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Published on: 07/06/2021
QB365 provides detailed and simple solution for every Book back Questions in class 11 Physics Subject. It will helps to get more idea about question pattern in every book back questions with solution.
Download Tamil Nadu 11th Standard Physics question papers, model tests, one-mark questions, important questions, and public exam papers in PDF format. Free study materials and answer keys for TN State Board students.
Questions + Answers key
Take MCQ Physics Test1.
A spring balance shows wrong readings after using for a long time. Why?
2.
State Hooke’s law of elasticity.
3.
A wire of length 2 m with the area of crosssection 10-6m2 is used to suspend a load of 980 N. Calculate
i) the stress developed in the wire
ii) the strain and
iii) the energy stored. Given: Y = 12 \(\times\) 1010N m−2.
4.
A metallic cube of side 100 cm is subjected to a uniform force acting normal to the whole surface of the cube. The pressure is 106 pascal. If the volume changes by 1.5\(\times\)10-5 m3, calculate the bulk modulus of the material.
5.
Explain elasticity using intermolecular forces.
1.
This is because of continuous usage, the wire loses its elasticity and does not regain it original dimension.
i.e: Length. Because of this, it shows wrong readings.
2.
Hooke's law of elasticity states that within the elastic limit, the strain produced in a body is directly proportional to the stress applied.
strain ∝ stress
3.
(i) stress = \(\frac { F }{ A } = \frac { 980 }{ { 10 }^{ -6 } } =98\times { 10 }^{ 7 }N \ m^{ -2 }\)
(ii) strain \(=\frac { stress }{ Y } =\frac { 98\times { 10 }^{ 7 } }{ 12\times { 10 }^{ 10 } } =8.17\times { 10 }^{ -3 }\)
(no units)
(iii) Since, volume = 2\(\times\)10-6m3
Energy = \(\frac{1}{2}\) (stress\(\times\)strain) \(\times\)volume \(=\frac{1}{2}\)(98 \(\times\)107 ) x (8.17\(\times\)10 3) 2 x 10−6 = 8 J
4.
By definition, K =\(\frac { \frac { F }{ A } }{ \frac { \Delta V }{ V } } =P\frac { V }{\Delta{V} }\)
\(K=\frac { { 10 }^{ 6 }\times 1 }{ 1.5\times { 10 }^{ -5 } } =6.67\times { 10 }^{ 10 }{ N \ m }^{ -2 }\)
5.
In a solid, interatomic forces bind two or more atoms together and the atoms occupy the positions of stable equilibrium. When a deforming force is applied on a body, its atoms are pulled apart or pushed closer. When the deforming force is removed, interatomic forces of attraction or repulsion restore the atoms to their equilibrium positions. If a body regains its original shape and size after the removal of deforming force, it is said to be elastic and the property is called elasticity. The force which changes the size or shape of a body is called a deforming force.
Examples: Rubber, metals, steel ropes.
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Tamilnadu Stateboard 11th Standard Subjects

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Computer Science

Physics

Chemistry

Maths

Biology

Economics

Physics

Chemistry

History

Business Maths and Statistics

Computer Science

Accountancy

Computer Applications

History

Computer Technology

Commerce

Computer Applications

Computer Technology

Tamil

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