$A$ metal wire of length $L_1$ and area of cross-section $A$ is attached to a rigid support. Another metal wire of length $L_2$ and of the same cross-sectional area is attached to the free end of the first wire. $A$ body of mass $M$ is then suspended from the free end of the second wire. If $Y_1$ and $Y_2$ are the Young's moduli of the wires respectively,the effective force constant of the system of the two wires is:
- A$\frac{Y_1 Y_2 A}{2(Y_1 L_2 + Y_2 L_1)}$
- B$\frac{Y_1 Y_2 A}{(L_1 + L_2)^{1/2}}$
- C$\frac{Y_1 Y_2 A}{Y_1 L_2 + Y_2 L_1}$
- D$\frac{(Y_1 Y_2)^{1/2} A}{(L_1 + L_2)^{1/2}}$
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Two masses $m_1$ and $m_2$ are connected by a light string passing over a pulley as shown in the figure. If the cross-sectional area of the string is $S$,its length is $L$,and its Young's modulus is $Y$,what is the elongation in the string?
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View SolutionIn an experiment to determine the Young's modulus of a wire of length exactly $1\;m$,the extension in the length of the wire is measured as $0.4\;mm$ with an uncertainty of $\pm 0.02\;mm$ when a load of $1\;kg$ is applied. The diameter of the wire is measured as $0.4\;mm$ with an uncertainty of $\pm 0.01\;mm$. The error in the measurement of Young's modulus $(\Delta Y)$ is found to be $x \times 10^{10}\;N/m^2$. The value of $x$ is (Take $g = 10\;m/s^2$)
DifficultJEE MAIN 2022
View SolutionIn steel,the Young's modulus and the strain at the breaking point are $2 \times 10^{11} \, N/m^2$ and $0.15$ respectively. The stress at the breaking point for steel is therefore:
Easy
View Solution$A$ copper wire of length $2.4 \ m$ and an aluminum wire of length $0.7 \ m$, both having diameter $2 \ mm$, are connected end to end. When stretched by a load, the total elongation is found to be $0.6 \ mm$. The applied load is (Young's modulus of copper $= 1.2 \times 10^{11} \ N/m^2$ and Young's modulus of aluminum $= 0.7 \times 10^{11} \ N/m^2$). (in $\pi \ N$)
On applying a stress of $20 \times 10^8 \ N/m^2$,the length of a perfectly elastic wire is doubled. Its Young's modulus will be:
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