May 2011 - Career Point

for this mirror,v = –a = – 60 cm, u = – 30 cm,Using mirror formula,1 1 1 + = , fm = – 20 cm or r = 40v u fwhen water is filled in space between glass pieces, anequi-convex lens of water is formed whose onesurface is silvered. This silvered convex glass may beassumed as a combination of a convex lens and aconcave mirror. Effective focal length of the1 2combination is given by = – where fl isF f1m flfocal length of the equiconvex lens of water.For the equiconvex lens, R 1 + r , R 2 = – r, µ = 4/31 ⎛ 1 1Using, = (µ–1) f ⎟ ⎞⎜ – , f = f 1 = 60 cm⎝ R1 R 2 ⎠Hence, effective focal length F is given by1 1 2= –F (–20)60or F = – 12 cmsince, a sharp image is again formed on the screentherefore, fore this effective mirror,F = – 12 cm, v = – a = – 60 cm, u = ?1 1 1Using, mirror formula, = + , u = – 15 cmF v uInitially the object was at a distance (–a/2) = – 30 cmfrom the mirror but now object must be at a distance(–15 cm). Hence, the object must be moved through[– 15 – (– 30)] cm = 15 cm towards the mirror(combination).3. A uniform rod of length l = 75 cm is hinged at one ofits ends and is free to rotate in vertical plane. It is areleased from rest when the rod is horizontal. Whenrod becomes vertical, it is broken at mid point andlower part now moves freely. Calculate distance ofthe centre of lower part from hinge, when it againbecomes vertical for the first time. (g = 10 ms –2 )Sol. When rod is released, it rotates about upper end itsgravitational potential energy converts into rotationalkinetic energy. Initially centre of mass of the rod is inlevel of O while its depth from O is l/2 when rodbecomes vertical as shown in figure. (1). If mass ofrod is m, then loss of gravitational potentialenergy = mg 2l .OFig. (1)ml2Moment of inertia of rod about O is I =3Let angular velocity or rod be ω when it becomesvertical. According to law of conservation energy,1 2 lI ω = mg2 2or ω =3gl= 40 rad/secNow the rod gets broken at its mid point as shown infigure (2). At this instant velocity of centre of lowerpart is v 0 = 43l ω (horizontally leftwards) and itsangular velocity is ω (clockwise).ωv 0Fig. (2)Now this part moves freely under gravity. Therefore,its angular velocity ω remains constant. This partagain becomes vertical for the first time aftercompleting half rotation. Therefore, time taken by theπ πlower part during this rotation is t = = sec.ω 40During this time. horizontal displacement centre of3πllower part is x = v 0 t = = 1.77 m4and vertical displacement = 21 gt2= 1.23 mBut initial depth of centre of lower part from hinge3lwas , therefore vertical depth of its centre from O4⎛ 3l 1 2 ⎞at this instant is y = ⎜ + gt ⎟ = 1.80 m.⎝ 4 2 ⎠Hence, distance of its centre from O isr =22x + y = 2.52 m4. Length of a horizontal arm of a U-tube is l = 21 cmand ends of both of the vertical arms are open tosurroundings of pressure 10500 Nm –2 . A liquid ofdensity ρ = 10 3 kg m –3 is poured into the tube suchthat liquid just fills horizontal part of the tube. Now,one of the open ends is sealed and the tube is thenrotated about a vertical axis passing through the othervertical arm with angular velocity ω 0 = 10radians/sec. If length of each vertical arm be a = 6cm. calculate the length of air column in the sealedarm.ωXtraEdge for IIT-JEE 17 MAY 2011