11 public questions tagged with this topic.
Pascal’s law ensures pressure is transmitted equally. A smaller piston area (A1) with force F1 creates pressure (P = F1/A1), which, applied to a larger area (A2), results in a larger force (F2 = P⋅A2), amplifying the input force.
Ref: NCERT Class 11 Physics, Thermal Properties and Gravitation.
F1 = A1A2F2, F2 = 2100×9.8 = 20580N. A1 = π(0.03)2, A2 = π(0.12)2, A1A2 = 0.00090.0144 = 116. F1 = 2058016 = 1286.25N≈1286N. As per NCERT, applying relevant law/formula with correct units and sign convention leads to 1286 N. This satisfies dimensional consistency and physical conditions given, so option C is scientifically correct.
Ref: NCERT Class 11 Physics, Thermal Properties and Gravitation.
F1 = A1A2F2, F2 = 1600×10 = 16000N. A1 = π(0.04)2, A2 = π(0.16)2, A1A2 = 0.00160.0256 = 116. F1 = 1600016 = 1000N. As per NCERT, applying relevant law/formula with correct units and sign convention leads to 1000 N. This satisfies dimensional consistency and physical conditions given, so option B is scientifically correct.
Ref: NCERT Class 11 Physics, Thermal Properties and Gravitation.
F1 = A1A2F2, F2 = 2300×9.8 = 22540N. A1 = π(0.05)2, A2 = π(0.2)2, A1A2 = 0.00250.04 = 116. F1 = 2254016 = 1408.75N≈1409N. As per NCERT, applying relevant law/formula with correct units and sign convention leads to 1409 N. This satisfies dimensional consistency and physical conditions given, so option D is scientifically correct.
Ref: NCERT Class 11 Physics, Thermal Properties and Gravitation.
F1 = A1A2F2, F2 = 1800×9.8 = 17640N. A1 = π(0.03)2, A2 = π(0.15)2, A1A2 = 0.00090.0225 = 125. F1 = 1764025 = 705.6N≈706N. As per NCERT, applying relevant law/formula with correct units and sign convention leads to 706 N. This satisfies dimensional consistency and physical conditions given, so option B is scientifically correct.
Ref: NCERT Class 11 Physics, Thermal Properties and Gravitation.
Pascal’s law: P = F1A1 = F2A2. F2 = F1×A2A1. F1 = 200N, A1 = 0.02m2, A2 = 0.1m2. F2 = 200×0.10.02 = 200×5 = 1000N. m = F2g = 100010 = 100kg.
Ref: NCERT Class 11 Physics, Thermal Properties and Gravitation.
By Pascal’s law, pressure is transmitted equally: P = F1A1 = F2A2. F2 = F1×A2A1. F1 = 100N, A1 = 0.01m2, A2 = 0.09m2. F2 = 100×0.090.01 = 100×9 = 900N.
Ref: NCERT Class 11 Physics, Thermal Properties and Gravitation.
F1 = A1A2F2, F2 = 2000×10 = 20000N. A1 = π(0.05)2, A2 = π(0.2)2, A1A2 = 0.00250.04 = 116. F1 = 2000016 = 1250N. As per NCERT, applying relevant law/formula with correct units and sign convention leads to 1250 N. This satisfies dimensional consistency and physical conditions given, so option B is scientifically correct.
Ref: NCERT Class 11 Physics, Thermal Properties and Gravitation.
F1 = A1A2F2, F2 = 1700×10 = 17000N. A1 = π(0.035)2, A2 = π(0.14)2, A1A2 = 0.0012250.0196 = 116. F1 = 1700016 = 1062.5N≈1063N. As per NCERT, applying relevant law/formula with correct units and sign convention leads to 1063 N. This satisfies dimensional consistency and physical conditions given, so option C is scientifically correct.
Ref: NCERT Class 11 Physics, Thermal Properties and Gravitation.
F1 = A1A2F2, F2 = 1400×9.8 = 13720N. A1 = π(0.025)2, A2 = π(0.1)2, A1A2 = 0.0006250.01 = 116. F1 = 1372016 = 857.5N≈858N. As per NCERT, applying relevant law/formula with correct units and sign convention leads to 858 N. This satisfies dimensional consistency and physical conditions given, so option B is scientifically correct.
Ref: NCERT Class 11 Physics, Thermal Properties and Gravitation.
F1 = A1A2F2, F2 = 1900×9.8 = 18620N. A1 = π(0.045)2, A2 = π(0.18)2, A1A2 = 0.0020260.0324 = 116. F1 = 1862016 = 1163.75N≈1164N. As per NCERT, applying relevant law/formula with correct units and sign convention leads to 1164 N. This satisfies dimensional consistency and physical conditions given, so option C is scientifically correct.
Ref: NCERT Class 11 Physics, Thermal Properties and Gravitation.
