Solution - Career Point

33. [B] For quarter revolution∆ V → = V → 2 – V→ 1angle between V → 1 andNWv 2a∴ ∆v =Also∴ ∆ → v =v 1–v 1S2 22 v1v + =tan –1 v = 45ºv34. [D] For vertical motionH = 21 gt 2 or t =→2V is 90ºE22 v south west2 H / g2v + v = 2 VFor horizontal motion, distance covered is givenby2πrn = utor 2πrn = uor u =42πr2 H / g2H / g35. [A] On descending(Mg – f) – Ma = 0(where f is the upthrust due to buoyancy)On ascending,f – (M – m)g – (M – m)a = 0⎛m =⎜⎝⎟ ⎞M+ g ⎠2aa36. [A] The masses will be lifted if the tension of thestring is more than the gravitational pull ofmass.50 N20 N 50 N50 – 2T = 0 or T = 25NSo, 5 kg weight cannot be lifted but 2 kg weightwill be lifted25 – 20 = 2a or a = 25 = 2.5 ms–237. [A] On cutting of string QR, the resultant force onm 1 remains zero because its weight mg isbalanced by the tension is the spring but onblock m 2 a resultant upward Force (m 1 – m 2 )g isden sped. The block m 1 will have no resultantacceleration where as m 2 does have an upward( m1− m2)gacceleration given bym⎛ 2pt ⎞38. [A] Here v = ⎜ ⎟⎝ M ⎠or1/ 21/ 2ds ⎛ 2pt ⎞ = ⎜ ⎟dt ⎝ M ⎠⎛ 2pt ⎞or ds = ⎜ ⎟⎝ M ⎠1/ 2dt⎛ 2p⎞integrating s = ⎜ ⎟⎠⎝ M1/ 2at t = 0, S = 0, so c = 01/ 2⎛ 8p⎞S = ⎜ ⎟⎠ t 3/2⎝ 9M22 t 3/2 + C339. [B] Let a small displacement be given to the systemin vertical plane of frame such that ST remainshorizontal then let vertical displacement ofcentres of rods up and QR be y then verticaldisplacment of centres of VT and RS will be 3yand that of TS will be 4y. Equating total verticalwork to zero we getWWPWTWQWSW(w + w)δy + (w + w)3 δy + w(4δy) – T(4δy) = 0or 2w + 6w + 4w = 4T or T = 3wWRyyyy40.[A] Normal reaction R = mg = 2 × 9.8 NFrictional force,F = µR = 0.2 × 2 × 9.8 = 3.92 NDistance traveled2 × 5 = 10 m∴ Work done = f × s = 3.92 × 10= 39.2 JXtraEdge for IIT-JEE 113APRIL 2010