DC Pandey Mechanics Volume 1

590 — Mechanics - I (b) The block will come to rest when ka = f The lower block will rebounce when or a = k …(A) x > m2g (kx = m2g) f k In the similar manner, we can write Substituting, x = m2g in Eq. (i), we get k 2f a1 − a2 = k …(i)  m2g  2  m2g  …(ii) k k …(n) 2m1 gH = k + 2m1g a2 − a3 = 2f …(B) or H = m2g  m2 + 2m1 k k  2m1  …… an − 1 − an = 2f Thus, H min = m2g  m2 + 2m1 Ans. k k  2m1  Adding Eqs. (i), (ii),… etc., we get x0 − an = n  2f  7. Ei = Ef k ∴ 1 1 m v2 2 1 or an = x0 − n  2f  2 mv2 = 2 + 2 kx2 k 3v2m Equating Eq. (A) and (B), we get or k = 4x2 Ans. k = x0 − n  2f  8. µmAg = 0.8 × 6 × 10 = 48 N f k k (mB + mC )g = (1 + 2) × 10 = 30 N f = x0 − = kx0 − 1 Since (mB + mC )g > µmAg, aA = aB = 0. 2f 2f 2 or n From conservation of energy principle we can prove k that maximum distance moved by C or maximum Number of cycles, extension in the spring would be xm = 2mCg = 2 × 1 × 10 k 1000 m = n = kx0 − 1 = 1  kx0 − 1 Ans. 2 4f 4 4 f  = 0.02 m 6. Conservation of mechanical energy gives, kxm EA = EB C ac or 1 m1v2 = 1 kx2 + m1 gx 2 2 mc g or 2m1 gH = kx2 + 2m1 gx …(i) v=0 At maximum extension B m1 aC = kxm − mCg mC x Substituting the values we have, aC = 10 m/s2. Ans. A m1 v = Ö2gH 9. Rate at which kinetic plus gravitational potential energy is dissipated at time t is actually the magnitude of power of frictional force at time t. m2 m2 |Pf | = f .v = (µmg cos α )(at) = (µmg cos α )[(g sin α − µg cos α )t ] = µmg2 cos α (sinα − µ cos α )t Ans.

Chapter 9 Work, Power and Energy — 591 10. From work-energy principle, W = ∆KE (c) 1 mv2 = mgR2 sin  Rl  + sin θ − sin θ + Rl   2l  Pt = 1 m (v2 − u2) ∴ 2 (P = power) 2gR2 sin  Rl  θ Rl   l  or t = m (v2 − u2) or v= + sin θ − sin + 2P …(i) Further v2 = 2gR2 sin  Rl  + sin θ − sin θ + Rl   l  F⋅v = P 2gR2 ∴ m ⋅ dv ⋅ v2 = P ∴ 2v ⋅ dv = l cos θ − cos θ + Rl   ⋅ dθ ds dt  dt v v2 dv = P x 2gR2 cos θ θ Rl   ∫ ∫or l − cos +  ds u m0 dv  dθ  ∴ (v3 − u3) = 3P ⋅ x = dt …(i) m dt 2v Ans. or m = 3x Here  ddθt  = ω = 1 P v3 − u3 v vR Substituting in Eq. (i) Substituting in Eq. (i), we get t = 2 3x (u + v) Hence proved. dv = gR cos θ − cos θ + Rl   (u2 + v2 + uv) dt l  11. (a) Mass per unit length = m At t = 0, θ = 0° l Hence, dv = gR 1 − cos  Rl   dt l  dm dα 12. From conservation of energy, α αh h= 0 m2 gh2 = m1 gh1 + 1 m1v12 + 1 m2v22 2 2 Here, v1 = v2 cos θ ∴ 2 × 10 × 1 = (0.5) (10) ( 5 − 1) + 1 × 0.5 2 dm = m R dα × v22 ×  2  2 1 l 5 2 + × 2 × v22 h = R cos α ∴ 20 = 6.18 + 0.2 v22 + v22 dU = (dm) gh = mgR2 cos α ⋅ dα 2m l ∫∴ l/ R = mgR2 sin  Rl  U= dU θ 0l 5m (b) KE = U i − U f 1m Here, Ui = mgR2 sin  Rl  v2 m2 l θ m1 v1 and ∫U f = l/ R + θ dU = mgR2 sin  l + θ − sin θ v2 = 3.39 m/s Ans. θl R ∴ KE = mgR2 sin  Rl  + sin θ − sin θ + Rl   and v1 = v2 cos θ = 2 × 3.39 l  5 Ans. or v1 = 3.03 m/s Ans.

592 — Mechanics - I 13. Net retarding force = kx + bMgx Substituting the values we have, ∴ Net retardation =  k + bMg  ⋅ x vf = 2 × 9.8 × 1.9 + 2300 (0.045)2 M 0.12 So, we can write = 8.72 m/s Ans. v⋅ dv = −  k + bMg  ⋅ x 15. (a) From work energy theorem, dx M Work done by all forces = Change in kinetic energy or 0 = −  k +MbMg  x ∴ Fx − 1 kx2 = 1 mv2 ∫ v ⋅ dv ∫ x dx 22 v0 0 ∴ v = 2Fx − kx2 or x= k M v0 m + bMg Substituting the values we have, Loss in mechanical energy v = 2 × 20 × 0.25 − 40 × 0.25 × 0.25 0.5 ∆E = 1 Mv02 − 1 kx2 2 2 = 15 m/s = 3.87 m/s Ans. 1 k  +MbMg  or ∆E = 2 Mv02 − 2 k v02 (b) From conservation of mechanical energy, v02   +MbMg   Ei = Ef 2 M k  or ∆E =  − k  or 1 mvi2 + 1 kxi2 = 1 kx2f 2 2 2 = v02bM 2g Ans. or xf = mvi2 + xi2 2(k + bMg) k 14. From conservation of mechanical energy, = 0.5 × 15 + (0.25)2 40 Ei = Ef = 0.5 m (compression) or 1 kxi2 + mghi = 1 mv 2 ∴ Distance of block from the wall 2 2 f = (0.6 − 0.5) m ∴ vf = 2ghi + k xi2 = 0.1 m Ans. m

10. Circular Motion 10.1INTRODUCTORY EXERCISE (b) ar = v2 ⇒ v= arR R 1. Direction of acceleration (acting towards centre) or v = (21.65) (2.5) continuously keeps on changing. So, it is variable acceleration. = 7.35 m/s Ans. Ans. 2. In uniform circular motion speed remains constant. (c) at = a sin 30° = (25)  21 = 12.5 m/s2 In projectile motion (which is a curved path) acceleration remains constant. 3. (a) ar = v2 = (2t)2 = 4t2 10.2INTRODUCTORY EXERCISE R 1.0 At t = 1 s, ar = 4 cm/s2 Ans. 1. v = 18 × 5 = 5 m/s Ans. (b) at = dv = 2 cm/s2 18 dt tan θ = v2 = (5)2 = 1 Rg 10 × 10 4 (c) a = ar2 + at2 = (4)2 + (2)2 = 2 5 cm/s2 Ans. ∴ θ = tan−1  41 Ans. 4. vav =s= 2R T/4 2. v = µRg t (T /4) ∴ = 4 2R RS µ = v2 = (5)2 (2πR )/ v R O gR 10 × 10 ∴ vav = 2 2 Ans. = 0.25 Ans. vπ 3. v = Rg tan θ = 50 × 10 × tan 30° 5. (Rω2) = Rα ≈ 17 m/s Ans. ∴ ω2 = α 4. a ≠ 0 ⇒ Fnet = ma ≠ 0 or (αt)2 = α Hence, particle is not in equilibrium. ∴ t= 1 = 1 5. If he applies the breaks, then to stop the car in a α4 distance r = 0.5 s Ans. Wall 6. at = dv = 8t at vθ dt a r ar = v2 = (4 t 2 )2 = 8 t4 R 54 27 ar At t = 3 s, v at = 24 m/s2 0 = v2 − 2a1 r ar = 24 m/s2 ∴ a1 = v2 = minimum retardation required. 2r tan θ = at = 1 ar (by friction) If he takes a turn of radius r, the centripetal ∴ θ = 45° Ans. acceleration required is 7. (a) ar = acos 30° = (25) 3 a2 = v2 (provided again by friction) 2 r = 21.65 m/s2 Ans. Since a1 < a2, it is better to apply brakes.

594 — Mechanics - I 6. T1 (c) N 2 − mg = mv2 R θ ∴ N2 = m  + v2  θ4 g   R T2 5 mg = 16 × 103 10 + 2500  θ 10 250  3 = 32 × 103 N = 32 kN Ans. T1 cos θ + T2 cos θ = mrω2 ...(i) 10.3INTRODUCTORY EXERCISE ω = (2nπ ) Here, n = number of revolutions per second. 1. (a) String will slack at height h1, discussed in article Substituting the proper values in Eq. (i), 10.5 where, 200 ×  35 + T2 ×  35 = (4)(3)(2nπ )2 h1 = u2 + gR = 5R (as u2 = 4gR) 3g 3 or 600 + 3T2 = 240 n2π 2 ...(ii) u2 − 2gh1 = 4gR − 2g × 5 R 3 Further, T1 sin θ = T2 sin θ + mg v= or 200 × 4 = T2 × 4 + 4 × 10 = 2 gR 5 5 3 or 800 = 4T2 + 200 ...(iii) 2. (a) Velocity becomes zero at height h2 discussed in Solving Eqs. (ii) and (iii) we get, article 10.5 where, T2 = 150 N and n = 0.66 rps u2 gR R = 39.6 rpm h2 = 2g = 2g = 2 7. (a) mg − N 1 = mv2 Now, h2 = R (1− cos θ) R ∴ R = R (1− cos θ) ⇒ θ = 60° 2 N1 N2 (b) T = mg cos θ = mg at θ = 60° mg 2 mg 3. (a) v = u2 − 2gh = 7gR − 2g(2R) = 3gR or mg − mg = mv2 (b) T + mg = mv2 = m (mgR) 2R RR ∴ mv2 = mg = 16kN = 8kN ∴ T = 2mg R2 2 (c) T − mg = mu2 = m (7gR) mv2 mv2 RR R R Now, N 2 − mg = or N2 = mg + ∴ T = 8mg = mg + mg = 3 mg 4. h = l − l cos 60° = l = 2.5 m 22 2 = 3 (16 kN) = 24 kN Ans. u=0 60° 2 (b) mg − N = mv2 or N = mg − mv2 h RR v0 0 = mg − mvm2 ax v0 = 2gh = 2 × 9.8 × 2.5 = 7 m/s Ans. R ∴ vmax = gR = 10 × 250 = 50 m /s Ans.

Exercises LEVEL 1 7. At A and C, v = 0. Therefore, radial acceleration v2 Assertion and Reason R is zero. But tangential acceleration g sin θ is 2. A to B S= 2R non-zero. and |∆v| = v2 + v2 − 2v ⋅ v cos 90° 8. For t < 3 sec, speed is negative which is not = 2v possible., Average acceleration = |∆v| = 2 v 9. In circular motion, a ≠ constant, so we cannot apply tt v = u + at directly. and average velocity = S = 2R In vertical circular motion, gravity plays an t t important role. The desired ratio is v = ω 10. R Nθ 3. Direction of acceleration keeps on changing. So, it a is variable accelration. Further, it is accelerated so it mg is non-inertial. θ 4. v v aa N cos θ = mg v = constant v is increasing ⇒ N = mg sec θ v a and N sin θ = mv2 R 11. Weight (plus normal reaction) provides the necessary centripetal force or weight is used in providing the centripetal force. v is decreasing Single Correct Option 1. Magnitude of tangential acceleration is constant but In first figure, v ⋅ a = 0 its direction keeps on changing. In second figure, v ⋅ a = positive and 2. At θ = 180°, |∆P| = 2 mv = maximum In third figure, v ⋅ a = negative At θ = 360° , |∆P | = 0 = minimum Further, ω and v are always perpendicular, so 3. h = l − l cos θ = l (1 − cos θ) ω⋅ v=0 lθ 5. Component of acceleration perpendicular to velocity is centripetal acceleration. ∴ ac = v2 or R = v2 = (2)2 = 2 m h R ac 2 6. At A v Tangential acceleration is g. Radial acceleration is v2 = 2gh = 2gl (1 − cos θ) = vm2 ax v2. R ∴ K max = 1 mv2 = mgl (1 − cos θ) Ans. 2  v2  2   ∴ anet = g2 +  R 4. Rod does not slack (like string). So, minimum velocity at topmost point may be zero also.

596 — Mechanics - I 5. T l θ u T mg T − mg = mu2 = m ( 5gR )2 r RR mg ∴ T = 6 mg Ans. Solving these two equations we get, 6. v2 = at = a (Here, at = asay) cos θ = g = l g n)2 R lω2 (2π or (at)2 = a = 10 2 (l = 1 m) R ×2  2π Ans. ∴ t = R = 20 = 2 s π Ans. a5 Ans. ∴ θ = cos−1 (5/8) 7. cos (dθ) components of T are cancelled and 10. T = mg = 0.1 × 10 = 8 N sin (dθ) components towards centre provide the cos θ 5/8 5 necessary centripetal force to small portion PQ. 11. f = ma = m at2 + ar2 dθ dθ TP QT = m (Rα )2 + (Rω2)2 = m (Rα )2 + [R (α t)2 ]2 dθ dθ 2  2 2 0.25  = 0.36 × 10−3  0.25 × 31 +   1 × 2  3 ∴ 2T sin (dθ) = (mPQ ) (R) ω2 = 50 × 10−6 N = 50 µ N Ans. For small angle, sin dθ ≈ dθ 12. v2 = 2gh ∴ 2T dθ =  m  (2θ) (R) (2nπ )2 T 2π ∴ T = 2π mn2R Substituting the values we get, v T = (2π ) (2π ) (300/60)2 (0.25) mg ≈ 250 N 0.3 × 300 × 10 Ans. T − mg = mv2 8. mv2 = µ mg Ans. l ...(i) R ...(ii) or T = mg + m (2gh) ∴ v = µ Rg = l = 30 m/s = 108 km/h = mg 1 + 2lh 9. T cos θ = mg Ans. T sin θ = m rω2 = m (l sin θ) ω2

Chapter 10 Circular Motion — 597 13. (ω1 − ω2) t = 2π Taking moment about C t = 2π = 2π No (x) = f (h) ω1 − ω2 (2π /T1) − (2π /T2)  mv 2    T1 T2 = 3 × 1 or (mg ) x =  R h T2 − T1 3 − 1 ∴ v = gRx = 1.5 min Ans. h Ans. 14. dB = vBt = 2.5 = 5 = 9.8 × 250 × 0.75 1.5 dC vCt 2 4 = 35 m/s Subjective Questions 5. Let ω be the angular speed of rotation of the 1. at = 8 m/s2 bowl.Two forces are acting on the ball. ar = v2 = (16)2 = 5.12 m/s2 R 50 ω a = at2 + ar2 = 9.5 m/s2 Ans. Rα N 2. µN A r mg N 1. normal reaction N 2. weight mg mg The ball is rotating in a circle of radius r (= R sin α ) with centre at A at an angular speed ω. Thus, N = mRω2 ...(i) µN = mg ...(ii) N sin α = mrω2 From these two equations we get, ω= g Ans. = mRω2 sinα …(i) µR Ans. and N cos α = mg …(ii) Dividing Eq. (i) by Eq. (ii), we get = 9.8 = 4.7 rad /s 0.15 × 3 1 = ω2R cos α g ⇒ R = v2 = (u cos θ)2 3. a = v2 ag ∴ ω= g R cos α R 6. R = L sin θ 4. L T C No θ Ni = 0 h f = mv 2/R x mg R mg T sin θ = mRω2 = m (L sin θ) ω2 In critical case, normal reaction on inner wheel N i ∴ T = mLω2 will become zero. Normal reaction on outer wheel Now, N o = mg. Friction will provide the necessary ∴ T cos θ = mg centripetal force. cos θ = mg = mg = g ∴ f = mv2 T mLω2 Lω2 R Hence proved.

598 — Mechanics - I 7. t = 2h = 2 × 2.9 = 0.77 s ∴ ω= g R cos θ g 9.8 Horizontal distance x = vt At lowermost point, θ = 0° ∴ v = x = 10 ∴ ω= g t 0.77 R ≈ 13 m/s Substituting ω = 2g / R in Eqs. (i), we have, a = v2 = (13)2 Ans. cos θ = 1 Ans. R 1.5 2 ≈ 113 m/s2 ∴ θ = 60° 8. (a) The frictional force provides the necessary 10. (a) 2 kg T centripetal force. T = m2 r2ω2 = (2) (2) (4)2 = 64 N ∴ mLω2 = µmg Centripetal force required to 1 kg block is or ω = µg L (b) Net force of circular motion will be provided by 1 kg the friction T = 64 N ω = αt …(i) f = 48 N Fnet = m at2 + an2 Fc = m1 r1 ω2 = (1) (1) (4)2 = 16 N ∴ µmg = m (Lα )2 + (Lω2)2 …(ii) But available tension is 64 N. So, the extra force of 48 N is balanced by friction acting radially Here, at = Lα and an = Lω2 outward from the centre. (b) fmax = µm1g = 0.8 × 1 × 10 = 8 N Substituting ω = αt in Eq. (ii) we have, µg = L2α 2 + L2α 4t4 2 kg T T 1 kg 8 N 1  µ 2g 2− L2α 2  4 (m2 r2 ω2) (m1 r1 ω2)  L2α  ∴ t =  4  T = m2 r2 ω2 = (2) (2) ω2 = 4 ω2 T − 8 = m1 r1 ω2 = (1) (1) (ω2) = ω2 Substituting the value of t in Eq. (i) we have, Solving Eqs. (i) and (ii) we get, ...(i) ...(ii) 1 ω = 1.63 rad /s (c) T = m2 r2 ω2 Ans.  µLg 2  4  ω = − α 2  9. N cos θ = mg ω 2 kg T m2 r2 ω2 R θN ∴ 100 = (2) (2) ω2 or ω = 5 rad /s mg 11. v2 = v02 + 2gh = (0.5 gr )2 + 2gr (1 − cos θ) r = (2.25 gr − 2gr cos θ) N sin θ = mrω2 At the time of leaving contact, N = 0 = m (R sin θ) ω2 ∴ mg cos θ = mv2 = 2.25 mg − 2 mg cos θ r Dividing these two equations, ∴ cos θ = 2.25 = 3 34 we get cos θ = g ...(i) R ω2 ∴ θ = cos−1 (3/4) Ans.

Chapter 10 Circular Motion — 599 h LEVEL 2 θ Single Correct Option v 1. EA = EB θ mg ∴ 1 mvA2 = 1 × 200 × (13 − 7)2 (m = 2 kg) 2 2 12. 2.5 mg − mg cos 30° = mv2 ∴ vA = 60 m/s Ans. At A, N = mvA2 = (2) (60)2 = 1440 N r RS θ 2. Let us see the FBD with respect to rotating cone T = 2.5 mg (non-inertial) v 150° N R F = mRω2 = centrifugal force mg 30° N , F and mg are balanced in the shown diagram. If displaced upwards, F will increase as R is increased. mg This will have a component up the plane. So, it will move upwards. Hence, the equilibrium is unstable. ∴ 1.63 g = v2 = v2 r2 3. µmg = mRω2 ∴ v = 5.66 m/s ∴ ω2 = µg = (1/3)g = 4 g R 5a/4 15a Fnet = (2.5 mg)2 + (mg)2 + (2) (2.5 mg) (mg) cos 150° Ans. = 1.7 mg 4. aA = g sin θ anet = Fnet = 1.7 g ≈ 16.75 m/s2 Ans. v2 2gh 2gR (1 − cos θ) m R R R aB = = = 13. v2 = v02 + 2 gh = v02 + 2 gR sin θ Given, aA = aB ∴ sin θ = 2 − 2 cos θ θ h 2 mg Squaring these two equations we have, sin2 θ = 4 + 4 cos2 θ − 8 cos θ v or 1 − cos2 θ = 4 + 4 cos2 θ − 8 cos θ 90° – θ On solving this equation, we get cos θ = 3 or θ = cos−1 (3/5) mg Ans. 5 = (5)2 + 2 × 10 × 2 sin θ 5. At the time of leaving contact = (25 + 40 sin θ) Now, 2 mg − mg sin θ = mv2 v R θθ or 2g − g sin θ = v2 R or 2 × 10 − 10 sin θ = 25 + 40 sin θ mg 2 N =0 Solving this equation we get, Ans. ∴ mg cos θ = mv2 = m (2gh) θ = sin−1  41 RR

600 — Mechanics - I 2h 2 R + R (1 − cos θ) 9. v = 2πr = (2π ) (0.5)  4 ∴ cos θ = = T 1.58 RR F1 F2 mg On solving this equation, we get cos θ = 5/6 or θ = cos−1  56 Ans. 6. hAB = R cos 37° − R cos 53° vB 37° ≈ 2 m/s B 37° F1 = mg = 100 N F2 = mv2 = 10 × 4 = 80 N r 0.5 g a^ = radial acceleration = 0.8R − 0.6R ∴ Net force by rod on ball = 0.2 R = F12 + F22 = 128 N Ans. Ans. ∴ vB = 2ghAB = 0.4 gR 10. t = T Ans. a⊥ =g cos 37° = (0.8g) = vB2 = 0.4 gR ∴ 2H = 2π r r gω ∴ r = radius of curvature at B ∴ ω = 2π g = π 2g 2H H =R 2 11. ω1t − ω2t = 2π 7. At the time of leaving contact at P, ∴ t = 2π ω1 − ω2 u = 2π a/4 P (2π /T1) − (2π /T2) v = T1T2 3a/4 θ θ T2 − T1 mg = (3600) (60) (3600) − (60) N =0 ∴ mg cos θ = mv2 = m (u2 + 2gh) = 3600 s 59 aa ∴ g (3a/4) = u2 + 2g (a/4) 12. Particle breaks off the sphere at cos θ = 2 a 3 θθ ∴ ag Ans. g u= Ans. 8. v2 = 4 r r2 2 The tangential acceleration at this instant is ∴ v= 2 g sin θ = g 1 − cos2 θ r = g 1− 4 = 5 g Ans. P = mv = 2m 93 r

Chapter 10 Circular Motion — 601 13. u = 5gR ∴ [cos θ 2gR cos θ ] + sin θ (−2gR sin θ) = 0 2 2gR cos θ ar v ∴ 4gR cos2θ − 2gR sin2θ = 0 at h=R or 2 cos2 θ − (1 − cos2 θ) = 0 u or cos θ = 1 3 v2 = u2 − 2gh = (5gR) − 2gR ∴ θ = cos−1  1  Ans. 3 = 3gR 16. at = ar ar = v2 = 3g dv = v2 (v = speed) R ∴ dt R ∴ at = g ∫ ∫v t or v−2 dv = ∴ a = ar2 + at2 or 1 dt ∴ = g 10 or R Ans. v0 0 Ans. 14. N = mRω2 1 − 1 = t  v0 v  R µN N 1 = 1 − t = R − v0t v v0 R Rv0 v = Rv0 R − v0t mg dx = Rv0 (x = distance travelled) dt R − v0t µN = mg From these two equations. we get ∫ ∫x t Rv0 dx = µ = g 0 0 R − v0t dt Rω2  1 = 10 = 0.2 x = Rv0  −  [ln (R − v0t )] t × (5)2 0 2 v0 15. h = R cos θ = − R ln 1 − vR0t v0 = 2gh = 2gR cos θ ∴ 1 − v0t = e−x/ R R θ h R θ or t = t = R (1 − e−x/ R ) v0 90°– θ Ans. Putting x = 2πR, we get v0 t = R (1 − e−2π ) v0 Vertical component of velocity is, v = v0 sin θ 17. If u > 5gl, then T and a are in same direction. = sin θ 2gR cos θ Hence, T ⋅ a is positive. For v to be maximum If u = 5gl, then T = 0 dv = 0 ∴ T⋅a=0 dθ 18. At lowest point T and a are always in same direction (towards centre). ∴ T ⋅ a is always positive.

602 — Mechanics - I = (h cot α ) (g tan α ) = gh More than One Correct Options 1. Radial acceleration is given by, C DB N α θ R P α A h mg α v2 Now, R ar = T = 2πR = 2π h cot α v gh At A, speed is maximum. Therefore, ar is maximum. = 2π h cot α At C, speed is minimum. g Therefore, ar is minimum. ∴ T∝ h Tangential acceleration is g sin θ. and T ∝ cot α At point B, θ = 90°. ...(i) 5. N cos θ = mg ...(ii) Therefore, tangential acceleration is maximum (= g). N sin θ = mRω2 2. T + mg = mu2 From Eq. (i), N = mg = constant l cosθ u as θ = constant Net force is the resultant of N and mg and both mg T h = 2l forces are constant. L Hence, net force is constant. ω v N ∴ 2 mg + mg = mu2 θ l R or u = 3gl θ v2 = u2 + 2 gh = 3gl + 2g (2l) = 7 gl h mg ∴ v = 7gl θ 3. R = h cot α ω = g tanθ R N cos α = mg N sin α = mv2 But R = h tan θ R Solving these two equations, we get ∴ ω = g tan θ h v = Rg tan α or ω∝ 1 (θ = constant) h

Chapter 10 Circular Motion — 603 Comprehension Based Questions (b) ar = v2 or v2 R l 1. ∆U = ∆U rod + ∆U ball ar = Mg  2l  + mgl = (10)2 = 100 m/s2 1 =  M + m at = g = 10 m/s2 2 gl Ans. ∴ a = ar2 + at2 at 2. ω = v = 100.49 m/s2 l (c) T = mv2 = (1) (10)2 = 100 N l1 Now, decrease in rotational kinetic energy = increase in potential energy (d) at = g = 10 m/s2 ∴ 1 I ω2 =  M + m gl v2 (2t)2 2 2 R 2 2. ar = = = 2t2 2 1  Ml2   vl   M m or 2  + ml 2  = 2 + gl at  3  45° v = 2 m/s  M + m gl ar a = Öar2 +at2 2 = 2Ö2 m/s2 ∴ v=  M m2  6 + 3. Maximum velocity is at bottommost point and dv m/s2 dt minimum velocity is at topmost point. um2 in + 2g (2L) = 2 at = = 2 umin 1 v = 2t On solving, we get ω = v = 2t = t R2 gL umin = 2 3 Ans. At 1 s, 4. umax = 2 umin = 4 gL ar = 2 m/s2, 3 at = 2 m/s2, ∴ K max = 1 m um2 ax = 8 mgL Ans. v = 2m/s 2 3 and ω = 1 rad /s 5. v = u 2 − 2g (L) (a) a ⋅ v = av cos 45° max = (2 2) (2) (1 2) = 4 m2/s3 (b) |a × ω | = aω sin 90° = (2 2) (1) (1) = 2 2 m/s3 h=L (c) v ⋅ ω = 0 as θ = 90° v umax (d) | v × a | = va sin 45° = 4 m2/s3 = 16 gL − 2gL = 10 gL Ans. 3. f = mv2 = F F =1 33 Ans. f R Match the Columns ∴ 1. (a) v = u2 − 2gh = 12gl − 2gl With increase in speed f will increase but F will = 10gl = 10 × 10 × 1 = 10 m/s f remain same.

604 — Mechanics - I (c) | aav | = | ∆v | t N1 C N2 = v 2 + vi2 − 2vf vi cos 90° f (T /4) f = mv 2/R = 4 2v (as vi = vf = v) T mg = (4 2) v= 2 2v2 Further, with increase in the value of v, friction f (2πR/v) πR will increase. Therefore, anticlockwise moment of f about C will also increase. Hence, clockwise = 2 m/s2 moment of N 2 should also increase. Thus, N 2 will increase. But N 1 + N 2 = mg. So, N 1 will decrease. (d) S= 2R = 2  2  π 4. r = |r | = (3)2 + (−4)2 = 5 m =2 2m π v = | v| = 16 + a2 Subjective Questions ar = |a | = 36 + b2 1. (a) Applying conservation of energy In uniform circular motion, v is always mgh = 1 m ( 3Lg )2 perpendicular to a. 2 ∴ v⋅a = 0 ∴ h = 3L Ans. 2 5Lg, the or −24 − ab = 0 (b) Since, 3Lg lies between 2Lg and ∴ ab = 24 ...(i) ar = v2 string will slack in upper half of the circle. r Assuming that string slacks when it makes an ∴ 36 + b2 = 16 + a2 ...(ii) angle θ with horizontal. We have 5 mg sin θ = mv2 …(i) L Solving these two equations, we can find the values of a and b. v2 = ( 3gL)2 − 2gL (1 + sin θ) …(ii) (d) r, v and a lie in same plane. But v × a is Solving Eq. (i) and Eq. (ii), we get sin θ = 1 perpendicular to this plane. 3 ∴ r⊥v×a and v2 = gL 3 or r ⋅ (v × a) = 0 Maximum height of the bob from starting point, 5. (a) Since, speed = constant v ar ∴ Average speed = this constant value = 1 m/s. (b) Average velocity = S T/4 t = 2R RS T /4 = 4 2R RO h = L (1 + sin θ) (2πR / v ) u = 4 2v 2π H = L (1 + sin θ) + v2 sin2 (90° − θ) 2g = 2 2 m/s π

Chapter 10 Circular Motion — 605 = 4L +  gL cos2θ = 4L + 4L Minimum Speed  3  6g 3 27 N = 40 L Ans. µN 27 θ θ Note Maximum height in part (b) is less than that in θ part (a), think why? mg 2. h = 0.8 sin 30° = 0.4 m Equation of motion are, ∴ v2 = 2gh N cos θ + µN sin θ = mg …(i) …(ii) (a) Just before, m R Ans. T1 − mg sin 30° = mv2 (R1 = 0.8 m) N sin θ − µN cos θ = vm2 in R1 Ans. …(iii) Solving these two equations, we get …(iv) T1 = mg + m (2g) (0.4) vmin = 4.2 m/s 2 0.8 Maximum Speed = 3mg 2 Equations of motion are, (b) Just after, N cos θ − µN sin θ = mg − 30° = mv2 (R2 = 0.4 m) N sin θ + µN cos θ = m vm2 ax R2 Ans. R T2 mg sin mg m (2g) (0.4) N 2 0.4 T2 = + θ θ or T2 = 5 mg 2 θ µN 3. h = l(1 − cos θ) mg v2 = v02 − 2gh = 3gl − 2gl(1 − cos θ) = gl(1 + 2 cos θ) Solving these two equations, we have At 45° means radial and tangential components of acceleration are equal. vmax = 15 m/s Ans. ∴ v2 = g sin θ 5. Let v be the velocity at that instant. Then, horizontal l component of velocity remains unchanged. or 1 + 2 cos θ = sin θ Ans. v Solving the equation we get, θ = 90° or π θ/2 2 tan −1  v 2    4. Banking angle, θ =  Rg g 36 km/h = 10 m/s ∴ θ = tan −1  100  = 27° ∴ v cos θ = u cos θ or v = u cos θ  20 × 9.8 2 cos θ Angle of repose, 2 θr = tan−1 (µ) = tan−1 (0.4) = 21.8° Tangential component of acceleration of this instant Since θ > θr , vehicle cannot remain in the given position with v = 0. At rest it will slide down. To will be, find minimum speed, so that vehicle does not slip down, maximum friction will act up the plane. at = g cos (π / 2 + θ/2) = − g sin θ/2 To find maximum speed, so that the vehicle does not an = a2 − at2 = g2 − g2 sin2 θ skid up, maximum friction will act down the plane. 2 = g cos θ 2

606 — Mechanics - I Since, an = v2 R  2 T′  u cos θ    cos θ  T sin 30°  2  or R = v2 = = u2 cos2 θ Ans. an g cos θ g cos3  θ2 30° 2 T cos 30° 6. After 1 s v = u + at = 20 $i + 10 $j, T v = 500 m/s = 10 5 m/s Now T − mg cos θ = mv2 a = − 10 $j r at = a cos θ T = 2 × 9.8 × cos 30° + 2 × (3.82)2 = a ⋅ v = − 100 (0.4 ) v 10 5 Ans. or T = 90 N Ans. = − 2 5 m/s2 ∴ R = T sin 30° = 45 N an = a2 − at2 = (10)2 − (2 5)3 T ′ = T cos 30° = 80 m/s2 = 4 5 m/s2 9. Speed of each particle at angle θ is, Ans. v = 2gh (from energy conservation) R = v2 = (10 5)2 = 25 5 m. Ans. where, h = R (1 − cos θ) an 4 5 ∴ v = 2gR(1 − cos θ) 7. (a) Force diagrams of m1 and m2 are as shown N + mg cos θ = mv2 R below or N + mg cos θ = 2mg(1 − cos θ) or N = 2mg − 3mg cos θ …(i) (Only horizontal forces have been shown) The tube breaks its contact with ground when Equations of motion are 2N cos θ > Mg T + µm1g = m1Rω2 Substituting, 2N cos θ = Mg T − µm1g = m2Rω2 …(i) or 4mg cos θ − 6mg cos2 θ = Mg Solving Eqs. (i) and (ii), we have …(ii) ω = 2m1 µg Substituting, θ = 60° (m1 – m2)R Ans. Ans. 2mg − 3mg = Mg 2 or M = 1 Ans. m2 Substituting the values, we have ωmin = 6.32 rad /s Note Initially normal reaction on each ball will be radially outward and later it will be radially (b) T = m2Rω2 + µm1 g inward, so that normal reactions on tube is = (1) (0.5) (6.32)2 + (0.5) (2) (10) radially outward to break it off from the ground. ≈ 30 N 10. At distance x from centre, 8. Speed of bob in the given position, Centrifugal force = mx ω2 v = 2gh ∴ Acceleration a = x ω2 Here, h = (400 + 400 cos 30° ) mm ∴ = 746 mm = 0.746 m or v. dv = x ω2 dx v = 2 × 9.8 × 0.746 = 3.82 m/s

Chapter 10 Circular Motion — 607 v = ω2 L Substituting in Eq. (i) ∫ ∫or dv v x dx 0a 1 91 2 =  2 × 3 − × = 3 or v2 = ω2 (L2 − a2) Fmax 2mg 3 mg 22 Fmax > Mg or v = ω L2 − a2 Ans. or 2 mg > Mg Ans. 11. N = mv2 3 R or m> 3 M Hence proved. 2 µmv2 fmax = µN = R 13. Minimum velocity of particle at the lowest position ∴ Retardation a = fmax to complete the circle should be 4gR inside a tube. m So, u = 4gR = µv2 h = R(1 − cos θ) R ∴ v2 = u2 − 2gh ∴  − ddvt  = µv2 or v2 = 4gR − 2gR(1 − cos θ) R = 2gR(1 + cos θ) ∫ ∫or v dv = − µ t or v2 = 2gR  2 cos2 θ2 v0 v2 R dt 0 or v = v0 + µv0t 1 R 12. Let R be the radius of the ring h = R(1 − cos θ) Rθ v v2 = 2gh = 2gR(1 − cos θ) u h mv2 = N + mg cos θ R or N = 2mg(1 − cos θ) − mg cos θ or v = 2 gR cos θ N = 2mg − 3mg cos θ 2 θ From ds = v ⋅ dt h We get R dθ = 2 gR cos θ ⋅ dt θθ v 2 t dt = 1 R π/ 2 sec  θ2 dθ 02 g 0 ∫ ∫or R ln  θ θ2 π / 2 g 2 0 or t= sec + tan In the critical condition, tension in the string is zero or t = R ln (1 + 2) Hence proved. and net upward force on the ring g F = 2N cos θ = 2mg(2 cos θ − 3 cos2 θ) …(i) 14. At position θ, F is maximum when dF = 0 v2 = v02 + 2gh dθ where, h = a(1 − cos θ) or −2 sin θ + 6 sin θ cos θ = 0 or cos θ = 1 ∴ v2 = ( 2ag )2 + 2ag(1 − cos θ) 3 or v2 = 2ag(2 − cos θ) …(i)

608 — Mechanics - I N + mg cos θ = mv2 Particle leaves contact with the track where N = 0 a sin θ (1 – cos θ) or N + mg cos θ = 2mg(2 − cos θ) vr or N = mg(4 − 3 cos θ) x Net vertical force, 1m θ F = N cos θ + mg 2g y = mg(4 cos θ − 3 cos2 θ + 1) 9 a = This force (or acceleration) will be maximum when or mg cos θ − m  2g  sin θ = mvr2 dF = 0 9 dθ or −4 sin θ + 6 sin θ cos θ = 0 or g cos θ − 2g sin θ = 2g(1 − cos θ) + 4g sin θ 99 So, either or 3 cos θ − 6 sin θ = 2 sin θ = 0, 9 θ = 0° Solving this, we get θ ≈ 37° Ans. or cos θ = 2 , 3 θ = cos−1  32 θ = 0°is unacceptable (b) From Eq. (i), we have Therefore, the desired position is at vr = 2g(1 − cos θ) + 4g sin θ θ = cos−1  32 9 Ans. 15. (a) Let vr be the velocity of mass relative to track at or vr = 2.58 m/s at θ = 37° Vertical component of its velocity is angular position θ. vy = vr sin θ From work energy theorem, KE of particle relative = 2.58 × 3 to track 5 = Work done by force of gravity + work done by = 1.55 m/s pseudo force Now, 1.3 = 1.55t + 5t2 Q s = ut + 1 gt2 1 mvr2  2g  2 ∴ 2 = mg(1 − cos θ) + m 9 sin θ 4g or 5t2 + 1.55t − 1.3 = 0 9 or vr2 = 2g(1 − cos θ) + sin θ …(i) or t = 0.38 s Ans.

JEE Main and Advanced Previous Years’ Questions (2018-13) JEE Main 1. The density of a material in the shape of a 4. A particle is moving in a circular path of cube is determined by measuring three radius a under the action of an attractive sides of the cube and its mass. If the potential U = − k. Its total energy is 2r 2 relative errors in measuring the mass and (2018) length are respectively 1.5% and 1% , the (a) − k (b) k 4a 2 2a 2 maximum error in determining the (c) zero (d) − 3 k density is (2018) 2 a2 (a) 2.5% (b) 3.5% (c) 4.5% (d) 6% 2. All the graphs below are intended to 5. In a collinear collision, a particle with an represent the same motion. One of them initial speed v0 strikes a stationary particle of the same mass. If the final does it incorrectly. Pick it up. (2018) total kinetic energy is 50% greater than Velocity Distance the original kinetic energy, the magnitude (a) Position (b) Time of the relative velocity between the two particles after collision, is (2018) Position Velocity (a) v0 (b) 2 v0 (c) v0 (d) v0 (c) (d) 4 2 2 Time Time 6. A particle is moving with a uniform speed in a circular orbit of radius R in a central 3. Two masses m1 = 5 kg and m2 = 10 kg force inversely proportional to the nth connected by an inextensible string over a power of R. If the period of rotation of the frictionless pulley, are moving as shown particle is T , then : (2018) in the figure. The coefficient of friction of (a) T ∝ R 3 / 2 for any n n +1 (b) T ∝ R 2 horizontal surface is 0.15. The minimum (c) T ∝ R (n + 1)/ 2 (d) T ∝ R n / 2 weight m that should be put on top of m2 7. A body is thrown vertically upwards. Which one of the following graphs to stop the motion is (2018) correctly represent the velocity vs time? mT (2017) m2 vv T (a) t (b) t m1 (c) v (d) v t m1g (a) 18.3 kg (b) 27.3 kg t (c) 43.3 kg (d) 10.3 kg

2 Mechanics Vol. 1 8. A body of mass m = 10−2kg is moving in a 12. The period of oscillation of a simple pendulum is T = 2π L/ g. Measured medium and experiences a frictional force F = − kv2. Its initial speed is v0 = 10 ms −1. value of L is 20.0 cm known to 1 mm If, after 10 s, its energy is 1/ 8 mv02, the accuracy and time for 100 oscillations of value of k will be (2017) the pendulum is found to be 90 s using a (a) 10−3 kgs−1 (b) 10−4 kgm−1 wrist watch of 1s resolution. The accuracy (c) 10−1 kgm−1s−1 (d) 10−3 kgm−1 in the determination of g is (2015) 9. A time dependent force F = 6t acts on a (a) 3% (b) 2% (c) 1% (d) 5% particle of mass 1 kg. If the particle starts 13. Two stones are thrown up simultaneously from the edge of a cliff 240 m high with from rest, the work done by the force during the first 1 s will be (2017) initial speed of 10 m/s and 40 m/s, (a) 22 J (b) 9 J (c) 18 J (d) 4.5 J respectively. Which of the following graph 10. A point particle of mass m, moves along best represents the time variation of the uniformly rough track PQR as shown relative position of the second stone with in the figure. The coefficient of friction, respect to the first? (Assume stones do not rebound after between the particle and the rough track hitting the ground and neglect air equals µ . The particle is released from resistance, take g = 10m /s2 ) (2015) rest, from the point P and it comes to rest 240 (y2 – y1)m 240 (y2 – y1)m at a point R. The energies lost by the ball, over the parts PQ and QR of the track, are (a) (b) equal to each other, and no energy is lost t 8 12 t(s) t(s) 8 12 when particle changes direction from PQ 240 (y2 – y1)m (c) 240 (y2 – y1)m to QR. The values of the coefficient of (d) friction µ and the distance x (= QR), are respectively close to (2016) 12 t(s) 8 12 t(s) P 14. Given in the figure are two blocks A and h = 2m B of weight 20 N and 100 N, respectively. These are being pressed against a wall by 30º R a force F as shown in figure. If the Q Horizontal coefficient of friction between the blocks is surface 0.1 and between block B and the wall is (a) 0.2 and 6.5 m (b) 0.2 and 3.5 m 0.15, the frictional force applied by the (c) 0.29 and 3.5 m (d) 0.29 and 6.5 m wall in block B is (2015) 11. A person trying to loose weight by burning f1 f2 fat lifts a mass of 10 kg upto a height of 1 m F RR B R A 1000 times. Assume that the potential energy lost each time he lowers the mass is dissipated. How much fat will he use up WA = 20 N f1 WB = 200 N considering the work done only when the (a) 100 N (b) 120 N (c) 80 N (d) 150 N weight is lifted up? Fat supplies 3.8 × 107 J of energy per kg which is converted to 15. The current voltage relation of diode is given by I = (e1000V/T − 1) mA, where the mechanical energy with a 20% efficiency applied voltage V is in volt and the rate. (Take, g = 9.8 ms−2) (2016) temperature T is in kelvin. If a student (a) 2.45 × 10−3 kg (b) 6.45 × 10−3 kg makes an error measuring ± 0.01 V while (c) 9.89 × 10−3 kg (d) 12.89 × 10−3 kg

Previous Years’ Questions (2018-13) 3 measuring the current of 5 mA at 300 K, the maximum height above the ground at what will be the error in the value of which the block can be placed without current in mA? (2014) slipping is (2014) (a) 0.2 mA (b) 0.02 mA (a) 1/6 m (b) 2/3 m (c) 0.5 mA (d) 0.05 mA (c) 1/3 m (d) 1/2 m 16. A student measured the length of a rod 19. When a rubber band is stretched by a and wrote it as 3.50 cm. Which instrument did he use to measure it? distance x, it exerts a restoring force of magnitude F = a x + bx2, where a and b are (a) A meter scale (2014) constants. The work done in stretching the (b) A vernier caliper where the 10 divisions in unstretched rubber band by L is (2014) vernier scale matches with 9 divisions in main scale and main scale has 10 divisions in 1 cm (a) aL2 + bL3 (b) 1/ 2(aL2 + bL3 ) (c) aL2 + bL3 (d) 1  aL2 + bL3  (c) A screw gauge having 100 divisions in the circular scale and pitch as 1 mm and pitch 23 2 2 3  as 1 mm 20. Let [ε0] denote the dimensional formula of (d) A screw gauge having 50 divisions in the circular scale the permittivity of vacuum. If M = mass, L = length, T = Time and 17. From a tower of height H, a particle is A = electric current, then (2013) thrown vertically upwards with a speed u. (a) [ε0 ] = [M−1 L−3 T 2 A] (b) [ε0 ] = [M−1 L−3 T 4 A2 ] The time taken by the particle to hit the (c) [ε0 ] = [M−2 L2 T −1 A−2 ] (d) [ε0 ] = [M−1 L2 T −1 A2 ] ground, is n times that taken by it to reach the highest point of its path. The relation between H, u and n is (2014) 21. A projectile is given an initial velocity of (a) 2gH = n2u2 (b) gH = (n − 2)2 u2 (i + 2j) m/s, where, i is along the ground (c) 2gH = nu2 (n − 2) (d) gH = (n − 2)2 u2 and j is along the vertical. If g = 10 m/s2, 18. A block of mass m is placed on a surface then the equation of its trajectory is (2013) with a vertical cross-section given by (a) y = x − 5x 2 (b) y = 2x − 5x 2 y = x3 / 6. If the coefficient of friction is 0.5, (c) 4y = 2x − 5x 2 (d) 4y = 2x − 25x 2 Answer with Explanations 1. (c)∴ Density, ρ= Mass = M or ρ = M 3. (b) Motion stops when pull due to m1 ≤ force of friction Volume L3 L3 between m and m2 and surface. ⇒ Error in density ∆ρ = ∆M + 3∆L ⇒ m1g ≤ µ(m2 + m)g ρM L ⇒ 5 × 10 ≤ 0.15(10 + m) × 10 ⇒ m ≥ 23.33 kg So, maximum % error in measurement of ρ is Here, nearest value is 27.3 kg ∆ρ × 100 = ∆M × 100 + 3∆ L × 100 ρM L So, mmin = 27.3 kg or % error in density = 1.5 + 3 × 1 % error = 4.5% 4. (c)∴Force = − dU dr 2. (b) If velocity versus time graph is a straight line with ⇒ F = − d  −k  = − k dr 2r 2 r3 negative slope, then acceleration is constant and negative. As particle is on circular path, this force must be centripetal force. With a negative slope distance-time graph will be ⇒ |F |= mv 2 parabolic  s = ut − 1 at 2  . r 2 So, option (b) will be incorrect.

4 Mechanics Vol. 1 So, k = mv 2 Now, with limits, we have r3 r ∫ ∫v dv = − k t ⇒ 1 mv 2 = k 10 v 2 2 2r2 dt m0 ∴ Total energy of particle = KE + PE = k − k =0 ⇒  − 1  v = − kt 2r2 2r2 v 10 m Total energy = 0 ⇒ 1 = 0.1 + kt vm 5. (b) Momentum is conserved in all type of collisions. ⇒ v= 1 kt =1 0.1 + 0.1 + 1000k Final kinetic energy is 50% more than initial kinetic m energy ⇒ 1 × m × v2 = 1 × v 2 ⇒ 2 8 0 1 1 150 1 2 mv 2 + 2 mv12 = 100 × 2 mv 2 …(i) ⇒ v = v0 = 5 2 0 2 m m ⇒ 1 =5 v0 m 0.1 + 1000 k Before collision v2 ⇒ 1 = 0.5 + 5000 k m ⇒ k = 0.5 v1 5000 ⇒ k = 10−4 kg/m After collision 9. (d) From Newton’s second law, ∆p = F ∆t Conservation of momentum gives, ⇒ ∆p = F∆t ⇒ p = ∫ dp= ∫1 F dt mv 0 = mv1 + mv 2 0 v 0 = v 2 + v1 …(ii) p= 1 6 t dt = 3 kg  m  From Eqs. (i) and (ii), we have ∫⇒ s 0 v12 + v 2 + 2 v1v 2 = v 2 Also, ∆k = ∆p2 = 32 = 4.5 2 0 2m 2 × 1 ⇒ 2 v1v 2 = − v 2 0 2 ∴ (v1 − v 2 )2 = (v1 + v 2 )2 − 4 v1v 2 So, work done = ∆k = 4.5 J = 2 v 2 10. (c) As energy loss is same, thus 0 µ mg cosθ ⋅ (PQ) = µmg ⋅ (QR ) or vrel = 2 v 0 ∴ QR = (PQ) cosθ 6. (c)∴Force = Mass × Acceleration = mω2R ⇒ QR = 4 × 3 = 2 3 ≈ 3.5 m and given, F ∝ 1 ⇒F= k 2 Rn Rn Further,decrease in potential energy = loss due to So, we have friction ∴ mgh = (µmg cosθ)d1 + (µmg )d 2 k m 2π  2 Rn T m × 10 × 2 = µ × m × 10 × 3 × 4 = R 2 + µ × m × 10 × 2 3 ⇒ T2 = 4π2m ⋅ Rn +1 k ⇒ 4 3µ =2 ⇒ µ = 1 = 0.288 = 0.29 n +1 23 ⇒ T∝R 2 7. (b) Initially velocity keeps on decreasing at a constant 11. (d) Work done in lifting mass = (10 × 9.8 × 1) × 1000 rate, then it increases in negative direction with same rate. If m is mass of fat burnt, then energy = m × 3.8 × 107 × 20 8. (b) Given, force, F = − kv 2 100 ∴ Acceleration, a = − k v 2 Equating the two, we get m ∴ m = 49 ≈ 12.89 × 10−3 kg or dv = − k v 2 ⇒ dv = − k . dt 3.8 dt m v2 m

Previous Years’ Questions (2018-13) 5 12. (a) Time period is given by, T = t ∴(y2 − y1) versus t graph is a straight line passing n through origin Further, T =2π L At t = 8 s, y2 − y1 = 240 m g From 8 s to 12 s y1 = 0 ⇒ g = (4 π 2 )L = (4 π 2 )(L) = (4π 2n2 )L ⇒ y2 = 240 + 40t − 1 × 10 × t2 T2 t2 2  t  2 n = 240 + 40t − 5t Percentage error in the value of ‘g’ will be ∴ ( y2 − y1 ) = 240 + 40t − 5t 2 ∆g × 100 =  ∆L  × 100 + 2  ∆t  × 100 Therefore, ( y2 − y1 ) versus t graph is parabolic, g L t substituting the values we can check that at t = 8 sec, 0.1  1  y2 − y1 is 240 m and at t = 12 sec, y2 − y1 is zero. 20 90 = × 100 + 2 × × 100 = 2.72% 14. (b) Note It is not given in the question, best assuming that both blocks are in equilibrium. The free body ∴The nearest answer is 3%. diagram of two blocks is as shown below, 13. (b) 10 m/s 40 m/s Reaction force, R = applied force F For vertical equilibrium of A; 12 +ve f1 = friction between two blocks = WA = 20 N For vertical equilibrium of B; 240 m –ve f2 = friction between block B and wall = WB + f1 = 100 + 20 = 120 N 15. (a) Given, I = (e1000V/T − 1) mA, dV = ± 0.01 V Let us first find, time of collision of two particles with T = 300 K So, I = e1000V/T − 1 ground in putting proper values in the equation ⇒ I + 1 = e1000V/T Taking log on both sides, we get s = ut + 1 at 2 2 log(I + 1) = 1000 V T ⇒ − 240 = 10t 1 − 1 × 10 × t 2 2 1 On differentiating, dI = 1000 dV I+1 T Solving, we get the position value of t1 = 8 sec dI = 1000 × (I + 1) dV Therefore, the first particle will strike the ground at 8 sec. T Similarly, − 240 = 40 t 2 − 1 × 10 × t 2 ⇒ dI = 1000 × (5 + 1) × 0.01 2 2 300 Solving this equation, we get positive value of t 2 = 0.2 mA =2s So, error in the value of current is 0.2 mA. Therefore, second particle strikes the ground at 16. (c) If student measures 3.50 cm, it means that there is 12 sec. If y is measured from ground. Then, an uncertainly of order 0.01 cm. For vernier scale with 1 MSD = 1 cm from 0 to 8 s y1 = 240 + s1 10 = 240 + u1t + 1 a1t 2 and 9 MSD = 10 VSD 2 LC of vernier caliper = 1MSD − 1VSD 1 or y1 = 240 + 10t − 2 × 10 × t2 = 1 1 − 190  10 1 Similarly, y2 = 240 + 40t − 2 × 10 × t2 = 1 cm = 0.01 cm 100 ⇒ t 2 − y1 = 30t

6 Mechanics Vol. 1 17. (c) Time taken to reach the maximum height, t1 = u So, the maximum height above the ground at which g the block can be placed without slipping is 1/6 m. t1 u 19. (c) Thinking Process We know that change in H potential energy of a system corresponding to a t2 conservative internal force as f ∫U f − U i = − W = − i F ⋅ d r If t 2 is the time taken to hit the ground, then Given, F = ax + bx2 i.e. 1 2 We know that work done in stretching the rubber band 2 2 −H = ut 2 − gt by L is|dW | = |Fdx| But t 2 = nt1 [Given] ∫|W | = L bx2 ) dx =  ax2 L +  bx3 L  2 0  3 0 (ax + 0 So, −H = u nu − 1 g n2u 2 =  aL2 − a × (0)2  + b × L3 − b × (0)3  g 2 g2  2 2   3 3  − H = nu 2 − 1 n2u 2 = |W | = aL2 + bL3 g 2g 23 ⇒ 2 gH = nu 2(n − 2 ) 20. (b) From Coulomb’s law, F = 1 q1q 2 4 πε0 R 2 18. (a) A block of mass m is placed on a surface with a vertical cross-section, then ε0 = q1q 2 4 πFR 2 y m Substituting the units, we have ε0 = C2 = [AT]2 N-m2 [MLT−2 ] [L2 ] y θx = [M−1L−3T4A 2 ] 21. (b) Initial velocity = ($i + 2 $j ) m/s dy d  x3  x2 Magnitude of initial velocity,  6  u = (1)2 + (2 )2 = 5 m/s tanθ = = dx dx 2 At limiting equilibrium, we get Equation of trajectory of projectile is µ = tanθ ⇒ 0. 5 = x2 2 y = x tanθ − gx2 (1+ tan2 θ)  tanθ = y =2 = 2  2u2  x 1  ⇒ x2 = 1 ⇒ x = ± 1 ∴ y = x × 2 − 10( x)2 [1+ (2 )2 ] Now, putting the value of x in y = x3 , we get 2( 5 )2 6 = 2 x − 10( x2 ) (1 + 4) 2× 5 When x = 1 When x = − 1 y = (1)3 = 1 y = (−1)3 = −1 = 2 x − 5x2 66 66

JEE Advanced 1. Two vectors A and B are defined as A = a$i Paragraph A (Q. Nos. 5-6) and B = a (cosωt$i + sinωt$j), where a is a constant and ω = π /6 rad s−1. If If the measurement errors in all the independent |aAt t+imB|e=t = 3τ|fAor −thBe|first time, the value of quantities are known, then it is possible to determine the error in any dependent quantity. This is done by the use τ, in seconds, is of series expansion and truncating the expansion at the ............. . [Numerical Value, 2018] first power of the error. For example, consider the 2. A spring block system is resting on a relation z = x/ y. If the errors in x, y and z are ∆x, ∆y and frictionless floor as shown in the figure. The spring constant is 2.0 Nm ∆z respectively, then −1and the mass of the b. lokcgk.iIsg2n0ore the mass of the spring. Initially, the spring is z ± ∆z = x ± ∆x = x 1± ∆xx 1± ∆y −1 in an unstretched condition. Another y ± ∆y y y  block of mass 1.0 kg moving with a speed of 2.0 ms−1 collides elastically with the The series expansion for 1± ∆yy −1 to first power in first block. The collision is such that the 2.0 kg block does not hit the wall. The , distance, in metres, between the two blocks when the spring returns to its un- ∆y/ y, is 1m (∆y/ y ). The relative errors in independent stretched position for the first time after variables are always added. So, the error in z will be the collision is ............ . ∆z =  ∆x + ∆yy [Numerical Value, 2018] z  x The above derivation makes the assumption that ∆x / x << 1, ∆ y/ y << 1. Therefore, the higher powers of these quantities are neglected. (2018) 5. Consider the ratio r = (1 − a) to be (1 + a) 1kg 2 ms–1 2 kg determined by measuring a dimensionless quantity a. If the error in the measurement of a is ∆ a (∆ a / a << 1), then Paragraph X (Q. Nos. 3-4) what is the error ∆r in determining r ? In electromagnetic theory, the electric and magnetic (a) ∆a (b) − 2∆a (c) 2∆a (d) 2a∆a (1+ a)2 (1+ a)2 (1− a)2 (1− a2 ) phenomena are related to each other. Therefore, the dimensions of electric and magnetic quantities must also 6. In an experiment, the initial number of be related to each other. In the questions below, [E] and radioactive nuclei is 3000. It is found that [B] stand for dimensions of electric and magnetic fields 1000 ± 40 nuclei decayed in the first 1.0s. For|x|<< 1, ln (1 + x) = x up to first power respectively, while [ε0 ] and [µ 0 ] stand for dimensions of in x. The error ∆λ, in the determination of the permittivity and permeability of free space, the decay constant λ in s−1, is respectively. [L ] and [T ] are dimensions of length and (a) 0.04 (b) 0.03 (c) 0.02 (d) 0.01 time, respectively. All the quantities are given in SI units. (2018) 7. A particle of mass m is initially at rest at the origin. It is subjected to a force and 3. The relation between [E] and [B] is starts moving along the X-axis. Its kinetic energy K changes with time as (a) [E ] = [B ] [L] [T] (b) [E] = [B][L]−1[T] dK / dt = γt, where γ is a positive constant (c) [E] = [B][L][T]−1 (d) [E] = [B][L]−1[T]−1 of appropriate dimensions. Which of the following statements is (are) true? 4. The relation between [ε0] and [µ 0] is (More than One Correct Option, 2018) (a) [µ 0 ] = [ε0 ][L]2 [T]−2 (b) [µ 0 ] = [ε0 ][L]−2 [T]2 (c) [µ 0 ] = [ε0 ]−1 [L]2 [T]−2 (d) [µ 0 ] = [ε0 ]−1[L]−2 [T]2

8 Mechanics Vol. 1 (a) The force applied on the particle is constant Y P b|R| (b) The speed of the particle is proportional to R =Q – P time S (c) The distance of the particle from the origin PS Q Q increases linearly with time OX (d) The force is conservative 8. A solid horizontal surface is covered with (a) S = (1− b2 )P + bQ (c) S = (1− b)P + bQ a thin layer of oil. A rectangular block of (b) S = (b − 1)P + bQ (d) S = (1− b)P + b2 Q mass m = 0.4 kg is at rest on this surface. An impulse of 1.0 N s is applied to the block 12. A flat plane is moving normal to its plane through a gas under the action of a at time t = 0, so that it starts moving constant force F. The gas is kept at a very low pressure. The speed of the plate v is along the X-axis with a velocity much less than the average speed u of the gas molecules. Which of the following v (t) = v0 e−t/ τ , where v0 is a constant and option(s) is/are true? τ = 4 s. The displacement of the block, in (More than One Correct Option, 2017) metres, at t = τ is .............. . (Take, (a)At a later time the external force F balances e−1 = 0.37). (Numerical Value, 2018) the resistive force 9. A ball is projected from the ground at an (b)The plate will continue to move with constant angle of 45° with the horizontal surface. It non-zero acceleration, at all time reaches a maximum height of 120 m and returns to the ground. Upon hitting the (c)The resistive force experienced by the plate ground for the first time, it loses half of its is proportional to v kinetic energy. Immediately after the bounce, the velocity of the ball makes an (d)The pressure differnce between the leading and angle of 30° with the horizontal surface. trailing faces of the plate is proportional to uv The maximum height it reaches after the bounce, in metres, is .......... . 13. A length-scale (l) depends on the permittivity (ε) of a dielectric material, (Numerical Value, 2018) Boltzmann’s constant (kB ), the absolute temperature (T ), the number per unit 10. A person measures the depth of a well by volume (n) of certain charged particles, measuring the time interval between and the charge (q) carried by each of the dropping a stone and receiving the sound particles. Which of the following of impact with the bottom of the well. The expression (s) for l is (are) dimensionally error in his measurement of time is correct? (More than One Correct Option, 2016) δT = 0. 01 s and he measures the depth of the well to be L = 20 m. Take the (a) l =  nq2  (b) l =  εkBT  acceleration due to gravity g = 10 ms−2 and  εkBT  nq 2 the velocity of sound is 300 ms−1. Then the fractional error in the measurement, (c) l =  q2  (d) l =  q2  δL, is closest to  εn2 / 3kBT   εn1/ 3kBT  L (Single Correct Option, 2017)     (a) 1% (b) 5% (c) 3% (d) 0.2% 11. Three vectors P,Q and R are shown in the 14. In an experiment to determine the figure. Let S be any point on the vector R. acceleration due to gravity g, the formula The distance between the points P and S used for the time period of a periodic motion is b [R]. The general relation among is T = 2π 7(R − r ). The values of R and r vectors P,Q and S is 5g (Single Correct Option, 2017) are measured to be (60 ± 1) mm and (10 ± 1) mm, respectively. In five successive

Previous Years’ Questions (2018-13) 9 measurements, the time period is found to ratio h/ l and the frictional force f at the bottom of the stick are (g = 10 ms−2) be 0.52 s, 0.56 , 0.57 s, 0.54 s and 0.59 s. (Single Correct Option, 2016) The least count of the watch used for the measurement of time period is 0.01 s. (a) h = 3 , f = 16 3 N (b) h = 3 , f = 16 3 N Which of the following statement(s) is (are) l 16 3 l 16 3 true? (More than One Correct Option, 2016) (c) h = 3 3 , f = 8 3 N (d) h = 3 3 , f = 16 3 N (a) The error in the measurement of r is 10% l 16 3 l 16 3 (b) The error in the measurement of T is 3.57% (c) The error in the measurement of T is 2% 17. The energy of a system as a function of (d) The error in the measurement of g is 11% time t is given as E(t) = A2 exp (−αt), where α = 0.2 s−1. The measurement of A 15. There are two vernier callipers both of has an error of 1.25%. If the error in the which have 1 cm divided into 10 equal measurement of time is 1.50%, the divisions on the main scale. The vernier percentage error in the value of E(t) at scale of one of the callipers (C1 ) has 10 t = 5 s is (Single Integer Type, 2015) equal divisions that correspond to 9 main 18. Planck’s constant h, speed of light c and scale divisions. The vernier scale of the gravitational constant G are used to form a unit of length L and a unit of mass M. other callipers (C2) has 10 equal divisions Then, the correct option(s) is/are that correspond to 11 main scale (More than One Correct Option, 2015) divisions. The readings of the two (a) M ∝ c (b) M ∝ G callipers are shown in the figure. The (c) L ∝ h (d) L ∝ G measured values (in cm) by callipers C1 19. Consider a vernier callipers in which each and C2 respectively, are 1 cm on the main scale is divided into 8 equal divisions and a screw gauge with (Single Correct Option, 2016) 100 divisions on its circular scale. In the vernier callipers, 5 divisions of the 234 vernier scale coincide with 4 divisions on the main scale and in the screw gauge, C1 one complete rotation of the circular scale moves it by two divisions on the linear 0 5 10 scale. Then, 234 (More than One Correct Option, 2015) C2 (a)if the pitch of the screw gauge is twice the 0 5 10 least count of the vernier callipers, the least count of the screw gauge is 0.01 mm (a) 2.87 and 2.87 (b) 2.87 and 2.83 (c) 2.85 and 2.82 (d) 2.87 and 2.86 (b)if the pitch of the screw gauge is twice the least count of the Vernier callipers, the least 16. A uniform wooden stick of mass 1.6 kg of count of the screw gauge is 0.05 mm length l rests in an inclined manner on a smooth, vertical wall of height h (< l) such (c)if the least count of the linear scale of the that a small portion of the stick extends screw gauge is twice the least count of the beyond the wall. The reaction force of the vernier callipers, the least count of the screw wall on the stick is perpendicular to gauge is 0.01 mm the stick. The stick makes an angle of 30° with the wall and the bottom of the stick (d) if the least count of the linear scale of the is on a rough floor. The reaction of the screw gauge is twice the least count of the wall on the stick is equal in magnitude to vernier callipers, the least count of the screw the reaction of the floor on the stick. The gauge is 0.005 mm.

10 Mechanics Vol. 1 20. A particle of unit mass is moving along the 22. A rocket is moving in a gravity free space with a constant acceleration of 2 ms−2 x-axis under the influence of a force and its along +x direction (see figure). The length total energy is conserved. Four possible of a chamber inside the rocket is 4 m. A forms of the potential energy of the particle ball is thrown from the left end of the are given in Column I (a and U 0 are chamber in +x direction with a speed of constants). Match the potential energies in 0.3 ms−1 relative to the rocket. At the Column I to the corresponding statements in same time, another ball is thrown in −x direction with a speed of 0.2 ms−1 Column II. (Matching Type, 2015) from its right end relative to the rocket. Column I Column II The time in seconds when the two balls  2  2 P. The force acting hit each other is (Single Integer Type, 2014) 1  on the particle is A. U1( x) = U0  −  x   zero at x = a 2 a a=2 m/s2 U0  x  2 The force acting 0.3 ms–1 0.2 ms–1 x 2 a Q. on the particle is B. U2( x) = zero at x = 0 4m U0  x  2 exp− x  2  R. The force acting 2 a a  on the particle is C. U3( x) =  zero at x = − a 23. A block of mass m1 = 1 kg another mass U0 x 1  x  3  S. The particle m2 = 2 kg are placed together (see figure) 2 3 a  experiences an on an inclined plane with angle of D. U4( x) =  a −  attractive force inclination θ. Various values of θ are given  towards x = 0 in in Column I. The coefficient of friction the region| x| < a between the block m1 and the plane is always zero. The coefficient of static and T. The particle with dynamic friction between the block m2 and total energy U0 the plane are equal to µ = 0.3. In Column II 4 expressions for the friction on the block m2 can oscillate are given. Match the correct expression of about the point the friction in Column II with the angles x = − a. given in Column I, and choose the correct option. The acceleration due to gravity is 21. Airplanes A and B are flying with denoted by g. constant velocity in the same vertical µ1 plane at angles 30° and 60° with respect m to the horizontal respectively as shown in 1m figure. The speed of A is 100 3 ms−1. At 2 time t = 0 s, an observer in A finds B at a θ θ µ2 distance of 500 m. This observer sees B moving with a constant velocity [Useful information tan (5.5° ) ≈ 0.1; perpendicular to the line of motion of A. tan (11.5° ) ≈ 0.2; tan (16.5° ) ≈ 0.3] If at t = t0, A just escapes being hit by B, t0 (Matching Type, 2014) in seconds is (Single Integer Type, 2014) Column I Column II A P. θ = 5° 1. m2g sin θ Q. θ = 10° 2. (m1 + m2 ) g sin θ B R. θ = 15° 3. µ m2g cos θ 30° 60° S. θ = 20° 4. µ (m1 + m2 ) g cos θ

Previous Years’ Questions (2018-13) 11 Codes PQRS (a) K (b) K (b) 2, 2, 2, 3 PQRS (d) 2, 2, 3, 3 t t (a) 1, 1, 1, 3 (c) 2, 2, 2, 4 24. In the figure, a ladder of mass m is shown (c) K (d) K leaning against a wall. It is in static tt equilibrium making an angle θ with the 27. A wire, which passes A horizontal floor. The coefficient of friction through the hole in a small between the wall and the ladder is µ1 and that between the floor and the ladder is bead, is bent in the form µ 2. The normal reaction of the wall on the of quarter of a circle. The 90° B ladder is N1 and that of the floor is N 2. If wire is fixed vertically on the ladder is about to slip, then (More than One Correct Option, 2014) ground as shown in the (a) µ1 = 0, µ 2 ≠ 0 and N2 tan θ = mg figure. The bead is released from near the 2 top of the wire and it slides along the wire mg (b) µ1 ≠ 0, µ 2 = 0 and N1 tan θ = 2 without friction. As the bead moves from mg A to B, the force it applies on the wire is 1 + µ1µ 2 (c) µ1 ≠ 0, µ 2 ≠ 0 and N2 = (Single Correct Option, 2014) (d) µ1 = 0, µ 2 ≠ 0 and N1 tan θ = mg (a) always radially outwards 2 (b) always radially inwards 25. Consider an elliptically Q (c) radially outwards initially and radially inwards shaped rail PQ in the later (d) radially inwards initially and radially outwards vertical plane with 4m later OP = 3 m and OQ = 4 m. 28. Using the expression 2d sinθ = λ, one calculates the values of d by measuring A block of mass 1 kg is 90° P the corresponding angles θ in the range 0 pulled along the rail from O 3m to 90°. The wavelength λ is exactly known P to Q with a force of 18 and the error in θ is constant for all values of θ. As θ increases from 0°, then N, which is always parallel to line PQ (Single Correct Option, 2013) (see figure). Assuming no frictional (a) the absolute error in d remains constant losses, the kinetic energy of the block (b) the absolute error in d increases (c) the fractional error in d remains constant when it reaches Q is (n × 10) J. The (d) the fractional error in d decreases value of n is (take acceleration due to gravity = 10 ms−2 ) (Single Integer Type, 2014) 26. A tennis ball is dropped on a horizontal 29. Match the Column I with Column II and smooth surface. It bounces back to its select the correct answer using the codes original position after hitting the surface. given below the column. (Matching Type, 2013) The force on the ball during the collision Column I Column II is proportional to the length of compression P. Boltzmann constant 1. [ML2T−1] of the ball. Which one of the following Q. Coefficient of viscosity 2. [ML−1T−1] sketches describes the variation of its R. Planck constant 3. [MLT−3K−1] kinetic energy K with time t most S. Thermal conductivity 4. [ML2T−2K−1] appropriately? The figures are only illustrative and not to the scale. (Single Correct Option, 2014)

12 Mechanics Vol. 1 Codes RS P QR S the initial speed (in ms−1) of the particle is 24 (b) 3 2 1 4 zero, the speed (in ms−1) after 5 s is PQ 13 (d) 4 1 2 3 (a) 3 1 (Single Integer Type, 2013) (c) 4 2 30. The diameter of a cylinder is measured Passage (Q. Nos. 33 & 34) using a vernier callipers with no zero A small block of mass 1 kg is released from rest at the top of a rough track. The track is a error. It is found that the zero of the circular arc of radius 40 m. The block slides along the track without toppling and a frictional vernier scale lies between 5.10 cm and force acts on it in the direction opposite to the instantaneous velocity. The work done in 5.15 cm of the main scale. The vernier overcoming the friction up to the point Q, as shown in the figure, is 150 J. scale has 50 divisions equivalent to 2.45 cm. The 24th division of the vernier scale exactly coincides with one of the main scale divisions. The diameter of the cylinder is (Single Correct Option, 2013) (Take the acceleration due to gravity, (a) 5.112 cm (b) 5.124 cm g = 10 ms−2) (Passage Type, 2013) (c) 5.136 cm (d) 5.148 cm y 31. The work done on a particle of mass m by R P 30°  x $i y $j a force, K (x2 + y2 )3/ 2 + (x2 + y2 )3/ 2  R Q (K being a constant of appropriate Ox dimensions), when the particle is taken from the point (a, 0) to the point (0, a) 33. The speed of the block when it reaches along a circular path of radius a about the origin in the x- y plane is the point Q, is (Single Correct Option 2013) (a) 5 ms−1 (b) 10 ms−1 (c) 10 3 ms−1 (d) 20 ms−1 (a) 2Kπ (b) Kπ (c) Kπ (d) 0 a a 2a 32. A particle of mass 0.2 kg is moving in one 34. The magnitude of the normal reaction dimension under a force that delivers a that acts on the block at the point Q is constant power 0.5 W to the particle. If (a) 7.5 N (b) 8.6 N (c) 11.5 N (d) 22.5 N Answer with Explanations 1. (2.0) A = a$i and B = acosω$i + asinωt$j π t = nπ ± π A + B = (a + acosωt )$i + asinωt $j 12 6 A − B = (a − acosωt )$i + asinωt $j t = (12 n ± 2 )s A+B = 3 A−B = 2 s ,10 s , 14 s and so on. (a + acosωt )2 + (asinωt )2 = (a − acosωt )2 + (asinωt )2 2. (2.09 m ) Just Before Collision, 3 2 m/s 1 kg 2 kg ⇒ 2 cos ωt = ± 3 × 2 sin ωt Just After Collision 22 tan ωt = ± 1 ⇒ ωt = nπ ± π 1 kg v1 2 kg v2 23 2 6

Previous Years’ Questions (2018-13) 13 Let velocities of 1 kg and 2 kg blocks just after collision 7. (a, b) K = 1 mv 2 ⇒ dK = mv dv be v1 and v2 respectively. 2 dt dt From momentum conservation principle, Given, ⇒ dK = γt ⇒ mv dv = γt 1 × 2 = 1v1 + 2 v2 …(i) dt dt ⇒ Collision is elastic. Hence e = 1 or relative velocity of ∴ ∫ ∫v t γ tdt ⇒ v2 = γ t2 ∴ separation = relation velocity of approach. v dv = 2 m2 0 0m v 2 − v1 = 2 …(ii) v = γ t ⇒ a = dv = γ From Eqs. (i) and (ii), m dt m v2 = 4 m/s, v1 = −2 m/s F = ma = γm = constant 3 3 V = ds = γ t ⇒ s = dt m γ t2 2 kg block will perform SHM after collision, m2 t = T = π m = 3.14 s NOTE Force is constant. In the website of IIT, option (d) is 2k given correct. In the opinion of author all constant forces are not necessarily conservative. For example : Distance = |v1|t = 2 × 3.14 = 2 . 093 = 2 . 09 m viscous force at terminal velocity is a constant force 3 but it is not conservative. 3. (c) In terms of dimension, Fe = Fm 8. (6.30) Linear impulse, J = mv0 ⇒ qE = qvB or E = vB ∴ v0 = J = 2.5 m/s [E] = [B][LT −1] m 4. (d) c = 1 ∴ v = v 0e −t / τ µ 0ε0 dx = v 0e −t / τ c2 = 1 dt µ 0ε0 xτ µ 0 = ε0−1 ⋅ c −2 [µ 0] = [ε0]−1[L−2 T 2] ∫ ∫dx = v0 e −t / τdt 5. (b) r = 1 − a 00 1+ a x = v 0  e −t / τ τ  −1  / τ 0 ln r = ln(1 − a) − ln(1 + a) x = 2.5 (− 4) (e −1 − e 0) Differentiating, we get = 2.5(− 4) (0.37 − 1) dr = − da − da x = 6.30 m r 1− a 1+ a 9. (30)Q H = u 2sin2 45° = 120 m 2g or, we can write ⇒ u 2 = 120 m 4g ∆r = −  ∆a + ∆a  r 1 − a 1 + a  ∆r = −2 ∆a If speed is v after the first collision, then speed should r 1 − a2 remain 1 times, as kinetic energy has reduced to or ∆r = −  2 ∆a  (r) = −2 ∆a 2  1 − a2  (1 + a)2   half. 6. (c) N = N0e −λt ⇒ v= u 2 In N = In N0 − λt Differentiating w.r.t λ , we get ∴ hmax = v 2 sin2 30° 2g 1 . dN = 0 − t N dλ = (u / 2 )2 sin2 30° ⇒ |dλ| = dN = 40 = 0.02 2g Nt 2000 × 1  u 2 / 4g  120   = = = 30 m  4 4

14 Mechanics Vol. 1 10. (a) Q t= L (a) RHS 5 = [L−3A 2T2] 1 1 L −1/2dL  1 dL [M−1L−3T4 A 2] [ML2T−2K −1] [K] dt = 5 2 + 300 dt = 1 1 dL + dL = 0. 01 = [L−3A 2T2] = [L−2] = [L−1] [L−1T2A 2] 2 5 20 300 dL  1 + 1  = 0. 01 Wrong 20 300 (b) RHS dL 15  = 0. 01, dL = 3 = [M−1L−3T4 A 2] [ML2T−2K −1] [K] 300  16 [L−3] [A 2T2] dL × 100 = 3 × 1 × 100 = 15 ~− 1% = [L−1T2A 2] = [L] Correct L 16 20 16 [L−3T2A 2] 11. (c) S = P + bR = P + b(Q − P) = P(1 − b) + bQ [A 2T2] [M−1L−3T4 A 2] [L−2] [ML2T−2K −1] [K] 12. (a,c,d) (c) RHS = v v = [L3] Wrong u u v1 v2 (d) RHS = [A 2T2] [M−1L−3T4 A 2] [L−1] [M1L2T−2K −1] = [A 2T2] = [L] Correct [L−2T2A 2] Just before the collision Just after the collision 14. (a,b,d) Mean time period = 0.52 + 0.56 + 0.57 + 0.54 + 0.59 v1 = u + 2 v ⇒ ∆v1 = (2u + 2 v ) 5 ⇒ F1 = dp1 = pA(u + v )(2u + 2v)= 2 pA(u + v )2 = 0.556 ≅ 0.56 sec as per significant figures dt Error in reading = |Tmean − T1| = 0.04 v2 = (u + 2 v ) ⇒ ∆v2 = (2u − 2 v ) |Tmean − T2| = 0.00 |Tmean − T3| = 0.01 F2 = dp2 = pA(u − v )(2u − 2v)= 2 pA(u − v )2 |Tmean − T4| = 0.02 dt |Tmean − T5| = 0.03 F2 Mean error = 0.1/ 5 = 0.02 % error in F1 T = ∆T × 100 = 0.02 × 100 = 3.57% [∆F is the net force due to the air T 0.56 molecules on the plate] % error in r = 1 × 100 = 10% 10 % error in R = 1 × 100 = 1.67% 60 ∆F = 2 pA(4uv ) = 8 pAuv % error in P = ∆F = 8 p(uv ) A ∆g × 100 = ∆(R ) + ∆(r) × 100 + 2 × ∆T g R−r T Fnet = (F − ∆F ) = ma [m is mass of the plate] = 2 × 100 + 2 × 3.57 F − (8ρAu )v = ma 50 13. (b, d) [n] = [L−3], [q ] = [AT] = 4% + 7% = 11% [ε] = [M−1 L−3A 2T4 ] 15. (b) For vernier C1 [T ] = [L] 10 VSD = 9 MSD = 9 mm [l] = [L] 1 VSD = 0.9 mm [kB ] = [M1L2T−2K −1] ⇒ LC = 1 MSD − 1 VSD = 1 mm − 0.9 mm = 0.1 mm

Previous Years’ Questions (2018-13) 15 Reading of C1 = MSR + (VSR) (LC) 18. (c, d) M ∝ hac bGc = 28 mm + (7 )(0.1) M ∝ (ML2T−1)a(LT−1)b(M−1L3T−2)c ∝ Ma − cL2a + b + 3c T − a − b − 2c Reading of C1 = 28.7 mm = 2.87 cm For vernier C 2 : the vernier C 2 is abnormal a−c =1 …(i) So, we have to find the reading form basics. 2 a + b + 3c = 0 …(ii) The point where both of the marks are matching : a + b + 2c = 0 …(iii) distance measured from main scale = distance measured from vernier scale On solving (i), (ii), (iii), a = 1, b = + 1, c = − 1 28 mm + (1 mm) (8) = (28 mm + x) + (1.1 mm) (7) 2 22 Solving we get, x = 0.3 mm ∴ M ∝ c only → (a) is correct. So, reading of C 2 = 28 mm + 0.3 mm In the same way we can find that, L ∝ h1/2c −3/2G1/2 = 2.83 cm 16. (d) L ∝ h, L ∝ G → (c), (d) are also correct. N1 19. (b,c) For vernier callipers 1 MSD = 1 cm 30° l N2 y 8 h x 5 VSD = 4 MSD mg ∴ 1 VSD = 4 MSD = 4 × 1 = 1 cm 60° 5 5 8 10 fO Least count of vernier callipers = 1 MSD − 1 VSD = 1 cm − 1 cm = 0.025 cm ∑ Fx = 0 8 10 N1cos 30° − f = 0 …(i) Fore screw gauge …(ii) ∑ Fy = 0 …(iii) Pitch of screw gauge = 2 × 0.025 = 0.05 cm N1 sin 30°+ N2 − mg = 0 Least count of screw gauge = 0.05 cm l ∑τ0 = 0 100 2 h mg cos 60° − N1 cos 30° = 0 = 0.005 mm Least count of linear scale of screw gauge = 0.05 mm Also, given N1 = N2 …(iv) Pitch = 0.05 × 2 = 0.1 cm Least count of screw gauge = 0.1 cm = 0.01 mm Solving Eqs. (i), (ii), (iii) and (iv) we have 100 h=3 3 l 16 20. A → P,Q,R,T, B → Q,S, C → P,Q,R,S, D → P,R,T and f = 16 3 (A) Fx = −dU = − 2 U0 [x − a] [x] [x + a] 3 dx a3 17. (4) E (t ) = A2e −αt …(i) F = 0 at x = 0, x = a, x = − a α = 0.2 s−1 and U = 0 at x = − a and x = a dA F(x) A   × 100 = 1.25%  dt  × 100 = 1.50 –a a t ⇒ (dt × 100) = 1. 5t = 1. 5 × 5 = 7.5 U(x) ∴  dE  × 100 = ± 2  dA  × 100 ± α (dt × 100) U0/ 2 E A U0 4 = ± 2 (1.25) ± 0.2 (7.5) a = ± 2.5 ± 1.5 = ± 4 % –a

16 Mechanics Vol. 1 U(X) 21. (5) Relative velocity of B with respect to A is X perpendicular to line PA. Therefore, parallel to PA, velocity components of A and B should be same. vA =100 3 m/s A 30° vB vB cos 30° vB sin30° 30° F(x) 60° B 60° –a a x 30° U(x) a P (X) ⇒ vA = 100 3 = vB cos 30° ⇒ vB = 200 m/s As A and B just hit x t0 = 500 = 500 = 5s X vB sin 30° 200 × 1/2 (B) Fx = − dU − U0  x  22. (2) dx a F(X) A B 0.3 m/s 0.2 m/s (C) Fx = − dU = U0 e−x2 /x2 [x] [x − a] [x + a] 4m dx a3 Relative to rocket neither of ball has any all elevation. So, if balls meet in time t then, 0.2 t + 0.3t = 4 ⇒ 0.5t = 4 ⇒ t =8s F(x) 23. (d) Block will not slip if (m1 + m2) g sin θ ≤ µm2g cos θ ⇒ 3 sin θ ≤  3  (2 ) cos θ 10 –a a tan θ ≤ 1 / 5 ⇒ θ ≤ 11.5° U0 U(x) (P) θ = 5° friction is static 3 f = (m1 + m2) g sin θ x (Q) θ = 10° friction is static f = (m1 + m2) g sin θ U0 3 (R) θ = 15° friction is kinetic f = µm2g cos θ (D) Fx = − dU = − U0 [( x − a)( x + a)] dx 2a3 (S) θ = 20° friction is kinetic ⇒ f = µm2g cos θ

Previous Years’ Questions (2018-13) 17 24. (c, d) µ 2 can never be zero for equilibrium. N µ1N1 h N1 A R cos θ θ θ v R N2 mg mg θ B µ2N2 When µ1 = 0, we have Radial force equation is N1 = µ 2N2 …(i) mg cos θ − N = mv 2 …(ii) R N2 = mg …(iii) Here, N = normal force on bead by wire τB = 0 ⇒ mg L cosθ = N1L sinθ …(iv) N = mg cos θ − mv 2 = mg(3 cos θ − 2 ) 2 R N = 0 at cos θ = 2 ⇒ N1 = mg cot θ 3 2 ⇒ N1 tanθ = mg ⇒ Normal force act radially outward on bead, if 2 cos θ > 2 and normal force act radially inward on When, µ1 ≠ 0 we have 3 µ1N1 + N2 = mg bead, if cos θ < 2 / 3 µ 2N2 = N1 ∴ Force on ring is opposite to normal force on bead. ⇒ N2 = 1 mg 28. (d) 2d sinθ = λ ⇒ d = λ + µ1µ 2 2 sinθ 25. (5) From work-energy theorem, Differentiate ∂(d ) = λ ∂(cosecθ) 2 Work done by all forces = change in kinetic energy or WF + Wmg = K f − K i ∂(d ) = λ (− cosec θ cot θ) ∂θ 2 18 × 5 + (1 × 10) (−4) = K f 90 − 40 = K f or K f = 50 ∂(d ) = − λ cosθ . ∂θ 2 sin2θ J = 5 × 10 J So, ∆d = − cosθ , ∆θ 26. (b) t=0 d sinθ (Before collision) As θ increases, cosθ, decreases t sinθ v = gt ⇒ ∆d decreases d K = 1 mg 2t 2 Alternate Solution 2 d= λ 2 sinθ K ∝ t 2 Therefore, K-t graph is parabola. ln d = ln λ − ln2 − ln sinθ During collision retarding force is just like the spring ∆(d ) = 0 − 0 − 1 × cosθ (∆θ) force (F ∝ x), therefore kinetic energy first decreases to elastic potential energy and then increases. d sinθ Most appropriate graph is therefore (b). Fractional error, | + (d )| = cot θ ∆θ| 27. (d) h = R − R cosθ Absolute error, ∆d = (d cot θ) ∆θ = λ × cosθ ⋅∆θ Using conservation of energy, mgR (1 − cos θ) = 1 mv 2 2 sinθ sinθ 2 ∆d = cosθ ⋅ λ ⋅ ∆θ sin2 θ

18 Mechanics Vol. 1 29. (c) (P) U = 1 kT r = OP = x $i + y$j ⇒ 2 F = ( x2 k [ML2T−2] = [k] K + y2)3/2 ⇒ [K] = [ML2T−2 K −1] ( x $i + y$j ) = k (r ) r3 (Q) F = ηA dv dx Since, F is along r or in radial direction. Therefore, work done is zero. ⇒ [η] = [MLT−2 ] = [ML−1 T−1 ] [L2 LT−1L−1 ] 32. (5) W = 1 mv 2 ⇒ Pt = 1 mv 2 22 (R) E = hν ⇒ [ML2T2] = [h] [T−1] ∴ v = 2 Pt = 2 × 0.5 × 5 = 5 m/s m 0.2 ⇒ [h] = [ML2T−1] 33. (b) Height fallen up to Q is R sin 30° = R = 20 m (S) dQ = k A∆θ 2 dt l Ei − Ef = work done against friction [ML2T−3 L] [MLT−3K −1] ⇒ [k] = [L2 K] =  1  ∴ 0−  mgh + 2 mv 2  = 150 30. (b) 1 MSD = 5.15 cm − 5.10 cm = 0.05 cm ∴ (1) (10) (20) − 1 × 1 × v 2 = 150 or v = 10 m/s 1 VSD = 2.45 cm = 0.049 cm 2 50 34. (a) N = mg cos 60° = mv 4 ∴ LC = 1 MSD − 1 VSD = 0.01 cm R Hence, diameter of cylinder = (Main scale reading) + 30° (Vernier scale reading) (LC) = 5.10 + (24) (0.001) = 5.124 cm µN N 60° 31. (d) r P mg Ox y ∴ N = mv 2 + mg R2 = (1)(10)2 + (1)(10) = 7.5 N 40 2