Length of a rod is 20, cm and area of cross-section 2,c{m^2} . Young's modulus of material of wi

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8.Mechanical Properties of Solids

medium

The length of a rod is $20\, cm$ and area of cross-section $2\,c{m^2}$. The Young's modulus of the material of wire is $1.4 \times {10^{11}}\,N/{m^2}$. If the rod is compressed by $5\, kg-wt$ along its length, then increase in the energy of the rod in joules will be

$8.57 \times {10^{ - 6}}$

$22.5 \times {10^{ - 4}}$

$9.8 \times {10^{ - 5}}$

$45.0 \times {10^{ - 5}}$

Solution

(a) Energy = $\frac{1}{2}Fl = \frac{1}{2} \times F \times \left( {\frac{{FL}}{{AY}}} \right) = \frac{1}{2} \times \frac{{{F^2}L}}{{AY}}$

$ = \frac{1}{2} \times \frac{{{{(50)}^2} \times 20 \times {{10}^{ – 2}}}}{{2 \times {{10}^{ – 4}} \times 1.4 \times {{10}^{11}}}}$$ = 8.57 \times {10^{ – 6}}J$

Standard 11

Physics

The work done in increasing the length of a $1$ $metre$ long wire of cross-section area $1\, mm^2$ through $1\, mm$ will be ……. $J$ $(Y = 2\times10^{11}\, Nm^{-2})$

medium

When a force is applied on a wire of uniform cross-sectional area $3 \times {10^{ – 6}}\,{m^2}$ and length $4m$, the increase in length is $1\, mm.$ Energy stored in it will be $(Y = 2 \times {10^{11}}\,N/{m^2})$

medium

A solid expands upon heating because

hard

(KVPY-2014)

If one end of a wire is fixed with a rigid support and the other end is stretched by a force of $10 \,N,$ then the increase in length is $0.5\, mm$. The ratio of the energy of the wire and the work done in displacing it through $1.5\, mm$ by the weight is

medium

$K$ is the force constant of a spring. The work done in increasing its extension from ${l_1}$ to ${l_2}$ will be

medium

The length of a rod is $20\, cm$ and area of cross-section $2\,c{m^2}$. The Young's modulus of the material of wire is $1.4 \times {10^{11}}\,N/{m^2}$. If the rod is compressed by $5\, kg-wt$ along its length, then increase in the energy of the rod in joules will be

Solution

Similar Questions

The work done in increasing the length of a $1$ $metre$ long wire of cross-section area $1\, mm^2$ through $1\, mm$ will be ……. $J$ $(Y = 2\times10^{11}\, Nm^{-2})$

When a force is applied on a wire of uniform cross-sectional area $3 \times {10^{ – 6}}\,{m^2}$ and length $4m$, the increase in length is $1\, mm.$ Energy stored in it will be $(Y = 2 \times {10^{11}}\,N/{m^2})$

A solid expands upon heating because

If one end of a wire is fixed with a rigid support and the other end is stretched by a force of $10 \,N,$ then the increase in length is $0.5\, mm$. The ratio of the energy of the wire and the work done in displacing it through $1.5\, mm$ by the weight is

$K$ is the force constant of a spring. The work done in increasing its extension from ${l_1}$ to ${l_2}$ will be

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