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Question: Three blocks, each of same mass m, are connected with wires W<sub>1</sub> and W<sub>2</sub> of same ...

Three blocks, each of same mass m, are connected with wires W1 and W2 of same cross-sectional area a and Young’s modulus Y. Neglecting friction the strain developed in wire W2 is

A

23mgaY\frac{2}{3}\frac{mg}{aY}

B

3mg2aY\frac{3mg}{2aY}

C

1mg3aY\frac{1mg}{3aY}

D

3mgaY\frac{3mg}{aY}

Answer

23mgaY\frac{2}{3}\frac{mg}{aY}

Explanation

Solution

If the system moves with acceleration a and T is the tension in the string W2W_{2} then by comparing this condition from standard case T=m1m2m1+m2g\mathbf{T =}\frac{\mathbf{m}_{\mathbf{1}}\mathbf{m}_{\mathbf{2}}}{\mathbf{m}_{\mathbf{1}}\mathbf{+}\mathbf{m}_{\mathbf{2}}}\mathbf{g}

In the given problem m1=(m+m)=2mm_{1} = (m + m) = 2m and m2=mm_{2} = m

\therefore Tension =m.2m.gm+2m=23mg= \frac{m.2m.g}{m + 2m} = \frac{2}{3}mg

\therefore Stress =Ta=23amg= \frac{T}{a} = \frac{2}{3a}mg and

Strain=StressYoung’s modulus=23mgaY\text{Strain} = \frac{\text{Stress}}{\text{Young's modulus}} = \frac{2}{3}\frac{mg}{aY}