Ratio of lengths of two wires A and B of same material is 1 : 2 and ratio of their diameter is 2 : 1

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8.Mechanical Properties of Solids

medium

The ratio of the lengths of two wires $A$ and $B$ of same material is $1 : 2$ and the ratio of their diameter is $2 : 1.$ They are stretched by the same force, then the ratio of increase in length will be

$2:1$

$1:4$

$1:8$

$8:1$

Solution

(c) $l = \frac{{FL}}{{AY}} \Rightarrow l \propto \frac{L}{{{d^2}}}$==>$\frac{{{l_1}}}{{{l_2}}} = \frac{{{L_1}}}{{{L_2}}} \times {\left( {\frac{{{d_2}}}{{{d_1}}}} \right)^2}$

$ = \frac{1}{2} \times {\left( {\frac{1}{2}} \right)^2} = \frac{1}{8}$

Standard 11

Physics

The temperature of a wire of length $1$ metre and area of cross-section $1\,c{m^2}$ is increased from $0°C$ to $100°C$. If the rod is not allowed to increase in length, the force required will be $(\alpha = {10^{ – 5}}/^\circ C$ and $Y = {10^{11}}\,N/{m^2})$

medium

Two similar wires under the same load yield elongation of $0.1$ $mm$ and $0.05$ $mm$ respectively. If the area of cross- section of the first wire is $4m{m^2},$ then the area of cross section of the second wire is….. $mm^2$

medium

A wire of length $L$ and radius $r$ is clamped rigidly at one end. When the other end of the wire is pulled by a force $F$, its length increases by $5\,cm$. Another wire of the same material of length $4 L$ and radius $4\,r$ is pulled by a force $4\,F$ under same conditions. The increase in length of this wire is $….cm$.

medium

(JEE MAIN-2022)

When a stress of $10^8\,Nm^{-2}$ is applied to a suspended wire, its length increases by $1 \,mm$. Calculate Young’s modulus of wire.

easy

A uniform rod of mass $m$, length $L$, area of cross-section $A$ and Young's modulus $Y$ hangs from the ceiling. Its elongation under its own weight will be

hard

The ratio of the lengths of two wires $A$ and $B$ of same material is $1 : 2$ and the ratio of their diameter is $2 : 1.$ They are stretched by the same force, then the ratio of increase in length will be

Solution

Similar Questions

The temperature of a wire of length $1$ metre and area of cross-section $1\,c{m^2}$ is increased from $0°C$ to $100°C$. If the rod is not allowed to increase in length, the force required will be $(\alpha = {10^{ – 5}}/^\circ C$ and $Y = {10^{11}}\,N/{m^2})$

Two similar wires under the same load yield elongation of $0.1$ $mm$ and $0.05$ $mm$ respectively. If the area of cross- section of the first wire is $4m{m^2},$ then the area of cross section of the second wire is….. $mm^2$

When a stress of $10^8\,Nm^{-2}$ is applied to a suspended wire, its length increases by $1 \,mm$. Calculate Young’s modulus of wire.

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