Page 1 CBSE | Physics Sample Paper 3 Solution CBSE Board Class XI Physics Sample Paper-3 Solution 1. Yes. Example is angle 2. Motion of a body thrown vertically/obliquely under constant g 3. - x-axis 4. Work 5. Since ? ? ? rF , larger arm means larger r which requires less F for same ? . 6. 3 7. 8. No process is possible whose sole result is the absorption of heat from a reservoir and the conversion of all of this heat into work. Page 2 CBSE | Physics Sample Paper 3 Solution CBSE Board Class XI Physics Sample Paper-3 Solution 1. Yes. Example is angle 2. Motion of a body thrown vertically/obliquely under constant g 3. - x-axis 4. Work 5. Since ? ? ? rF , larger arm means larger r which requires less F for same ? . 6. 3 7. 8. No process is possible whose sole result is the absorption of heat from a reservoir and the conversion of all of this heat into work. CBSE | Physics Sample Paper 3 Solution 9. Systematic Errors Random Errors 1. Errors in which the deviation from true value tends to have fixed size and sign. 2. They can be attributed to a fixed cause and can be eliminated. Deviation from true value is irregular in size as well as sign. Irregular pattern does not allow them to be attributed to any fixed cause and hence cannot be eliminated, only minimized. 10. For 3 kg: F – F 32 = m 3a (i) For 2 kg: F 23 = m 2a (ii) But from Newton’s third law 23 32 FF ?? Therefore, from (i) and (ii), F – m 2a = m 3a F = (m 2 + m 3)a 2 5 a 1 m/s 5 ? ? ? Therefore, F 32 = m 2a = 2.1 = 2N (1/2) Page 3 CBSE | Physics Sample Paper 3 Solution CBSE Board Class XI Physics Sample Paper-3 Solution 1. Yes. Example is angle 2. Motion of a body thrown vertically/obliquely under constant g 3. - x-axis 4. Work 5. Since ? ? ? rF , larger arm means larger r which requires less F for same ? . 6. 3 7. 8. No process is possible whose sole result is the absorption of heat from a reservoir and the conversion of all of this heat into work. CBSE | Physics Sample Paper 3 Solution 9. Systematic Errors Random Errors 1. Errors in which the deviation from true value tends to have fixed size and sign. 2. They can be attributed to a fixed cause and can be eliminated. Deviation from true value is irregular in size as well as sign. Irregular pattern does not allow them to be attributed to any fixed cause and hence cannot be eliminated, only minimized. 10. For 3 kg: F – F 32 = m 3a (i) For 2 kg: F 23 = m 2a (ii) But from Newton’s third law 23 32 FF ?? Therefore, from (i) and (ii), F – m 2a = m 3a F = (m 2 + m 3)a 2 5 a 1 m/s 5 ? ? ? Therefore, F 32 = m 2a = 2.1 = 2N (1/2) CBSE | Physics Sample Paper 3 Solution OR The two ways are: 1. Friction adjusts its direction to be always opposite to applied force. 2. Friction adjusts its magnitude up to a certain limit, to be equal to the applied force. F ms = ? s N = ? ? ? ? ? s mg 0.2 2 10 4N Since, applied force < F ms, the static friction acting = f s = 2 N. 11. We know that p 2mk ? ?? ? ? ? ? ? ?? ?? 21 11 11 and p' 2mk' 2m k k 2mk p 100 10 10 Therefore, 11 pp p1 10 p p 10 ? ? ?? p 100 10% p ? ?? 12. ? ? ? rpsin mv Direction of ? rp ? direction of 13. Polar satellites - Their orbit is perpendicular to the orbit of geostationary satellites. These are used for communication purpose. Also, the height above the Earth’s surface is lower. Negative sign of total energy indicates attractive nature of force between the satellite and the Earth. Page 4 CBSE | Physics Sample Paper 3 Solution CBSE Board Class XI Physics Sample Paper-3 Solution 1. Yes. Example is angle 2. Motion of a body thrown vertically/obliquely under constant g 3. - x-axis 4. Work 5. Since ? ? ? rF , larger arm means larger r which requires less F for same ? . 6. 3 7. 8. No process is possible whose sole result is the absorption of heat from a reservoir and the conversion of all of this heat into work. CBSE | Physics Sample Paper 3 Solution 9. Systematic Errors Random Errors 1. Errors in which the deviation from true value tends to have fixed size and sign. 2. They can be attributed to a fixed cause and can be eliminated. Deviation from true value is irregular in size as well as sign. Irregular pattern does not allow them to be attributed to any fixed cause and hence cannot be eliminated, only minimized. 10. For 3 kg: F – F 32 = m 3a (i) For 2 kg: F 23 = m 2a (ii) But from Newton’s third law 23 32 FF ?? Therefore, from (i) and (ii), F – m 2a = m 3a F = (m 2 + m 3)a 2 5 a 1 m/s 5 ? ? ? Therefore, F 32 = m 2a = 2.1 = 2N (1/2) CBSE | Physics Sample Paper 3 Solution OR The two ways are: 1. Friction adjusts its direction to be always opposite to applied force. 2. Friction adjusts its magnitude up to a certain limit, to be equal to the applied force. F ms = ? s N = ? ? ? ? ? s mg 0.2 2 10 4N Since, applied force < F ms, the static friction acting = f s = 2 N. 11. We know that p 2mk ? ?? ? ? ? ? ? ?? ?? 21 11 11 and p' 2mk' 2m k k 2mk p 100 10 10 Therefore, 11 pp p1 10 p p 10 ? ? ?? p 100 10% p ? ?? 12. ? ? ? rpsin mv Direction of ? rp ? direction of 13. Polar satellites - Their orbit is perpendicular to the orbit of geostationary satellites. These are used for communication purpose. Also, the height above the Earth’s surface is lower. Negative sign of total energy indicates attractive nature of force between the satellite and the Earth. CBSE | Physics Sample Paper 3 Solution 14. The stress required to fracture a material whether by compression, tension, or shear is called breaking stress. Yes, the wire is under stress as its own weight acts as load. 15. For adiabatic expression PV ? = const. Therefore, PV ? = '' PV ? ? 5/3 5/3 5 5/3 1600 '(8 ) 2 ' V P V PV ?? Or ?? a 1600 P' 50P 32 Therefore, fall in pressure = 1600 – 500 = 1550 P a 16. (i) For isothermal expansion ?T = 0, hence ?U = 0 (ii) For adiabatic expansion ?U = ?Q – ?W = - ?W = -P ?V as ?Q = 0. 17. ? ? ? ? ? ? A 36 1000 36 km/h 10 m/s 60 60 AC 54 km/h 15 m/s ? ? ? ? ? BA B A 15 10 5 m/s ? ? ? ? ? ? ? ? CA C A 15 ( 10) 25 m/s ? ? ? ? ? ? ? ? ? Time taken by C to cover 1 km = ? 1000 40s 25 To avoid accident, B should cover 1 km is less than 40s. ?? 2 1 s ut at 2 2 1 1000 5 40 a.(40) 2 ? ? ? ? 200 800a ?? 800a = 1000 – 200 = 800 ? a = 1 m/s 2 Page 5 CBSE | Physics Sample Paper 3 Solution CBSE Board Class XI Physics Sample Paper-3 Solution 1. Yes. Example is angle 2. Motion of a body thrown vertically/obliquely under constant g 3. - x-axis 4. Work 5. Since ? ? ? rF , larger arm means larger r which requires less F for same ? . 6. 3 7. 8. No process is possible whose sole result is the absorption of heat from a reservoir and the conversion of all of this heat into work. CBSE | Physics Sample Paper 3 Solution 9. Systematic Errors Random Errors 1. Errors in which the deviation from true value tends to have fixed size and sign. 2. They can be attributed to a fixed cause and can be eliminated. Deviation from true value is irregular in size as well as sign. Irregular pattern does not allow them to be attributed to any fixed cause and hence cannot be eliminated, only minimized. 10. For 3 kg: F – F 32 = m 3a (i) For 2 kg: F 23 = m 2a (ii) But from Newton’s third law 23 32 FF ?? Therefore, from (i) and (ii), F – m 2a = m 3a F = (m 2 + m 3)a 2 5 a 1 m/s 5 ? ? ? Therefore, F 32 = m 2a = 2.1 = 2N (1/2) CBSE | Physics Sample Paper 3 Solution OR The two ways are: 1. Friction adjusts its direction to be always opposite to applied force. 2. Friction adjusts its magnitude up to a certain limit, to be equal to the applied force. F ms = ? s N = ? ? ? ? ? s mg 0.2 2 10 4N Since, applied force < F ms, the static friction acting = f s = 2 N. 11. We know that p 2mk ? ?? ? ? ? ? ? ?? ?? 21 11 11 and p' 2mk' 2m k k 2mk p 100 10 10 Therefore, 11 pp p1 10 p p 10 ? ? ?? p 100 10% p ? ?? 12. ? ? ? rpsin mv Direction of ? rp ? direction of 13. Polar satellites - Their orbit is perpendicular to the orbit of geostationary satellites. These are used for communication purpose. Also, the height above the Earth’s surface is lower. Negative sign of total energy indicates attractive nature of force between the satellite and the Earth. CBSE | Physics Sample Paper 3 Solution 14. The stress required to fracture a material whether by compression, tension, or shear is called breaking stress. Yes, the wire is under stress as its own weight acts as load. 15. For adiabatic expression PV ? = const. Therefore, PV ? = '' PV ? ? 5/3 5/3 5 5/3 1600 '(8 ) 2 ' V P V PV ?? Or ?? a 1600 P' 50P 32 Therefore, fall in pressure = 1600 – 500 = 1550 P a 16. (i) For isothermal expansion ?T = 0, hence ?U = 0 (ii) For adiabatic expansion ?U = ?Q – ?W = - ?W = -P ?V as ?Q = 0. 17. ? ? ? ? ? ? A 36 1000 36 km/h 10 m/s 60 60 AC 54 km/h 15 m/s ? ? ? ? ? BA B A 15 10 5 m/s ? ? ? ? ? ? ? ? CA C A 15 ( 10) 25 m/s ? ? ? ? ? ? ? ? ? Time taken by C to cover 1 km = ? 1000 40s 25 To avoid accident, B should cover 1 km is less than 40s. ?? 2 1 s ut at 2 2 1 1000 5 40 a.(40) 2 ? ? ? ? 200 800a ?? 800a = 1000 – 200 = 800 ? a = 1 m/s 2 CBSE | Physics Sample Paper 3 Solution 18. a = -kx dv a kx dx ? ? ? ? dv kx dx ? ? ? Integrating both sides, we get x uo dv kx dx ? ? ? ? ?? 2 2 2 11 ( u ) kx 22 ? ? ? ? or 2 2 2 11 m( u ) m kx 22 ? ? ? ? Therefore, loss in K.E. = 2 1 m kx 2 19. (a) During free fall acceleration of thief = g = acceleration of load So that load is unable to apply any force. Let the force by load be N. mg – N ? N = 0 = force applied by load on man (b) Along horizontal direction, ? ? ext F0 . Net linear momentum is conserved. Before firing, the system is at rest. Therefore, 0 = b b g g mm ? ? ? b gb g m m ? ? ? ? So, to conserve linear momentum, the gun recoils. (c) The sand yields but the cemented floor doesn’t. Hence, the time taken by man to come to rest increases in case of sand. Since, p F t ? ? ? , force on man is less. 20. We know that 12 KA(T T )t Q x ? ? A = area of 6 faces = 6 × (3 × 10 -1 ) 2 = 54 × 10 -3 m L = 12 KA(T T )t x ? or 12 KA(T T )t m xL ? ?Read More

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