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**Previous Years JEE Advanced Questions**

**[IIT-JEE 2010]**

**Sol.**4

$\mathrm{F}=6 \mathrm{A}=\mathrm{YSA}=\frac{\mathrm{YA}}{\ell} \cdot \mathrm{x}$

$140=\sqrt{\frac{\mathrm{n} \times 10^{9} \times 4.9 \times 10^{-7}}{1 \times 0.1}}=70 \times \sqrt{\mathrm{n}}$

$\therefore \mathrm{n}=4$

(A) 0.25 (B) 0.50 (C) 2.00 (D) 4.00

**[JEE-Advance-2013]**

**Sol.**(C)

Force same = F

$\mathrm{Y}=\frac{F / A}{\Delta L / L}$

$F=Y A_{1} \frac{\Delta L_{1}}{L_{1}}=Y A_{2} \frac{\Delta L_{2}}{L_{2}}$

$\frac{\Delta L_{2}}{\Delta L_{1}}=\frac{L_{2}}{L_{1}} \times \frac{A_{1}}{A_{2}}$

(A) P has more tensile strength than Q

(B) P is more ductile than Q

(C) P is more brittle than Q

(D) The Young’s modulus of P is more than that of Q

**[JEE-Advance-2015]**

**Sol.**(A,B)

Slope of this graph represents the reciprocal of Young’s modulus.

since Slope of P > Slope of Q

Hence Y of P < Y of Q