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Page 1 MARKING SCHEME PHYSICS Subject Code – 042 CLASS – XII Academic Session 2024 – 25 Maximum Marks:70 Time Allowed: 3hours [SECTION – A] Ans.1- (B) (1 mark) VA> VB [VA = VC] In the direction of electric field, the electric potential decreases. Ans.2- (B) In the state of equilibrium, (1 mark) The potential on the surface of bigger sphere = the potential at the surface of the smaller sphere 12 12 = kq kq rr ? 11 22 = qr qr ? 22 1 1 2 1 2 2 22 2 2 2 1 11 E E = = ? = q r r r r q r r rr Ans.3 - (C) (1 mark) AtP2, B2 = 00 II 3 3 2 2 ?? = ? ?? ? ?? ?? a a AtP1, B1 = ( ) 00 (I 4) I 2 2 4 ?? = ?? aa ? 0 2 0 1 I B 3 I B 4 ? ?? ?? ? ?? = ? ?? ?? ? ?? a a ? 2 1 B 4 B3 = Ans.4 - (D)Sound waves as well as light waves (1 mark) Ans.5 -(A) (1 mark) Ans.6 - (C)When all the given components are connected(1 mark) Page 2 MARKING SCHEME PHYSICS Subject Code – 042 CLASS – XII Academic Session 2024 – 25 Maximum Marks:70 Time Allowed: 3hours [SECTION – A] Ans.1- (B) (1 mark) VA> VB [VA = VC] In the direction of electric field, the electric potential decreases. Ans.2- (B) In the state of equilibrium, (1 mark) The potential on the surface of bigger sphere = the potential at the surface of the smaller sphere 12 12 = kq kq rr ? 11 22 = qr qr ? 22 1 1 2 1 2 2 22 2 2 2 1 11 E E = = ? = q r r r r q r r rr Ans.3 - (C) (1 mark) AtP2, B2 = 00 II 3 3 2 2 ?? = ? ?? ? ?? ?? a a AtP1, B1 = ( ) 00 (I 4) I 2 2 4 ?? = ?? aa ? 0 2 0 1 I B 3 I B 4 ? ?? ?? ? ?? = ? ?? ?? ? ?? a a ? 2 1 B 4 B3 = Ans.4 - (D)Sound waves as well as light waves (1 mark) Ans.5 -(A) (1 mark) Ans.6 - (C)When all the given components are connected(1 mark) IR = IXC = IXL = 10V XC = XL = R Z = 22 CL R (X X ) +- Z = 22 R (R R) +- Z = R VS = IZ = IR = 10 V So, the source voltage is also 10 V When the capacitor is short circuited then Z = 22 L R (X ) + = v?? 2 + ?? 2 = ?? v2 VL = I ? XL = 10 R 5 2 2R ?= V Ans.7 - (B)(1 mark) Ans.8 - (B) The distance of closest approach(1 mark) 2 1 const V = d ...(1) 2 2 const 2 V = d ...(2) From equations (1) and (2), 2 2 2 1 V 2 V = ?V2 = 2 V1 ? V2 = 2 V Given, (V1 = V) Ans.9 - (C)(1 mark) ?? 2 ?? - ?? 1 ?? = ?? 2 - ?? 1 ?? 1 3 [1 3 2] 2[ 6] 6 v - -= -- Page 3 MARKING SCHEME PHYSICS Subject Code – 042 CLASS – XII Academic Session 2024 – 25 Maximum Marks:70 Time Allowed: 3hours [SECTION – A] Ans.1- (B) (1 mark) VA> VB [VA = VC] In the direction of electric field, the electric potential decreases. Ans.2- (B) In the state of equilibrium, (1 mark) The potential on the surface of bigger sphere = the potential at the surface of the smaller sphere 12 12 = kq kq rr ? 11 22 = qr qr ? 22 1 1 2 1 2 2 22 2 2 2 1 11 E E = = ? = q r r r r q r r rr Ans.3 - (C) (1 mark) AtP2, B2 = 00 II 3 3 2 2 ?? = ? ?? ? ?? ?? a a AtP1, B1 = ( ) 00 (I 4) I 2 2 4 ?? = ?? aa ? 0 2 0 1 I B 3 I B 4 ? ?? ?? ? ?? = ? ?? ?? ? ?? a a ? 2 1 B 4 B3 = Ans.4 - (D)Sound waves as well as light waves (1 mark) Ans.5 -(A) (1 mark) Ans.6 - (C)When all the given components are connected(1 mark) IR = IXC = IXL = 10V XC = XL = R Z = 22 CL R (X X ) +- Z = 22 R (R R) +- Z = R VS = IZ = IR = 10 V So, the source voltage is also 10 V When the capacitor is short circuited then Z = 22 L R (X ) + = v?? 2 + ?? 2 = ?? v2 VL = I ? XL = 10 R 5 2 2R ?= V Ans.7 - (B)(1 mark) Ans.8 - (B) The distance of closest approach(1 mark) 2 1 const V = d ...(1) 2 2 const 2 V = d ...(2) From equations (1) and (2), 2 2 2 1 V 2 V = ?V2 = 2 V1 ? V2 = 2 V Given, (V1 = V) Ans.9 - (C)(1 mark) ?? 2 ?? - ?? 1 ?? = ?? 2 - ?? 1 ?? 1 3 [1 3 2] 2[ 6] 6 v - -= -- 1 3 1 2 1 12 12 12 6 v - - - = + = = ?? = –6 cm Ans.10 - (B)Diffraction (1 mark) Ans.11- (A)doping level (1 mark) Ans.12- (C)+0.4% (1 mark) Ans.13- (A) (1 mark) Ans.14- (A)(1 mark) Ans.15- (D)(1 mark) Ans.16- (A)(1 mark) [SECTION – B] Ans.17– Given Ø ?? = ?? .???????? = ?? .???? × ?? .?? × ???? -???? ?? ?? = ?? .?? × ???? ???? ???? ?? .?? .= ???? - Ø ?? = ???? ?? ½ ?? = ???? ???? -Ø ?? ½ = ?? .???? × ???? -???? ×?? ×???? ?? ?? .???? × ???? -???? ×?? .?? ×???? ???? - ?? .???? ×?? .?? ×???? -???? ½ = ???? .???? × ???? -???? ?? .?? × ???? -???? (?? .???? - ?? .???? ) = ???? .???? × ???? -???? ?? .?? ×???? ???? = ???? .?? × ???? -?? ?? ½ Ans.18 - ?? 1 = 4 × 10 -7 ?? ?? 2 = 6× 10 -7 ?? Distance at which dark fringe is observed ?? = (?? + 1 2 ) ???? ?? ½ First Dark fringe for ?? 1 ?? 1 = 1 2 4×10 -7 10 -2 ?? = 2 × 10 -5 ?? ½ Page 4 MARKING SCHEME PHYSICS Subject Code – 042 CLASS – XII Academic Session 2024 – 25 Maximum Marks:70 Time Allowed: 3hours [SECTION – A] Ans.1- (B) (1 mark) VA> VB [VA = VC] In the direction of electric field, the electric potential decreases. Ans.2- (B) In the state of equilibrium, (1 mark) The potential on the surface of bigger sphere = the potential at the surface of the smaller sphere 12 12 = kq kq rr ? 11 22 = qr qr ? 22 1 1 2 1 2 2 22 2 2 2 1 11 E E = = ? = q r r r r q r r rr Ans.3 - (C) (1 mark) AtP2, B2 = 00 II 3 3 2 2 ?? = ? ?? ? ?? ?? a a AtP1, B1 = ( ) 00 (I 4) I 2 2 4 ?? = ?? aa ? 0 2 0 1 I B 3 I B 4 ? ?? ?? ? ?? = ? ?? ?? ? ?? a a ? 2 1 B 4 B3 = Ans.4 - (D)Sound waves as well as light waves (1 mark) Ans.5 -(A) (1 mark) Ans.6 - (C)When all the given components are connected(1 mark) IR = IXC = IXL = 10V XC = XL = R Z = 22 CL R (X X ) +- Z = 22 R (R R) +- Z = R VS = IZ = IR = 10 V So, the source voltage is also 10 V When the capacitor is short circuited then Z = 22 L R (X ) + = v?? 2 + ?? 2 = ?? v2 VL = I ? XL = 10 R 5 2 2R ?= V Ans.7 - (B)(1 mark) Ans.8 - (B) The distance of closest approach(1 mark) 2 1 const V = d ...(1) 2 2 const 2 V = d ...(2) From equations (1) and (2), 2 2 2 1 V 2 V = ?V2 = 2 V1 ? V2 = 2 V Given, (V1 = V) Ans.9 - (C)(1 mark) ?? 2 ?? - ?? 1 ?? = ?? 2 - ?? 1 ?? 1 3 [1 3 2] 2[ 6] 6 v - -= -- 1 3 1 2 1 12 12 12 6 v - - - = + = = ?? = –6 cm Ans.10 - (B)Diffraction (1 mark) Ans.11- (A)doping level (1 mark) Ans.12- (C)+0.4% (1 mark) Ans.13- (A) (1 mark) Ans.14- (A)(1 mark) Ans.15- (D)(1 mark) Ans.16- (A)(1 mark) [SECTION – B] Ans.17– Given Ø ?? = ?? .???????? = ?? .???? × ?? .?? × ???? -???? ?? ?? = ?? .?? × ???? ???? ???? ?? .?? .= ???? - Ø ?? = ???? ?? ½ ?? = ???? ???? -Ø ?? ½ = ?? .???? × ???? -???? ×?? ×???? ?? ?? .???? × ???? -???? ×?? .?? ×???? ???? - ?? .???? ×?? .?? ×???? -???? ½ = ???? .???? × ???? -???? ?? .?? × ???? -???? (?? .???? - ?? .???? ) = ???? .???? × ???? -???? ?? .?? ×???? ???? = ???? .?? × ???? -?? ?? ½ Ans.18 - ?? 1 = 4 × 10 -7 ?? ?? 2 = 6× 10 -7 ?? Distance at which dark fringe is observed ?? = (?? + 1 2 ) ???? ?? ½ First Dark fringe for ?? 1 ?? 1 = 1 2 4×10 -7 10 -2 ?? = 2 × 10 -5 ?? ½ First Dark fringe for ?? 2 ?? 2 = 1 2 6×10 -7 10 -2 ?? = 3 × 10 -5 ?? First dark fringe will be the distance where both dark fringes will coincide i.e LCM of ?? 1 &?? 1 ½ i.e. 2 × 10 -5 ?? × 3 × 10 -5 ?? = 6 × 10 -5 ?? ½ OR (II) ?????? ?? = ?? 1 + ?? 2 + 2v?? 1 v?? 2 cos? 0.5 M Since, ?? 1 = ?? 2 = ?? Net I = I + I + 2 I cos? = 2I (1 + cos?) = 2I (2 cos 2 Ø 2 ) 0.5 M For path difference ?/4 , phase difference is p/2 0.5 M Net I = 4 I cos 2 ?? 4 Net I = 2 I 0.5 M (2 Marks) Ans.19 - (I)The direction of the magnetic field is perpendicular and inward into the plane of thepaper 0.5M (II) For a head-on collision to take place, the radius of the path of each ion should be equal to 0.5 m. 0.5 mv r qB == m 0.5M 26 5 19 4 10 2.4 10 B 4.8 10 0.5 mv qr - - ? ? ? == ?? 0.5M B = 0.04 T 0.5M For VI Candidate (a) As Pitch (p)= 2????? ?????? O ???? 0.5M Or, p= 2 ?? 3.14 ?? 1.7?? 10 -27 ?? 2 ?? 10 5 ?????? 30 0 1.6 ?? 10 -19 ?? 1.5 m Or, P=7.7X10 -3 m 0.5M (b)As, done by magnetic field is always zero K.E=1/2mv 2 0.5M KE=3.4 X 10 -17 J 0.5M Ans.20 –(i) Nuclear fission –W 0.5M Reason: As W has binding energy per nucleon less then Y and X and nucleus is larger in size. 0.5M Page 5 MARKING SCHEME PHYSICS Subject Code – 042 CLASS – XII Academic Session 2024 – 25 Maximum Marks:70 Time Allowed: 3hours [SECTION – A] Ans.1- (B) (1 mark) VA> VB [VA = VC] In the direction of electric field, the electric potential decreases. Ans.2- (B) In the state of equilibrium, (1 mark) The potential on the surface of bigger sphere = the potential at the surface of the smaller sphere 12 12 = kq kq rr ? 11 22 = qr qr ? 22 1 1 2 1 2 2 22 2 2 2 1 11 E E = = ? = q r r r r q r r rr Ans.3 - (C) (1 mark) AtP2, B2 = 00 II 3 3 2 2 ?? = ? ?? ? ?? ?? a a AtP1, B1 = ( ) 00 (I 4) I 2 2 4 ?? = ?? aa ? 0 2 0 1 I B 3 I B 4 ? ?? ?? ? ?? = ? ?? ?? ? ?? a a ? 2 1 B 4 B3 = Ans.4 - (D)Sound waves as well as light waves (1 mark) Ans.5 -(A) (1 mark) Ans.6 - (C)When all the given components are connected(1 mark) IR = IXC = IXL = 10V XC = XL = R Z = 22 CL R (X X ) +- Z = 22 R (R R) +- Z = R VS = IZ = IR = 10 V So, the source voltage is also 10 V When the capacitor is short circuited then Z = 22 L R (X ) + = v?? 2 + ?? 2 = ?? v2 VL = I ? XL = 10 R 5 2 2R ?= V Ans.7 - (B)(1 mark) Ans.8 - (B) The distance of closest approach(1 mark) 2 1 const V = d ...(1) 2 2 const 2 V = d ...(2) From equations (1) and (2), 2 2 2 1 V 2 V = ?V2 = 2 V1 ? V2 = 2 V Given, (V1 = V) Ans.9 - (C)(1 mark) ?? 2 ?? - ?? 1 ?? = ?? 2 - ?? 1 ?? 1 3 [1 3 2] 2[ 6] 6 v - -= -- 1 3 1 2 1 12 12 12 6 v - - - = + = = ?? = –6 cm Ans.10 - (B)Diffraction (1 mark) Ans.11- (A)doping level (1 mark) Ans.12- (C)+0.4% (1 mark) Ans.13- (A) (1 mark) Ans.14- (A)(1 mark) Ans.15- (D)(1 mark) Ans.16- (A)(1 mark) [SECTION – B] Ans.17– Given Ø ?? = ?? .???????? = ?? .???? × ?? .?? × ???? -???? ?? ?? = ?? .?? × ???? ???? ???? ?? .?? .= ???? - Ø ?? = ???? ?? ½ ?? = ???? ???? -Ø ?? ½ = ?? .???? × ???? -???? ×?? ×???? ?? ?? .???? × ???? -???? ×?? .?? ×???? ???? - ?? .???? ×?? .?? ×???? -???? ½ = ???? .???? × ???? -???? ?? .?? × ???? -???? (?? .???? - ?? .???? ) = ???? .???? × ???? -???? ?? .?? ×???? ???? = ???? .?? × ???? -?? ?? ½ Ans.18 - ?? 1 = 4 × 10 -7 ?? ?? 2 = 6× 10 -7 ?? Distance at which dark fringe is observed ?? = (?? + 1 2 ) ???? ?? ½ First Dark fringe for ?? 1 ?? 1 = 1 2 4×10 -7 10 -2 ?? = 2 × 10 -5 ?? ½ First Dark fringe for ?? 2 ?? 2 = 1 2 6×10 -7 10 -2 ?? = 3 × 10 -5 ?? First dark fringe will be the distance where both dark fringes will coincide i.e LCM of ?? 1 &?? 1 ½ i.e. 2 × 10 -5 ?? × 3 × 10 -5 ?? = 6 × 10 -5 ?? ½ OR (II) ?????? ?? = ?? 1 + ?? 2 + 2v?? 1 v?? 2 cos? 0.5 M Since, ?? 1 = ?? 2 = ?? Net I = I + I + 2 I cos? = 2I (1 + cos?) = 2I (2 cos 2 Ø 2 ) 0.5 M For path difference ?/4 , phase difference is p/2 0.5 M Net I = 4 I cos 2 ?? 4 Net I = 2 I 0.5 M (2 Marks) Ans.19 - (I)The direction of the magnetic field is perpendicular and inward into the plane of thepaper 0.5M (II) For a head-on collision to take place, the radius of the path of each ion should be equal to 0.5 m. 0.5 mv r qB == m 0.5M 26 5 19 4 10 2.4 10 B 4.8 10 0.5 mv qr - - ? ? ? == ?? 0.5M B = 0.04 T 0.5M For VI Candidate (a) As Pitch (p)= 2????? ?????? O ???? 0.5M Or, p= 2 ?? 3.14 ?? 1.7?? 10 -27 ?? 2 ?? 10 5 ?????? 30 0 1.6 ?? 10 -19 ?? 1.5 m Or, P=7.7X10 -3 m 0.5M (b)As, done by magnetic field is always zero K.E=1/2mv 2 0.5M KE=3.4 X 10 -17 J 0.5M Ans.20 –(i) Nuclear fission –W 0.5M Reason: As W has binding energy per nucleon less then Y and X and nucleus is larger in size. 0.5M (ii) Nuclear fusion-Z 0.5M Reason: As Z has binding energy per nucleon more then Y and X and nucleus is smaller in size. 0.5M Ans. 21 - (I) ?????????? ???????????????? ? ?? ???????????????????? ???????? 1 M (II) Alternating current changes direction every half cycle. 0.5 M So average drift velocity is zero 0.5 M [SECTION – C] (3 Marks) Ans.22 -(I) X = Full wave rectifier½ Y = Filter½ (Output Waveform for X) ½ (Output Waveform for Y) ½ (ii) 1Read More
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