1
MHT CET 2024 9th May Evening Shift
MCQ (Single Correct Answer)
+1
-0

An e.m.f. $E=E_0 \cos \omega t$ is applied to circuit containing L and R in series. If $\mathrm{X}_{\mathrm{L}}=2 \mathrm{R}$, then the power dissipated in the circuit is

A
$\frac{\mathrm{E}_0{ }^2}{12 \mathrm{R}}$
B
$\frac{\mathrm{E}_0{ }^2}{10 \mathrm{R}}$
C
$\frac{\mathrm{E}_0{ }^2}{8 \mathrm{R}}$
D
$\frac{\mathrm{E}_0{ }^2}{6 \mathrm{R}}$
2
MHT CET 2024 9th May Evening Shift
MCQ (Single Correct Answer)
+1
-0

In the following circuit, the reading in the ammeter is

MHT CET 2024 9th May Evening Shift Physics - Current Electricity Question 46 English

A
25.1 mA
B
22.5 mA
C
21.5 mA
D
21.25 mA
3
MHT CET 2024 9th May Evening Shift
MCQ (Single Correct Answer)
+1
-0

A black body radiates power ' P ' and maximum energy is radiated by it at a wavelength $\lambda_0$. The temperature of the black body is now so changed that it radiates maximum energy at the wavelength $\frac{\lambda_0}{4}$. The power radiated by it at new temperature is

A
64 P
B
256 P
C
4 P
D
16 P
4
MHT CET 2024 9th May Evening Shift
MCQ (Single Correct Answer)
+1
-0

In Young's double slit experiment using monochromatic light of wavelength ' $\lambda$ ', the maximum intensity of light at a point on the screen is ' K ' units. The intensity of light at a point where the path difference is $\frac{\lambda}{6}$ ' is $\left(\cos 60^{\circ}=\sin 30^{\circ}=0.5, \sin 60^{\circ}=\cos 30^{\circ}=\sqrt{3} / 2\right)$

A
$\frac{3 \mathrm{~K}}{4}$
B
$\frac{\mathrm{K}}{4}$
C
$\frac{\mathrm{K}}{2}$
D
$\mathrm{K}$
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