Solids - NEET Physics Chapterwise MCQs & PYQs

NEET Solids MCQs & PYQs

Practice NEET Solids Questions

Question 21:

easy

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)

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 \(L + \Delta L = 2L\).

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Question 22:

easy

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 \(1.25 \times 10^{11}\text{ N/m}^2\))

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 \(\frac{\Delta V}{V} = \frac{\Delta P}{B} = \frac{10^5}{1.25 \times 10^{11}} = 8 \times 10^{-7}\).

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Question 23:

easy

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 :

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 \(W \approx 16 \times 10^{-3}\text{ J}\).

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Question 24:

easy

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).

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Question 25:

easy

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 \(L = \frac{B}{\rho g}\).

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Question 26:

easy

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}\.

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Question 27:

easy

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: \(Y = \frac{FL}{\pi R^2 \Delta L} ⇒ R^2 \propto \frac{1}{Y} ⇒ \frac{R_B}{R_S} = \sqrt{\frac{Y_S}{Y_B}} = \sqrt{\frac{2 \times 10^{11}}{1 \times 10^{11}}} = \sqrt{2} : 1\).

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Question 28:

easy

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).

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Question 29:

easy

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, \(\frac{R_B}{R_S} = \sqrt{\frac{Y_S}{Y_B}} = \sqrt{\frac{2 \times 10^{11}}{1 \times 10^{11}}} = \sqrt{2} : 1\).

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Question 30:

easy

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).

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