At what angle should the two forces $2 \overrightarrow{\mathrm{P}}$ and $\sqrt{2} \overrightarrow{\mathrm{P}}$ act so that the resultant force is $\sqrt{10} \overrightarrow{\mathrm{P}}$ ?
$\cos ^{-1}\left(\frac{1}{\sqrt{2}}\right)$
$\cos ^{-1}\left(\frac{1}{2}\right)$
$\cos ^{-1}\left(\frac{\sqrt{3}}{2}\right)$
$\cos ^{-1}\left(\frac{1}{\sqrt{3}}\right)$
If the r.m.s. velocity of gas is V at temperature T , then
$\mathrm{VT}^2=$ constant
$\mathrm{V}^2 \mathrm{~T}=$ constant
$\mathrm{V}^2 / \mathrm{T}=$ constant
$\quad \mathrm{VT}=$ constant
Light propagates 2 cm distance in glass of refractive index 1.5 in certain time. In the same time same light propogates a distance of 2.25 cm in a medium. The refractive index of the medium is
$\frac{4}{3}$
$\frac{3}{2}$
$\frac{8}{3}$
$\frac{3}{4}$
In photoelectric effect experiment, the stopping potential for a given metal is $\mathrm{V}_0$ (in volt) when radiation of wavelength $\lambda_0$ is used. If radiation of wavelength $5 \lambda_0$ is used with the same metal, then the stopping potential is (in volt) ( $\mathrm{h}=$ Planck's constant, $\mathrm{c}=$ velocity of light, $\mathrm{e}=$ electronic charge)
$\quad \mathrm{V}_0-\frac{4 \mathrm{hc}}{5 \mathrm{e} \lambda_0}$
$\quad \mathrm{V}_0+\frac{4 \mathrm{hc}}{5 \mathrm{e} \lambda_0}$
$\mathrm{V}_0+\frac{2 \mathrm{hc}}{3 \mathrm{e} \lambda_0}$
$\quad \mathrm{V}_0-\frac{2 \mathrm{hc}}{3 \mathrm{e} \lambda_0}$
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