2-Solution combined Test 1 (Paper 2)

SOLUTIONS TO COMBINED TEST - 1 (PAPER - 2) 1. Mass defect m = 4m – mHe = 4(1.007825) – 4.002603 = 0.028697 u MeV = 0.028697 u × 932 u = 26.7 MeV 2. A concave or convex mirror is to be placed left of the object. The object and the image both will be real for concave mirror and virtual for convex mirror. 3. f max = µN = µB𝑙 = µB 100 5  0.5t 𝑙 = 3 × 2 × 1 × 4 100 5  0.5 t 150 = 5  0.5 t at t = 0 fmax = 30N but mg = 20 for f = 20 at t = 0 when rod start to move fmax = 20 , t = 5 sec. 5 The potential of the centre is positive. From symmetry of the problem B and D will move away on the line BD and A and C will move away on the line AC. Since masses are different, their speeds will be different. 6. Putting k = 1, → → v2  v1 for oblique collision and putting k = 1, →  , →  → for headon collision v1 0 v2 v 7. R + Distance travelled by A = Distance travelled by B 1 R + vt = vt + 2 t = vBA = VB – VA = at2 . 8. (A) when circuit is not under resonance. (B) when circuit under resonance. (C) There will always be a phase difference between current and voltage across WY 13. Time period of revolution 2m T = qB 14. The charged particle touches the x-axis at times t = T, 2T, 3T, ..... 15. Let the area of the plates be A and the charge on capacitor be Q Q = KCV (1) The electrostatic force on lower capacitor plate is Q2 Fe = 2A 0 ..........(2) For rod to be in equilibrium, net moment of force on the rod about O should be zero. mg 𝑙 2 – F 𝑙 = 0 or, mg e 2 Q2 = Fe = 2A 0 A0 mg = K2C2V2 d m g = K2 C V2 16. After the removal of dielectric, the force on lower metal plate does not change because charge on capacitor remains same. Hence the rod remains in equilibrium and the angular acceleration of rod is zero. mV 18 r = qB 4mV rB = eB mV rD = eB  r = 2mV A eB  r = 2mV C eB 19. By snell law n = sin i sin r Since for 1st case angular incedence is same for all ray. So r will be less for red.