Solution - Career Point

5100−j(D) 95 C 4 + ∑j=0C 3= 95 C 4 + 99 C 3 + 98 C 3 + 97 C 3 + 96 C 3 + 95 C 3= ( 95 C 4 + 95 C 3 ) + 96 C 3 + 97 C 3 + 98 C 3 + 99 C 3= ( 96 C 4 + 96 C 3 ) + 97 C 3 + 98 C 3 + 99 C 3= ( 97 C 4 + 97 C 3 ) + 98 C 3 + 99 C 3= ( 98 C 4 + 98 C 3 ) + 99 C 3= 99 C 4 + 99 C 3= 100 C 420. [A] → p,q; [B] → q,t; [C] → q; [D] → s(A) Given lines intersect if2 −13 − 4 4 − 51 1 λ = 0λ 2 1⇒ λ = 0, – 1⎛ x + 1 ⎞⎜ 1−(B) lim 4x ⎜−tan 1 x + 2x→∞⎜ x + 11+⎟ ⎟⎟⎟ =2 = y 2 + 4y + 5⎝ x + 2 ⎠⇒ y = –1, – 3(C) y 2 – ax (– x – y) = 0⇒ for perpendicular lines a + b = 0⇒ 1 + a = 0 ⇒ a = – 1(D) ( a × b ) × a = ( ĵ– kˆ ) × a⇒ ( a . a ) b – ( a . b ) a = ( ĵ– kˆ ) × aon solving,we get b = îPHYSICS1.[B] T = m 1 r 1 ω 2 2 1 also T = m 2 g + ω2r22.[A]∴ m 1 r 1 ω 1 2 = m 2or, 0.1 × ω 1 2 =222 2 2g + ( ω2r2)1 2 2( 10) + (10)20.1 ω 2 1 = 1ω 1 = 10 rad/s v 1 = r 1 ω 1 = 10 m/sAs f = µN = mgor, µmlω 2 = mg ⇒ ω =gµl5.[D]Net force acting on container due to liquidcoming out from the holes is given by⎡ 3H H ⎤F = ρA ⎢2 g × − 2g × ⎥ = ρgAH towads left⎣ 4 4 ⎦∴ F = f = ρgAH towards right.Now,6.[C] T A =7.[A]8.[A]τ F = ρgAH × 2H into the plane of paper.τ f = ρgAH × 2H out of plane of papers∴ τ F = τ f hence τ N = 0PA VAand T B =n RAPBVBn RGiven, P A = P B , V A = V B and n A = 2n B∴ T A =Now,VVABT B2ABEquivalent circuit1 1'22'Middle plateABT MB= × = 2T M3 3'11' 33'ε 1C 1 V 0V 0ε 2 C 2 V 02, 2'⎡ε1ε0Aε2ε0A⎤Total charge on 2 & 2' plate = ⎢ + ⎥V⎣ d1d2⎦⎡ εσ = ε 0 V ⎢⎣dC1130º C 15º45º15Oε+dA22⎥⎦⎤30B3.[C]4.[C]At terminal velocity net force is zero.6πη(r 1 + r 2 ) V T + 34 π (r1 3 + r 2 3 ) ρg = 34 π (r1 3 + r 2 3 ) σgAs supporting plane is lowered slowly∴ N = mg – kxA I30ºl120ºOSine ruleRBXtraEdge for IIT-JEE 92APRIL 2010