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#Pascal's law

18 public questions tagged with this topic.

A hydraulic press uses a small piston of area 0.008m2 to exert 120N. What force is produced by a large piston of area 0.

Pascal’s law: P = F1A1 = F2A2. F2 = F1×A2A1. F1 = 120N, A1 = 0.008m2, A2 = 0.032m2. F2 = 120×0.0320.008 = 120×4 = 480N. As per NCERT, applying relevant law/formula with correct units and sign convention leads to 480 N. This satisfies dimensional consistency and physical conditions given, so option C is scientifically correct.

Ref: NCERT Class 11 Physics, Thermal Properties and Gravitation.

A 1500kg car is lifted by a hydraulic system with a small piston of radius 6cm and a large piston of radius 18cm. What f

F1 = A1A2F2, F2 = 1500×9.8 = 14700N. A1 = π(0.06)2, A2 = π(0.18)2, A1A2 = 0.00360.0324 = 19. F1 = 147009 = 1633.33N≈1633N. As per NCERT, applying relevant law/formula with correct units and sign convention leads to 1633 N. This satisfies dimensional consistency and physical conditions given, so option C is scientifically correct.

Ref: NCERT Class 11 Physics, Thermal Properties and Gravitation.

A hydraulic press applies 50N on a piston of area 0.02m2. What force is exerted by a piston of area 0.1m2?

Pressure transmitted: P = F1A1 = F2A2. F2 = F1×A2A1. F1 = 50N, A1 = 0.02m2, A2 = 0.1m2. F2 = 50×0.10.02 = 50×5 = 250N. As per NCERT, applying relevant law/formula with correct units and sign convention leads to 250 N. This satisfies dimensional consistency and physical conditions given, so option C is scientifically correct.

Ref: NCERT Class 11 Physics, Thermal Properties and Gravitation.

A hydraulic lift raises a 1600kg load using a small piston of radius 4cm and a large piston of radius 16cm. What force i

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.

A hydraulic system has a small piston (0.005m2) and a large piston (0.025m2). If 80N is applied on the small piston, wha

Pascal’s law: P = F1A1 = F2A2. F2 = F1×A2A1. F1 = 80N, A1 = 0.005m2, A2 = 0.025m2. F2 = 80×0.0250.005 = 80×5 = 400N. As per NCERT, applying relevant law/formula with correct units and sign convention leads to 400 N. This satisfies dimensional consistency and physical conditions given, so option C is scientifically correct.

Ref: NCERT Class 11 Physics, Thermal Properties and Gravitation.

A hydraulic lift raises a 2300kg load using a small piston of radius 5.0cm and a large piston of radius 20cm. What force

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.

A car lift uses a small piston of radius 4cm to lift a 1200kg car on a piston of radius 12cm. What force is applied on t

F1 = A1A2F2, F2 = mg = 1200×10 = 12000N. A1 = π(0.04)2, A2 = π(0.12)2. A1A2 = (0.04)2(0.12)2 = 0.00160.0144 = 19. F1 = 120009 = 1333.33N≈1333N. As per NCERT, applying relevant law/formula with correct units and sign convention leads to 1333 N. This satisfies dimensional consistency and physical conditions given, so option B is scientifically correct.

Ref: NCERT Class 11 Physics, Thermal Properties and Gravitation.

A hydraulic system applies 180N on a small piston of area 0.012m2. What force is exerted by a large piston of area 0.06m

Pascal’s law: P = F1A1 = F2A2. F2 = F1×A2A1. F1 = 180N, A1 = 0.012m2, A2 = 0.06m2. F2 = 180×0.060.012 = 180×5 = 900N. As per NCERT, applying relevant law/formula with correct units and sign convention leads to 900 N. This satisfies dimensional consistency and physical conditions given, so option C is scientifically correct.

Ref: NCERT Class 11 Physics, Thermal Properties and Gravitation.

A hydraulic lift raises a 2000kg load using a small piston of radius 5cm and a large piston of radius 20cm. What force i

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.