1.Solution combined Test 1 ( Paper 1)

SOLUTIONS TO COMBINED TEST - 1 (PAPER - 1) 1. The loop rotates anticlockwise due to magnetic torque and as a result north and south poles will attract each other. 2. Net torque on particles.  = – (E q – E q).𝑙 = 2m𝑙2.   q q  = –  2 – (r – x) 20 (r  x)  𝑙 = 2m𝑙2.    = q 2 r 2m (x = 𝑙)  = 2 3. As the magnet-1 falls into solenoid A, the magnetic flux associated with solenoid A increases. From Lenz's law, induced current in solenoid A will oppose this increase in magnetic flux. Hence direction of induced current in solenoids is as shown.The nature of magnetic field produced by solenoid B is as shown. Therefore magnet 2 will be attracted by magnetic field due to solenoid B. 2k 4. Electric field between the two cylinders = r  Force on charge q = This force is centripetal force 2kq r  2kq r mv2 = r   v = = 5. Energy of photon is given by mc2 now the maximum energy of photon is equal to the maximum energy of electron = eV eV hence , mc2 = eV  m = 2 1.6  1019 18 103 = 8 2 = 3.2 × 10–32 kg c (3 10 ) 6. v = v0 sin (t + /4) = v0cos (t– /4) Since V lags current, an inductor can bring it in phase with current. → ˆ ˆ 7. E  Ex i  Ey j , V = –Ex x – Ey y for A and B 16 – 4 = – Ex (– 2 – 2) – Ey (2 – 2) Ex = 3 V/m for B and C 12 – 16 = –Ex {2– (– 2)} – Ey (4 – 2) Ey = –4 V/m.  → ˆ ˆ E  (3i – 4 j)V / m 8. AP = AQ e–lt = A  1 t e T AQ \ t = T ln AP 9. Nx = N0 .e1t 2 = 0.2 N0  N = N0 .e2t 2 = 0.8 N0 e(12 )  4  t = 8 × 109 years 10. Solve dimensionaly  12   Q2  [F] =  2  rn   [F] =  2  r 2rn   0   0  Comparing with coulmb's law , f = 1 40 q1q2 r 2  n = 0 11. In the initial state, charge on each capacitor is shown in figure-1. Let charge q flow anticlockwise in the circuit before it achieves steady state as shown in figure-2. Applying KVL to figure 2. 360  q  3 q 1.5  300  q  0 2 or 180 C  final charge on 1.5 F capacitor is q = 180 C and final charge on 2 F capacitor is 300 – q = 120 C. 12. E = 3x2 + 0.4 N/C |V| = E dx = (3x2  0.4)dx =  3 0.2 0.2  x 0 x  0.4x 0 |V| = (0.2)3 + 0.4 × 0.2 = 0.088 volt C = Q = V 0.88 0.088 = 10 µF 13. For Balmer series, n1 = 2, n2 = 3,4,........ (lower) (higher)  In transition (V), Photon of Balmer series is absorbed. In transition  E2 = – 3.4 eV, E4 = – 0.85 eV E = 2.55 eV E = hc   = hc  E  = 486 nm. Wavelength of radiation = 103 nm = 1030 Å 12400  E = 1030 Å ~– 12.0 eV So difference of energy should be 12.0 eV (approx) Hence n1 = 1 and n2 = 3 (–13.6)eV (–1.51)eV  Transition is V. For longest wavelength, energy difference should be minimum. So in visible portion of hydrogen atom, minimum energy emitted is in transition V. 14. By principal of energy conservation. PB = PR + PL Near the starting of the circuit PR Ldi By KVL,  = iR + dt Initially i is small so di = i2 R and P = L i di dt . dt has greater value at the starting of the circuit, PL > PR 15. Charge on a1 Charge on a2 = (r1 = (r2 )  )  r1 Ratio of charges = r2 K [ (r1  )] KQ E1, Field produced by a1 = 2 1 = r1 E2, Field produced by a2 = as r2 > r1 Therefore E1 > E2 E1 > E2 KQ r2 i.e. Net field at A is towards a2. K.(r1) V1 = V2 = r1 K.(r2 ) r2 = K  = K   V = V2 . 2k 16. Electric field due to infinitely long charged wire at distance r from it = r Now, we take small element dr on wire (2) which is having distance r from wire (1) Now, force on this element = ( dr). 2k1 r or dF = 2k1 2dr r 2𝑙 2k   2𝑙  Total force = dF =  1 2 dr = 2k  𝑙n   r 1 2 𝑙 1  𝑙  Total force = 2 × 9 × 109 × 1 × 10–6 × 𝑙n4 × 10–6 × 𝑙n2 = 9 × 10–3 N. 17. Electric field inside the cavity = →    Here a  along line joining  30 → a centers of sphere and cavity  Force on the e– inside cavity = 30 (e) Cavity ——  acceleration of electron a = ae 3om. Now for distance Cavity —— d = = 2 r v2 = u2 + 2ad 2 2rea v = 30m After solving, we get v = 4 × 105 m/sec. v = 4. 18. x + y = 1.8 Magnification = x = 0.6, y = 1.2 xy y = – 2  y = 2x  x f = x  y = 0.4 m 19. → → →     3ˆi  Ey ˆj  2kˆ =  (ˆi  ˆj)  (2ˆi  3kˆ) =  3(ˆj)  2(kˆ)  3ˆi = Ey = 3 Aliter : → →  E = 3 v . E  0 y 20. 1 i1 = 3 1 i2 = 3 Vx + 3i1 + 4 – 3i2 = Vy Vx – Vy = – 4 + 3 × 1 – 3 × 1 3 3 = – 4 + 1.5 – 1.2 = –4 V  potential difference = 4 V.