Question

Gravitational potential in a region is given by v=(3x+4y+12z)J/kg . the modulus of gravitational field at (x=1, y=0 ,z=3 ) is_
(A) 20 N/kg (B) 13 N/kg
(C) 12 N/kg (D) 5 N/kg

Answer 1

Answer:

(B) 13 N/kg

Explanation:

1) We have,

Gravitational Potential in a region , $V(x,y,z)=(3x+4y+12z) J/kg$

Gravitational Field,  

$F= -\nabla V=-[\frac{\partial V}{\partial x}i+\frac{\partial V}{\partial y}j+\frac{\partial V}{\partial x}k] \\ \\ =-[3i+4j+12k]$

Hence, Gravitational field is constant.

It is same at all points.

Hence, Magnitude of Gravitational Field

$|F|=\sqrt{3^2+4^2+12^2}=\sqrt{9+16+144}\\ \\=\sqrt{169}=13 N/kg$

Answer 2

SOLUTION.

V = 3x + 4y + 12z

In such question we need some differential knowledge and I prefer you to break such question into axis like.....

V(X) = 3x....(1)

V(Y) = 4y.....(2)

V(Z) = 12z.....(3)

We know that dv/dx = - E

Differentiate (1) (2) or (3) with respect to their corresponding variables.

E(X) = -3

E(y) = -4

E(z) = -12

In vector form we can write this notation as...

E = -(3i+4j+12k)

where i,j,k are the unit vector along X,y,z axis.

it's magnitude can be written as.

|E| = √3²+4²+12²

|E| = √169

|E| = 13

#answerwithquality

#BAL