Explanation
Hint: Use Torricelli's equation for velocity
$$\textbf{Step1: Torricelli's equation for velocity}$$
Torricelli's theorem states that speed v by which water comes out from the tank is proportional to the square toot of twice the acceleration created by gravity and vertical height between surface and center of the opening.
$$V = \sqrt{2\ *\ g\ *\ h}$$
$$\textbf{Step2: Calculation of velocity of water coming out}$$
On putting the values, g = 10 $$m/s^{2}$$ and h = 20 cm = 0.2 m.
$$V = \sqrt{2\ *\ 10\ *\ 0.2}$$
= 2 m/s
Answer:
Hence, option B (2 m/s) is the correct answer.
Hint:
Use Torricelli’s equation for velocity and projectile motion’s horizontal range formula
Torricelli's theorem stats that speed v by which water come out from tank is proportional to square toot of twice the acceleration created by gravity and vertical height between surface and center of opening.
Hence, option B is the correct answer.
The manometer here gives the stagnation pressure i.e. the velocity of liquid is zero at point B.$$ \\ $$Hence the pressure drop is given by $$ \\ \Delta P=\rho gh={ P}_{ B }-{ P }_{ A }\\ \rho =0.8{ gm }/{ { cm }^{ 3 } }=800kg/{ m }^{ 3 } ,h=0.1m , { \rho }_{ air }=1.3\times { 10 }^{ 3 }{ gm }/{ { cm }^{ 3 }}=1.3{ kg }/{ { m }^{ 3 } }\\ \Delta P=800\times 9.81\times 0.1\\ \DeltaP=784.8 N/{ m }^{ 2 }\\ $$
From Bernoulli's equation for stagnation pressure
$$\\ { P}_{ A }+\dfrac { { \rho }_{ air }{ { V } }^{ 2 } }{ 2 } ={ P }_{ B }\\V=\sqrt { \dfrac { 2\Delta P }{ { \rho }_{ air } } } \\ V=\sqrt {\dfrac { 2\times 784.8 }{ 1.3 } } \\ V=34.74{ m }/{ s }\\ V\cong 3470{ cm}/{ s } $$
Applying Bernoulli's principle on a horizontal pipe, we have $$P_{ 1 }+\dfrac { 1 }{ 2 } \rho { v_{ 1 } }^{ 2 }=P_{ 2 }+\dfrac { 1 }{ 2 } \rho { v_{ 2 } }^{ 2 }\\ \Rightarrow \dfrac { 1 }{ 2 } \rho { v_{ 2 } }^{ 2 }-\dfrac { 1 }{ 2 } \rho { v_{ 1 } }^{ 2 }=P_{ 1 }-P_{ 2 }\\ \Rightarrow \dfrac { 1 }{ 2 } { v_{ 2 } }^{ 2 }-\dfrac { 1 }{ 2 } { v_{ 1 } }^{ 2 }=\dfrac { P_{ 1 }-P_{ 2 } }{ \rho } \\ \Rightarrow \text{Change in kinetic energy per unit mass } (\Delta K.E.)=\dfrac { P_{ 1 }-P_{ 2 } }{ \rho } \\\Rightarrow \Delta K.E.=\dfrac{8}{800}=10^{-2}\mathrm{J/Kg}$$
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