Department of Pre-University Education, KarnatakaPUC Karnataka Science Class 11

# A Vertical Metal Cylinder of Radius 2 Cm and Length 2 M is Fixed at the Lower End and a Load of 100 Kg is Put on It. Find (A) the Stress (B) the Strain and (C) the Compression of the Cylinder. - Physics

Short Note

A vertical metal cylinder of radius 2 cm and length 2 m is fixed at the lower end and a load of 100 kg is put on it. Find (a) the stress (b) the strain and (c) the compression of the cylinder. Young modulus of the metal = 2 × 1011 N m−2.

#### Solution

Given:
Radius of cylinder (r) = 2 cm = $2 \times {10}^{- 2} \text{ m}$

Length of cylinder (L) = 2 m
Mass of the load = 100 kg
Young's modulus of the metal = $2 \times {10}^{11} \text{ N/ m}^2$

(a) Stress(ρ) is given by :  $\frac{F}{A}$

Here, F is the force given by mg = $100 \times 10 = 1000 N$ ( Taking g = 10 m/s2)

A is the area of cross-section = πr2  = 4π $\times {10}^{- 4} m^2$
$\Rightarrow \text{ Stress } \rho = \frac{\text{ mg }}{A}$
$= \frac{\left( 100 \times 10 \right)}{\left( 4\pi \times {10}^{- 4} \right)}$
$= 7 . 96 \times {10}^5 \text{ N/ m}^2$
(b) Strain is given by:

$\text{ Strain } = \frac{\rho}{Y} = \frac{\left( 7 . 96 \times {10}^5 \right)}{\left( 2 \times {10}^{11} \right)}$

$= 4 \times {10}^{- 6}$

(c) Compression of the cylinder:
ΔL = strain × L
= 4 × 10−6 × 2 = 8 × 10−6 m

Concept: Elastic Moduli - Determination of Young’s Modulus of the Material of a Wire
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#### APPEARS IN

HC Verma Class 11, Class 12 Concepts of Physics Vol. 1
Chapter 14 Some Mechanical Properties of Matter
Q 2 | Page 300