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IIT JEE 2009 Physics Paper2 Code 1 Solutions

14.    A solid sphere of radius R has a charge Q distributed in its volume with a charge density r = kra, where k and a are constants and r is the distance from its centre. If the electric field at r = R/2 is 1/8 times that at r = R, find the value of a.

Sol.   2


        We have ,

        ρ = kra

        E(r=R/2)=1/8 E(r=R)

       qenclosed/(4π ε0 (R/2)2 ) = 1/8 q/(4π ε0 R2 )

       32qenclosed = Q

      qenclosed = ∫0R/2 kra 4π r2 dr= 4πk/((a+3) ) (R/2)(a+3)

       Q = 4πk/((a+3) ) (R/2)(a+3)

       Q/qenclosed = 2a+3

       2a+3 = 32

       a = 2

15.    A metal rod AB of length 10x has its one end A in ice at 0oC and the other end B in water at 100oC. If a point P on the rod is maintained at 400oC, then it is found that equal amounts of water and ice evaporate and melt per unit time. The latent heat of evaporation of water is 540 cal/g and latent heat of melting of ice is 80 cal/g. If the point P is at a distance of lx from the ice end A, find the value of l. [Neglect any heat loss to the surrounding.]

Sol.   9

              metal rod

       dmice/dt = dmvapour/dt 

       dmice/λxLice = 300kS/(100 - π )xLvapour

      λ = 9


16.    Two soap bubbles A and B are kept in a closed chamber where the air is maintained at pressure 8 N/m2. The radii of bubbles A and B are 2 cm and 4 cm, respectively. Surface tension of the soap-water used to make bubbles is 0.04 N/m. Find the ratio nB/nA, where nA and nB are the number of moles of air in bubbles A and B, respectively. [Neglect the effect of gravity].

Sol.   6


        PA = P0 + 4T/RA  = 16 N/m2

        PB = P0 +  4T/RA = 12 N/m2

       nB/nA  = PB/PA (RB/RA)3 6

17.    A 20 cm long string, having a mass of 1.0 g, is fixed at both the ends. The tension in the string is 0.5 N. The string is set into vibrations using an external vibrator of frequency 100 Hz. Find the separation (in cm) between the successive nodes on the string.

Sol.   5


        v = √T/µ = 10 m/s

       λ = v/f = 10/100 = 10 cm

        Distance between the successive nodes = λ/2 = 5 cm

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