Question

Length of a metallic rod of mass m and cross-sectional area A is L. If mass M is suspended at the lower end of this rod suspended vertically stress at the cross-section situated at 3l/4 distance from its lower end is ........ . (A) Mg/A (B) (M + m/4) g/A (C) (M + 3/4 m) g/A (D) (M + m) g/A

Answer 1

Answer:

$(C) (M + \dfrac{3}{4}m) g/A$

Explanation:

Given,

Area of the cross section = A

length of rod = L

mass of the rod = m

mass of the of object suspended to the rod = M

since the cross is at 3L/4 from its lower end so the total mass suspend to the cross section is (mass of the object +3/4 of mass of object) = $M +\dfrac{3}{4}m$.

Since the force acting on a body due to gravity, F = mass x acceleration due to gravity

                                                                                $= (M+\dfrac{3}{4}m)g$

Hence, stress at the cross section is,

                                      $S\ =\ \dfrac{F}{A}$

                                          $=\ \dfrac{(M+\dfrac{3}{4}m)g}{A}$