# 11 - Elasticity Questions Answers

**Asked By: KANDUKURI ASHISH**

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**Joshi sir comment**

Due to inner sphere attraction, outer sphere contracts so it is compressive stress.$GravitationalforceisG\left(\rho 4\pi {R}^{2}t\right)\left[\pi {(R\theta /2)}^{2}\right]/4{R}^{2}\phantom{\rule{0ex}{0ex}}Outwardforceduetostress=S\left[2\pi \right(R\theta /2\left)t\right](\theta /2)\phantom{\rule{0ex}{0ex}}Nowcompare.\phantom{\rule{0ex}{0ex}}Informforanyissueandwaitforvideo$

A solid sphere is subjected to uniform pressure from all direction. The ratio of volumetric strain to lateral strain produced in it, is

**Asked By: SWATI KAPOOR**

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**Joshi sir comment**

V = 4/3 πr^{3}

so dV = 4πr^{2}

now calculate dV/V

What is the length of a wire whose density is 8 g/cm^{3}, which suspended vertically, would break due to its own weight? [Breaking stress of wire is 2.4 x 10^{8} N/m^{2}]

**Asked By: SWATI KAPOOR**

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**Joshi sir comment**

use stress = force/area

When a load of 8 kg is hung on a wire, then extension of 3 cm takes place, the work done by internal forces of wire is

**Asked By: SWATI KAPOOR**

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**Joshi sir comment**

The work done per unit volume to stretch the length of area of cross-section 2 mm^{2} by 2% will be [ Y = 8 x 10^{10} N/m^{2} ]

**Asked By: SWATI KAPOOR**

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**Joshi sir comment**

To break a wire a stress of 9 x 10^{5} N/m^{2} is required. If density of the material of wire is 3 g/cc, then the minimum length of wire which will break by its own weight will be

**Asked By: SWATI KAPOOR**

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**Joshi sir comment**

use stress = force /area

A steel wire is 1m long and 1mm^{2} in area of cross-section. If it takes 200 N to stretch this wire by 1mm, how much force will be required to stretch a wire of the same material as well as diameter from its normal length of 10m to a length of 1002 cm?

**Asked By: SWATI KAPOOR**

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**Joshi sir comment**

by first set of data calculate Y then by using this Y solve the question

The length of wire, when M_{1} is hung from it, is l_{1} and is l_{2} with both M_{1} and M_{2} hanging. The natural length of wire is [ M_{2 }is on M_{1} ]

**Asked By: SWATI KAPOOR**

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**Joshi sir comment**

Y = M_{1}gL/A(l_{1}-L) = (M_{1}+M_{2})gL/(l_{2}-L)

solve for L

A nylon rope 3 cm in diameter has a breaking strength of 1.5 x 10^{5} N. The breaking strength for a similar rope 1.5cm in diameter is

**Asked By: SWATI KAPOOR**

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**Joshi sir comment**

breaking stress will remain same