1
MHT CET 2026 17th April Evening Shift
MCQ (Single Correct Answer)
+1
-0
In an a.c. circuit with pure capacitance 'C' and a.c. source $E = E_0\sin\omega t$, the equation of instantaneous current is given by
A
$I = E_0\,\omega C \cdot \sin(\omega t)$
B
$I = E_0\omega C\sin\left(\omega t + \dfrac{\pi}{2}\right)$
C
$I = \dfrac{E_0}{\omega C}\sin(\omega t)$
D
$I = \dfrac{E_0}{\omega C}\sin\left(\omega t + \dfrac{\pi}{2}\right)$
2
MHT CET 2026 17th April Evening Shift
MCQ (Single Correct Answer)
+1
-0
In a series LCR circuit alternating e.m.f. and current are given by the equations $V = V_0\sin(\omega t)$ and $I = I_0\sin\left(\omega t + \dfrac{\pi}{3}\right)$ respectively. The average power dissipated in the circuit over one cycle of a.c. is $(\cos 60^\circ = 0.5)$
A
zero
B
$\dfrac{V_0 I_0}{2}$
C
$\dfrac{\sqrt{3}}{2}V_0 I_0$
D
$\dfrac{V_0 I_0}{4}$
3
MHT CET 2026 17th April Evening Shift
MCQ (Single Correct Answer)
+1
-0
A ray of light incident on one face of an equilateral glass prism having refractive index $\sqrt{2}$, produces the emergent ray which just grazes along the adjacent face. The value of angle of incidence is $(\sin 90^\circ = 1)$
$\left(\sin 30^\circ = \dfrac{1}{2}\right)\left(\sin 45^\circ = \dfrac{1}{\sqrt{2}}\right)$
A
$\sin^{-1}\left(\sqrt{2}\sin 15^\circ\right)$
B
$\sin^{-1}\left(\dfrac{1}{\sqrt{2}}\sin 15^\circ\right)$
C
$\sin^{-1}\left(\sqrt{2}\sin 30^\circ\right)$
D
$\sin^{-1}\left(\dfrac{1}{\sqrt{2}}\sin 45^\circ\right)$
4
MHT CET 2026 17th April Evening Shift
MCQ (Single Correct Answer)
+1
-0
An unpolarised light of intensity $64 \text{ Wm}^{-2}$ passes through three polarizers successively such that the transmission axis of last polarizer is crossed with the first. The intensity of the emerging light is $6 \text{ Wm}^{-2}$. The angle between the transmission axis of the first two polarizers is
A
$\dfrac{1}{2}\sin^{-1}\left(\dfrac{\sqrt{3}}{2}\right)$
B
$\dfrac{1}{2}\cos^{-1}\left(\dfrac{\sqrt{3}}{2}\right)$
C
$\dfrac{1}{2}\sin^{-1}\left(\dfrac{1}{\sqrt{3}}\right)$
D
$\dfrac{1}{2}\cos^{-1}\left(\dfrac{1}{\sqrt{3}}\right)$

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