Ncert Exercises & Solutions

Question 9.10:
A rigid bar of mass 15 kg is supported symmetrically by three wires each 2.0 m long. Those at each end are of copper and the middle one is of iron. Determine the ratio of their diameters if each is to have the same tension.

Young's modulus of iron, Y $_{1}$ $= 190 \times 10^{9} Pa$

Young's modulus of copper, $Y_{2} = 110 \times 10^{9} Pa$

The extension in each case is the same. Since the wires are of the same length, the strain will also be the same. The Young’s modulus is given by:

$Y = \frac{Stress}{Strain} = \frac{\frac{F}{A}}{Strain} = \frac{\frac{4 F}{{πd}^{2}}}{Strain} . . . . . . . . . . . . . . . . (i)$

Where,
F = Tension force
A = Area of cross-section

d=Diameter of the wire

So, Y $\propto \frac{1}{d^{2}}$

Therefore,

$\frac{d_{2}}{d_{1}} = \sqrt{\frac{Y_{1}}{Y_{2}}} = \sqrt{\frac{190 \times 10^{9}}{110 \times 10^{9}}} = \sqrt{\frac{19}{11}} = 1.31 : 1$

NEET Information

courses

Company

Botany Questions

Chemistry Questions

Physics Questions

Zoology Questions