May 2011 - Career Point

KNOW IIT-JEEBy Previous Exam QuestionsPHYSICS1. A non-conducting disc of radius a and uniform positivesurface charge density σ is placed on the ground, withits axis vertical. A particle of mass m and positivecharge q is dropped, along the axis of the disc, from aheight H with zero initial velocity, the particle hasq/m = ε 0 g/σ.[IIT-1990](a) Find the value of H if the particle just reaches thedisc(b) Sketch the potential energy of the particle as afunction of its height and find its equilibriumposition.Sol. (A) Given that : a = radius of disc, σ = surfacecharge density, q/m = 4ε 0 g/sThe K.E. of the particle, which it react reaches thedisc can be taken as zero.Potential due to a charge disc at any axial pointsituated at a distance x from 0.σ 2 2V(x) = [ a + x – x]2ε0σ 2 2Hence, V(H) = [ a + H – H]2ε0σaand V(O) =2ε0According to law of conservation energy, Loss ofgravitation potential energy = gain in electricpotential energy.H(m,q)or H = 2[a + H – (a2 2+ H ) ]2 +or H = 2a + 2H – 2 (a H )or 2 (a2+ H ) = H + 2aor 4a 2 + 4H 2 = H 2 + 4a 2 + 4aHor 3H 2 = 4aH or4aH =3[Q H = O is not valid](B) Total potential energy of the particle at height hU(x) = mgx + qV(x) = mgx + σ 2( a + x– x)]2ε0= mgx + 2mg [ (a2+ x ) – x]= mg [2 (a2 2+ x ) – x]...(ii)dUFrom equilibrium : = 0 dxaThis gives : x =3From equation (ii) graph between U(x) and x is andshown aboveU2 mga3 mgaO a/ 3 H = 4a/3X2HO amgH = q∆H = q[V(0) – V(H)]mgH = q[a – { (a2 2 σ+ H ) – H }]2ε 0σqFrom the given relation : = 2 mg (given)2ε0Putting this is equation (i), we get,mgH = 2mg [a – { (a2 2+ H ) – H }]...(i)2. Light is incident at an angle α on one planar end of atransparent cylindrical rod of refractive index.Determine the least value of n so that the light enteringthe rod does not emerge from the curved surface of rodirrespective of the value α [IIT- 1992]αβ90 – βSol. The light entering the rod does not emerge from thecurved surface of the rod when the angle 90 – r isgreater than the critical angle.nXtraEdge for IIT-JEE 7 MAY 2011