Mechanical Properties of Solids JEE Advanced previous year questions with solutions

3 solved JEE Advanced questions on Mechanical Properties of Solids, free to read — no sign-in needed. The full chapter has 5 questions; sign in to attempt the remaining 2 in the exam simulator.

1. Q1JEE Advanced Adv 2020 (Paper 2)
   A cubical solid aluminum $\left(bulkmodulus=-V\dfrac{dP}{dV}=70GPa\right)$ block has an edge length of $1m$ on the surface of the earth. It is kept on the floor of a $5km$ deep ocean. Taking the average density of water and the acceleration due to gravity to be ${10}^{3}kg{m}^{-3}$ and $10m{s}^{-2}$, respectively, the change in the edge length of the block in $mm$ is ________.
   Show answer & solution

   Answer: 0.24

   $B=\dfrac{P}{\left(-\dfrac{dV}{V}\right)}\Rightarrow \dfrac{dV}{V}=-\dfrac{P}{B}\Rightarrow 3\dfrac{d\ell }{\ell }=\dfrac{P}{B}$ $d\ell =\left(\dfrac{-P}{B}\right)\dfrac{\ell }{3}=\dfrac{{10}^{3}\dfrac{kg}{{m}^{3}}\times \dfrac{10m}{{s}^{2}}\times 5\times {10}^{3}}{70\times {10}^{9}}\times \dfrac{1m}{3}$ $d\ell =0.238\times {10}^{-3}$ $\Rightarrow d\ell =0.238mm$ $=0.24$
2. Q2JEE Advanced Adv 2018 (Paper 2)
   A steel wire of diameter $0.5mm$ and Young's modulus $2\times {10}^{11}N{m}^{-2}$ carries a load of mass M. The length of the wire with the load is 1.0 m. A vernier scale with 10 divisions is attached to the end of this wire. Next to the steel wire is a reference wire to which a main scale, of least count 1.0 mm, is attached. The 10 divisions of the vernier scale correspond to 9 divisions of the main scale. Initially, the zero of vernier scale coincides with the zero of main scale. If the load on the steel wire is increased by $1.2kg$ , the vernier scale division which coincides with a main scale division is __________. Take $g=10m{s}^{-1}and\pi =3.2.$
   Show answer & solution

   Answer: 3

   $d=0.5mm$ $Y=2\times {10}^{11}$ $l=1m$ $∆l=\dfrac{Fl}{Ay}=\dfrac{mgl}{\dfrac{\pi {d}^{2}}{4}y}=\dfrac{1.2\times 10\times 1}{\dfrac{\pi }{4}\times {\left(5\times {10}^{-4}\right)}^{2}\times 2\times {10}^{11}}$ $∆l=\dfrac{1.2\times 10}{\dfrac{3.2}{4}\times 25\times {10}^{-8}\times 2\times {10}^{11}}$ $=\dfrac{12}{1.6\times 25\times {10}^{3}}=\dfrac{12}{40\times {10}^{3}}=0.03mm$ So 3rd division of vernier scale will coincicle with main scale
3. Q3JEE Advanced Adv 2010 (Paper 1)
   A $0.1 \mathrm{~kg}$ mass is suspended from a wire of negligible mass. The length of the wire is $1 \mathrm{~m}$ and its cross-sectional area is $4.9 \times 10^{-7} \mathrm{~m}^2$. If the mass is pulled a little in the vertically downward direction and released, it performs simple harmonic motion of angular frequency $140 \mathrm{rad} \mathrm{s}^{-1}$. If the Young's modulus of the material of the wire is $n \times 10^9 \mathrm{Nm}^{-2}$, the value of $n$ is
   Show answer & solution

   Answer: 4

   $\omega=\sqrt{\frac{K}{m}}=\sqrt{\frac{Y A}{l m}}=\sqrt{\frac{\left(n \times 10^9\right)\left(4.9 \times 10^{-7}\right)}{1 \times 0.1}}$ Putting, $\omega=140 \operatorname{rad} \mathrm{s}^{-1}$ in above equation we get, $$n=4$$ $\therefore$ Answer is 4 .