One end of a uniform rod of mass $M$ and cross-sectional area $A$ is suspended from the other end. The stress at the mid-point of the rod will be :

\frac{2 M g}{A}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

\frac{3 M g}{2 A}

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

\frac{M g}{A}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Zero

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is B $\frac{3 M g}{2 A}$
The weight of suspended mass is given as,

$W_{1} = m g$

The weight of the rod acting at the midpoint is given as,

$W_{2} = \frac{m g}{2}$

The stress at the midpoint is given as,

$σ = \frac{W_{1} + W_{2}}{A}$

$σ = \frac{m g + \frac{m g}{2}}{A}$

$σ = \frac{3 m g}{2 A}$

Suggest Corrections

Similar questions

One end of a uniform rod of mass M and cross-sectional area A is suspended from a rigid support and an equal mass M is suspended from the other end. The stress at the mid-point of the rod will be

Q. One end of a uniform bar of weight

w_{1}

is suspended from the roof and a weight

w_{2}

is suspended from the other end, the area of cross-section is A. What is the stress at the mid point of the rod?

Q. One a uniform rod of mass

m_{1}

and cross-sectional area A is hung from a ceiling. The other end of bar is supporting mass

m_{2}

The stress at the midpoint is

Join BYJU'S Learning Program

One end of a uniform rod of mass M and cross-sectional area A is suspended from the other end. The stress at the mid-point of the rod will be :

The correct option is B 3Mg2AThe weight of suspended mass is given as, W1=mg The weight of the rod acting at the midpoint is given as, W2=mg2 The stress at the midpoint is given as, σ=W1+W2A σ=mg+mg2A σ=3mg2A

One end of a uniform rod of mass $M$ and cross-sectional area $A$ is suspended from the other end. The stress at the mid-point of the rod will be :