A uniform rod of length l and mass M is suspended from two inextensible string as shown in the figure. What will be the tension in the left string when right string snaps?

Question

A uniform rod of length l and mass M is suspended from two inextensible string as shown in the figure. What will be the tension in the left string when right string snaps? <img class="img-fluid question-image" alt="Question" src="https://cdn.tardigrade.in/q/nta/p-wrqzx8tpmshmgcwt.jpg" /> (A) (m g/2) (B) mg (C) (m g/4) (D) (m g/8)

Tardigrade Admin · Accepted Answer

Using Newton's second law on rod, mg-T=ma...(i) <img class="img-fluid question-image" alt="Solution" src="https://cdn.tardigrade.in/q/nta/p-wrqzx8tpmshmhwvp.jpg" /> Using the torque equation about the left string, (m g l/2)=Iα =(m l2/3)× (a/(l/2)) ⇒ a=(3 g/4) . Putting the value of a in equation (i) , T=(m g/4) .

Q. A uniform rod of length $l$ and mass $M$ is suspended from two inextensible string as shown in the figure. What will be the tension in the left string when right string snaps?

$\frac{m g}{2}$

$m g$

$\frac{m g}{4}$

$\frac{m g}{8}$

Using Newton's second law on rod,
$m g - T = ma ... (i)$

Using the torque equation about the left string,
$\frac{m g l}{2} = I α = \frac{m l ^{2}}{3} \times \frac{a}{( l /2 )}$
$\Rightarrow a = \frac{3 g}{4}$ .
Putting the value of $a$ in equation $(i)$ ,
$T = \frac{m g}{4}$ .

Q. A uniform rod of length l and mass M is suspended from two inextensible string as shown in the figure. What will be the tension in the left string when right string snaps?

2mg​

mg

4mg​

8mg​