Tiêu đề 4-rotation-and-gravitation-562a4785-6d34-4b22-8c7f-72fc93657e6f.pdf ✅

ROTATION AND GRAVITATION 1. (C) Since all the particles on a helix are equidistant from the axis, we can use I = mR2 , where m is the total mass of wire. Length of helical wire can be found by unrolling the helix into a straight line. 2. (A) Use the formula for moment of inertia of a triangular plate about its base (I = mh2 /6). Note that the two diagonals of a rectangle will not be in general perpendicular to each other. Hence perpendicular axis theorem cannot be used. 3. (A) F ⋅ l 2 = ml 2 12 α ⇒ α = 6F ml F = mac ⇒ ac = F m aB = lα 2 − ac = 3F m − F m = 2 m/s 2 (right) 4. (B) 40 100 [ Iω2 2 + 1 2 m(ωr) 2 ] = Iω2 2 ⇒ 3 5 Iω2 2 = 1 5 mω 2 r 2 I = 2 3 mr 2 5. (D) Relative acceleration 2μg 6. (D) ω L 2 cos θ = v0 ω L 2 sin θ = v ∴ v = v0tan θ 7. (B) mg L 2 = 1 3 mL2α ∴ α = 3g 2L and αx = g ⇒ 3g 2L x = g ∴ x = 2L 3 ∴ Distance from B = L 3 8. (A) Impulse = mV − (−mV0 ) = m(V + V0 ) and about centre of mass angular impulse = m(V + V0 ) l 2 cos θ = mV 2 12 ⋅ ω( = change in angular momentum ) ⇒ ω = 6(V + V0 )cos θ l 9. (A)
mg − T = ma ⇒ mg = 3 2 ma Tr = 1 2 mr 2α a = 2g/3 a = rα T = mg − 2mg 3 = mg 3 T = f N = mg f ≤ Nμ ⇒ mg 3 ≤ mgμ ⇒ μ ≥ 1 3 10. (C) About centre of earth, m√ 5 6 veR = mvr (∵ ΔL = 0) Where v is the velocity at maximum distance r and 1 2 h̸(√ 5 6 ve) 2 − GMh̸ R = 1 2 h̸v 2 = GMη̸̸ r ⇒ 5 6 GM R − GM R = 5 6 GM R ⋅ R 2 r 2 − GM r Let r R = x ⇒ x 2 − 6x + 5 = 0 ⇒ x = 5 or 1 11. (C) ω = | Vs2 −Vs1 r2−r2 | = | Vs1 /2−Vs1 4Rs1 −Rs1 | = Vs1 6Rs1 = 2π 6TS1 ⇒ T 2 ∝ R 3 ⇒ ( 1 8 ) 2 = ( 104 Rs2 ) 3 ⇒ Rs2 = 40,000 km ⇒ Rs2 = 4Rs1 as V0 = √ GM R ⇒ Vs2 = Vs1 2 ⇒ ω = 2π 6Ts1 = π/3rad/hr 12. (A) W = ΔV = V∞ − Vp = −Vp To find Vp we considering of radius x and thickness dx. dVp = − GdM √x 2 + 16R2 , | dM = M(2πxdx) π(4R) 2 − π(3R) 2 = 2Mxdx 7R2 ⇒ VP = −∫ GdM x 2 + 16R2 = −∫3R 4R 2MGxdx 7R2√x 2 + 16R2 = 2GM 7R (4√2 − 5) 13. (C) Work done for rotation is minimum when, moment of inertia is minimum and MI is minimum when axis passes through the centre of mass. Let it be at a distance x from 0.3 kg. ⇒ x = (0.3)(0) + (0.7) + (1.4) 0.3 + 0.7 = 0.98 14. (A) moon ⇒ GM x 2 = G(81M) (60R − x) 2 ⇒ x = 6R 15. (B) 16. (A) If another hemisphere(identical) is added so that it becomes a complete sphere then total intensity at both point P and Q becomes same = IP + IQ where Ip and IQ are intensities at P and Q respectively due to only given hemisphere ⇒ IP + IQ = intensity due to complete sphere = G(2m) (2R) 2 = Gm 2R2 ⇒ IQ = Gm 2R2 − IP 17. (B) 1 2 h̸u 2 − GMh̸ R = −GMh̸ R+h ⇒ (4000) 2 2 − 10(6.4 × 106 ) = −10 (6.4 × 106 ) 2 6.4 × 106 + h ⇒ 8 − 64 = −64 × 106 × 6.4 6.4 × 106 + h ⇒ 6.4 × 106 × 7 + 7 h = 8 × 6.4 × 106 ⇒ h = 64 7 × 105 m ≈ 914.3 km
18. (A) 19. (A) V = √ GM r m × √ 2 3 v × r = mv1r1 (r1 = min distance ) Also 1 2 m × 2 3 v 2 − GMm r = 1 2 MV1 2 − GMm r1 . . . . (ii) Solving r1 = r 2 20. (B) For earth, Ve = √ 2GM R For Sun + Earth, Potential energy = (− GM R − 3×105 2.5×104 GM R ) m = − GMm R (1 + 12) = − 13GMm R v = √ 2GM R ⋅ √13 = √13ve (S) 2.5 × 104 3 × 105M ∴ 1 2 mv 2 = 13GMm R ∴ ve = 40.4 km/s ≃ 42 km/s. 21. (BC) N = Fsin θ = GMmx R3 × R 2x = const. a = Fcos θ m = GMx R3 × √x 2 − R2/4 x = GM R3 √x 2 − R2/4 22. (AC) V0 = √ GM r ,K0 = 1 2 m × GM r , E = − GMm 2R ; 1 2 mVe 2 − GMm r = 0 1 2 mVe 2 = GMm r ⇒ K = GMm r ⇒ Ve = √ 2GM r = √2V0 = 1.41V0 ⇒ speed increases nearly 41% 23. (ABCD) Work done by kinetic friction on ONE body may be positive/negative/zero. Direction of frictional force in B,C,D is correct for providing the necessary torque. 24. (BC) I = 1 3 Ml 2 1 2 Iω 2 = Mg ⇒ ω = √ 3g 1 25. (AC) NA + fB = 2mg;NB = fA ; mgR + −fAR − fBR = 0 ; mg = fA + fB 26. (BC) The angular momentum of the system is conserved. Kinetic energy will not be conserved because friction is there. 27. (ABC) Parallel Axis theorem, check the distance carefully. ID = IB (symmetric)
28. (BCD)Since normal is impulsive, friction will also be impulsive and it will reduce ω and give some horizontal velocity to C.M. v ≤ ω friction cannot act when there is no tendency of relative motion. 29. (AC) mgr = fR N1sin θ + f = mg N1cos θ = N2 f ≤ μN2 30. (AD) Due to torque of friction about CMω eventually decreases to zero, initially there is no translation. Friction is sufficient for pure rolling therefore after sometime pure rolling begins. There is no external force in × direction therefore momentum is conserved along × direction. 31. (AD) Linear impulse = mv0 Angular impulse = (2R/3) Linear impulse. This will give the angular speed of sphere just after collision. Impulse of friction DURING collision is negligible. 32. (CD) 33. (ABC) ( T1 T2 ) 2 = ( r1 r2 ) 3 = ( 1 4 ) 4 ⇒ T1: T2 = 1: 8 V0 = √ GM r ⇒ V1: V2 = √ 1 R ÷ 1 4R = 2: 1 L = mvr ⇒ L1 = m(2v)r, L2 = mv(4r) ⇒ L1: L2 = 1: 2 34. (CD) Field at P = ρG 4 3 πPC1 ̅̅̅̅̅ − ρG 4 3 πPC2 ̅̅̅̅̅; ρ : density of massive sphere. 35. (AC) AB is a⃗ AP is R⃗⃗ At point P, velocity is out of the plane Angular momentum L0 ⃗ ⃗ = L⃗⃗ × mv ⃗⃗⃗⃗⃗⃗⃗ = (a⃗ + R⃗⃗) × mv ⃗⃗⃗⃗⃗⃗⃗ L0 ⃗ ⃗ = (⏟R⃗⃗ × mv ⃗⃗ ⃗⃗⃗⃗ ⃗) Constant in direction magnitude + (⏟a⃗ × mv ⃗⃗ ⃗⃗⃗⃗ ⃗) Constant in magnitude only 36. (AC) I0α = τ0 [ ml 2 12 + mx 2 ] α = mgx α = 12gx l 2 + 12x 2 For α maximum dα dt = 0 ; d [ 12gx l 2 + 12x 2 ] dx = 0 12g 12x 2 + l 2 − 288gx 2 (12x 2 + l 2) 2 − 12g(12x 2 − l 2 ) (12x 2 + l 2) 2 = 0 ; 12x 2 − l 2 = 0 ; x = l 2√3 Now put x in (i) ; α = g√3 l