A gardener water the plants by a pipe of radius 3.5 cm. The water comes out at rate of 10 cm³/s. The reactionary force exerted on the hand of gardener is
1. 22/7 x 10-³ N
2. 3.5 × 10-3 N
3. 22/7 x 10-6 N
4. 3.5 × 10-4 N
Answers
Answer:
A gardener water the plants by a pipe of radius 3.5 cm. The water comes out at rate of 10 cm³/s. The reactionary force exerted on the hand of gardener is
Explanation:
22/7 x 10-³ N
Answer:
The reactionary force exerted on the hands of the Gardner is 2.6×10⁻⁵ N
Explanation:
The radius (r) of the pipe = 3.5 cm = 3.5×10⁻² m
The rate of water coming out of the pipe in terms of volume (v) and time (t),
dv/dt = 10 cm³/s = 10×10⁻⁶ m³/s
The density (D) of water = 10³ kg/m³
Since density, D = m/v
m = D×v
The rate of change will be: dm/dt = D (dv/dt)
= 10³ ˣ 10 ˣ 10⁻⁶
dm/dt = 10⁻² kg/s → (1)
'A' is the area of cross-section of the pipe.
The velocity (V) of the water flowing out,
V = (dv/dt) × (1/A)
V = (10 ˣ 10⁻⁶) × (1÷((3.5×10⁻²)²))
On solving, V = 0.002598 m/s → (2)
From Newton's 2nd law of motion, F = dP/dt
Where P is the momentum.
∴ Reactionary force, F = d(mV)/dt
= V × (dm/dt)
From equations (1) & (2)
F = 0.002598 × 10⁻²
F = 0.00002598
∴ F = 2.6×10⁻⁵ N