(C) Given:Length $L = 6\,m$Area $A = 3\,mm^2 = 3 \times 10^{-6}\,m^2$Young's modulus $Y = 2 \times 10^{11}\,N/m^2$Mass $m = 4\,kg$Gravity on Earth $g_

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(C) Given:Length $L = 6\,m$Area $A = 3\,mm^2 = 3 \times 10^{-6}\,m^2$Young's modulus $Y = 2 \times 10^{11}\,N/m^2$Mass $m = 4\,kg$Gravity on Earth $g_e = 10\,m/s^2$Gravity on planet $g_p = \frac{1}{4} g_e = \frac{10}{4} = 2.5\,m/s^2$The tension $F$ in the wire is equal to the weight of the block on the planet:$F = m \times g_p = 4 \times 2.5 = 10\,N$The formula for elongation $\Delta L$ is:$\Delta L = \frac{FL}{AY}$Substituting the values:$\Delta L = \frac{10 \times 6}{3 \times 10^{-6} \times 2 \times 10^{11}}$$\Delta L = \frac{60}{6 \times 10^5} = 10 \times 10^{-5} = 10^{-4}\,m$Converting to millimeters:$\Delta L = 10^{-4} \times 10^3\,mm = 0.1\,mm$

Class 11 Physics Mechanical Properties of Solids Young’s Modulus

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The Young's modulus of a steel wire of length $6\,m$ and cross-sectional area $3\,mm^2$ is $2 \times 10^{11}\,N/m^2$. The wire is suspended from its support on a given planet. $A$ block of mass $4\,kg$ is attached to the free end of the wire. The acceleration due to gravity on the planet is $\frac{1}{4}$ of its value on the Earth. The elongation of the wire is (Take $g$ on the Earth $= 10\,m/s^2$):

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A
$1\,cm$
B
$1\,mm$
C
$0.1\,mm$
D
$0.1\,cm$

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Mechanical Properties of Solids

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Young’s Modulus

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An aluminum rod (Young's modulus $Y = 7 \times 10^9 \, N/m^2$) has a breaking strain of $0.2\%$. The minimum cross-sectional area of the rod required to support a load of $10^4 \, N$ is:

Medium

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Two wires $A$ and $B$ made of the same material and areas of cross-section in the ratio $1: 2$ are stretched by the same force. If the masses of the wires $A$ and $B$ are in the ratio $2: 3$, then the ratio of the elongations of the wires $A$ and $B$ is

MediumTS EAMCET 2025

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$A$ metallic rod breaks when the strain produced is $0.2 \%$. The Young's modulus of the material of the rod is $7 \times 10^9 \,N/m^2$. The area of cross-section required to support a load of $10^4 \,N$ is:

MediumKCET 2022

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To break a wire of $1 \, m$ length, a minimum weight of $40 \, kg \, wt$ is required. Then, the wire of the same material with double the radius and $6 \, m$ length will require a breaking weight of ....... $kg \, wt$.

Medium

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$A$ steel wire of length $3.2 \, m$ $(Y_{S} = 2.0 \times 10^{11} \, N/m^{2})$ and a copper wire of length $4.4 \, m$ $(Y_{C} = 1.1 \times 10^{11} \, N/m^{2})$,both of radius $1.4 \, mm$,are connected end to end. When stretched by a load,the net elongation is found to be $1.4 \, mm$. The load applied,in Newtons,is. (Given $\pi = \frac{22}{7}$)

MediumJEE MAIN 2022

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An aluminum rod (Young's modulus $Y = 7 \times 10^9 \, N/m^2$) has a breaking strain of $0.2\%$. The minimum cross-sectional area of the rod required to support a load of $10^4 \, N$ is:

Two wires $A$ and $B$ made of the same material and areas of cross-section in the ratio $1: 2$ are stretched by the same force. If the masses of the wires $A$ and $B$ are in the ratio $2: 3$, then the ratio of the elongations of the wires $A$ and $B$ is

$A$ metallic rod breaks when the strain produced is $0.2 \%$. The Young's modulus of the material of the rod is $7 \times 10^9 \,N/m^2$. The area of cross-section required to support a load of $10^4 \,N$ is:

To break a wire of $1 \, m$ length, a minimum weight of $40 \, kg \, wt$ is required. Then, the wire of the same material with double the radius and $6 \, m$ length will require a breaking weight of ....... $kg \, wt$.

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