if the density of earth increase by 20% and the radius decrease by 20% and then the new value of the 'g' on the earth surface will be
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g = GM / R²
= G × (4/3 × π × R³ × d) / R²
= 4πGdR/3
From above
g ∝ dR
g₁/g₂ = (d₁/d₂) × (R₁/R₂)
g₁/g₂ = (100 / 120) × (100 / 80)
g₁/g₂ = 25/24
g₂ = 24g₁/25
= 0.96 g₁
= 0.96 × 9.8 m/s²
= 9.408 m/s²
Value of g on surface of earth will be 9.408 m/s²
= G × (4/3 × π × R³ × d) / R²
= 4πGdR/3
From above
g ∝ dR
g₁/g₂ = (d₁/d₂) × (R₁/R₂)
g₁/g₂ = (100 / 120) × (100 / 80)
g₁/g₂ = 25/24
g₂ = 24g₁/25
= 0.96 g₁
= 0.96 × 9.8 m/s²
= 9.408 m/s²
Value of g on surface of earth will be 9.408 m/s²
sanidhyasingla:
the options are-----> A 0.8g B 0.90g c 0.96g d g
now please edit your answer
please edit your answer
Read my whole answer.
c) 0.96 g is correct answer
as 9.8m/s^2 is nothing but g therefore 0.96*9.8=0.96*g=0.96g
Answered by
11
0.96g....... ..........
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