May 2011 - Career Point

k 21[Where L is the relative distance between A and B]i.e., µ ≤ where C is the critical angle.sin Cor t 2 2L 2L= =Here C = 90 – ra A / B a A – a B1L = 2m ⇒ L = aA/B t 2 = 0.785 + 0.5 + 1.05 + 0.52 = 2.83 (app.)⇒11Putting values we get, t 2 = 4 or t = 2s.]µ ≤⇒ µ ≤sin(90 – r)cos rDistance moved by B during that time is given by11As a limiting case µ =...(i)S = 2Bcos r2t1 0.79 2× 0.7Applying Snell's law at A= × × 4 =2sin αsin α2 2µ =⇒ sin r = ...(ii)sin rµSimilarly for A = 8 2 m.× 10 = 7 2 mThe smallest angle of incident on the curved surface 4. Two light springs of force constants k 1 and k 2 and aπ block of mass m are in one line AB on a smoothis when α = . This can be taken as a limiting case2 horizontal table such that one end of each spring isfor angle of incidence on plane surfacefixed on rigid on rigid supports and the other end is freesin π / 2 1shown in the figure. The distance CD between the freeFrom (ii) sin r = ⇒ µ =...(iii)µ sin rends of the springs is 60 cms. If the block moves alongFrom (i) and (ii) sin r = cos rAB with a velocity 120 cm/sec in between the springs,⇒ r = 45ºcalculate the period of oscillation of the block.⇒1 1(k 1 = 1.8 N/m, k 2 = 3.2 N/m. m = 200 gm)µ = =cos 45º 1/ 2[IIT-1985]60 cm⇒ µ = 2k 1This is the least value of the refractive index of rodVfor light entering the rod and not leaving it from thecurved surface.A C m D B3. Two block A and B of equal masses are placed onrough inclined plane as shown in figure. When andwhere will the two blocks come on the same line on theinclined plane if they are released simultaneously?Initially the block A is 2 m behind the block B.Sol. The mass will strike the right spring, compress it.The K.E. of the mass will convert into P.E. of thespring. Again the spring will return to its natural sizethereby converting its P.E. to K.E. of the block. TheTtime taken for this process will be . where 2Co-efficient of kinetic friction for the blocks A andmT = 2π .B are 0.2 and 0.3 respectively (g = 10 m/s 2 ).k2 m[IIT-2005]T m 0.2∴ t 1 = = π = π 2 K 2 1. 8= 0.785AThe block will move from A to B without anyacceleration. The time taken will beB A60t 2 = = 0.5 120BNow the block will compress the left spring and then45ºthe spring again attains ite natural length. The timemgsinθ – µ kmg cosθtaken will beSol. a =mm 0.2∴ a A = g sin θ – µ kA g cos θ ...(i)t 3 = π = π = 1.05K 1 1. 8and a B = g sin θ – m kB g cos θ ...(ii)Putting values we getAgain the block moves from B to A, completing oneoscillation. the time taken for doing so0.89 0.79a A = and α B =6022t 4 = = 0.5 120a AB is relative acceleration of A' w.r.t.∴ The complete time of oscillation will beB = a A – a B= t 1 + t 2 + t 3 + t 4XtraEdge for IIT-JEE 8 MAY 2011