Two blocks of mass 2 ~kg and 4 ~kg are connected by a metal wire going over a smooth pulley as shown in figure. The radius of wire is 4.0 \times 10^-5 ~m and Young's modulus of the metal is 2.0 \times 10^11 ~N / m^2 . The longitudinal strain developed in the wire is \frac1\alpha \pi . The value of \alpha is \underline\hspace2cm . [Use g=10 ~m / s^2 ]

Properties of Matter PYQ — JEE Main 2024 · 31 Jan · Shift 2 · Q81 with Solution

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Correct answer: 12

Since the pulley is smooth and the wire is light, this is an Atwood machine with masses $m_1=2\text{ kg},\quad m_2=4\text{ kg}.$

The acceleration is $a=\frac{m_2-m_1}{m_1+m_2}g=\frac{4-2}{4+2}\cdot 10=\frac{10}{3}\text{ m/s}^2.$

Now tension in the wire: $T=m_1(g+a)=2\left(10+\frac{10}{3}\right)=2\cdot \frac{40}{3}=\frac{80}{3}\text{ N}.$

(Equivalently, $T=m_2(g-a)$ gives the same result.)

Radius of wire: $r=4.0\times 10^{-5}\text{ m}.$

So area is

\pi\cdot 16\times 10^{-10}=1.6\times 10^{-9}\pi\text{ m}^2.$$ --- 3. **Use Young's modulus relation** Young's modulus is $$Y=\frac{\text{stress}}{\text{strain}}$$ so $$\text{strain}=\frac{\text{stress}}{Y}=\frac{T/A}{Y}=\frac{T}{AY}.$$ Given $$Y=2.0\times 10^{11}\text{ N/m}^2.$$ Thus $$\text{strain}=\frac{\frac{80}{3}}{\left(1.6\times 10^{-9}\pi\right)\left(2.0\times 10^{11}\right)}.$$ First simplify denominator: $$1.6\times 10^{-9}\times 2.0\times 10^{11}=3.2\times 10^2=320.$$ Hence $$\text{strain}=\frac{80/3}{320\pi}=\frac{80}{960\pi}=\frac{1}{12\pi}.$$ --- 4. **Compare with the given form** Given strain is $$\frac{1}{\alpha\pi}.$$ So, $$\frac{1}{\alpha\pi}=\frac{1}{12\pi}$$ which gives $$\alpha=12.$$ --- 5. **Comparison with stored answer** Derived answer: $12$ Stored correct answer: $12$ They agree.