The two femurs each of cross-sectional area 10 cm2 support the upper part of a human body of mass 40 kg. The average pressure sustained by the femurs is (Take g = 10 m s-2)

Question

The two femurs each of cross-sectional area 10 cm2 support the upper part of a human body of mass 40 kg. The average pressure sustained by the femurs is (Take g = 10 m s-2) (A) 2 × 103 N m-2 (B) 2 × 104 N m-2 (C) 2 × 105 N m-2 (D) 2 × 106 N m-2

Tardigrade Admin · Accepted Answer

Total cross-sectional area of the femurs is,

A = 2 \times 10 c m^{2} = 2 \times 10 \times 1 0^{- 4} m^{2} = 20 \times 1 0^{- 4} m^{2}

Force acting on them is

F = m g = 40 k g \times 10 m s^{- 2} = 400 N

∴

Average pressure sustained by them is

P = \frac{F}{A} = \frac{400 N}{20 \times 1 0 ^{- 4} m ^{2}} = 2 \times 1 0^{5} N m^{- 2}

Q. The two femurs each of cross-sectional area $10 c m^{2}$ support the upper part of a human body of mass $40 k g$ . The average pressure sustained by the femurs is $(T ak e g = 10 m s^{- 2})$

$2 \times 1 0^{3} N m^{- 2}$

$2 \times 1 0^{4} N m^{- 2}$

$2 \times 1 0^{5} N m^{- 2}$

$2 \times 1 0^{6} N m^{- 2}$

Q. The two femurs each of cross-sectional area 10cm2 support the upper part of a human body of mass 40kg. The average pressure sustained by the femurs is (Takeg=10ms−2)

2×103Nm−2

2×104Nm−2

2×105Nm−2

2×106Nm−2

Total cross-sectional area of the femurs is, A=2×10cm2=2×10×10−4m2=20×10−4m2 Force acting on them is F=mg=40kg×10ms−2=400N ∴ Average pressure sustained by them is P=AF​=20×10−4m2400N​=2×105Nm−2

Q. The two femurs each of cross-sectional area $10 c m^{2}$ support the upper part of a human body of mass $40 k g$ . The average pressure sustained by the femurs is $(T ak e g = 10 m s^{- 2})$

$2 \times 1 0^{3} N m^{- 2}$

$2 \times 1 0^{4} N m^{- 2}$

$2 \times 1 0^{5} N m^{- 2}$

$2 \times 1 0^{6} N m^{- 2}$