Question

The Young's modulus of steel is twice that of brass. Two wires of same length and of same area of cross section, one of steel and another of brass are suspended from the same roof. If we want the lower ends of the wires to be at the same level, then the weights added to the steel and brass wires must be in the ratio of

• A. 4 : 1
• B. 1 : 1
• C. 1 : 2
• D. 2 : 1

Answer 1

Answer: (D)

2 : 1

Answer 2

The correct option is(D): 2:1

Let L and A be length and area of cross section of
each wire. In order to have the lower ends of the

wires to be at the same level (i.e. same elongation
is produced in both wires), let weights $W_s \, \, and W_b$
are added to steel and brass wires respectively.
Then
By definition of Young's modulus, the elongation
produced in the steel wire is
$\Delta L_s = \frac{ W_s L}{ Y_s A}$ \hspace20mm $\bigg( as \, \, Y+ \frac{W/A}{\Delta L/L} \bigg)$
and that in the brass wire is
$\Delta L_b = \frac{ W_b L}{ Y_b A}$
But $\Delta L_s = \Delta L_b \, \, \, \, \,$ (given)
$\therefore \, \, \, \, \frac{W_sL}{Y_s A}= \frac{W_b L}{ Y_b A } \, \, \, or \, \, \, \frac{W_s}{W_b} =\frac{Y_s}{Y_b}$
As $\frac{Y_s}{Y_b} = 2 \, \, \, \, \, \, (given)$
$\frac{W_s}{W_b} = \frac{2}{1}$