Answer:
The pressure is 
Explanation:
From the question we are told that
The gauge pressure at the mouth is 
The radius of the column is 
The speed of the liquid outside the body is 
The area of the column is 
The area inside the mouth 
Generally according to continuity equation

=> 
=> 
=> 
So

=> 
=> 
substituting values


Now the height of inside the mouth is 
Now the height of the column is 
Generally according to Bernoulli's equation
![p_1 = [\frac{1}{2} \rho v_2^2 + h_2 \rho g +p_2] -[\frac{1}{2} \rho * v_1^2 + h_1 \rho g ]](https://tex.z-dn.net/?f=p_1%20%3D%20%20%5B%5Cfrac%7B1%7D%7B2%7D%20%20%5Crho%20v_2%5E2%20%2B%20h_2%20%5Crho%20g%20%2Bp_2%5D%20-%5B%5Cfrac%7B1%7D%7B2%7D%20%5Crho%20%2A%20v_1%5E2%20%2B%20h_1%20%5Crho%20g%20%5D)
Now
which is the density of water
is the gauge pressure of the atmosphere which is zero
So
![p_1 = [(0.5 * 1000 * (3.1)^2) +(0.008 * 1000 * 9.8) + 0]-](https://tex.z-dn.net/?f=p_1%20%3D%20%20%5B%280.5%20%2A%201000%20%2A%20%283.1%29%5E2%29%20%2B%280.008%20%2A%201000%20%2A%209.8%29%20%2B%200%5D-)

Answer:
1. E x 4πr² = ( Q x r³) / ( R³ x ε₀ )
Explanation:
According to the problem, Q is the charge on the non conducting sphere of radius R. Let ρ be the volume charge density of the non conducting sphere.
As shown in the figure, let r be the radius of the sphere inside the bigger non conducting sphere. Hence, the charge on the sphere of radius r is :
Q₁ = ∫ ρ dV
Here dV is the volume element of sphere of radius r.
Q₁ = ρ x 4π x ∫ r² dr
The limit of integration is from 0 to r as r is less than R.
Q₁ = (4π x ρ x r³ )/3
But volume charge density, ρ = 
So, 
Applying Gauss law of electrostatics ;
∫ E ds = Q₁/ε₀
Here E is electric field inside the sphere and ds is surface element of sphere of radius r.
Substitute the value of Q₁ in the above equation. Hence,
E x 4πr² = ( Q x r³) / ( R³ x ε₀ )
Answer:
B) Yes, but only those electrons with energy greater than the potential difference established between the grid and the collector will reach the collector.
Explanation:
In the case when the collector would held at a negative voltage i.e. small with regard to grid So yes the accelerated electrons would be reach to the collecting plate as the kinetic energy would be more than the potential energy that because of negative potential
so according to the given situation, the option b is correct
And, the rest of the options are wrong
What do you mean? Is there a picture ?