Home
Class 12
PHYSICS
For an infinitely long solid cylinder of...

For an infinitely long solid cylinder of radius R having a uniform density , the gravitational field at a distance `r_(1)` is `E_(1)` and at a distance `r_(2)` is `E_(2)` , then

A

`E_(1) lt E_(2)` for `r_(1) lt r_(2) lt R `

B

`E_(1) gt E_(2)` for `R lt r_(1) lt r_(2)`

C

both 1 and 2

D

none of the above

Text Solution

AI Generated Solution

The correct Answer is:
To solve the problem regarding the gravitational field due to an infinitely long solid cylinder, we need to analyze the gravitational field at different distances from the center of the cylinder. ### Step-by-step Solution: 1. **Understanding the Gravitational Field**: The gravitational field \( E \) due to an infinitely long solid cylinder varies depending on the distance from the axis of the cylinder. The gravitational field inside the cylinder (for \( r < R \)) and outside the cylinder (for \( r > R \)) behaves differently. 2. **Gravitational Field Inside the Cylinder**: For a point inside the cylinder (where \( r < R \)), the gravitational field \( E \) is given by: \[ E = \frac{G M(r)}{r^2} \] where \( M(r) \) is the mass enclosed within radius \( r \). Since the mass is uniformly distributed, \( M(r) \) can be calculated based on the volume of the cylinder up to radius \( r \). 3. **Gravitational Field Outside the Cylinder**: For a point outside the cylinder (where \( r > R \)), the gravitational field \( E \) behaves as if all the mass of the cylinder were concentrated at its axis. Thus, it is given by: \[ E = \frac{G M}{r^2} \] where \( M \) is the total mass of the cylinder. 4. **Comparing \( E_1 \) and \( E_2 \)**: - **Case A**: If \( r_1 < r_2 < R \) (both points are inside the cylinder): - Since \( E \) increases as we move closer to the surface of the cylinder, we have \( E_1 < E_2 \). Thus, this statement is correct. - **Case B**: If \( r_1 > R \) and \( r_1 < r_2 \) (both points are outside the cylinder): - Here, \( E \) decreases as we move away from the cylinder. Therefore, \( E_1 > E_2 \) for \( r_1 < r_2 \). This statement is also correct. 5. **Conclusion**: Both statements regarding the gravitational fields \( E_1 \) and \( E_2 \) are correct. Therefore, the final answer is that both options A and B are correct. ### Final Answer: Both options A and B are correct.
Promotional Banner

Similar Questions

Explore conceptually related problems

An infinitely long solid cylinder of radius R has a uniform volume charge density rho . It has a spherical cavity of radius R//2 with its centre on the axis of cylinder, as shown in the figure. The magnitude of the electric field at the point P , which is at a distance 2 R form the axis of the cylinder, is given by the expression ( 23 r R)/( 16 k e_0) . The value of k is . .

Inside a uniformly charged infinitely long cylinder of radius 'R' and volume charge density 'rho' there is a spherical cavity of radius 'R//2'. A point 'P' is located at a dsintance 2 R from the axis of the cylinder as shown. Then the electric field strength at the point 'P' is

An infinitely long cylinderical wire of radius R is carrying a current with current density j=alphar^(3) (where alpha is constant and r is the distance from the axis of the wire). If the magnetic fixed at r=(R)/(2) is B_(1) and at r=2R is B_(2) then the ratio (B_(2))/(B_(1)) is

A non-conducting solid sphere of radius R, has a uniform charge density. The graph representing variation of magnitude of electric field (E) as a function of distance (x) from the centre of the sphere is

An infinite cylinder of radius r_(o), carrying linear charge density lamda. The equation of the equipotential surface for the cylinder is

The gravitational potential at a point outside the solid sphere of radius "R" and at a distance of "r" will be :

There is a solid sphere of radius R having uniformly distributed charge throughout it. What is the relation between electric field E and distance r from the centre (r is less than R) ?

A solid non-conduction cylinder of radius R is charge such that volume charge density is proporation to r where r is distance from axis. The electric field E at a distance r(r lt R) well depend on r as.

A uniform solid of valume mass density rho and radius R is shown in figure. (a) Find the gravitational field at a point P inside the sphere at a distance r from the centre of the sphere. Represent the gravitational field vector vec(l) in terms of radius vector vec(r ) of point P. (b) Now a spherical cavity is made inside the solid sphere in such a way that the point P comes inside the cavity. The centre is at a distance a from the centre of solid sphere and point P is a distance of b from the centre of the cavity. Find the gravitational field vec(E ) at point P in vector formulationand interpret the result.

A current l is flowing through a long straight solid conductor of radius R = 4 cm. Magnetic field at a distance 2 cm from the axis of conductor is B_0 The distance from the axis, where magnetic field is 1.5 B_0 is (current density is uniform)