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$X_C=\frac{1}{\omega C}$

On increasing $\omega$ $$ XC decreases so Z decreases, current in circuit increases.

$\frac{H_{D.C.}}{H_{A.C.}}=\frac{I^{2}R}{I_{rms}^{2}R}=1$

$B_0=\frac{E_0}{v}v=\frac{\omega }{k}$

$\Rightarrow \mu =\frac{c}{v}=\frac{3\times 10^8}{\frac{9}{4}\times 10^8}=\frac{4}{3}$

$\frac{\Delta \lambda }{\lambda }=\frac{v}{c}$

$\Rightarrow v=\frac{0.6\times 10^{-9}\times 3\times 10^8}{589\times 10^{-9}}$ = 3.06 × 10⁵ m/s

Given $F\propto \frac{1}{r^n}$

$m{\omega }^{2}R=\frac{K}{R^n}$

$?\frac{m.4{\pi }^{2}R}{T^2}=\frac{K}{R^n}$

$?T^2\propto R^{n+1}$

$T\propto R^{\frac{n+1}{2}}$

^{$e=\left|\frac{d?}{dt}\right|=4t+8$}

^{$\Rightarrow i=\frac{e}{R}=\frac{4t+8}{4}=t+2$}

^{$\Rightarrow$ Heat produced (H)= ${\displaystyle \underset{0}{\overset{2}{\int }}{i}^{2}Rdt}$}

^{$\Rightarrow H=4\lfloor \frac{8}{3}+8+8\rfloor =\frac{224}{3}J$}

No, the CUEE exam centres may not allow candidates to take printouts from their offices. They need to ensure that they are carrying the hard copy of the CUEE admit card. The candidates must take a printout and take it to the exam centre. 

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$\frac{\Delta T}{T}=\frac{\alpha \Delta \theta }{2}$

$\Rightarrow \frac{\Delta T}{T}\times 100=10^{-3}\mathrm{\%}$

No, the CUEE centre authorities will not allow the candidates to carry eatables or drinks inside the exam hall. The candidates are not allowed to eat anything during exam hours. 

$v=\frac{\omega }{k}=\frac{5\pi }{\pi /4}=20cm/s$

No, the CUEE centres do not have any provision to accommodate the candidates' parents. They need to wait outside the exam centre or come back later after the exam is over. 

$f\text{'}=f\lfloor \frac{v+\frac{v}{4}}{v}\rfloor =\frac{5f}{4}$

$\Rightarrow \frac{f\text{'}}{f}=\frac{5}{4}$

$\Rightarrow \frac{\Delta f}{f}=\frac{1}{4}$

$\eta =1-\frac{T_2}{T_1}=1-\frac{500}{1000}=\frac{1}{2}$

$\eta =\frac{W}{Q_1}$

$\Rightarrow Q_1=\frac{W}{\eta }=\frac{1000}{1/2}$

$\Rightarrow Q_1=2000J$

As such, a candidate can make a request to the CUEE centre authorities to deposit their personal belongings for the test duration. But the authorities will not be responsible for any loss of belongings. It is advisable not to carry any valuables to the exam centre. 

During the process are under P – V graph continuously increases but rate of increment varies.

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

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

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

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

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