JEE MAIN - Physics (2020 - 9th January Morning Slot - No. 6)
Water flows in a horizontal tube (see figure).
The pressure of water changes by 700 Nm–2
between A and B where the area of cross section
are 40 cm2 and 20 cm2, respectively. Find the
rate of flow of water through the tube.
(density of water = 1000 kgm–3)_9th_January_Morning_Slot_en_6_1.png)
The pressure of water changes by 700 Nm–2
between A and B where the area of cross section
are 40 cm2 and 20 cm2, respectively. Find the
rate of flow of water through the tube.
(density of water = 1000 kgm–3)
1810 cm3/s
2420 cm3/s
3020 cm3/s
2720 cm3/s
Explanation
Rate of flow of water is same at both A and B
AAVA = ABVB
(40)VA = (20)VB
VB = 2VA .......(1)
Using Bernoulli's theorem
PA + $${1 \over 2}\rho V_A^2$$ = PB + $${1 \over 2}\rho V_B^2$$
$$ \Rightarrow $$ PA - PB = $${1 \over 2}\rho V_B^2$$ - $${1 \over 2}\rho V_A^2$$
$$ \Rightarrow $$ 700 = $${1 \over 2} \times 1000\left( {4V_A^2 - V_A^2} \right)$$
$$ \Rightarrow $$ VA = 0.68 m/s = 68 cm/s
$$ \therefore $$ Flow rate = AAVA = (40)(68) = 2720 cm3/s
AAVA = ABVB
(40)VA = (20)VB
VB = 2VA .......(1)
Using Bernoulli's theorem
PA + $${1 \over 2}\rho V_A^2$$ = PB + $${1 \over 2}\rho V_B^2$$
$$ \Rightarrow $$ PA - PB = $${1 \over 2}\rho V_B^2$$ - $${1 \over 2}\rho V_A^2$$
$$ \Rightarrow $$ 700 = $${1 \over 2} \times 1000\left( {4V_A^2 - V_A^2} \right)$$
$$ \Rightarrow $$ VA = 0.68 m/s = 68 cm/s
$$ \therefore $$ Flow rate = AAVA = (40)(68) = 2720 cm3/s
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