The bulk modulus for an incompressible liquid is :
For an incompressible liquid, the volume change \(\Delta V = 0\) for any pressure change \(\Delta P\). Since bulk modulus is given by \(B = -V \frac{\Delta P}{\Delta V}\), dividing by zero results in \(B = \infty\) (infinity).
If \(\rho\) is the density of the material of a wire and \(B\) is the breaking stress, the greatest length of the wire that can hang freely without breaking is:
Breaking stress is \(B = \frac{\text{Maximum Tension}}{\text{Area}}\). For a wire of length \(L\) hanging freely, the maximum tension is at the support: \(T = mg = A L \rho g\). Hence, \(B = L \rho g\), which gives \(L = \frac{B}{\rho g}\).
A steel rod has a radius of 10 mm and a length of 1 m. A 100 kN force stretches it along its length. The elongation in rod is (Assume young’s modulus \(Y = 2 times 10^{11}\text{ N/m}^2\))
Formula: \(Delta L = frac{FL}{AY}\). Area \(A = pi r^2 = 3.14 times 10^{-4}\text{ m}^2\. Substituting the given values yields \(Delta L = frac{10^5 times 1}{3.14 times 10^{-4} times 2 times 10^{11}} approx 1.59\text{ mm}\.
The Young’s modulus of brass and steel are \(1 \times 10^{11}\text{ N/m}^2\) and \(2 \times 10^{11}\text{ N/m}^2\) respectively. If wires of both materials, having same length, are loaded with same weight, then they both extend by 4 mm. Ratio of the radii of two wires \(R_B : R_S\) is
Since length, load, and extension are the same: \(Y = \frac{FL}{\pi R^2 \Delta L} ⇒ R^2 \propto \frac{1}{Y} ⇒ \frac{R_B}{R_S} = \sqrt{\frac{Y_S}{Y_B}} = \sqrt{\frac{2 \times 10^{11}}{1 \times 10^{11}}} = \sqrt{2} : 1\).
Given below are two statements: one is labelled as Assertion A and the other is labelled as Reason R.
Assertion (A): The unit of stress is same as that of pressure.
Reason (R): Stress is a vector quantity.
In the light of above statements, select the correct option.
Assertion (A) is true as both stress and pressure are measured in \( \text{N/m}^2 \) (or Pa). Reason (R) is false because stress is a tensor quantity (neither a scalar nor a simple vector).
The Young’s modulus of brass and steel are \(1 \times 10^{11}\text{ N/m}^2\) and \(2 \times 10^{11}\text{ N/m}^2\) respectively. If wires of both materials, having same length, are loaded with same weight, then they both extend by 4 mm. Ratio of the radii of two wires \(R_B : R_S\) is
Using \(Y = \frac{FL}{\pi R^2 \Delta L}\), for constant force, length, and extension, \(R^2 \propto \frac{1}{Y}\). Thus, \(\frac{R_B}{R_S} = \sqrt{\frac{Y_S}{Y_B}} = \sqrt{\frac{2 \times 10^{11}}{1 \times 10^{11}}} = \sqrt{2} : 1\).
Assertion (A): Identical springs of steel and copper are equally stretched. More work will be done on the steel spring.
Reason (R): Steel is more elastic than copper.
Assertion (A) is true. Work done to stretch a spring is \(W = \frac{1}{2} k x^2\). Steel has a higher Young's modulus than copper, implying a higher spring constant \(k\) for identical dimensions.
Thus, more work is done on the steel spring.
Reason (R) is true. Steel is indeed more elastic than copper (possesses a higher Young's modulus).
Reason (R) correctly explains Assertion (A).
Given below are two statements: one is labelled as Assertion A and the other is labelled as Reason R.
Assertion A: Strain is a dimensionless quantity.
Reason R: Unit of Young’s modulus is same as that of stress.
In the light of the above statements, choose the correct answer from the option given below.
Strain is indeed dimensionless since it's the ratio of two similar physical quantities (change in dimension over original dimension). The unit of Young's modulus is indeed identical to stress (both are \(\text{N/m}^2\)). However, the second statement is not the reason for the first.
The experiment which is used to determine Young’s modulus of the material of a given wire, is
Searle's apparatus/experiment is specifically designed and used to determine the Young's modulus of elasticity of a metal wire by measuring elongation under load.
A wire of length $L$ area of cross section $A$ is hanging from a fixed support. The length of the wire changes to $L_1$ when mass $M$ is suspended from its free end. The expression for Young’s modulus is:
(2020)
Young's modulus $Y = \frac{\text{Stress}}{\text{Strain}} = \frac{Mg/A}{\Delta L/L}$. Here, the change in length $\Delta L = L_1 - L$. Substituting this gives $Y = \frac{MgL}{A(L_1 - L)}$.