A truck is pulling a car out of a ditch by means of a steel cable that is $9.1\,m$ long and has a radius of $5\,mm$, when the car just begins to move the tension in the cable is $800\,N$. How much has the cable stretched ? (Young’s modulus for steel is $ 2 \times 10^{11}\,Nm^{-2}$)
A truck is pulling a car out of a ditch by means of a steel cable that is $9.1\,m$ long and has a radius of $5\,mm$, when the car just begins to move the tension in the cable is $800\,N$. How much has the cable stretched ? (Young’s modulus for steel is $ 2 \times 10^{11}\,Nm^{-2}$)
$\mathrm{Y}=\frac{\mathrm{F} l}{\mathrm{~A} \Delta l}$
$\Delta l =\frac{\mathrm{F} l}{\mathrm{AY}}=\frac{\mathrm{F} l}{\pi r^{2} \mathrm{Y}}$
$\Delta l =\frac{800 \times 9.1}{3.14 \times\left(5 \times 10^{-3}\right)^{2} \times 2 \times 10^{11}}$
$=46.369 \times 10^{-5}$ $\approx 4.64 \times 10^{-4} \mathrm{~m}$
Similar Questions
Two wires are made of the same material and have the same volume. The first wire has cross-sectional area $A$ and the second wire has cross-sectional area $3A$. If the length of the first wire is increased by $\Delta l$ on applying a force $F$, how much force is needed to stretch the second wire by the same amount?
Two wires are made of the same material and have the same volume. The first wire has cross-sectional area $A$ and the second wire has cross-sectional area $3A$. If the length of the first wire is increased by $\Delta l$ on applying a force $F$, how much force is needed to stretch the second wire by the same amount?
- [AIEEE 2009]
A copper wire of length $4.0m$ and area of cross-section $1.2\,c{m^2}$ is stretched with a force of $4.8 \times {10^3}$ $N.$ If Young’s modulus for copper is $1.2 \times {10^{11}}\,N/{m^2},$ the increase in the length of the wire will be
A copper wire of length $4.0m$ and area of cross-section $1.2\,c{m^2}$ is stretched with a force of $4.8 \times {10^3}$ $N.$ If Young’s modulus for copper is $1.2 \times {10^{11}}\,N/{m^2},$ the increase in the length of the wire will be
An aluminum rod (Young's modulus $ = 7 \times {10^9}\,N/{m^2})$ has a breaking strain of $0.2\%$. The minimum cross-sectional area of the rod in order to support a load of ${10^4}$Newton's is
An aluminum rod (Young's modulus $ = 7 \times {10^9}\,N/{m^2})$ has a breaking strain of $0.2\%$. The minimum cross-sectional area of the rod in order to support a load of ${10^4}$Newton's is
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.
A composite rod made up of two rods $AB$ and $BC$ are joined at $B$ . The rods are of equal length at room temperature and have equal masses. The coefficient of linear expansion a of $AB$ is more than that of $BC$. The composite rod is suspended horizontal by means of a thread at $B$. When the rod is heated
A composite rod made up of two rods $AB$ and $BC$ are joined at $B$ . The rods are of equal length at room temperature and have equal masses. The coefficient of linear expansion a of $AB$ is more than that of $BC$. The composite rod is suspended horizontal by means of a thread at $B$. When the rod is heated