Intensity of gravitational field due to a uniform rod of mass M and length L at a point P as shown in figure is

Question

Intensity of gravitational field due to a uniform rod of mass M and length L at a point P as shown in figure is (A)  (GM/2L2) (B)  (4GM/9L2) (C)  (GM/L2) (D)  (2GM/L2)

Tardigrade Admin · Accepted Answer

dl = (Gdm/x2) ∫ dl = G∫(M/L)(dx/x2) ⇒ I = (GM/L)∫2LL×-2dx=(GM/L)×(1/2L)=(GM/2L2)

Q. Intensity of gravitational field due to a uniform rod of mass M and length L at a point P as shown in figure is

$\frac{GM}{2 L ^{2}}$

$\frac{4 GM}{9 L ^{2}}$

$\frac{GM}{L ^{2}}$

$\frac{2 GM}{L ^{2}}$

$d l = \frac{G d m}{x ^{2}}$
$\int d l = G \int \frac{M}{L} \frac{d x}{x ^{2}}$
$\Rightarrow I = \frac{GM}{L} \int_{L}^{2 L} \times^{- 2 d x} = \frac{GM}{L} \times \frac{1}{2 L} = \frac{GM}{2 L ^{2}}$

Q. Intensity of gravitational field due to a uniform rod of mass M and length L at a point P as shown in figure is

2L2GM​

9L24GM​

L2GM​

L22GM​

dl=x2Gdm​ ∫dl=G∫LM​x2dx​ ⇒I=LGM​∫L2L​×−2dx=LGM​×2L1​=2L2GM​

$\frac{GM}{2 L ^{2}}$

$\frac{4 GM}{9 L ^{2}}$

$\frac{GM}{L ^{2}}$

$\frac{2 GM}{L ^{2}}$

$d l = \frac{G d m}{x ^{2}}$
$\int d l = G \int \frac{M}{L} \frac{d x}{x ^{2}}$
$\Rightarrow I = \frac{GM}{L} \int_{L}^{2 L} \times^{- 2 d x} = \frac{GM}{L} \times \frac{1}{2 L} = \frac{GM}{2 L ^{2}}$