MHT CET 2021 22th September Morning Shift | MHT CET Year Wise Previous Years Questions

MHT CET 2021 22th September Morning Shift

MCQ (Single Correct Answer)

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Kinetic energy of a proton is equal to energy '$$E$$' of a photon. Let '$$\lambda_1$$' be the de-Broglie wavelength of proton and '$$\lambda_2$$' is the wavelength of photon. If $$\frac{\lambda_1}{\lambda_2} \alpha E^n$$, then the value of '$$n$$' is

$$\frac{1}{2}$$

$$\frac{1}{4}$$

MHT CET 2021 22th September Morning Shift

MCQ (Single Correct Answer)

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A drop of liquid of density '$$\rho$$' is floating half immersed in a liquid of density '$$d$$'. If '$$T$$' is the surface tension, then the diameter of the drop of the liquid is

$$\sqrt{\frac{6 \mathrm{~T}}{\mathrm{~g}(2 p-\mathrm{d})}}$$

$$\sqrt{\frac{T}{g(2 \rho-d)}}$$

$$\sqrt{\frac{2 T}{g(2 p-d)}}$$

$$\sqrt{\frac{12 T}{g(2 \rho-d)}}$$

MHT CET 2021 22th September Morning Shift

MCQ (Single Correct Answer)

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Assuming the atom is in the ground state, the expression for the magnetic field at a point nucleus in hydrogen atom due to circular motion of electron is [$$\mu_0=$$ permeability of free space, $$\mathrm{m}=$$ mass of electron, $$\epsilon_0=$$ permittivity of free space, $$\mathrm{h}=$$ Planck's constant ]

$$\frac{\mu_0 \mathrm{e}^7 \pi \mathrm{m}^2}{8 \in_0^3 \mathrm{~h}^5}$$

$$\frac{\mu_0 \mathrm{e}^5 \pi \mathrm{m}^3}{8 \epsilon_0^3 \mathrm{~h}^5}$$

$$\frac{\mu_0 \mathrm{e}^5 \pi^2 \mathrm{~m}^2}{8 \epsilon_0^2 \mathrm{~h}^4}$$

$$\frac{\mu_0 \mathrm{e}^7 \pi^2 \mathrm{~m}^2}{8 \epsilon_0^3 \mathrm{~h}^5}$$