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Calculating Suspended Mass from Tensile Stress in a Wire

Q: Why did the problem specify π = 22/7?

Because the diameter is 14mm and the stress has a factor of 7, the factor of 7 in the denominator of π cancels cleanly with the given values, leading to an integer result.

A metal block of mass m is suspended from a rigid support through a metal wire of diameter 14mm. The tensile stress developed in the wire under equilibrium state is 7× 10^5Nm^(-2). The value of mass m is kg. (Take, g=9.8ms^(-2) and π =22/7)

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Question type: Numerical
Exam relevance: JEE Main · Physics
Academic status: Answer verified against the official NTA key
Source: Previous-year question
Editorial review: 19 September 2026

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Students also ask

Why is the tension

T

equal to

m g

The mass is suspended in static equilibrium from a rigid support, meaning its acceleration is zero, so $Σ F_{y} = T - m g = 0 ⟹ T = m g$ .

Why did the problem specify

π = 22 / 7

Because the diameter is $14 mm$ and the stress has a factor of $7$ , the factor of $7$ in the denominator of $π$ cancels cleanly with the given values, leading to an integer result.

Answer

Accepted answer: 11 kg

The value of mass $m$ is $11 kg$ .

Why each option works or fails

Step-by-step solution

Given: Wire diameter $D = 14 mm = 14 \times 10^{- 3} m$ , tensile stress $σ = 7 \times 10^{5} N \cdot m^{- 2}$ , acceleration due to gravity $g = 9.8 m \cdot s^{- 2}$ , and $π = \frac{22}{7}$ .
Find: The suspended mass $m$ in kilograms.
Visualise: A vertical wire of circular cross-section supports a hanging block of mass $m$ . In static equilibrium, the tension $T$ in the wire balances the weight: $T = m g$ .
Strategise: Tensile stress is defined as force per unit cross-sectional area: $σ = \frac{F}{A} = \frac{m g}{π (D / 2)^{2}} = \frac{4 m g}{π D^{2}}$ . Rearrange to isolate $m$ : $m = \frac{π D^{2} σ}{4 g}$ .
Execute: Substitute the values: $m = \frac{\frac{22}{7} \times (14 \times 10^{- 3})^{2} \times 7 \times 10^{5}}{4 \times 9.8}$ . Evaluating numerator and denominator: $\frac{22}{7} \times 7 \times 10^{5} = 22 \times 10^{5}$ . Then $(14 \times 10^{- 3})^{2} = 196 \times 10^{- 6}$ . The product is $22 \times 10^{5} \times 196 \times 10^{- 6} = 431.2$ . Dividing by $4 \times 9.8 = 39.2$ : $\frac{431.2}{39.2} = 11 kg$ .
Verify: Check dimensions: $[e x t {N m}^{- 2}] [m^{2}] / [{m s}^{- 2}] = N / [{m s}^{- 2}] = kg$ . Magnitude $11 kg$ is physically reasonable for a thick $14 mm$ wire under moderate stress.

Calculating Suspended Mass from Tensile Stress in a Wire

What feels right?

The value of mass $m$ is $11 kg$ .

Why each option works or fails

Solution

01Given

02Find

03Visualise

04Strategise

05Execute

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