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JEE Main 2023 (Online) 30th January Morning Shift
Numerical
+4
-1
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The general displacement of a simple harmonic oscillator is $$x = A\sin \omega t$$. Let T be its time period. The slope of its potential energy (U) - time (t) curve will be maximum when $$t = {T \over \beta }$$. The value of $$\beta$$ is ______________.

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2
JEE Main 2023 (Online) 30th January Morning Shift
Numerical
+4
-1
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In Young's double slit experiment, two slits $$S_{1}$$ and $$S_{2}$$ are '$$d$$' distance apart and the separation from slits to screen is $$\mathrm{D}$$ (as shown in figure). Now if two transparent slabs of equal thickness $$0.1 \mathrm{~mm}$$ but refractive index $$1.51$$ and $$1.55$$ are introduced in the path of beam $$(\lambda=4000$$ $$\mathop A\limits^o $$) from $$\mathrm{S}_{1}$$ and $$\mathrm{S}_{2}$$ respectively. The central bright fringe spot will shift by ___________ number of fringes.

JEE Main 2023 (Online) 30th January Morning Shift Physics - Wave Optics Question 35 English

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3
JEE Main 2023 (Online) 30th January Morning Shift
Numerical
+4
-1
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A thin uniform rod of length $$2 \mathrm{~m}$$, cross sectional area '$$A$$' and density '$$\mathrm{d}$$' is rotated about an axis passing through the centre and perpendicular to its length with angular velocity $$\omega$$. If value of $$\omega$$ in terms of its rotational kinetic energy $$E$$ is $$\sqrt{\frac{\alpha E}{A d}}$$ then value of $$\alpha$$ is ______________.

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4
JEE Main 2023 (Online) 30th January Morning Shift
Numerical
+4
-1
Change Language

A horse rider covers half the distance with $$5 \mathrm{~m} / \mathrm{s}$$ speed. The remaining part of the distance was travelled with speed $$10 \mathrm{~m} / \mathrm{s}$$ for half the time and with speed $$15 \mathrm{~m} / \mathrm{s}$$ for other half of the time. The mean speed of the rider averaged over the whole time of motion is $$\frac{x}{7} \mathrm{~m} / \mathrm{s}$$. The value of $$x$$ is ___________.

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