07 NEET EXAMINATION SOLUTION (6) Avail Vide Lectures f Exerienced Faculty Page
Sl. The lean exressin which satisfies the utut f this lgic gate is C = A., Whichindicates fr AND gate. We can see, utut C is zer when either f A r (inuts) is zer. Sl. Sl. When articles are emitted, the mass number decreases by 4 units and atmic number decreases by units and fr article, atmic number is increased by and mass number remains the same. Given reactin is 9 e 4 88 Y 8 Number f articles = 4 8 6 4 4 4 Decreases in atmic number = 4 = 8 Frm atmic number 88, number f articles emitted = 88 84 4 Hence, 4 and 4 articles are emitted. Particle mmentum is inversely rrtinal t the assciated de rglie wavelength h Hence, grah (d) is reresenting crrect relatin. Sl.4 The central maxima lies between the first minima n bth sides. The angular width f central maxima = a Sl.5 We knw that, E 0 i cs t L...(i) and E 0 i cs t L Nw i E E i L L 0 0 cs t cs t E cs t L L L L E cs t L L 0 0 Sl.6 Sl.7 Efficiency f transfrmer, Given, Hence, utut wer Inut wer V s I s = 00 W, V = 0 V, I = 0.5 A 00 0.90 r 90 % 0 0.5 Magnetic di r magnetic inclinatin is given by tan v H...(i) V I V I where, V and H are vertical and hrizntal cmnents f earth`s magnetic field resectively. s s Page
Given, H Frm Eqs. (i) and (ii), we get tan 60...(ii) V V H Sl.8 Magnetic Lrentz frce is given as, F = q( v ) Sl.9 Sl.0 Sl. Sl. Sl. Fr the balance int f meter bridge, Case I When balancing length is 55 cm Case II P l 55 45 Q 00 l Q 45 55 Q When an unknwn resistance x is nvlved P x l x 75 Q 00 l Q 5 45 P x l x 75 8 48 + x = = 55 Q 00 l Q 5 All the caacitrs are cnnected in arallel. Therefre, the effective caacitance between A and = 6 C = 6 = F x In the case (a), the fields due t charges at the site crners cancle each ther. S, in this case net electric field at the centre will be zer. t x Given, equatin f wave, y 5sin 0.04 4 The standard equatin f a wave in the given frm is x y a sin t Cmaring the given equatin with the standard equatin, we get a = 5 and 5 0.04 Therefre, maximum velcity f articles f the medium. max a 5 5 5 cms.5 ms Ttal energy in SHM, E m a, (where, a = amlitude) K m a y K E m Kinetic energy, When y K E m E 4 4 a a E E E 4 Page
Sl.4 Rt mean square seed f mlecules in a gas is defined as the square rt f mean f the square f the seed f different mlecules, i.e., rms... N RT r rms (accrding t kinetic thery f gases) M while mst rbable seed is the seed which maximum number f mlecules in a gas have at cnstant temerature and is gives by m RT M It is bvius that, rms m. Als, Maxwell distributin fr the seed f mlecules in a gas is asymmetrical. Sl.5 Fr an adiabatic rcess, ressure () and vlume (V) f the system are related as V = cnstant...(i) where, is the rati f secific heat caacities at cnstant ressure and cnstant vlume. Fr an ideal gas, V = nrt (ideal gas equatin) Frm Eqs. (i) and (ii), we get V = nrt nrt = cnstant T = cnstant [ number f mles (n), gas cnstant (R) are cnstant] Sl.6 It is clear frm the grah, ab > cd. and S A < S Sl.7 Fig. is streamlined, s air resistance f it will be minimum. Fr Fig. surface area is maximum, s air resistance fr it is maximum. Hence, crrect sequence is < <. Sl.8 As, F / A Y strain = strain F AY Sl.9 The acceleratin due t gravity n the new lant can be fund using the relatin GM g R 4 ut M R, being density. Thus, Eq. (i) becmes 4 G R 4 g G R g R R g R g R g R g g g g R g Page 4
Sl.0 Accrding t arallel axes therem, IX Y IXY MR 6 kg-m Sl. Kinetic energy is directly rrtinal t square f velcity. K Sl. Sl. The area under F-t grah gives change in mmentum. Fr 0 t s, 6 6 kg-m/s 6 kg-m/s Fr t 4 s, Fr 4 t 8 s, 4 kg-m/s S, ttal change in mmentum fr 0 t 8 s net 6 6 kg-m/s = N-s The bdy is describing a vertical circle, m m T mg cs T mg cs l l Tensin is maximum when cs = and velcity is maximum. th cnditins are satisfied at = 0, i.e. at lwest int. Sl.4 Sl.5 As, 0 na a Sn an n Distance travelled in (n ) secnd is s n a n S, distance travelled in the last s is a sn sn an a n n n a n n n n n a n n Dimensins f = Unit f Unit f A metre newtn newtn metre amere metre amere kg ms m kgm s A ML T A A N0 Sl.6 Frm the questin, N = N and t / = T yr 00 S, N 0 0.69 T.0.0T t lg N lg 00 = 6.65 T 0.69 Page 5
Sl.7 Crrect answer is. Sl.8 Cnditin fr nth secndary minima is that ath difference a sin n = n nth secndary maxima is ath difference = n a sin n Fr st minima, 5500 A, 0 n a sin 0...(i) Fr nd maxima, ath difference = a sin n Dividing Eq. (i) Eq. (ii), we get...(ii) sin n sin 4 n 4 n sin Sl.9 Sl.0 Sl. Resnance frequency, 0 0 6 6 0 LC 000 0 0 Nw, given = 70 k-rad/s = 70000 rad/s Here, 0 Thus, the circuit is caacitive dminant circuit. Sft irn is used fr making transfrmer cres. Time erid f susended magnet 5 rad/s T I Mcs Frequency Mcs I cs r 400 cs 60 cs cs0 5 6 6 :9 9 Sl. r m q 4 r m q Page 6
Sl. Fr a tentimeter, the internal resistance (r) is given by l l r R Given, R 0, l 75 cm and l 65 cm 75 r R 0 0.54.54 65 Sl.4 If a surce is cnnected between int P and R, same charge will flw thrugh tw caacitrs in arm PQ and QR. Similarly, same charge will flw thrugh caacitrs in arms PT, TS and SR. S, equivalent caacitance f left side C C C C C C and equivalent caacitance f right side C C C C C C C Nw, C' and C" will in arallel cmbinatin, C C 5C Hence, C C C 6 Similarly, if a surce is cnnected between ints P and Q, then equivalent caacitance C 5C C C 4 4 Hence, the required rati is given by Sl.5 Ptential, (V) = x + 5 C 5C / 6 C 5C / 4 dv Intensity f the electric field = 6x 6 dx (E at x = ) = Vm Sl.6 Sl.7 Amlitude f ressure variatin m ka and intensity f wave I A S, m i.e., ressure variatin (intensity) / In SHM, velcity I a y 60 0 mms Page 7
Sl.8 Pressure due t an ideal gas is given by M V On utting, M V = as the density f gas. Sl.9 C As, we knw C V then value f mlar secific heat f an ideal gas at cnstant vlume, i.e., C V R Sl.40 Sl.4 Sl.4 At 0 C frm b t c, temerature f matter des nt changes but its state changes. Similarly, frm d t e state f matter changes withut changing temerature. Hence, b t c and d t e shw hase changes. all bearing als knwn as anti-frictin bearings, are small metallic r ceramic sheres used t reduce frictin between shafts and axles in a number f alicatins. all bearings are ftenused in a series t absrb the weight laced n a mving art r in individual cases t reduce frictin in axle assemblies. Nst ball bearings are manufactured t meet very exacting standards f rundness, since any defrmatin can cause the mving arts t fail unexectedly. Yung s mdulus f material f the bdy is given by lngitudinal stress MgL Y lngitudinal strain A l Putting the numerical values, we have L = m, A = 50 mm = 50 0 6 m l = 0.5 mm = 0.5 0 m, M = 50 kg 509.8 Y 9.60 Nm 6 500 0.50 0 Sl.4 Let the each side f square lamina is d. S, I EF = I GH (due t synnetry) I AC = I D (due t synnetry) Nw, accrding t the erendicular axes therem, I AC + I D = I 0 r I AC = I 0...(i) and I EF + I GH = I 0 r I EF = I 0...(ii) Frm Eqs. (i) and (ii), we get I AC = I EF I AD = I EF + S, I AD = md md md md md as IEF 4 4 = 4I EF Page 8
Sl.44 Frm the questin, we see that the mass f the bdy is 6 kg and the dislacement is given by x. S, the velcity 4 dx t wuld be equal t. dt t Thus, velcity at time t = s, () = =. Hence, the initial kinetic energy wuld be equal t wuld be KEi m 6 S, we see that the wrk dne by the frce is equal t W = KE f KE i = 0 = J KEi m 0 0 and similarly, the final kinetic energy Sl.45 Linear mmentum gained = imulse frm 0 t 4 s = area enclsed by grah frm 0 t 4 s r N-s ANSWER-KEY Q ue. 4 5 6 7 8 9 0 Ans. A A D C C A C A Q ue. 4 5 6 7 8 9 0 Ans. A D C D D D C A A D Q ue. 4 5 6 7 8 9 0 Ans. C C D A C C A Q ue. 4 5 6 7 8 9 40 Ans. D C C D A D D Q ue. 4 4 4 44 45 Ans. C C C D C Page 9