May 2011 - Career Point
May 2011 - Career Point May 2011 - Career Point
Ans.[D]m 1 = 0.01 kgvH = 5mm 2 = 0.2 kgAns.[B](A)(C)3π2A ,4(B) A,5π3A ,6(D)A4π3A , π320 mBallBulletθ30º120ºA100 m2HT = = 1 secgLet v 1 & v 2 be velocity of bullet & ballrespectively just after collision.v 2 × 1 = 20 ⇒ v 2 = 20& v 1 = 100From conservation of momentum0.01 × v = (0.01 × 100) + (0.2 × 20)0.01 v = 1 + 4 = 55v = = 500 m/sec.−21025. Which of the field patterns given below is validfor electric field as well as for magnetic field ?(A)(B)BHere φ = π + θA cos 30º = B sin θ ⇒ B sin θ =30º + B cosθ = A ⇒ B cos θ = 2A3A2and A sinSolving above, B = A and θ = 60º = 3π .Hence φ = 240º =4π327. A long insulated copper wire is closely wound asspiral of ‘N’ turns. The spiral has inner radius ‘a’and outer radius ‘b’. The spiral lies in the X-Yplane and a steady current ‘I’ flows through thewire. The Z-component of the magnetic field atthe centre of the spiral is -y(C)(D)IabxAns.[C] Electric lines of force for induced electric field isclosed loop.26. A point mass is subjected to two simultaneoussinusoidal displacements in x-direction, x 1 (t) = A⎛ 2π⎞sin ωt and x 2 (t) = A sin ⎜ωt+ ⎟ .Adding a third⎝ 3 ⎠sinusoidal displacement x 3 (t) = B sin(ωt + φ)brings the mass to a complete rest. The values ofB and φ are :(A)(C)µ 0NI⎛ b ⎞ln⎜⎟2( b − a)⎝ a ⎠µ 0NI ⎛ b ⎞ ln ⎜ ⎟2b⎝ a ⎠(B)(D)µ 0NI⎛ b + a ⎞ln⎜⎟2( b − a)⎝ b − a ⎠µ 0NI⎛ b + a ⎞ ln ⎜ ⎟2b⎝ b − a ⎠XtraEdge for IIT-JEE 86 MAY 2011
SECTION – IIy⇒⎛ GM ⎞Ans.[C,D]K = m ⎜ e⎟ = mv 2⎝ r ⎠Multiple correct Choice TypedrThis section contains 4 multiple choice questions. EachAns.[A]question has four choices (A), (B), (C) and (D), out ofwhich ONE OR MORE may be correct.a rx29. Two solid spheres A and B of equal volumes butof different densities d A and d B are connected by astring. They are fully immersed in a fluid ofbdensity d F . They get arranged into an equilibriumstate as shown in the figure with a tension in thestring. The arrangement is possible only if –NNo. of turns per unit thickness =b − aµmagnetic field at centre due to element = 0 (dN)iA2rµdB = 0 i ⎛ N ⎞⎜ ⎟ dr2r ⎝ b − a ⎠bBµB =0iN drµ2(b − a) ∫ =0iNln 32(b − a)a(A) d A < d F(B) d B > d F(C) d A > d F(D) d A + d B = 2d F28. A satellite is moving with a constant speed ‘V’ in Ans.[A,B,D]a circular orbit about the earth. An object of mass‘m’ is ejected from the satellite such that it justVd f gescapes from the gravitational pull of the earth. AtAthe time of its ejection, the kinetic energy of theobject is -Vd A g1 2(A) mV (B) mV2Vd f g2B3 2(C) mV (D) 2mV2Vd B g2system will be in equilibrium with tension inAns.[B]string only if d f > d A and d B > d f . If both A & BV M•are considered as a system then2Vd f g = V (d A + d b )gr⇒ d A + d B = 2d f30. Which of the following statement(s) is/arecorrect?(A) If the electric field due to a point chargevaries as r –2.5 instead of r –2 , then the Gaussmv 2 GmM law will still be valid.⇒= er2(B) The Gauss law can be used to calculate therfield distribution around an electric dipole.GM⇒ r = e...(1)(C) If the electric field between two point charges2Vis zero somewhere, then the sign of the twoIf K.E. of mass m = was k then fromcharges is the same.(D) The work done by the external force inGmME = K – e moving a unit positive charge from point A at= 0rpotential V A to point B at potential V B is (V B– V A ).XtraEdge for IIT-JEE 87 MAY 2011
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SECTION – IIy⇒⎛ GM ⎞Ans.[C,D]K = m ⎜ e⎟ = mv 2⎝ r ⎠Multiple correct Choice TypedrThis section contains 4 multiple choice questions. EachAns.[A]question has four choices (A), (B), (C) and (D), out ofwhich ONE OR MORE may be correct.a rx29. Two solid spheres A and B of equal volumes butof different densities d A and d B are connected by astring. They are fully immersed in a fluid ofbdensity d F . They get arranged into an equilibriumstate as shown in the figure with a tension in thestring. The arrangement is possible only if –NNo. of turns per unit thickness =b − aµmagnetic field at centre due to element = 0 (dN)iA2rµdB = 0 i ⎛ N ⎞⎜ ⎟ dr2r ⎝ b − a ⎠bBµB =0iN drµ2(b − a) ∫ =0iNln 32(b − a)a(A) d A < d F(B) d B > d F(C) d A > d F(D) d A + d B = 2d F28. A satellite is moving with a constant speed ‘V’ in Ans.[A,B,D]a circular orbit about the earth. An object of mass‘m’ is ejected from the satellite such that it justVd f gescapes from the gravitational pull of the earth. AtAthe time of its ejection, the kinetic energy of theobject is -Vd A g1 2(A) mV (B) mV2Vd f g2B3 2(C) mV (D) 2mV2Vd B g2system will be in equilibrium with tension inAns.[B]string only if d f > d A and d B > d f . If both A & BV M•are considered as a system then2Vd f g = V (d A + d b )gr⇒ d A + d B = 2d f30. Which of the following statement(s) is/arecorrect?(A) If the electric field due to a point chargevaries as r –2.5 instead of r –2 , then the Gaussmv 2 GmM law will still be valid.⇒= er2(B) The Gauss law can be used to calculate therfield distribution around an electric dipole.GM⇒ r = e...(1)(C) If the electric field between two point charges2Vis zero somewhere, then the sign of the twoIf K.E. of mass m = was k then fromcharges is the same.(D) The work done by the external force inGmME = K – e moving a unit positive charge from point A at= 0rpotential V A to point B at potential V B is (V B– V A ).XtraEdge for IIT-JEE 87 MAY <strong>2011</strong>
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