A steel wire can withstand a load up to $2940\, N$. A load of $150\, kg$ is suspended from a rigid support. The maximum angle through which the wire can be displaced from the mean position, so that the wire does not break when the load passes through the position of equilibrium, is
- $30^{\circ} $
- $60^{\circ} $
- $80^{\circ} $
- $85^{\circ} $
The Correct Option is B
Solution and Explanation
$T \cos \theta=m g$
$\Rightarrow \cos \theta=\frac{150 \times 9.8}{2940}$
$\Rightarrow \cos \theta=0.5$
$\Rightarrow \theta=60^{\circ}$
Top Questions on Stress and Strain
- Two blocks of mass \(2 \, \text{kg}\) and \(4 \, \text{kg}\) are connected by a metal wire going over a smooth pulley as shown in the figure. The radius of the wire is \(4.0 \times 10^{-5} \, \text{m}\) and Young's modulus of the metal is \(2.0 \times 10^{11} \, \text{N/m}^2\). The longitudinal strain developed in the wire is \(\frac{1}{\alpha \pi}\). The value of \(\alpha\) is _____. [Use \(g = 10 \, \text{m/s}^2\)]
- JEE Main - 2024
- Physics
- Stress and Strain
- The equivalent resistance between A and B is _________.
- JEE Main - 2023
- Physics
- Stress and Strain
- The Young's modulus of a steel wire of length $6 m$ and cross-sectional area $3 \,mm ^2$, is $2 \times 10^{11}$ $N / m ^2$ The wire is suspended from its support on a given planet A block of mass $4 kg$ is attached to the free end of the wire The acceleration due to gravity on the planet is $\frac{1}{4}$ of its value on the earth The elongation of wire is (Take $g$ on the earth $=10\, m / s ^2$ ) :
- JEE Main - 2023
- Physics
- Stress and Strain
- A force is applied to a steel wire ' $A$ ', rigidly clamped at one end As a result elongation in the wire is $02\, mm$ If same force is applied to another steel wire ' $B$ ' of double the length and a diameter $24$ times that of the wire ' $A$ ', the elongation in the wire ' $B$ ' will be (wires having uniform circular cross sections)
- JEE Main - 2023
- Physics
- Stress and Strain
A steel wire of length 3.2 m (Ys = 2.0 × 1011 Nm-2) and a copper wire of length 4.4 m (Yc = 1.1 × 1011 Nm-2), both of radius 1.4 mm are connected end to end. When stretched by a load, the net elongation is found to be 1.4 mm. The load applied, in Newton, will be:
\((Given: π = \frac{22}{7})\)- JEE Main - 2022
- Physics
- Stress and Strain
Questions Asked in BITSAT exam
- Two capacitors, of capacitance C, are connected in series. If one of them is filled with a dielectric substance K, what is the effective capacitance?
- BITSAT - 2023
- electrostatic potential and capacitance
- What is the dimension of the Gravitational constant?
- BITSAT - 2023
- Gravitation
- Minimum value of 5cos(2x) + 5sin(2x)
- BITSAT - 2023
- Trigonometric Functions
NaOH is deliquescent
- BITSAT - 2023
- States of matter
- The freezing point of the 0.05 molal solution of non - electrolyte in water is? (kf = 1.86)
- BITSAT - 2023
- Solutions
Concepts Used:
Stress and Strain
Stress and Strain are the terms in physics, which are used to explain deformation of solids.
What is Stress?
Force applied per unit area is known as stress.
- σ=F/A
- σ is stress applied
- F is force applied
- A is that the area of force applied
- Stress is measured by unit N/m2
What is Strain?
As a result of stress, change of shape is observed in the body. The change or deformity consequential to the stress acting on the body is called strain. Strain can be defined as the amount or measure of deformity that takes place due to the force applied on the object.
Strain is denoted with (ε). It has no units.
Longitudinal Strain = Δ L/L
Relation Between Stress and Strain
The English scientist Robert Hooke, while studying spring and elasticity, noticed that many materials displayed an identical property when the stress-strain relationship was studied. There exists a linear region where the force required to stretch the material was proportional to the extension of the material; this is called Hooke’s law. Mathematically, the law is presented as:
F = -k.x
Where, F = the force
x = the extension length
k = spring constant in N/m
SUBSCRIBE TO OUR NEWS LETTER
