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JEE Advanced
Exam
JEE Advanced
Subject
Physics
Chapter
Mechanical Properties of Solids

JEE Advanced · Physics · Mechanical Properties of Solids

In the adjacent figure, ABCD are four points in a smooth horizontal...

In the adjacent figure, ABCD are four points in a smooth horizontal plane representing the four corners of a square of diagonal 20 cm . A very small object of mass 14 gm placed at the centre of the square connected to points A, B, C and D by four elastic strings of same dimensions but of different moduli of elasticity as shown in the figure. Natural length and cross-sectional area of each string are 10 cm and π2×10−6 m2\pi^2 \times 10^{-6} \mathrm{~m}^2 respectively. The mass is slightly displaced from equilibrium and then released to perform SHM along the diagonal AC. (Y1=9Y,Y2=4Y,Y3=16Y\left(Y_1=9 \mathrm{Y}, \mathrm{Y}_2=4 \mathrm{Y}, \mathrm{Y}_3=16 \mathrm{Y}\right. and Y4=\mathrm{Y}_4= 4 Y , where Y=4.9×109 N/m2\mathrm{Y}=4.9 \times 10^9 \mathrm{~N} / \mathrm{m}^2 ). The time period of oscillation of object is k millisecond. The value of K is In the adjacent figure, ABCD are four points in a smooth horizontal plane representing the four corners of a square of

Correct answer

1

Step-by-step solution

T=π[mℓ9YA+mℓ16YA]=π[mℓYA][13+14]=7π12mℓYAm=144×10−3 kg\begin{aligned} & \mathrm{T}=\pi\left[\sqrt{\frac{\mathrm{m} \ell}{9 \mathrm{YA}}}+\sqrt{\frac{\mathrm{m} \ell}{16 \mathrm{YA}}}\right] \\ & =\pi\left[\sqrt{\frac{\mathrm{m} \ell}{\mathrm{YA}}}\right]\left[\frac{1}{3}+\frac{1}{4}\right] \\ & =\frac{7 \pi}{12} \sqrt{\frac{\mathrm{m} \ell}{\mathrm{YA}}} \\ & \mathrm{m}=144 \times 10^{-3} \mathrm{~kg}\end{aligned} ℓ=AO=BO=CO=DO=10×10−2 m A=π2×10−6 m2Y=4.9×109 T=7π12144×10−3×10×10−2π2×10−6×49×108=7π12×12π×1710−2108=1×10−3=1 ms\begin{aligned} & \ell=A O=B O=C O=D O=10 \times 10^{-2} \mathrm{~m} \\ & \mathrm{~A}=\pi^2 \times 10^{-6} \mathrm{~m}^2 \\ & \mathrm{Y}=4.9 \times 10^9 \\ & \mathrm{~T}=\frac{7 \pi}{12} \sqrt{\frac{144 \times 10^{-3} \times 10 \times 10^{-2}}{\pi^2 \times 10^{-6} \times 49 \times 10^8}}=\frac{7 \pi}{12} \times \frac{12}{\pi} \times \frac{1}{7} \sqrt{\frac{10^{-2}}{10^8}} \\ & =1 \times 10^{-3} \\ & =1 \mathrm{~ms} \end{aligned} In the adjacent figure, ABCD are four points in a smooth horizontal plane representing the four corners of a square of
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