Two steel wires of same length but radii $r$ and $2r$ are connected together end to end and tied to a wall as shown. The force stretches the combination by $10\ mm$ . How far does the midpoint $A$ move ......... $mm$
Two steel wires of same length but radii $r$ and $2r$ are connected together end to end and tied to a wall as shown. The force stretches the combination by $10\ mm$ . How far does the midpoint $A$ move ......... $mm$
- A
$2$
- B
$4$
- C
$6$
- D
$8$
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In which case there is maximum extension in the wire, if same force is applied on each wire
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A uniform wire (Young's modulus $2 \times 10^{11}\, Nm^{-2}$ ) is subjected to longitudinal tensile stress of $5 \times 10^7\,Nm^{-2}$ . If the over all volume change in the wire is $0.02\%,$ the fractional decrease in the radius of the wire is close to
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- [JEE MAIN 2013]
In nature the failure of structural members usually result from large torque because of twisting or bending rather than due to tensile or compressive strains. This process of structural breakdown is called buckling and in cases of tall cylindrical structures like trees, the torque is caused by its own weight bending the structure. Thus, the vertical through the centre of gravity does not fall withinthe base. The elastic torque caused because of this bending about the central axis of the tree is given by $\frac{{Y\pi {r^4}}}{{4R}}$ $Y$ is the Young’s modulus, $r$ is the radius of the trunk and $R$ is the radius of curvature of the bent surface along the height of the tree containing the centre of gravity (the neutral surface). Estimate the critical height of a tree for a given radius of the trunk.
In nature the failure of structural members usually result from large torque because of twisting or bending rather than due to tensile or compressive strains. This process of structural breakdown is called buckling and in cases of tall cylindrical structures like trees, the torque is caused by its own weight bending the structure. Thus, the vertical through the centre of gravity does not fall withinthe base. The elastic torque caused because of this bending about the central axis of the tree is given by $\frac{{Y\pi {r^4}}}{{4R}}$ $Y$ is the Young’s modulus, $r$ is the radius of the trunk and $R$ is the radius of curvature of the bent surface along the height of the tree containing the centre of gravity (the neutral surface). Estimate the critical height of a tree for a given radius of the trunk.
The length of wire, when $M_1$ is hung from it, is $I_1$ and is $I_2$ with both $M_1$ and $M_2$ hanging. The natural length of wire is ........
The length of wire, when $M_1$ is hung from it, is $I_1$ and is $I_2$ with both $M_1$ and $M_2$ hanging. The natural length of wire is ........
A brass rod of length $2\,m$ and cross-sectional area $2.0\,cm^2$ is attached end to end to a steel rod of length $L$ and cross-sectional area $1.0\,cm^2$ . The compound rod is subjected to equal and opposite pulls of magnitude $5 \times 10^4\,N$ at its ends. If the elongations of the two rods are equal, then length of the steel rod $(L)$ is ........... $m$ $(Y_{Brass}=1.0\times 10^{11}\,N/m^2$ and $Y_{Steel} = 2.0 \times 10^{11}\,N/m^2)$
A brass rod of length $2\,m$ and cross-sectional area $2.0\,cm^2$ is attached end to end to a steel rod of length $L$ and cross-sectional area $1.0\,cm^2$ . The compound rod is subjected to equal and opposite pulls of magnitude $5 \times 10^4\,N$ at its ends. If the elongations of the two rods are equal, then length of the steel rod $(L)$ is ........... $m$ $(Y_{Brass}=1.0\times 10^{11}\,N/m^2$ and $Y_{Steel} = 2.0 \times 10^{11}\,N/m^2)$