JEE Advanced 2019 · Paper 2

A thin and uniform rod of mass M and length L is held vertical on a floor with large friction. The rod is released from rest so that it falls by rotating about its contact-point with the floor without slipping. Which of the following statement(s) is/are correct, when the rod makes an angle 60° with vertical? [g is the acceleration due to gravity]

1. AThe angular speed of the rod will be √(3g/2L)
2. BThe angular acceleration of the rod will be 2g/L
3. CThe radial acceleration of the rod’s center of mass will be 3g/4
4. DThe normal reaction force from the floor on the rod will be Mg/16

A mixture of ideal gas containing 5 moles of monatomic gas and 1 mole of rigid diatomic gas is initially at pressure P₀, volume V₀, and temperature T₀. If the gas mixture is adiabatically compressed to a volume V₀/4, then the correct statement(s) is/are, (Given 2^1.2 = 2.3; 2^3.2 = 9.2, R is gas constant)

1. AThe work |W| done during the process is 13RT₀
2. BThe average kinetic energy of the gas mixture after compression is in between 18RT₀ and 19RT₀
3. CThe final pressure of the gas mixture after compression is in between 9P₀ and 10P₀
4. DAdiabatic constant of the gas mixture is 1.6

A free hydrogen atom after absorbing a photon of wavelength λₐ gets excited from the state n = 1 to the state n = 4. Immediately after that the electron jumps to n = m state by emitting a photon of wavelength λₑ. Let the change in momentum of atom due to the absorption and the emission are Δpₐ and Δpₑ, respectively. If λₐ/λₑ = 1/5, which of the option(s) is/are correct? [Use hc = 1242 eV nm ; 1 nm = 10⁻⁹ m, h and c are Planck’s constant and speed of light, respectively]

1. Am = 2
2. Bλₑ = 418 nm
3. CΔpₐ/Δpₑ = 1/2
4. DThe ratio of kinetic energy of the electron in the state n = m to the state n = 1 is 1/4

If the tension in each string is T₀, the correct match for the highest fundamental frequency in f₀ units will be,

1. AI → P, II → Q, III → T, IV → S
2. BI → P, II → R, III → S, IV → Q
3. CI → Q, II → S, III → R, IV → P
4. DI → Q, II → P, III → R, IV → T

The length of the strings 1, 2, 3 and 4 are kept fixed at L₀, 3L₀/2, 5L₀/2, and 7L₀/4, respectively. Strings 1, 2, 3, and 4 are vibrated at their 1st, 3rd, 5th, and 14th harmonics, respectively, such that all the strings have same frequency. The correct match for the tension in the four strings in the units of T₀ will be,

1. AI → P, II → R, III → T, IV → U
2. BI → P, II → Q, III → T, IV → U
3. CI → P, II → Q, III → R, IV → T
4. DI → T, II → Q, III → R, IV → U