Solution - Career Point

K TFor a cylinder = 2 in case ofK Rpure rolling upto B :andK T = 32 mghK R = 31 mghAfter B : rotational kinetic energy is constanthowever translational kinetic energy increases.At C : K T = 32 mgh + mgh = 35 mghwhile K R = 31 mgh∴K T = 5KR14. [2]According to the formula, the position of theparticle is at the centre of the path. It goes by 7 unitsto the right and then by another 7 units back to theinitial position and then it goes to the negative sideby 7/2 units for the first time.So the minimum time is 2 × 4T plus the additionaltime (t) to cover 7/2 units on the negative side.π 2π πHere ω = or = or T = 4s2 T 24∴ t min = 2 × + t = 2 + t 47 π π π 1Now = 7sin t ⇒ t = sin 30º =sin ⇒t = s2 2 2 6 31∴ t min = 2 + = 2.33s ≈ 2315. [6] Let ω be the actual angular velocity of thesatellite from east to west and ω e be the angularspeed of the earth (west to east).Then ω relative = ω – (–w e ) = ω + ω e ⇒ ω = ω rel – w eBy the dynamics of circular motion2GMm2 = mω 2 R or ω 2 gR e=3 (Q GM 2 =gRe )RR⇒ ω =2e3gRR2∴ ω rel =122e3gRR+ ωe10×6.4 × 10⇒ ω rel =+ 7.27 × 10 –53 212 × 102π(Q ω e = = 7.27 × 10 –5 )86400⇒ ω rel =22.6×10 –5 +7.27×10 –5 = 30×10 –5 rad s –12π 2π∴ τ = = = 2.09×10 4 s = 5hr 48 min.−530×10w rel16. [1]444πPrπPrπPrV 1 = , V 2 = , V 3 =8ηl18ηl28ηl3πPr 4and V =8ηlNow V = V 1 + V 2 + V 3Substituting the values, we getl1l2l3l =l1l2 + l1l3+ l2l31×2×3 6== m1×2 + 1×3 + 2×3 1117. [1]π3πAt x 1 = and x 2 =3k2kSin k x 1 or sin k x 2 is not zero.Therefore, neither of x 1 or x 2 is a node⎛ 3 1 ⎞ π 7π∆x = x 2 – x 1 = ⎜ − ⎟ =⎝ 2 3 ⎠ k 6kSince2π π > ∆x >k kλλ > ∆x > 2Therefore, φ 1 = π7πand φ 2 = k.∆x = 6∴φ 1 6 =φ 72≈ 1⎛ 2π⎞⎜k= ⎟⎝ k ⎠18. [1]With an AC source, current in the circuit ismaximum whenZ = Z min = R (the resistance of coil)∴ R = 624 = 4ΩWhen connected with an dc source of emf 12V.12i =(r internal resistance of source)r + R12= = 1.5 A4 + 419. [1] For incident electronh h hλ 1 = = =p 2Km2mVehcand shortest wavelength of X-rays is λ 2 = Ve∴λ 1 1 =λ c2⎛ V ⎞⎛⎜ ⎟⎜⎝ 2 ⎠⎝emλ 1Substituting the values, we get = 1λ⎞⎟⎠2XtraEdge for IIT-JEE 101APRIL 2010