2 = 5 cm × 4 cm = 20 cm 2DA = 2(A 2 – A 1 ) = 24 cm 2 = 24 × 10 –4 m 2W = 3 × 10 –4 J, T = ?−×−∴ T = W 43 10 11= = = 0. 125 N m∆A−424 × 10 8T12. (b) : Capillary rise, h = 2 cosθrρgFor given value of T and r, h ∝ cosθρcosθcos cosAlso, h 1 = h 2 = h 3 or1 θ2θ3= =ρ1ρ2ρ3Since, r 1 > r 2 > r 3 , so cosq 1 > cosq 2 > cosq 3For 0 ≤ q < p/2, q 1 < q 2 < q 3Hence, 0 ≤ q 1 < q 2 < q 3 < p/213. (d) : Water will not overflow but will change itsradius of curvature.14. (c) : Let n droplets each of radius r coalesce to forma big drop of radius R.\ Volume of n droplets = Volume of big dropn r R n R 3× 4 3 = 4 3π π3 3⇒ = 3r... (i)Volume of big drop, V = 4 3π R3...(ii)Initial surface area of n droplets,3RAi n 2r r3r2π π (Using (i))3R= = ⎛ ⎝ ⎜ 4 ⎞43 3 3Vπ πR ⎠⎟ =r 3 r r (Using (ii))Final surface area of big dropAf 2 V= R = ⎛ R⎝ ⎜ 4 ⎞ 3 34π3π⎠⎟ =3 R R (Using (ii))Decrease in surface area∆A= Ai− Af3V3V⎛1 1⎞= − = 3V −r R ⎝⎜ r R ⎠⎟\ Energy released = Surface tension× Decrease in surface area⎛1 1⎞= T × ∆A= 3VT −⎝⎜ r R ⎠⎟15. (b) : The wettability of a surface by a liquid dependsprimarily on angle of contact between the surface and theliquid.vvv

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