1
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
$q : p$
2
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
$\dfrac{K\nu_1 - 3\nu_2}{K-3}$
3
MHT CET 2026 17th April Morning Shift
MCQ (Single Correct Answer)
+1
-0
The kinetic energy of a free electron increases to 3 times the previous K.E. The ratio of new de-Broglie wavelength to previous de-Broglie wavelength is
A
$\dfrac{1}{\sqrt{3}}$
B
$\dfrac{1}{3}$
C
$3$
D
$\sqrt{3}$
4
MHT CET 2026 17th April Morning Shift
MCQ (Single Correct Answer)
+1
-0
For wavelength of visible radiation of Hydrogen spectrum, Balmer gave an equation as $\lambda = \dfrac{xm^2}{m^2-4}$, where m is the integer value. The value of x in terms of Rydberg's constant R is
A
$\dfrac{R}{4}$
B
$\dfrac{4}{R}$
C
$2R$
D
$4R$

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