Question

Each of the three blocks P, Q and R shown in figure have a mass of 3 kg. Each of the wires A and B has cross-sectional area 0.005 cm² and Young modulus 2 x 10¹¹ N m^-2. Neglect friction. Find the longitudinal strain developed in each of the wires. Take g = 10 m s^-2.

• A. ${10}^{-4},{10}^{-4}$
• B. $2\times {10}^{-4},4\times {10}^{-4}$
• C. ${10}^{-4},2\times {10}^{-4}$
• D. ${10}^{-4},4\times {10}^{-4}$

Answer 1

The correct option is C ${10}^{-4},2\times {10}^{-4}$

The block R will descend vertically and the blocks P and Q will move on the frictionless horizontal table. Let the common magnitude of the acceleration be a. Let the tensions in the wires A and B be $T_A$ and $T_B$ respectively.
Writing the equations of motion of the blocks P, Q and R, we get
$T_A$=(3kg) a...(i)
$T_B-T_A$=(3 kg ) a....(ii)
And (3 kg)g-$T_B$=(3 kg ) a....(iii)
By (i) and (ii),
$T_B$=2$T_A$
By (i) and (iii),
$T_A+T_B$=(3 kg) g=30 N
or, 3$T_A$=30 N
or, $T_A=10Nand{T}_B=20N$
Longitudinal strain=$\frac{Longitudinalstress}{Youngmodulus}$
Strain in wire A$\frac{\frac{10N}{0.005c{m}^2}}{2\times {10}^{11}N{m}^{-2}}={10}^{-4}$
And stain in wire B= $\frac{\frac{20N}{0.005c{m}^2}}{2\times {10}^{11}N{m}^{-2}}=2\times {10}^{-4}$