Tiêu đề 03. CURRENT ELECTRICITY(H).pdf ✅

NEET REVISION 03. CURRENT ELECTRICITY(H) NEET REVISION Date: March 18, 2025 Dura on: 1:00:00 Total Marks: 180 INSTRUCTIONS INSTRUCTIONS PHYSICS 1. () : Explana on 2. () : Explana on If then poten al drop across resistor present between and is so, cur‐ rent flow from to . . By junc on rule . 3. () : Explana on By applying wheatstone bridge principle 4. () : Explana on Resistance of any bulb is Net resistance of the circuit is Current in each branch Reading of ammeter 5. () : Explana on Circuit can be redrawn as The upper and lower branches can be replaced with a ba ery of and internal resistance , which are given by Total so 6. () : Explana on The power (heat produced per unit me) in a heater is given by , where is the voltage and is the resistance. If the resistance is halved by using only one-half of the coil, the power is doubled . Since power is directly propor onal to the square of the current , using half of the coil will result in the same amount of heat being produced in half the me, as stated in the asser on. So cor‐ rect op on is (2). I = P. D across R2 = IR2 = E r + R2 ER2 r + R2 Q = CER2 r + R2 VB − VC = 12 2Ω B C 2V B C = = 1 A 2 2 I1 = 2 + 1 = 3 A RAB = and I = = 3A 10 3 10 10/3 R = = = 484 Ω V 2 P (220) 2 100 Req = = = 121 Ω R 4 484 4 V = i Req ⇒ 220 = i × 121 i = 1.81 Amp = amp = 0.45 amp 1.81 4 = 3 × 0.45 = 1.35 amp emf ε0 r0 ε0 = = = 2 V + ε1 r1 ε2 r2 + 1 r1 1 r2 + 2 4 2 4 + 1 4 1 4 = + = + = = 1 r0 1 r1 1 r2 1 4 1 4 2 4 1 2 ⇒ r0 = 2Ω emf = 2 + 2 = 4 V I = = 2A 4 2 P = V 2/R V R (P = V 2/(R/2) = 2V 2/R) (PαI 2)
NEET REVISION 7. () : Explana on At When capacitor is fully charged it will behave as open circuit, Net resistance of circuit, Finally current will reduce to 1 mA . So correct op on is (4) 8. () : Explana on 9. () : Explana on Net . net internal resistance . . 10. () : Explana on Let resistor be connected between & . Even if we remove all the units, then s ll should remain same. for to be independent of no. of unit cell. 11. () : Explana on 12. () : Explana on For a balanced meter bridge When & are interchanged and will also interchange 13. () : Explana on For balanced meter bridge when when t = 0, I = = 2 mA 2 V 1kΩ R = 1kΩ + 1kΩ = 2kΩ ∴ Current = = 1 mA 2 V 2 k Ω x = (2R + x)R 3R + x 3Rx + x 2 = 2R2 + Rx x 2 + 2Rx − 2R2 = 0 x = = R(√3 − 1) −2R±√4R2+8R2 2 = (√3 + 1)(√3 − 1) = 3 − 1 = 2 Ω emf ∑eq = (5 − 2)2 = 6 V req = 5Ω i = = A 6V (5 + 7)Ω 1 2 ′x ′ C D RAB ⇒ RAB = RCD = x RAB = = R(R+x+R) R+(R+x+R) 2R2+Rx 3R+x RAB x = 2R2+Rx 3R+x x 2 + 2Rx − 2R 2 = 0 x = R ⇒ x = (√3 − 1) R VG − 2 × 2 + 3 − 2 × 2 + 2 × 1 = VH VG − 3 = VH ⇒ VG − VH = 3V y × 39.5 = x × (100 − 39.5) x = = 8.16 Ω 12.5×39.5 60.5 x y l1 (100 − l1) l2 = 60.5 cm = R l S 100 − l ⇒ R = Sl 100 − l R1 = 4 Ω t1 = 0 ∘C R2 = 6 Ω t2 = 100 ∘C α = = = R2 − R1 R1t2 − R2t1 6 − 4 4 × 100 − 0 2 400 α = 0.005 ∘C −1
NEET REVISION 14. () : Explana on In metre-bridge, In first case, unknown resis‐ tance in right gap ) So, In second case, is shunted with Effec ve resistance in the right gap becomes For new balancing length , a er solving we get, Change in balance length, 15. () : Explana on Resistance of each bulb, In parallel, Power of circuit at , Number of bulbs , (nearest lowest integer) So, correct op on is (3). 16. () : Explana on Let and be the emf and the internal resistance of the ba ery. The poten al difference across the terminals of the ba ery when a current is drawn from it is As per ques on Subtrac ng eqn. (1) from eqn. (2), we get Subs tu ng this value of in eqn. (1), we get 17. () : Explana on Consider a thin spherical shell of radius r and thick‐ ness . The resistance of the shell is 18. () : Explana on In the figure, Ra o of heat developed per second across , and = P Q l 100 − l = (X = 2 X 40 60 X = = 3 Ω 2×60 40 X 2 Ω X′ = = Ω 2 × 3 5 6 5 = l ′ ⇒ = ⇒ = R X′ l ′ 100 − l ′ 2 6/5 l ′ 100 − l ′ = 5 3 l ′ 100 − l ′ l ′ = 62.5 cm ∴ l ′ − l = 62.5 − 40 cm = 22.5 cm R = = = 200Ω V 2 P (100) 2 50 R′ = = 100Ω 200 × 200 400 120 V P ′ = = 144W (120) 2 100 = = 2.88 144 50 i. e. n = 2 ε r V I V = ε − Ir 50 V = ε − (11 A)r (1) and 60 V = ε − (1 A)r (2) 10 V = (10 A)r r = = 1 Ω 10 V 10 A r 50 V = ε − (11 A)(1 Ω) 50 V = ε − 11 V ε = 50 V + 11 V = 61 V dr dR = ρ dr 4πr 2 R = ∫ ρ = b ∫ a dr dr 4πr 2 ρ 4π 1 r 2 R = [ − ] = ρ 4π 1 a 1 b ρ(b−a) 4πab P = 100 Ω, Q = 10 Ω, R = 300 Ω, S = 30 Ω I1 = = I 330I 330 + 110 3 4 I2 = I − I1 = I − I = 3 4 I 4 P, Q R S, HP : HQ : HR : HS = ( I) 2 × 100 : ( I) 2 × 10 : ( ) 2 × 300 3 4 3 4 I 4 : ( ) 2 × 30 I 4 = 90 : 9 : 30 : 3 = 30 : 3 : 10 : 1
NEET REVISION 19. () : Explana on Resistance, or In parallel combina on of such four resistances, 20. () : Explana on Wheatstone bridge principle is not valid for the given circuit we can apply delta star transforma‐ on. 21. () : Explana on For balancing the meter bridge, case : case :Length per meter of wire, 22. () : Explana on The equivalent circuit is as shown below: 23. () : Explana on 24. () : Explana on and So and 25. () : Explana on R = ρl A R ∝ l ∴ = = = 4 R1 R2 l1 l2 L L/4 ⇒ R2 = (∵ R1 = R) R 4 = + + + 1 R′ 1 R1 1 R2 1 R3 1 R4 or = + + + 1 R′ 1 R/4 1 R/4 1 R/4 1 R/4 ⇒ = + + + 1 R′ 4 R 4 R 4 R 4 R ∴ = ⇒ R ′ = 1 R′ 16 R R 16 ⇒ RAB = Ω 13 11 1 st = P Q R S = ; r×40 r(100−40) X 25 X = 50 3 2 nd r ′ = 2r ⇒ = 2r × l 2r(100 − l) 50/3 25 ⇒ = l 100 − l 2 3 ⇒ l = 40 cm Req = 6 + 5 + 8 = 19 Ω P = i 2R ∴ imax = √ = √ = √15 A Pmax R 36 2.4 Total maximum power = (imax) 2 ( ) 3R 2 = (15)(1.5)(2.4) = 54 W = R1R2 R1 + R2 8 3 R1 + R2 = 12 ⇒ R1R2 = 32 ⇒ R2 − R1 = √(R1 + R2) 2 − 4R1R2 = √12 2 − 4 × 32 = 4Ω R1 = 4Ω R2 = 8 Ω ⇒ = = l1 l2 4 8 1 2 εeq = = = V ∑ ε i r i ∑ 1 ri + − 2 1 1 1 2 1 + + 1 1 1 1 1 1 1 3