Title 1--introduction-to-vectors--forces-.pdf ✅

Introduction to Vectors & Forces 1. (C) A⃗ × B⃗⃗ lies in plane ⊥ to A⃗, B⃗⃗, C⃗ (A⃗ × B⃗⃗) × C⃗ will be in plane containing A, B and C 2. (C) VT ̅̅̅ = VE ̅̅̅ + VN ̅̅̅̅ [ VT = actual velocity of the train VE = velocity of train along east VN = velocity of train along north First Condition VE − 10 VN = 1 Second Condition VE − 30 VN = 0 ∴ VE = 30VN = 20; V0 (my velocity) = 50 3. (B) ⇒ R = F, N = mg ⇒ R = 0, N = F + mg ⇒ R = Fsin θ N = mg + Fcos θ 4. (D) A⃗ and B⃗⃗ may be anti-parallel with magnitude of A twice of B.A 2 + B 2 + ABcos θ = A 2 ⇒ B(B + 2cos θ) = 0 ⇒ B = 0 or cos θ = − B 2A ∴ A ̅ ⋅ B⃗⃗ = ABcos θ = − B 2 2 ≤ 0 5. (C) a⃗ and b⃗⃗ must be || or anti-parallel with |a⃗| > |b⃗⃗| 6. (B) a⃗ × c⃗ = sin θpˆ b⃗⃗ × c⃗ = sin αpˆ pˆ is unit vector prep to plane of incidence (outward) 7. (C) sin 45∘ = Asin 75∘ C and sin 30∘ = Bsin 75∘ C , sin 75∘ = sin (30 + 45) = sin 30∘ cos 45 + cos 30∘ sin 45∘ = 1 2√2 + √3 2√2 = √3 + 1 2√2
⇒ 1 √2 = B(√3 + 1) 2√2C ⇒ B = √2C √3 + 1 = C(√3 − 1) √2 8. (BD) |A⃗ + B⃗⃗| = |A⃗ − B⃗⃗| A 2 + B 2 + 2ABcos θ = A 2 + θ 2 − 2ABcos θ ABcos θ = 0 9. (BD) a⃗ is anti-parallel to b⃗⃗ and |a⃗| = |b⃗⃗| ⇒ a⃗ + b⃗⃗ = 0⃗⃗ 10. (A) N + F = mg R = N + Mg F = Rμ } ⇒ F = (m + M)gμ 1 + μ 11. (ABD) If μ = 0.5 fmax = 20 N which will not be needed so block will remain at rest at f = 10 N If μ = 0.2, f = 8N hence block will move down. 12. (D) DB̅̅̅̅ = a ̅ − b ̅ DB̅̅̅̅ ⋅ AB̅̅̅̅ = (a ̅ − b ̅) ⋅ a⃗ = a 2 − b ̅ ⋅ a ̅ Also a ̅ + b ̅ = c ̅ a 2 + b 2 + 2a ̅ ⋅ b ̅ = c 2From (i) and (ii) DB̅̅̅̅ ⋅ AB̅̅̅̅ = 3a 2+b 2−c 2 2 13. (D) tan α = Qsin θ P+Qcos θ ⇒ √3 2 = 2Psin θ P+2Pcos θ 14. (ABD) If |F1 ̅ | = 2, |F ̅ 2 | = 3, and |F ̅ 3 | = 4 then maximum value of F1 ̅ + F2 ̅̅̅ + F3 ̅̅̅ can be 9 and minimum value can be 0 and particle can take accelerations in between 9 m = 4.5 m/s 2 and 0 m = 0 m/s 2 15. (AB) RA = 10N and RB = 0 in case 1 Case - II RAsin 60∘ = RBsin 60∘ RAcos 60∘ + RBcos 60∘ = mg ⇒ RA = RB = 10N
⇒ RB = RAcos 60∘ ⇒ RA = 2mg √3 = 20/√3 RAsin 60∘ = mg RB = mg/√3 = 10/√3 16. (C) b = xiˆ + yjˆ + zk a⃗ ⋅ b⃗⃗ = 1 ⇒ x + y + z = 1 a⃗ × b⃗⃗ = jˆ − kˆ ⇒ z − y = 0; x − z = 1 ; y − x = −1 17. (AB) As AB = BC ⇒ VSRcos θ + VSRsin θ ⇒ 5cos θ + 1 = 5sin θ ⇒ 5√2sin (θ − 45∘ ) = 1 ⇒ θ = 45∘ + sin−1 1 5√2 ≈ 53∘ time = 0.4 km 5sin θkm/hr = 0.1hr = 6 min 18. (B) b⃗⃗ × c⃗ = b⃗⃗ × (−(a⃗ + b⃗⃗)) = a⃗ × b⃗⃗; c⃗ × a⃗ = (−(a⃗ + b⃗⃗) × a⃗) = a⃗ × b⃗⃗ 19. (CD) Let velocity components (horizontal and vertical) be V each as for stationary observer it is falling at angle of 45∘ , let u; velocity of man in the direction of horizontal component of rain. ⇒ v = 2 m/s( i.e. 2√2cos 45∘ )Now relative horizontal components relative velocity of rain in opposite direction if u > v and |v − u| = 2 (i.e. (2√2sin 45∘ ) ⇒ u = 4 m/s 20. (D) (a⃗ + b⃗⃗ + c⃗) 2 = a 2 + b 2 + c 2 + 2(a⃗ ⋅ b⃗⃗ + b⃗⃗ ⋅ c⃗ + c⃗ ⋅ a⃗) 21. (D) Magnitude of vector should not change V 22. (ABCD) (i) Maximum value of |a⃗ + b⃗⃗ + c⃗| in case 1 will be 14 and minimum value will be zero because a⃗. b⃗⃗ and c⃗ can form a triangle. (ii) Maximum value of |a⃗ + b⃗⃗ + c⃗| in Case II will be 12 and minimum value will be |1 + 4 − 7| = 2 23. (C) Sum of magnitude of any two vector is greater or equal to magnitude of the third, if three vector adding to zero. 24. (C) For Parallel 8t t 2 = −t 2 2 (t 3 = −16( Not possible as t ≥ 0) For perpendicular a⃗ ⋅ b⃗⃗ = 0 = 8t 3 − 2t 2 t = 1 4
25. (B) 26. (B) x 2 = (1500 − 10t) 2 + (1800 − 15t) 2 , for x to be minimum, its first derivative should be zero. Thus dx dt = 0. = −20(1500 − 10t) − 30(1800 − 15t) or t = 129.23 s 27. (D) 28. (A) cos θ = A ̅⋅B ̅ |A ̅||B ̅| = √3 2 , θ = π 6 29. (A) nˆ = A ̅×B ̅ |A ̅×B ̅| , A ̅ × B ̅ = | iˆ jˆ kˆ 2 3 1 1 −1 1 | = 4iˆ − jˆ − 5kˆ ,|A ̅ × B ̅| = √42, nˆ = 1 √42 (4iˆ − jˆ − 5kˆ ) 30. (C) aav = Vf−Vi t = 5jˆ−4iˆ 10 or |aav| = 1 √2 North west. 31. (B) (2iˆ + 3jˆ + 8kˆ ) ⋅ (4jˆ − 4iˆ + αkˆ ) = 0 or −8 + 12 + 8α = 0, α = −1 2 32. (D) vssin θ = vriver sin θ = 1 2 θ = 30∘ 33. (D) A ̅ × B ̅ = | iˆ jˆ kˆ 3 −2 1 4 3 −2 | nˆ = A ̅ × B ̅ |A ̅ × B ̅| = iˆ + 10jˆ + 17kˆ √390 x ̅ = |B ̅|nˆ = √29(iˆ + 10jˆ + 17kˆ ) √390 34. (C) V ̅ rm = V ̅ r − V ̅m − 3kˆ − 4jˆ; tan β = 4 3 ⇒ β = 53∘ North vertical 35. (A) A regular hexagon will be formed if we continue. After 5th turn displacement = OE = 500 m After 3rd turn displacement = OC = 1000 m ( OC is diameter of the circle circumscribing regular hexagon). 36. (D) Use parallelogram law. 37. (A) V ̅ cm = (Vrx − Vm)iˆ + Vryjˆ Case 1 : tan 90∘ = Vry Vrx−Vm = Vry Vrx−8 or Vrx = 8 ms−1 Case 2: tan 30∘ = Vrx−Vm Vry = 8−12 Vry or Vry = −4√3 ms−1 Vr = 8iˆ − 4√3jˆ = 4√2 2 + 3 = 4√7 ms−1 tan θ Vry Vrx = 4√3 8 = √3 2 or θ = tan−1 √3 2 with respect to road (horizontally). 38. (C) As 13N > 8 + 4[∵ Rmax = A + B] 39. (A) Vws = Vw − Vs ⇒ (Vwxiˆ + Vwyjˆ) − 10iˆ = 5jˆ or (Vwxiˆ + Vwyjˆ) = 5jˆ + 10iˆ or |Vw| = 5√5 and tan θ = 1 2 or θ = tan−1 1 2 North of east. 40. (C) d ̅ = 5iˆ − 2jˆ + 4kˆ W = F ̅ ⋅ d ̅ = (6iˆ + 3jˆ + kˆ ) ⋅ (5iˆ − 2jˆ + 4kˆ ) = 28J