May 2011 - Career Point

PHYSICSStudents'ForumExpert’s Solution for Question asked by IIT-JEE Aspirants1. Two circular rings A and B, each of radius a = 30 cmare placed coaxially with their axes horizontal in auniform electric field E = 10 5 NC –1 directed verticallyupwards as shown in figure. Distance betweencentres of these rings A and B is h = 40 cm. Ring Ahas a positive charge q 1 = 10 µC while ring B has anegative charge of magnitude q 2 = 20 µC. A particleof mass m = 100 gm and carrying a positive chargeq = 10 µC is released from rest at the centre of thering A. Calculate its velocity when it has moved adistance of 40 cm.B E AahSol. Weight of the particle is W = mg = 1 newtonForce on particle, due to electric field E is F = qE = 1newton (upwards)It means weight of the particle is balanced by theforce F. Hence, net force on the particle is due tocharge on rings.Since particle is at centre of ring A therefore initiallyno force is exerted due to charge of this ring. But ringB is negatively charged. Therefore particleexperiences a resultant force towards centre of ringB.Separation between centres of rings is h = 40 cm anddistance moved the particle is also 40 cm. It meansvelocity of particle is to be calculated when it reachesthe centre of ring B.According to law of conservation of energy, kinetic⎛ 1 2 ⎞energy ⎜ mv ⎟ of particle at centre of B = Loss of⎝ 2 ⎠its electrical potential energy. Potential energy ofparticle at centre of A is1 qq11 q(–q 2)U 1 = U 0 + +4πε 0 a 4πε2 20 a + hWhere U 0 is potential energy due to electric field E.Potential energy at centre of B is1 qq1U 2 = U 0 ++1 q(–q ).24πε2 20 a + h 4πε 0 aaBut1mv22= U 1 – U 2∴ v = 6 2 ms –12. Two thin similar convex glass pieces are joinedtogether, front to front, with rear portion silvered andthe combination of glass pieces is placed at a distancea = 60 cm from a screen. A point object is placed onoptical axis of the combination such that its m = 2times the magnified image formed on the screen. Ifair between the glass pieces is replaced by water (µ =4/3), calculate the distance through which the objectmust be displaced so that a sharp image is againformed on the screen.Sol. A thin convex glass piece has both the surfacesequally curved, therefore, it works as a glass plate.When two thin similar convex glass pieces are joinedfront to front, though they form shape of an equiconvexlens as shown in figure(1), but its opticalpower is zero because medium inside this lens is air.AirFig. (1)When its rear surface is silvered, it works as aconcave mirror. This concave mirror has focal lengthf m = – r/2 where r is the radius of curvature of eachglass piece. Since, this concave mirror forms animage on screen, therefore the image is real. Hencescreen and object both are on same side of the mirroras shown figure (2).ScreenObjectaAirFig. (2)Since, image is m = 2 times magnified, therefore,distance of object from the mirror is equal toa/m = 30 cmXtraEdge for IIT-JEE 16 MAY 2011