MHT CET 2025 26th April Evening Shift | MHT CET Year Wise Previous Years Questions

MHT CET 2025 26th April Evening Shift

MCQ (Single Correct Answer)

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A long rectangular conducting loop of width ' $l$ ', mass ' $m$ ' and resistance ' $R$ ' is placed partly in a perpendicular magnetic field ' B '. It is pushed downwards with velocity ' $V$ ' so that it may continue to fall freely. The velocity ' $V$ ' is

$\frac{\mathrm{mgR}^2}{\mathrm{~B} l}$

$\frac{\mathrm{B}^2 l^2 \mathrm{R}}{\mathrm{mg}}$

$\frac{\mathrm{mgR}}{\mathrm{B}^2 l^2}$

$\frac{\mathrm{mgl}}{\mathrm{B}^2 \mathrm{R}^2}$

MHT CET 2025 26th April Evening Shift

MCQ (Single Correct Answer)

-0

The depth at which acceleration due to gravity becomes $\frac{g^{\prime}}{n}$ is ( $R=$ radius of earth, $\mathrm{g}=$ acceleration due to gravity) ( $\mathrm{n}=$ integer)

$\frac{R(n-1)}{n}$

$\frac{R(n+1)}{n}$

$\frac{R(n-1)^2}{n}$

$\frac{R(n+1)^2}{n}$

MHT CET 2025 26th April Evening Shift

MCQ (Single Correct Answer)

-0

The polarising angle of transparent medium is ' $\theta$ '. Let the speed of light in the medium be ' v '. Then the relation between ' $\theta$ ' and ' $\mathbf{v}$ ' is [ $\mathrm{c}=$ velocity of light in air]

$\quad \theta=\sin ^{-1}\left(\frac{\mathrm{v}}{\mathrm{c}}\right)$

$\theta=\tan ^{-1}\left(\frac{\mathrm{v}}{\mathrm{c}}\right)$

$\theta=\cot ^{-1}\left(\frac{\mathrm{v}}{\mathrm{c}}\right)$

$\quad \theta=\cos ^{-1}\left(\frac{\mathrm{v}}{\mathrm{c}}\right)$

MHT CET 2025 26th April Evening Shift

MCQ (Single Correct Answer)

-0

A black sphere has radius $R$ whose rate of radiation is E at temperature T . If radius is made R / 2 and temperature 3T, the rate of radiation will be

$\frac{3 \mathrm{E}}{2}$

$\frac{27 \mathrm{E}}{8}$

$\frac{81 \mathrm{E}}{4}$

$\frac{9 \mathrm{E}}{4}$

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