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Applying conservation of momentum for collision of blocks

$2mv+0=3m{v}^{\text{'}}$

$?v^{\text{'}}=\frac{2v}{3}\dots \left(1\right)$

Now, from conservation of energy

$\frac{1}{2}\times 3m\times {v^{\text{'}}}^2=\frac{1}{2}k{x}_0^2$

$x_0=\sqrt[]{\frac{3m{v}^{\text{'}2}}{k}}=v\text{'}\sqrt[]{\frac{3m}{k}}$

$=\frac{2v}{3}\sqrt[]{\frac{3m}{k}}=v\sqrt[]{\frac{4m}{3k}}$

$v_T\propto r^2$

$\frac{v_{T_1}}{v_{T_2}}=\frac{r_1^2}{r_2^2}=\frac{r^2}{9r^2}$

Angular momentum imparted to sprinkler per

$second=vr\frac{dm}{dt}=\frac{0.6}{60}\times 0.5\times 1=0.005N.m$

$?l\frac{d\omega }{dt}=0.005$

$?\frac{d\omega }{dt}=\frac{0.005}{500\times {10}^{-7}}$

$=100\frac{rad}{s^2}$

$\frac{l_1}{l_2}=\frac{M_1{R}_1^2}{M_2{R}_2^2}=\frac{\rho \times 2\pi R_1\times R_1^2}{\rho \times 2\pi R_2\times R_2^2}=\frac{R_1^3}{R_2^3}=\frac{8}{1}$

In rotational motion, angular velocity of every particle on rigid body will be same. But linear velocity is given as v =   $r\omega .$

The CUEE centre authorities will likely close the gates half an hour to 15 minutes before the commencement of the exam. Hence, candidates must reach well in advance to avoid a last-minute rush. 

$K.E=\frac{1}{2}l{\omega }^2$

$K.E=\frac{1}{2}\times 4\times 16=32=4\times 8J$

For normal emergence

$r=A$

From Snell’s Law

$\frac{\sin ?i}{\sin ?r}=\frac{\mu }{1}$

$?\sin ?i=\sqrt[]{2}\sin ?{30}^0$

$?i={45}^0$

$f?mg$

$??mr{?}^2?mg$

$??=\frac{g}{r{?}^2}=\frac{10}{1*100}=0.1$

$F=mr{\omega }^2=\frac{k}{r}$

$\begin{array}{l}\Rightarrow \omega \propto \frac{1}{r}\\ \Rightarrow T\propto r\end{array}$

Once the CUEE centre is allotted to the candidate, it will not be possible to change the centre location. Once the CUEE admit card is sent to the candidate, they cannot change the exam centre. Candidates must make the necessary travel arrangements and attend the exam in the allotted exam centre. 

Candidates will be able to know about the VTUEEEE exam dates through this page, we will ll also share the dates table here as and when it is released.

$\begin{array}{l}\Rightarrow x={\left(t-2\right)}^2\\ \Rightarrow v=2t-4\end{array}$

$v_{t=0}=-4m/s$

$v_{t=2}=0m/s$

Work done $=\Delta K$ $$

$\Rightarrow W=\frac{1}{2}m\times 0-\frac{1}{2}\times 2\times 16$

$$ => W = 16 J

Let is the angle made by the wire with the vertical.

Here,  .

Balancing torque about COM

$T_A\left(\frac{l}{2}\right)=T_B\left(\frac{l}{6}\right)$

$?\frac{T_A}{T_B}=\frac{1}{3}$

As per Newton third law, the horse exerts a force on the ground, the ground will also exert a force on the horse.

$f\propto \frac{\sqrt[]{T}}{l}$

$\Rightarrow T\propto l^2$

$\Rightarrow T\propto l^2$ for same frequency

l’ = 1.5 l

$$ $\Rightarrow T\text{'}=2.25T$  0020

l = m (2v) = 1 * 2 * 15 = 30 kg m s^-1

$\theta =\omega t$

$2\pi =\left(\frac{V}{3R}\right)T$

$?t=\frac{6\pi R}{V}$

The distance travelled by car in this time is

$s=vt$

$?s=v\left(\frac{6\pi R}{V}\right)$

$=6\pi R$

$\begin{array}{l}t=\frac{T}{2}\\ \Rightarrow N=N_0{\left(\frac{1}{2}\right)}^{\frac{t}{T}}\end{array}$

$\Rightarrow N=N_0{\left(\frac{1}{2}\right)}^{\frac{1}{2}}=\frac{N_0}{\sqrt[]{2}}$