1
MHT CET 2026 17th April Morning Shift
MCQ (Single Correct Answer)
+1
-0
A lens of refractive index '$\mu$' has focal length '$f$'. When the lens is immersed in a liquid of refractive index '$\mu_0$', its focal length becomes '$f_0$'. Then '$f_0$' is given by
A
$\dfrac{(\mu_0-\mu)f}{\mu(\mu_0-1)}$
B
$\dfrac{\mu(\mu_0-1)f}{(\mu_0-\mu)}$
C
$\dfrac{(\mu-\mu_0)f}{\mu_0(\mu-1)}$
D
$\dfrac{\mu_0(\mu-1)f}{(\mu-\mu_0)}$
2
MHT CET 2026 17th April Morning Shift
MCQ (Single Correct Answer)
+1
-0
In Young's double slit experiment, width of the second slit is double the width of first slit, consequently the amplitude of the light from two slits. '$I_m$' is the maximum intensity. The resultant intensity '$I$' when they interfere with the phase difference of $\phi$ is given by
A
$\dfrac{I_m}{9}\left(1 + 8\cos^2\dfrac{\phi}{2}\right)$
B
$\dfrac{I_m}{7}\left(3 + 5\cos^2\dfrac{\phi}{2}\right)$
C
$\dfrac{I_m}{5}\left(1 + 2\cos^2\dfrac{\phi}{2}\right)$
D
$\dfrac{I_m}{3}\left(1 + 6\cos^2\dfrac{\phi}{2}\right)$
3
MHT CET 2026 17th April Morning Shift
MCQ (Single Correct Answer)
+1
-0
In a single slit diffraction experiment, for wavelength '$\lambda$', half angular width of the principal maxima is '$\theta$'. Also for wavelength of light '$p\lambda$', the half angular width of the principal maxima is '$q\theta$'. The ratio of the half angular widths of the first secondary maxima in the first case to second case will be
A
$p : 1$
B
$1 : q$
C
$p : q$
D
$q : p$
4
MHT CET 2026 17th April Morning Shift
MCQ (Single Correct Answer)
+1
-0
Photoelectric emission is observed from a metallic surface for frequencies $\nu_1$ and $\nu_2$ of the incident light rays ($\nu_1 > \nu_2$). If the ratio of maximum value of kinetic energy of the photoelectron emitted in first case to that in second case 3 : K, then the threshold frequency of the metallic surface is
A
$\dfrac{K\nu_1 - \nu_2}{K-1}$
B
$\dfrac{K-1}{K\nu_1 - \nu_2}$
C
$\dfrac{K-3}{K\nu_1 - 3\nu_2}$
D
$\dfrac{K\nu_1 - 3\nu_2}{K-3}$

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