A steel rod of length 1 m and radius 10 mm is stretched by a force 100 kN along its length. Th

Name: Problem 41, Chapter 9: Objective NCERT at your FINGERTIPS: Physics
Uploaded: 2021-09-25T18:29:03-08:00
Duration: 1 min 28 s
Channel: Kamlesh Goyal
Description: A steel rod of length 1 m and radius 10 mm is stretched by a force 100 kN along its length. The stress produced in the rod is (YSteel =2 ×10^11 N m^-2. ) (a) 3.18 ×10^6 N m^-2 (b) 3.18 ×10^7 N m^-2 (c) 3.18 ×10^8 N m^-2 (d) 3.18 ×10^9 N m^-2.

step 1 So hello students in this question we have given that we have a steel roll of length 1 meter rig

Question

A steel rod of length $1 \mathrm{~m}$ and radius $10 \mathrm{~mm}$ is stretched by a force $100 \mathrm{kN}$ along its length. The stress produced in the rod is $\left(Y_{\text {Steel }}=2 \times 10^{11} \mathrm{~N} \mathrm{~m}^{-2}\right.$ ) (a) $3.18 \times 10^{6} \mathrm{~N} \mathrm{~m}^{-2}$ (b) $3.18 \times 10^{7} \mathrm{~N} \mathrm{~m}^{-2}$ (c) $3.18 \times 10^{8} \mathrm{~N} \mathrm{~m}^{-2}$ (d) $3.18 \times 10^{9} \mathrm{~N} \mathrm{~m}^{-2}$.

   A steel rod of length $1 \mathrm{~m}$ and radius $10 \mathrm{~mm}$ is stretched by a force $100 \mathrm{kN}$ along its length. The stress produced in the rod is $\left(Y_{\text {Steel }}=2 \times 10^{11} \mathrm{~N} \mathrm{~m}^{-2}\right.$ )
(a) $3.18 \times 10^{6} \mathrm{~N} \mathrm{~m}^{-2}$
(b) $3.18 \times 10^{7} \mathrm{~N} \mathrm{~m}^{-2}$
(c) $3.18 \times 10^{8} \mathrm{~N} \mathrm{~m}^{-2}$
(d) $3.18 \times 10^{9} \mathrm{~N} \mathrm{~m}^{-2}$.

Objective NCERT at your FINGERTIPS: Physics

NCERT 11th Edition

Chapter 9, Problem 41

Instant Answer

Solved by Expert Kamlesh Goyal

09/25/2021

Step 1

The formula for the area of a circle is given by $A = \pi r^{2}$, where $r$ is the radius of the circle. Here, the radius of the rod is given as $10 \, \text{mm}$, which is $10 \times 10^{-3} \, \text{m}$. So, the area $A$ is given by: \[A = \pi (10 \times Show more…

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A steel rod of length $1 \mathrm{~m}$ and radius $10 \mathrm{~mm}$ is stretched by a force $100 \mathrm{kN}$ along its length. The stress produced in the rod is $\left(Y_{\text {Steel }}=2 \times 10^{11} \mathrm{~N} \mathrm{~m}^{-2}\right.$ ) (a) $3.18 \times 10^{6} \mathrm{~N} \mathrm{~m}^{-2}$ (b) $3.18 \times 10^{7} \mathrm{~N} \mathrm{~m}^{-2}$ (c) $3.18 \times 10^{8} \mathrm{~N} \mathrm{~m}^{-2}$ (d) $3.18 \times 10^{9} \mathrm{~N} \mathrm{~m}^{-2}$.

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Key Concepts

Cross-sectional Area Calculation

When dealing with cylindrical objects, the cross-sectional area is determined using the formula A = ?r², where r is the radius of the cylinder. Correctly calculating this area is crucial for accurately determining the stress experienced by the material.

Stress

Stress is the internal force experienced per unit area within a material when an external force is applied. It is computed by dividing the applied force by the cross-sectional area over which the force acts, and is typically measured in pascals (N/m²).

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