JEE Physics Mechanical Properties Of Solids MCQs Set 03

Answer: (d) \(5\times10^{-4}\) J

Explanation:
1. \(W=\tfrac12Fx=\frac{YA}{2L}x^2\)
2. \(=\frac{2\times10^{11}\times10^{-6}}{2\times2}\times(10^{-4})^2=5\times10^{-4}\) J

Teacher's Note:
1. Work done in stretching is the area under the force-extension line, \(\tfrac12\times\text{force}\times\text{extension}\).

Answer: (c) \(1/\gamma\)

Explanation:
1. Isothermal bulk modulus \(E_T=P\)
2. Adiabatic bulk modulus \(E_S=\gamma P\)
3. Ratio \(E_T/E_S=1/\gamma\)

Teacher's Note:
1. For a gas the isothermal bulk modulus equals the pressure and the adiabatic one is \(\gamma\) times as large.

Answer: (c) \(1\times10^{12}\)

Explanation:
1. \(B=\frac{\Delta P}{\Delta V/V}\)
2. \(\Delta P=100\times1.013\times10^6\approx10^8\) dyne/cm\(^2\) and \(\Delta V/V=10^{-4}\)
3. \(B=\frac{10^8}{10^{-4}}=10^{12}\) dyne/cm\(^2\)

Teacher's Note:
1. 0.01 percent means a fractional change of \(10^{-4}\). Take care with CGS units.

Answer: (c) 0.5%

Explanation:
1. Stress \(=\frac{1000}{10^{-6}}=10^9\) Pa
2. Strain \(=\frac{10^9}{2\times10^{11}}=5\times10^{-3}\)
3. Percentage increase = 0.5%

Teacher's Note:
1. Convert mm\(^2\) to m\(^2\) before dividing and remember that GPa is \(10^9\) Pa.

Answer: (a) 70 degrees C

Explanation:
1. Strain in the wire \(=\frac{F}{AY}=\frac{200}{2\times10^{-6}\times2\times10^{11}}=5\times10^{-4}\)
2. The tension vanishes when the thermal expansion equals this strain: \(\alpha\Delta T=5\times10^{-4}\)
3. \(\Delta T=50\), so the temperature is 20 + 50 = 70 degrees C

Teacher's Note:
1. Heating a wire fixed between rigid supports reduces the tension as the wire tries to expand.

Answer: (a) \(\tfrac12YX^2\)

Explanation:
1. Energy per unit volume \(=\tfrac12\times\text{stress}\times\text{strain}\)
2. Stress \(=YX\)
3. So the energy density is \(\tfrac12YX^2\)

Teacher's Note:
1. Energy density is the area under the stress-strain line.

Answer: (a) 9

Explanation:
1. The spring constant is inversely proportional to the number of turns (length)
2. Three-fourths of the length has spring constant \(k'=\tfrac43k\)
3. \(x'=\frac{mg}{k'}=\frac34x=\frac34\times12=9\) cm

Teacher's Note:
1. A shorter piece of the same spring is stiffer. Extension of the same load falls in proportion to the length.

Answer: (b) \(\frac{3a}{2b^2c}\)

Explanation:
1. The steel wire supports \(3m\) and the brass wire supports \(2m\)
2. \(\Delta l=\frac{FL}{\pi r^2Y}\)
3. \(\frac{\Delta l_s}{\Delta l_b}=\frac{3mg}{2mg}\cdot\frac{L_s}{L_b}\cdot\frac{r_b^2}{r_s^2}\cdot\frac{Y_b}{Y_s}=\frac32\cdot a\cdot\frac1{b^2}\cdot\frac1c=\frac{3a}{2b^2c}\)

Teacher's Note:
1. The upper wire carries the weight of everything hanging below it. Write the ratio of each factor of \(FL/AY\) separately.

Answer: (a) Length = 50 cm and diameter = 0.5 mm

Explanation:
1. \(\Delta l=\frac{4FL}{\pi d^2Y}\propto\frac{L}{d^2}\) for the same force and material
2. The values of \(L/d^2\) (cm per mm\(^2\)) are 50/0.25 = 200, 100/1 = 100, 200/4 = 50 and 300/9 = 33
3. The first wire has the largest extension

Teacher's Note:
1. Extension grows with length but falls with the square of the diameter, so thin wires stretch most.