September 2011 - Career Point

KNOW IIT-JEEBy Previous Exam QuestionsPHYSICS1. A transverse harmonic disturbance is produced in astring. The maximum transverse velocity is 3 m/s andmaximum transverse acceleration is 90 m/s 2 . If thewave velocity is 20 m/s then find the waveform.[IIT-2005]Sol. The wave form of a transverse harmonic disturbancey = a sin (ωt ± kx ± φ)Given v max = aω = 3 m/s...(i)A max = aω 2 = 90 m/s 2 ....(ii)Velocity of wave v = 20 m/s ...(iii)Dividing (ii) by (i)aω2 90= ⇒ ω = 30 rad/s ...(iv)aω3Substituting the value of ω in (i) we geta = 303 = 0.1 m ...(v)Now2 π 2π 2π ω 30 3k = = = = = =λ v / v vvv 20 2From (iv), (v) and (vi) the wave form is⎡ 3 ⎤y = 0.1 sin ⎢30t ± x ± φ⎥ ⎣ 2 ⎦...(vi)2. A 5m long cylindrical steel wire with radius 2 × 10 –3m is suspended vertically from a rigid support andcarries a bob of mass 100 kg at the other end. If thebob gets snapped, calculate the change in temperatureof the wire ignoring radiation losses. (For the steelwire : Young's modulus = 2.1 × 10 11 Pa; Density= 7860 kg/m 3 ; Specific heat = 420 J/kg-K).[IIT-2001]Sol. When the mass of 100 kg is attached, the string isunder tension and hence in the deformed state.Therefore it has potential energy (U) which is givenby the formula.U = 21 × stress × stain × volume21 (Stress)= × 2 Y22× πr 2 l1 (Mg / πr)= × πr 2 1 M g ll = ...(i)2 Y22πrYThis energy is released in the form of heat, therebyraising the temperature of the wireQ = mc ∆T...(iii)22From (i) and (iii) Since U = Q Therefore2 21 M g l∴ mc∆T =22 πrY2 21 M g l∴ ∆T =22 πrYcmHerem = mass of string = density × volume of string= ρ × πr 2 l2 21 M g∴ ∆T =2 22 ( πr) Ycρ1 (100×10)= × 2−3211(3.14×2×10 ) × 2.1×10 × 420×7860= 0.00457ºC3. The x – y plane is the boundary between twotransparent media. Medium –1 with z ≥ 0 has arefractive index 2 and medium –2 with z ≤ 0 has arefractive index 3 . A ray of light in medium –1given by the vector A = 6 3 ^i + 8 3^j – 10 ^k isincident on the plane of separation. Find the unitvector in the direction of the refracted ray inmedium –2.[IIT-2003]Sol.Y6^3 i + 8^3 jZ^ ^6 3 i + 8 3 j8 3^jOM ' XOX–10 ^KMFig(1)M '6 3^iFigure 1 shows vector2→^ ^ ^A = 6 3 i+8 3 j–10k6 3i+8 3^jFig(2)Figure 2 shows vector A → = 6 3^i + 8 3^j – 10^kThe perpendicular to line MOM' is Z-Axis which hasa unit vector of ^k . Angle between vector IO →and→ZO can be found by dot product→IO . ZO →= (IO) (ZO) cos i(6(63)2^+ (83)^3 i+8 3 j–10 k).(– k)2^+ (–10)2^(–1)2= cos i⇒ i = 60Unit vector in the direction MOM' from figure (1) isXtraEdge for IIT-JEE 6 SEPTEMBER 2011