BK

Bhavesh Kumar

Numerade Educator
Teaching Assistant

Biography

I have done my masters in Mathematics form prestigious Ramjas college from Delhi University. I have been teaching Mathematics since my graduation time. After Masters I am a full time teacher of Mathematics. I have a total experience of 8 years of teaching Mathematics.

Education

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Educator Statistics

Numerade tutor for 5 years
70 Students Helped

Topics Covered

Applications of the Derivative
Discover the Power of Gravitation: Exploring the Science Behind It

Bhavesh's Textbook Answer Videos

03:59
Master Resource Book in JEE Main Physics

Moment of inertia of cylinder about an axis through the centre and perpendicular to its axis is
$$
I_{c}=M\left(\frac{R^{2}}{4}+\frac{L^{2}}{12}\right)
$$
Using theorem of parallel axes, moment of inertia of the cylinder about an axis through its edge would be
$$
I=I_{c}+M\left(\frac{L}{2}\right)^{2}=M\left(\frac{R^{2}}{4}+\frac{L^{2}}{12}+\frac{L^{2}}{4}\right)=M\left(\frac{R^{2}}{4}+\frac{L^{2}}{3}\right)
$$
When $L=6 R, \quad I_{h}=\frac{49}{4} M R^{2}$

Chapter 10: Gravitation
Section 2: Round 2
Bhavesh Kumar
02:12
Master Resource Book in JEE Main Physics

$\mathrm{As}, I_{s}=\frac{2}{5} M R_{s}^{2}, I_{h}=\frac{2}{3} M R_{h}^{2}$
As $\quad I_{s}=I_{h}$
$\therefore \quad \frac{2}{5} M R_{s}^{2}=\frac{2}{3} M R_{h}^{2}$
$\therefore$
$\frac{R_{s}}{R_{h}}=\frac{\sqrt{5}}{\sqrt{3}}$

Chapter 10: Gravitation
Section 2: Round 2
Bhavesh Kumar
01:44
Master Resource Book in JEE Main Physics

Here, $\pi r=l \therefore r=1 / \pi$
Moment of inertia of a ring about its diameter $=\frac{1}{2} M r^{2}$ $\therefore$ Moment of inertia of semicircle $=\frac{1}{2}\left[m\left(\frac{l}{\pi}\right)^{2}\right]=\frac{m l^{2}}{2 \pi^{2}}$.

Chapter 10: Gravitation
Section 2: Round 2
Bhavesh Kumar
03:24
Master Resource Book in JEE Main Physics

As, $\omega_{2}=\omega_{1}+\alpha t$
$\therefore \quad 40 \pi=20 \pi+\alpha \times 10$
or $\alpha=2 \pi \mathrm{rad} \mathrm{s}^{-2}$
$\begin{aligned}&\text { From, } \quad \omega_{2}^{2}-\omega_{1}^{2} & =2 \alpha \theta \\& & (40 \pi)^{2}-(20 \pi)^{2} & =2 \times 2 \pi \theta \\&\Rightarrow \quad \theta & =\frac{1200 \pi^{2}}{4 \pi}=300 \pi\end{aligned}$
Number of rotations completed $=\frac{\theta}{2 \pi}=\frac{300}{2 \pi}=150$

Chapter 10: Gravitation
Section 2: Round 2
Bhavesh Kumar
02:34
Master Resource Book in JEE Main Physics

From conservation of angular momentum,
$$l_{1} \omega_{1}=I_{2} \omega_{2}$$
$\therefore \quad \frac{\omega_{1}}{\omega_{2}}=\frac{l_{2}}{l_{1}}$
Now, $\begin{aligned} \frac{E_{1}}{E_{2}} &=\frac{\frac{1}{2} l_{1} \omega_{1}^{2}}{\frac{1}{2} I_{2} \omega_{2}^{2}} \\ &=\frac{l_{1}}{l_{2}} \times\left(\frac{l_{2}}{l_{1}}\right)^{2}=\frac{l_{2}}{l_{1}} \end{aligned}$
As $\quad I_{1}>I_{2}$
$\therefore \quad E_{1}<E_{2}$

Chapter 10: Gravitation
Section 2: Round 2
Bhavesh Kumar
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