A uniform rope of mass M and length L is fixed at its upper end vertically from a rigid support. Then the tension. in the rope at the distance I from the rigid support is x

Question

A uniform rope of mass M and length L is fixed at its upper end vertically from a rigid support. Then the tension. in the rope at the distance I from the rigid support is x (A) M g (L/L+I) (B) (M g/L)(L-I) (C) Mg (D) (I/L) M g

Tardigrade Admin · Accepted Answer

For the lower part <img class="img-fluid question-image" alt="image" src="https://cdn.tardigrade.in/img/question/physics/896656346af33e8ca9689ee25c341f58-.png" /> Mass of the lower part is m' m'= Mass per unit length × length of lower part =(M/L)(L-I) So, Using vecF text net =m veca here veca=0 T-(M/L)(L-I) g=0 T=(M/L)(L-I) g

Q. A uniform rope of mass $M$ and length $L$ is fixed at its upper end vertically from a rigid support. Then the tension. in the rope at the distance $I$ from the rigid support is $x$

$M g \frac{L}{L + I}$

$\frac{M g}{L} (L - I)$

$M g$

$\frac{I}{L} M g$

For the lower part

Mass of the lower part is $m^{'}$
$m^{'} =$ Mass per unit length $\times$ length of lower part
$= \frac{M}{L} (L - I)$
So, Using $F_{net} = m a$
here $a = 0$
$T - \frac{M}{L} (L - I) g = 0$
$T = \frac{M}{L} (L - I) g$

Q. A uniform rope of mass M and length L is fixed at its upper end vertically from a rigid support. Then the tension. in the rope at the distance I from the rigid support is x

MgL+IL​

LMg​(L−I)

Mg

LI​Mg