A metal block of mass m is suspended from a rigid support through a metal wire of diameter 14 mm. The tensile stress developed in the wire under equilibrium state is 7 10^5 N m^-2. The value of mass…

Given: tensile stress S = 7 10^5 N m^-2, wire diameter = 14 mm, so radius r = 7 10^-3 m. Find: mass m of the metal block. Using stress, S = force area = mgA So, m = S Ag Area of cross-section of the wire is A = r^2 = (7 10^-3)^2 Substituting the given values, m = 7 10^5 (7 10^-3)^29.8 Using = 227, m = 7 10^5 227 (7 10^-3)^29.8 Thus, m = 11 kg Therefore, the mass of the block is 11 kg. Note: 14 mm is the diameter, so the radius is half of it.

A metal block of mass m is suspended from a rigid support through… — Solution | JEE Physics

NVAEasyJEE 2023Elasticity & Stress-Strain Curve

A metal block of mass $m$ is suspended from a rigid support through a metal wire of diameter $14 \, \text{mm}$ . The tensile stress developed in the wire under equilibrium state is $7 \times 10^5 \, \text{N m}^{-2}$ . The value of mass $m$ is _____ kg.

Answer

Correct answer:11

Step-by-step solution

Standard Method

Given: tensile stress $S = 7 \times 10^5 \, \text{N m}^{-2}$ , wire diameter $= 14 \, \text{mm}$ , so radius $r = 7 \times 10^{-3} \, \text{m}$ .

Find: mass $m$ of the metal block.

Using stress,

S = \frac{\text{force}}{\text{area}} = \frac{mg}{A}

So,

m = \frac{S \times A}{g}

Area of cross-section of the wire is

A = \pi r^2 = \pi \left(7 \times 10^{-3}\right)^2

Substituting the given values,

m = \frac{7 \times 10^5 \times \pi \times \left(7 \times 10^{-3}\right)^2}{9.8}

Using $\pi = \frac{22}{7}$ ,

m = \frac{7 \times 10^5 \times \frac{22}{7} \times \left(7 \times 10^{-3}\right)^2}{9.8}

Thus,

m = 11 \, \text{kg}

Therefore, the mass of the block is $11 \, \text{kg}$ .

Note: $14 \, \text{mm}$ is the diameter, so the radius is half of it.

Common mistakes

Using $14 \, \text{mm}$ directly as the radius. This is wrong because $14 \, \text{mm}$ is the diameter. Always divide by $2$ first to get the radius before using $A = \pi r^2$ .
Not converting mm to m before substituting in SI units. This is wrong because stress is given in $\text{N m}^{-2}$ , so radius must be in metres. Use $7 \, \text{mm} = 7 \times 10^{-3} \, \text{m}$ .
Taking area as proportional to diameter instead of radius squared. This is wrong because cross-sectional area of a wire is $A = \pi r^2$ , not $\pi d^2$ without adjustment. First find $r$ , then square it.

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