A horizontal fire hose with a nozzle of cross-sectional area (5/√21)×10-3 m2 delivers a cubic metre of water in 10s. What will be the maximum possible increase in the temperature of water while it hits a rigid wall (neglecting the effect of gravity)?

Question

A horizontal fire hose with a nozzle of cross-sectional area (5/√21)×10-3 m2 delivers a cubic metre of water in 10s. What will be the maximum possible increase in the temperature of water while it hits a rigid wall (neglecting the effect of gravity)? (A) 1°C (B) 0.1°C (C) 10°C (D) 0.01°C

Tardigrade Admin · Accepted Answer

A=(5/√21)×10-3m2 (dv/dt)=Av =(1/10), v=(1/10A),(1/2)mv2 =msΔθ Δθ =(v2/2s)

Q. A horizontal fire hose with a nozzle of cross-sectional area $\frac{5}{21} \times 1 0^{- 3} m^{2}$ delivers a cubic metre of water in 10s. What will be the maximum possible increase in the temperature of water while it hits a rigid wall (neglecting the effect of gravity)?

1°C

0.1°C

10°C

0.01°C

$A = \frac{5}{21} \times 1 0^{- 3} m^{2}$
$\frac{d v}{d t} = A v = \frac{1}{10}, v = \frac{1}{10 A}, \frac{1}{2} m v^{2} = m s Δ θ$
$Δ θ = \frac{v ^{2}}{2 s}$

Q. A horizontal fire hose with a nozzle of cross-sectional area 21​5​×10−3m2 delivers a cubic metre of water in 10s. What will be the maximum possible increase in the temperature of water while it hits a rigid wall (neglecting the effect of gravity)?

1°C

0.1°C

10°C

0.01°C

A=21​5​×10−3m2 dtdv​=Av=101​,v=10A1​,21​mv2=msΔθ Δθ=2sv2​

Q. A horizontal fire hose with a nozzle of cross-sectional area $\frac{5}{21} \times 1 0^{- 3} m^{2}$ delivers a cubic metre of water in 10s. What will be the maximum possible increase in the temperature of water while it hits a rigid wall (neglecting the effect of gravity)?

$A = \frac{5}{21} \times 1 0^{- 3} m^{2}$
$\frac{d v}{d t} = A v = \frac{1}{10}, v = \frac{1}{10 A}, \frac{1}{2} m v^{2} = m s Δ θ$
$Δ θ = \frac{v ^{2}}{2 s}$