Question

Two L-shaped wires are kept over each other as shown. The wire on the left is fixed and the wire on the right is movable on the left wire without any friction. The whole system is in a horizontal plane. Now we make a soap film of surface tension 0.1 N/m in between the wires such that it covers the common quadrilateral area.
If the right wire has a mass of 125 g, then its initial acceleration on being released is

• A. $4m/s^2$
• B. $3m/s^2$
• C. $2m/s^2$
• D. $1m/s^2$

Answer 1

The correct option is C $2m/s^2$


$F_1=T{l}_1=0.1\times 1.5=15\times {10}^{-2}N$

$F_2=T{l}_2=0.1\times 2=20\times {10}^{-2}N$

Net force on the wire, $F_{net}=\sqrt{F_1^2+F_2^2}=\left(\sqrt{{20}^2+{15}^2}\right)\times {10}^{-2}$

$=25\times {10}^{-2}N$

$\therefore$ acceleration of the wire, $a=\frac{F_{net}}{m}=\frac{25\times {10}^{-2}}{125\times {10}^{-3}}=\frac{25}{12.5}=2m/s^2$