Step 1: Power Dissipation Formula
The power dissipation across a resistor is given by: \[ P = \frac{V^2}{R}. \]
Step 2: Resistance of Each Half
When the wire is cut into two equal halves, the resistance of each half becomes: \[ R_{\text{half}} = \frac{R}{2}. \]
Step 3: Power Dissipation Across Each Half
When \( V_0 \) is applied across each half: \[ P_{\text{half}} = \frac{V_0^2}{R_{\text{half}}} = \frac{V_0^2}{\frac{R}{2}} = \frac{2V_0^2}{R}. \]
Step 4: Total Power Dissipation Across Two Halves
The total power dissipation across the two halves is: \[ P_2 = 2 \times P_{\text{half}} = 2 \times \frac{2V_0^2}{R} = \frac{4V_0^2}{R}. \]
Step 5: Ratio of Power Dissipation
The original power dissipation \( P_1 \) is: \[ P_1 = \frac{V_0^2}{R}. \] The ratio \( P_2 : P_1 \) is: \[ \frac{P_2}{P_1} = \frac{\frac{4V_0^2}{R}}{\frac{V_0^2}{R}} = 4. \]
Step 6: Relation to \( \sqrt{x} \)
The ratio \( P_2 : P_1 \) is given as \( \sqrt{x} : 1 \), so: \[ \sqrt{x} = 4 \quad \Rightarrow \quad x = 16. \]

The equivalent resistance between the points \(A\) and \(B\) in the given circuit is \[ \frac{x}{5}\,\Omega. \] Find the value of \(x\). 
A Wheatstone bridge is initially at room temperature and all arms of the bridge have same value of resistances \[ (R_1=R_2=R_3=R_4). \] When \(R_3\) resistance is heated, its resistance value increases by \(10%\). The potential difference \((V_a-V_b)\) after \(R_3\) is heated is _______ V. 
Match the LIST-I with LIST-II for an isothermal process of an ideal gas system. 
Choose the correct answer from the options given below:
Which one of the following graphs accurately represents the plot of partial pressure of CS₂ vs its mole fraction in a mixture of acetone and CS₂ at constant temperature?

In the given figure, the blocks $A$, $B$ and $C$ weigh $4\,\text{kg}$, $6\,\text{kg}$ and $8\,\text{kg}$ respectively. The coefficient of sliding friction between any two surfaces is $0.5$. The force $\vec{F}$ required to slide the block $C$ with constant speed is ___ N.
(Given: $g = 10\,\text{m s}^{-2}$) 
Kirchhoffs Circuit Laws allow us to solve complex circuit problems.

It states that the “total current or charge entering a junction or node is exactly equal to the charge leaving the node as it has no other place to go except to leave, as no charge is lost within the node“.
It states that “in any closed loop network, the total voltage around the loop is equal to the sum of all the voltage drops within the same loop” which is also equal to zero.