A 3~m long wire of radius 3~mm shows an extension of 0.1~mm when loaded vertically by a mass of 50~kg in an experiment…

Solution & Explanation

Related Formula

Y = StressStrain = (F / A)/(Δ L / L) = (mg L)/(π r² Δ L)

where, $Y$ $Y$ = Young's modulus $F = mg$ $F = mg$ = stretching force (load) $A = π r²$ $A = \pi r^2$ = cross-sectional area of the wire $L$ $L$ = original length of the wire $Δ L$ $\Delta L$ = extension produced

Core Logic

Given parameters:

Original length of wire, $L = 3~m$ $L = 3\mathrm{~m}$
Radius of wire, $r = 3~mm = 3 \times 10⁻³~m$
Mass loaded, $m = 50~kg$ $m = 50\mathrm{~kg}$
Acceleration due to gravity, $g = 3π~m/s²$ $g = 3\pi\mathrm{~m/s}^{2}$
Extension, $Δ L = 0.1~mm = 10⁻⁴~m$

Substitute the values into the Young's modulus formula:

Y = 50 \times (3π) \times 3π \times (3 \times 10⁻³)² \times 10⁻⁴

Step 1: Detailed Computation

Simplify the equation:

Y = 450ππ \times 9 \times 10⁻⁶ \times 10⁻⁴

Cancelling $π$ $\pi$ from numerator and denominator:

Y = 4509 \times 10⁻¹⁰ = 50 \times 10¹⁰ = 5 \times 10¹¹~N/m²

Comparing this with $P \times 10¹¹~N/m²$ $P \times 10^{11}\mathrm{~N/m}^{2}$:

$P = 5$ $P = 5$

Pattern Recognition

Sees: Standard Young's modulus measurement. Trap: Don't forget that the radius is in millimeters and the extension is in millimeters. Convert all units to SI systematically first. Calculation Tip: The variable $g = 3π$ $g = 3\pi$ is chosen deliberately to cancel the $π$ $\pi$ in the area formula. Keep your eye open for these mathematical shortcuts! ✓

Chapter Mix

Class 11 Physics: Elasticity

More Elasticity Previous-Year Questions — Page 5

Q58 jee_main_2024_31_jan_morning Bulk Modulus

The depth below the surface of sea to which a rubber ball be taken so as to decrease its volume by

0.02%

is ________

m

. (Take density of sea water

= 10³ kgm⁻³

, Bulk modulus of rubber

= 9 \times 10⁸ Nm⁻²

, and

g = 10 ms⁻²

)

Numerical Answer. Answer: 18 to 18

Solution

Related Formula

β = (-Δ P)/((Δ V)/(V))

Δ P = ρ g h

Core Logic

The change in pressure $Δ P$ $\Delta P$ is the hydrostatic pressure at depth $h$ $h$.

Δ P = -β (Δ V)/(V)

ρ g h = -β (Δ V)/(V)

Step 2: Calculation

Given values: $ρ = 10³ kg/m³$ $\rho = 10^3\mathrm{\,kg/m^3}$ $g = 10 m/s²$ $g = 10\mathrm{\,m/s^2}$ $β = 9 \times 10⁸ N/m²$ $\beta = 9 \times 10^8\mathrm{\,N/m^2}$ $(Δ V)/(V) = -0.02% = -(0.02)/(100)$

Substitute into the equation:

10³ \times 10 \times h = - (9 \times 10⁸) \times (-(0.02)/(100))

10⁴ \times h = 9 \times 10⁸ \times 2 \times 10⁻⁴

10⁴ h = 18 \times 10⁴

h = 18 m

Chapter Mix

Class 11 Physics: Mechanical Properties Of Solids Class 11 Physics: Mechanical Properties Of Fluids

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Related Formula

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Pattern Recognition

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More Elasticity Previous-Year Questions — Page 5

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Solution & Explanation

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Reference Study Guides

More Elasticity Previous-Year Questions — Page 5

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Core Logic

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