the pressure at the base of tank is 2×10^5 n/ m^2 . given atmospheric pressure is 1×10^5 n/m^2 and g= 10 m/s . a block of wood having specific gravity 0.8 is released from the base of beaker . find the total vertical distance covered by the block of wood before it comes momentarily to stop.
Answers
Answer:
h = 12.5x10^5 m
Explanation:
The general formula for pressure is;
p = ρgh
Here p is the pressure, g is the gravity, ρ is the specific gravity and h is the height.
Now we are given two pressures and net pressure inside the system would be;
Net pressure = 2×10^5 - 1×10^5 = 1×10^5
Putting the other values in formula;
1×10^5 = 0.8 x 10 x h
h = 12.5x10^5 m
The total vertical distance covered by the block of wood before it comes momentarily to stop is 12755.1 meters.
Given:
Pressure at base = 2 × 10⁵ N/m² = P₁
Pressure at surface = 1 × 10⁵ N/m² = P₂
Acceleration due to gravity = 10 m/s
Specific gravity = 0.8
Explanation:
The pressure is given by the formula:
P = ρgh
Where,
P = P₁ - P₂ = 1 × 10⁵ N/m²
ρ = Density (Here, the specific density is ratio of density of a substance to density of a given reference material)
g = Acceleration due to gravity
h = Height (vertical distance)
On substituting the values, we get,
1 × 10⁵ = 0.8 × 9.8 × h
h = (1 × 10⁵)/(0.8 × 9.8)
h = (1 × 10⁵)/7.84
∴ h = 12755.1 m