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

A step down transformer is used to reduce the main supply from '$$V_1$$' volt to '$$V_2$$' volt. The primary coil draws a current '$$\mathrm{I}_1$$' $$\mathrm{A}$$ and the secondary coil draws '$$\mathrm{I}_2$$' A. $$(\mathrm{I}_1<\mathrm{I}_2)$$. The ratio of input power to output power is

$$\frac{V_1 V_2}{I_1 I_2}$$

$$\frac{V_1 I_1}{V_2 I_2}$$

$$\frac{\mathrm{I}_1 \mathrm{I}_2}{\mathrm{~V}_1 \mathrm{~V}_2}$$

$$\frac{\mathrm{V}_1 \mathrm{I}_2}{\mathrm{~V}_2 \mathrm{I}_1}$$

MHT CET 2021 22th September Evening Shift

MCQ (Single Correct Answer)

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For the circuit shown below, instantaneous current through inductor '$$\mathrm{L}$$' and capacitor '$$\mathrm{C}$$' is respectively.

MHT CET 2021 22th September Evening Shift Physics - Alternating Current Question 21 English

$$\frac{-\mathrm{e}_0}{\omega \mathrm{L}} \cos \omega \mathrm{t} ; \mathrm{e}_0 \omega \mathrm{c} \cos \omega \mathrm{t}$$

$$\frac{-\mathrm{e}_0}{\omega \mathrm{L}} \sin \omega t ; \frac{\mathrm{e}_0}{\omega \mathrm{C}} \cos \omega \mathrm{t}$$

$$\frac{\mathrm{e}_0 \mathrm{C}}{\mathrm{L}} \cos \omega \mathrm{t}; \frac{\mathrm{e}_0 \mathrm{~L}}{\mathrm{C}} \sin \omega \mathrm{t}$$

$$\frac{-\mathrm{e}_0 \mathrm{C}}{\mathrm{L}} \sin \omega \mathrm{t} ; \frac{\mathrm{e}_0 \mathrm{~L}}{\mathrm{C}} \cos \omega \mathrm{t}$$

MHT CET 2021 22th September Evening Shift

MCQ (Single Correct Answer)

-0

The light of wavelength '$$\lambda$$' is incident on the surface of metal of work function $$\phi$$ and emits the electron. The maximum velocity of electron emitted is [$$\mathrm{m}=$$ mass of electron and $$\mathrm{h}=$$ Planck's constant, $$\mathrm{c}=$$ velocity of light]

$$\left[\frac{2(h c-\lambda)}{\mathrm{m} \lambda}\right]^{\frac{1}{2}}$$

$$\left[\frac{2(h c-\phi) \lambda}{m c}\right]$$

$$\left[\frac{2(h c-\lambda)}{m \lambda}\right]$$

$$\left[\frac{2(\mathrm{hc}-\lambda \phi)}{\mathrm{m} \lambda}\right]^{\frac{1}{2}}$$

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