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6 tan θ = or tan θ = 3 or θ = 60°2 1/ 323.[B] Magnetic field of solenoid, B 1 =µ 0N1i1lMagnet flux of coil, φ 2 = N 2 B 1 A 2 = N 2⎛ µ 0 N1i1⎞⎜ ⎟ A 2⎝ l ⎠φ2µ 0 N1N 2A2As φ 2 = M i 1 , so M = =i ldi∴ induced emf, |e| = M 1dtµ 0 N1N2A2dior |e| =× 1l dt– 7−34π×10 × 2000×300 × 1.2×10=0.30= 4.8 × 10 –2 Volt224.[A] Z = R + ( X − X ) 225.[D]26.[C]27.[B]28.[D]LC1×40.25Here X L = 2πfL = 2 × 3.14 × 500 × (8.1 × 10 –3 )= 25.4 Ω11and X C = =2πfC−62×3.14×500×12.5×10= 25.4 Ω2 25.4 25. 4∴ Z = ( ) ( ) 210 + − = 10 ΩENow i rms = rms 100= = 10 AZ 10∴ V R = i rms × R = 10 × 10= 100 Vu 2 sin 2 θ 2gh = ug2(14) sin 29.8( 45)=Loss in P.E. = gain in K.E.mg r = 21 mv 2 ⇒ v 2 = 2g ra c =v 2r= 2gT – mg cos θ =2(h)14 ⇒ h = 10 m9.8m v2r⇒ T = 3 mg29.[C] Loss in P.E. =gain in K.E. + work done against frictionmg R = 21 m (1.4 gR) + WfW f = 0.3 mgR30.[C]Now,W B → C = mg R + 0.3 mgR= 1.3 mgRdm dv × u = Mdt dt0.5 × 400 = 2000 ×∆ v5⇒ ∆v = 0.5 ms –1CHEMISTRY31.[C] HCO 3 – can donate a proton to CO 3 2– and it canaccept a proton to form H 2 CO 3 .32.[C] O F 2 ⇒ O = +2O 2 F 2 ⇒ O = +1O 2 [PtF 6 ] ⇒ O 2 + + [PtF 6 ] –1O 2 + ⇒ 2x = +1x = +1/233.[A] ∆G = ∆H –T ∆Sif ∆H = ⊕ & ∆S = ⊕& T ∆S > ∆H than∆G = Θ & process is spontaneous.34.[C] Calcined Gypsum is calcium sulphate35.[A] Inter particles forces between CH 3 COCH 3 &CHCl 3 are strong H-bonding. Thus solutionshows negative deviation. ∆V mixing = Negative.36.[C] Addition of inert gas at constant volume doesnot cause any effect on the equilibrium.37.[A] For I order Reaction t 1/2 is constant10gm⎤⎥ t1/2 = 24↓ ⎦5g⎤⎥ t1/2 = 24↓ ⎦= 96 hours.2.5g ⎤⎥↓ ⎥ t1/2 = 241.25g⎥⎦38.[D]↓ ⎤⎥ t0.625g⎦1/ 2= 24Eq. wt. of Ag wt. of Ag=Eq. wt. of O wt. of108 W =8 1.6WAg = 21.6 gm,2 O 2XtraEdge for IIT-JEE 104APRIL 2010