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).