Tiêu đề 03. CURRENT ELECTRICITY.pdf ✅

03. CURRENT ELECTRICITY NEET PREPARATION (MEDIUM PHYSICS PAPER) Date: March 12, 2025 Dura on: 1:00:00 Total Marks: 180 INSTRUCTIONS INSTRUCTIONS PHYSICS 1. () : Explana on The resistance of a metallic wire at tempera‐ ture is given by where, is the coefficient of expansion. The resistance of wire increases on increasing the temperature. Also, from ohm's law, the ra‐ o of is equal to . Hence, on increasing the temperature, the resistance of metallic wire is increased, so the ra o also increases. 2. () : Explana on Correct op on is (2). 3. () : Explana on Resistance of the lamp, Current in the lamp, When lamp is operated on a 120 , For the same current 3 , total resistance R' For proper glow, the resistance required to be put in a series will be 4. () : Explana on Since current through the both the rods is same 5. () : Explana on When poten ometer is connected between and , then it measures only and when connected between and , then it mea‐ sures 6. () : Explana on In a poten ometer experiment, the emf of a cell can be measured if the poten al drop along the poten ometer wire is more than the emf of the cell. t ∘C Rt = R0(1 + αt) α V /i R V /i P = ⇒ I 2R = W t W t ⇒ W = I 2Rt R0 = V 2 P = = 10 Ω (30) 2 90 I = 30/10 = 3A V A = = = 40 Ω V I 120 3 ∴ R = R′ = R0 − 40 − 10 = 30Ω I = neAvd 2(n) eAvL = ne(2A)vR ⇒ = 1 vL vR A B E1 A C E1 − E2 ∴ = E1 E1−E2 l1 l2 = E1−E2 E1 l2 l1 ⇒ 1 − = E2 E1 100 300 ⇒ = 1 − E2 E1 1 3 ⇒ = E2 E1 2 3 ∴ = E1 E2 3 2
7. () : Explana on Let be the resistance of the wire (1) The heat generated is . (2) Resistance of each part will be . When they are connected in parallel, the resistance will be . Hence, . (3) In case of four wires connected in parallel, the resistance will be . (4) Hence, largest amount of heat will be generated in case of four parts connected in parallel. 8. () : Explana on If the poten al fall per is then the re‐ sistance should become 9. () : Explana on Resistance of metallic wire increases with rise of temperature. So standard resistance should be increased to get the null point at same place. So op on (4) is correct. 10. () : Explana on Resistance of voltmeter should be high. 11. () : Explana on Total power ( ) consumed As we know, Power i.e., Minimum capacity should be 12 . 12. () : Explana on 13. () : Explana on When 3 cells are connected correctly and one incorrectly, then equivalent emf of the combi‐ na on is and effec ve internal resistance of the combi‐ na on is 14. () : Explana on Let cells out of cells be wrongly connnected to the external resistor . Then: Total emf of combina on where of each cell. Total internal resistance . Required . 15. () : Explana on As both resistors are in parallel combina on so poten al drop ( ) across both are same. We know Power Time(t) is same for both Note: We are asked to find ra o of thermal energies 16. () : Explana on Here : Put the value in equa on at R H1 = V 2t R R/2 R/4 H2 = 4V 2t/R R/8 ∴ H3 = 8V 2t R H4 = = V 2t R/2 2V 2 ⋅ t R cm V /4 40Ω P = (15 × 40) + (5 × 100) + (5 × 80) + (1 × 1000) = 2500 W P = V I ⇒ I = A = = 11.3A 2500 220 125 11 A V = xl ⇒ iR = xl ⇒ i × 10 = ( ) × 50 × 10 −2 = 0.1 2×10 −3 10 −2 ⇒ i = 10 × 10 −3A = 10mA Eeq = (E + E + E − E) = 2E reff = (r + r + r + r) = 4r (m) (n) R = (n − 2m)E E = emf = nr I = ( ) E n − 2m R + nr I = = = = 0.6 A (n−2m)ε R+nr [10−2(2)]×2 10+10(1) 6×2 20 V P = ⇒ P ∝ V 2 R 1 R P = = Energy time E t R1 , R2 = = = = E1 E2 P1 P2 R2 R1 200 100 2 1 = 2 : 1 R0 = 8 Ω R100 = 10 Ω Rt =?, T = 400 ∘C α = (1) RT−R0 R0t α = 10 − 8 8 × 100 ⇒ α = / ∘C 1 400 Rt = R0(1 + αΔT) t = 400 ∘C, Rt = 8(1 + 400α) = 8 (1 + 4 × ) 1 4 Rt = 16 Ω
17. () : Explana on Given ohm-metre . 18. () : Explana on 19. () : Explana on Electric appliances with metallic bodies like heaters and presses o en have three-pin con‐ nec ons for safety purposes. The third pin, known as the Earth or Ground pin, is meant to provide an addi onal path for electrical cur‐ rent to safely dissipate into the ground in the event of a fault or short circuit, preven ng electrical shocks or fires. It reduces hea ng of connec ng cables, and contributes to overall electrical safety. So correct op on is (3). 20. () : Explana on Charge on capacitor At (switch closed) resistance of‐ fered by capacitor is zeroSo correct op on is (1). 21. () : Explana on Applying Kirchoff's loop law, The current in the circuit is The terminal poten al difference is 22. () : Explana on . 23. () : Explana on Dri velocity current charge of electron cross-sec on area number density of electrons 24. () : Explana on . across any cell, 25. () : Explana on Here, Number of electrons, Number of protons, The current due to electrons is from le to right in the direc on of op‐ posite to the mo on of electrons). The current due to protons is from le to right in the direc on of mo‐ on of protons). The total current in the discharge tube is from le to right. Hence the current in the discharge tube is and its direc on will be towards right. ρ = 2.2 × 10 −8 l = 314m, D = 1.2 × 10 −2m, D1 = 1.0 × 10 −2m. R = ρl (D2−D2 1 ) π 4 ⇒ R = 4×2.2×10 −8×314 π(1.2 2−1.0 2)×10−4 = 4 × 5 × 10 −2Ω = 2 × 10 −1Ω R = ⇒ 0.7 = ρL A ρ×1 (1×10−3) 22 2 7 ρ = 2.2 × 10 −6Ω. m q = CE (1 − e−tlCReq) ∴ I = = e−tlCR dq dt E Req t = 0 I = = ⇒ E Req E R + r −IR − Ir2 − E2 + E1 − Ir1 = 0 ∴ I = E1−E2 r1+r2+R V = IR = × R E1−E2 r1+r2+R Rt = R0 [1 + α(t − t0)] ⇒ 6.8 = 2 [1 + α(80 − 0)] ⇒ α = = 0.03/ ∘C = 3 × 10 −2/ ∘C 2.4 80 vd = I eAn I = e = A = n = ∴ vd = ms 5 −1 (1.6×10 −19)×(4×10 −6)×(5×10 26) = ms−1 = 0.0156 ms 1 1 64 = 1.56 × 10 −2ms−1 I = = NE Nr E r P D ΔV = E − Ir = E − ( ) r = 0 E r ne = 3.31 × 10 15 np = 3.12 × 10 15 Ie = = nee t (3.31×10 15)(1.6×10 −19C) 1 s = 5.296 × 10 −4 A ≈ 0.5 mA (i. e. Ip = = npe t (3.12×10 15)(1.6×10 −19C) 1 s = 4.992 × 10 −4 A ≈ 0.5 mA (i. e. ∴ I = Ie + Ip = 0.5 mA + 0.5 mA = 1 mA 1 mA