A proton and an alpha particle of the same velocity enter in a uniform magnetic field which is acting perpendicular to their direction of motion. The ratio of the radii of the circular paths described by the alpha particle and proton is

Process-AB $\Rightarrow$ Isobaric,                                               &n

from newtan’s law of cooling :

$\frac{dT}{dt}=-k\left(T-T_0\right)$

Using average value :

$\frac{75-65}{5}=-k\left[\frac{\left(75+65\right)2}{2}-25\right]$

$\frac{\left(65-T\right)}{5}=-k\left[\frac{\left(65+T\right)}{2}-25\right]$

$\Rightarrow \frac{10}{65-T}=\frac{45}{\frac{65+T}{2}-25}$

$\Rightarrow T=57°C$

Efficiency $\eta =50\mathrm{\%}$ $$

$\Rightarrow \eta =\left(1-\frac{T_2}{T_1}\right)\times 100=50$

$\frac{{}_{}}{}$ $\Rightarrow 1-\frac{T_2}{T_1}=\frac{50}{100}\Rightarrow 1-\frac{T_2}{T_1}=\frac{1}{2}$  -(i)

After the change :- $1-\frac{T_2^1}{T_1}=\frac{50\times \left(1.3\right)}{100}=\frac{65}{100}=.65$ $\frac{{}_{}^{}}{{}_{}}\frac{}{}\frac{}{}$ -(ii)

$\Rightarrow 1-\left(\frac{T_2-40}{T_1}\right)=.65$ --(ii)

$\Rightarrow \frac{40}{T_1}=.65-.50=.15\Rightarrow T_1=\frac{4000}{15}=266.7k$

At Resonance

X_L = X_C

then l_L = l_C

Now phasor diagram

for L & C

So, Net current = zero

$=\left(l_L+l_C\right)$

Therefore current through R circuit at resonance will be zero

T₁ = 227°C                                                &n