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Energy required to melt

Q = $MS\Delta T+ML$  

$\Rightarrow 10^{-1}\times 2\times 10^3\times 10+10^{-1}\times 3.33\times 10^5$      

->3.53 × 10⁴ J

Heat produce in wire

H = l²RT

  $Q=3.53\times 10^4={\left(\frac{1}{2}\right)}^2\times \left(4\times 10^3\right)\times t$

$t=\frac{3.53\times 10^4\times 4}{4\times 10^3}=35.3sec$             

$\overrightarrow{x}=\left|\overrightarrow{P}+\overrightarrow{Q}\right|=\sqrt[]{P^2+Q^2}$          

=> $\overrightarrow{y}=\left|\overrightarrow{P}-\overrightarrow{Q}\right|=\sqrt[]{P^2+Q^2}$

$\left|\overrightarrow{x}+\overrightarrow{y}\right|=\sqrt[]{x^2+y^2+2xy}$

$\Rightarrow \sqrt[]{3\left(P^2+Q^2\right)}=\sqrt[]{{\left(\sqrt[]{P^2+Q^2}\right)}^2+{\left(\sqrt[]{P^2+Q^2}\right)}^2+2\left(P^2+Q^2\right)cos{\theta }_1}$

$\Rightarrow {\theta }_1=60°$

=>Using same formula : θ₂ = 90°

Since the line $\frac{x}{5}cos\theta +\frac{y}{12}sin\theta =1$ is the equation of tangent to the ellipse $\frac{x^2}{5^2}+\frac{y^2}{12^2}=1$ at (5 cos q, 12 sin q). Hence option (A) is correct option.

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Let $?\frac{a}{r}$ , a, ar are in G.P. $\therefore \frac{a}{r},$ 2a, ar are in A.P.

$2a-\frac{a}{r}=ar-2a\Rightarrow 2-\frac{1}{r}=r-2\Rightarrow r^2-4r+1=0\Rightarrow r=2+\sqrt[]{3}$

$A/q,a{r}^2=3r^2\Rightarrow a=3$

$\begin{array}{l}\therefore d=ar-2a=3\sqrt[]{3}\\ \therefore r^2-d=7+\sqrt[]{3}\end{array}$

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Using conservation of linear momentum:

=> 40 × 3m = 60 × m + v × 2m

=> v = 30 m/s

$K{E}_i=\frac{1}{2}\times 3m\times {\left(40\right)}^2$      

$K{E}_f=\frac{1}{2}\times m\times {\left(60\right)}^2+\frac{1}{2}\times 2m\times {\left(30\right)}^2$     

$\Rightarrow \frac{K{E}_f}{K{E}_i}=\frac{54}{48}$

Fractional change in kinetic energy = 1 $-\frac{K{E}_f}{K{E}_i}=\frac{1}{8}$

Suppose acceleration of wedge is a and acceleration of block w. r.t. wedge is a₁ then N cos60° = Ma = 16 a -> N = 32 a

For block w.r.t. wedge

N + 8a sin 30° = 8g cos 30°

N = 8g cos 30° - 8a sin 30°

->32a = 8g cos 30° - 8a sin 30°

->a = $\frac{\sqrt[]{3}}{9}g$  

Now for 8 kg,

$8gsin30°+8acos30°=m{a}_1$  

$a_1=g\times \frac{1}{2}+\frac{\sqrt[]{3}}{9}g\times \frac{\sqrt[]{3}}{2}$  

$=\frac{2}{3}g$  

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Energy of electron in first existed state will be -3.4 eV.

So total energy difference will be (2.6 + 3.4) eV.

Wavelength $\left(\lambda \right)=\frac{1242eV-nm}{6eV}=207nm$

$Frequency=\frac{c}{\lambda }=\frac{3\times 10^8}{207\times 10^{-9}}=1.45\times 10^{9}MHz$

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De Broglie wavelength

$\lambda =\frac{h}{mV}=\frac{h}{\sqrt[]{2mE}}E\to K.E.$              

$E=\frac{3}{2}$  KT for gas

$So\lambda =\frac{h}{\sqrt[]{3mKT}}=\frac{6.6\times 10^{-34}}{\sqrt[]{3\times 9\times 10^{-31}\times 1.38\times 10^{-23}\times 300}}$

$\begin{array}{l}\lambda =6.26\times 10^{-9}m\\ \lambda =6.26nm\end{array}$                

For equilibrium net force acting on the system should be zero.