Young’s modulus of rubber is \(10^4\text{ N/m}^2\) and area of cross-section is \(2\text{ cm}^2\). If force of \(2 \times 10^5\text{ dynes}\) is applied along its length, then length of wire becomes how much times of its initial length \(L\):- (Assume stress \(\propto\) strain)
1. three times
2. four times
3. two times
4. No change in length
View Answer
Converting Young's modulus to CGS units gives \(Y = 10^5\text{ dyne/cm}^2\). Stress is \(F/A = 2 \times 10^5 / 2 = 10^5\text{ dyne/cm}^2\). Since \(\text{Strain} = \text{Stress}/Y = 1\), we have \(\Delta L = L\). Thus, the final length becomes \(L + \Delta L = 2L\).
A metal block is experiencing an atmospheric pressure of \(1 \times 10^5\text{ N/m}^2\). When the same block is placed in a vacuum chamber, the fractional change in its volume is (the bulk modulus of metal is \(1.25 \times 10^{11}\text{ N/m}^2\))
1. \(4 \times 10^{-7}\)
2. \(2 \times 10^{-7}\)
3. \(8 \times 10^{-7}\)
4. \(1 \times 10^{-7}\)
View Answer
The bulk modulus is defined as \(B = \frac{\Delta P}{\Delta V/V}\). Moving to vacuum causes a pressure change of \(\Delta P = 10^5\text{ N/m}^2\). Thus, the fractional volume change is \(\frac{\Delta V}{V} = \frac{\Delta P}{B} = \frac{10^5}{1.25 \times 10^{11}} = 8 \times 10^{-7}\).
When the load on a wire is increased from \(3\text{ kg wt}\) to \(5\text{ kg wt}\) the elongation increases from \(0.61\text{ mm}\) to \(1.02\text{ mm}\). The required work done during the extension of the wire is :
1. \(16 \times 10^{-3}\text{ J}\)
2. \(8 \times 10^{-2}\text{ J}\)
3. \(20 \times 10^{-2}\text{ J}\)
4. \(11 \times 10^{-3}\text{ J}\)
View Answer
The work done during the extension is \(W = \frac{1}{2} (F_2 x_2 - F_1 x_1)\). Converting values: \(F_1 = 3 \times 9.8\text{ N}\), \(F_2 = 5 \times 9.8\text{ N}\), \(x_1 = 0.61 \times 10^{-3}\text{ m}\), and \(x_2 = 1.02 \times 10^{-3}\text{ m}\) yields \(W \approx 16 \times 10^{-3}\text{ J}\).
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:
1. \(\frac{2B}{rho g}\)
2. \(\frac{rho}{Bg}\)
3. \(\frac{B}{rho g}\)
4. \(\frac{rho g}{2B}\)
View Answer
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 \(L = \frac{B}{\rho g}\).
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
1. \(\sqrt{2} : 1\)
2. \(1 : \sqrt{2}\)
3. 4 : 1
4. 1 : 4
View Answer
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.
1. Both (A) and (R) are true and (R) is the correct explanation of (A)
2. Both (A) and (R) are true but (R) is not the correct explanation of (A)
3. (A) is true but (R) is false
4. Both (A) and (R) are false
View Answer
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
1. \(\sqrt{2} : 1\)
2. \(1 : \sqrt{2}\)
3. 4 : 1
4. 1 : 4
View Answer
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.
1. Both (A) & (R) are true and the (R) is the correct explanation of the (A)
2. Both (A) & (R) are true but the (R) is not the correct explanation of the (A)
3. (A) is true but (R) is false
4. Both (A) and (R) are false
View Answer
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).