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**JEE main Previous Year Topic Wise Questions of Physics with Solutions**

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*Simulator***Previous Years JEE Advanced Questions**

Q. A 0.1 kg mass is suspended from a wire of negligible mass. The length of the wire is 1m 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 rad s–1. If the Young’s modulus of the material of the wire is $\mathrm{n} \times 10^{9} \mathrm{Nm}^{-2}$, the value of n is ?

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**Sol.**4 $\mathrm{F}=6 \mathrm{A}=\mathrm{YSA}=\frac{\mathrm{YA}}{\ell} \cdot \mathrm{x}$ $140=\sqrt{\frac{\mathrm{n} \times 10^{9} \times 4.9 \times 10^{-7}}{1 \times 0.1}}=70 \times \sqrt{\mathrm{n}}$ $\therefore \mathrm{n}=4$

Q. One end of a horizontal thick copper wire of length 2L and radius 2R is welded to an end of another horizontal thin copper wire of length L and radius R. When the arrangement is stretched by applying forces at two ends, the ratio of the elongation in the thin wire to that in the thick wire is :-
(A) 0.25 (B) 0.50 (C) 2.00 (D) 4.00

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**Sol.**(C) Force same = F $\mathrm{Y}=\frac{F / A}{\Delta L / L}$ $F=Y A_{1} \frac{\Delta L_{1}}{L_{1}}=Y A_{2} \frac{\Delta L_{2}}{L_{2}}$ $\frac{\Delta L_{2}}{\Delta L_{1}}=\frac{L_{2}}{L_{1}} \times \frac{A_{1}}{A_{2}}$

Q. In plotting stress versus strain curves for two materials P and Q, a student by mistake puts strain on the y-axis and stress on the x-axis as shown in the figure. Then the correct statement(s) is (are):-
(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

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**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

only 3 qn??

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