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#force calculation

29 public questions tagged with this topic.

Surface Tension, Capillarity and Excess Pressure

What is the force on a submarine window (0.06m2) at 400m depth in seawater (ρ\=1.03×103kg/m3), interior at atmospheric p

Gauge pressure: Pg = ρgh = 1.03×103×9.8×400 = 4.0376×106Pa. F = PgA = 4.0376×106×0.06 = 2.42256×105N. As per NCERT, applying relevant law/formula with correct units and sign convention leads to 2.4 × 10⁵ N. This satisfies dimensional consistency and physical conditions given, so option B is scientifically correct.

Ref: NCBI Bookshelf, Ideal Gas Law: P1V1/T1 = P2V2/T2.

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Surface Tension, Capillarity and Excess Pressure

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.

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Viscosity, Stokes' Law and Reynolds Number

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.

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Viscosity, Stokes' Law and Reynolds Number

A hydraulic press applies 350N on a small piston of area 0.025m2. What force is exerted by a large piston of area 0.1m2?

Pascal’s law: P = F1A1 = F2A2. F2 = F1×A2A1. F1 = 350N, A1 = 0.025m2, A2 = 0.1m2. F2 = 350×0.10.025 = 350×4 = 1400N. As per NCERT, applying relevant law/formula with correct units and sign convention leads to 1400 N. This satisfies dimensional consistency and physical conditions given, so option C is scientifically correct.

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

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Viscosity, Stokes' Law and Reynolds Number

A hydraulic press applies 250N on a small piston of area 0.02m2. What force is exerted by a large piston of area 0.08m2?

Pascal’s law: P = F1A1 = F2A2. F2 = F1×A2A1. F1 = 250N, A1 = 0.02m2, A2 = 0.08m2. F2 = 250×0.080.02 = 250×4 = 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 C is scientifically correct.

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

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Equation of Continuity and Flow Rate

A plate of area 0.05m2 moves at 0.7m/s over a 0.8mm thick water film (η\=1.0×10−3Pa s). What force is required?

Viscous force: F = ηvAl. η = 1.0×10−3Pa s, v = 0.7m/s, A = 0.05m2, l = 0.8×10−3m. F = 1.0×10−3×0.7×0.050.8×10−3 = 1.0×10−3×43.75 = 0.04375N. As per NCERT, applying relevant law/formula with correct units and sign convention leads to 0.044 N. This satisfies dimensional consistency and physical conditions given, so option C is scientifically correct.

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

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Fluid Pressure, Pascal's Law and Manometers

A plate of area 0.07m2 moves at 0.6m/s over a 0.9mm thick machine oil film (η\=0.113Pa s). What force is required?

Viscous force: F = ηvAl. η = 0.113Pa s, v = 0.6m/s, A = 0.07m2, l = 0.9×10−3m. F = 0.113×0.6×0.070.9×10−3 = 0.113×46.67 = 5.27N. As per NCERT, applying relevant law/formula with correct units and sign convention leads to 5.3 N. This satisfies dimensional consistency and physical conditions given, so option C is scientifically correct.

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

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Surface Tension, Capillarity and Excess Pressure

A plate of area 0.09m2 moves at 0.5m/s over a 1.0mm thick water film (η\=1.0×10−3Pa s). What force is required?

Viscous force: F = ηvAl. η = 1.0×10−3Pa s, v = 0.5m/s, A = 0.09m2, l = 1.0×10−3m. F = 1.0×10−3×0.5×0.091.0×10−3 = 1.0×10−3×45 = 0.045N. As per NCERT, applying relevant law/formula with correct units and sign convention leads to 0.045 N. This satisfies dimensional consistency and physical conditions given, so option B is scientifically correct.

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

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Viscosity, Stokes' Law and Reynolds Number

A plate of area 0.08m2 moves at 0.4m/s over a 0.25mm thick oil film (η\=0.034Pa s). What force is required?

Viscous force: F = ηvAl. η = 0.034Pa s, v = 0.4m/s, A = 0.08m2, l = 0.25×10−3m. F = 0.034×0.4×0.080.25×10−3 = 0.034×128 = 4.352N. As per NCERT, applying relevant law/formula with correct units and sign convention leads to 4.35 N. This satisfies dimensional consistency and physical conditions given, so option C is scientifically correct.

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

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Equation of Continuity and Flow Rate

A hydraulic system applies 300N on a small piston of area 0.015m2. What force is exerted by a large piston of area 0.045

Pascal’s law: P = F1A1 = F2A2. F2 = F1×A2A1. F1 = 300N, A1 = 0.015m2, A2 = 0.045m2. F2 = 300×0.0450.015 = 300×3 = 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.

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