A Uniform rod AB is hinged at its end A and the other end of the rod is connected to a block through a massless string as shown in the figure. The pulley is smooth and massless. Masses of the block and the rod are same and are equal to m . The acceleration of the block just after the release from this position is

Question

A Uniform rod AB is hinged at its end A and the other end of the rod is connected to a block through a massless string as shown in the figure. The pulley is smooth and massless. Masses of the block and the rod are same and are equal to m . The acceleration of the block just after the release from this position is <img class="img-fluid question-image" alt="Question" src="https://cdn.tardigrade.in/q/nta/p-2hjmqeu4huusjavr.jpg" /> (A) (6 g/13) (B) (g/4) (C) (3 g/8) (D) None

Tardigrade Admin · Accepted Answer

For the rod: textTℓ - textmg((ℓ /2))=( textm ℓ 2/3)× ( texta/ℓ ) ...(i) For the: mg-T=ma ...(ii) Solving (i) and (ii) we get: a=(3 g/8) <img class="img-fluid question-image" alt="Solution" src="https://cdn.tardigrade.in/q/nta/p-2hjmqeu4huusibpv.jpg" />

$\frac{6 g}{13}$

$\frac{g}{4}$

$\frac{3 g}{8}$

$N o n e$

For the rod: $T ℓ - mg (\frac{ℓ}{2}) = \frac{m ℓ ^{2}}{3} \times \frac{a}{ℓ}$ ...(i)
For the: $m g - T = ma$ ...(ii)
Solving (i) and (ii) we get: $a = \frac{3 g}{8}$

136g​

4g​

83g​

None

For the rod: Tℓ−mg(2ℓ​)=3mℓ2​×ℓa​ ...(i) For the: mg−T=ma ...(ii) Solving (i) and (ii) we get: a=83g​

$\frac{6 g}{13}$

$\frac{g}{4}$

$\frac{3 g}{8}$

$N o n e$

For the rod: $T ℓ - mg (\frac{ℓ}{2}) = \frac{m ℓ ^{2}}{3} \times \frac{a}{ℓ}$ ...(i)
For the: $m g - T = ma$ ...(ii)
Solving (i) and (ii) we get: $a = \frac{3 g}{8}$