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: \[ \sqrt{x} = 4 \quad \Rightarrow \quad x = 16. \]
Given below are two statements. One is labelled as Assertion (A) and the other is labelled as Reason (R):
Assertion (A): In an insulated container, a gas is adiabatically shrunk to half of its initial volume. The temperature of the gas decreases.
Reason (R): Free expansion of an ideal gas is an irreversible and an adiabatic process. \text{In the light of the above statements, choose the correct answer from the options given below:}
In the figure shown below, a resistance of 150.4 $ \Omega $ is connected in series to an ammeter A of resistance 240 $ \Omega $. A shunt resistance of 10 $ \Omega $ is connected in parallel with the ammeter. The reading of the ammeter is ______ mA.

Two cells of emf 1V and 2V and internal resistance 2 \( \Omega \) and 1 \( \Omega \), respectively, are connected in series with an external resistance of 6 \( \Omega \). The total current in the circuit is \( I_1 \). Now the same two cells in parallel configuration are connected to the same external resistance. In this case, the total current drawn is \( I_2 \). The value of \( \left( \frac{I_1}{I_2} \right) \) is \( \frac{x}{3} \). The value of x is 1cm.
Let \( a \in \mathbb{R} \) and \( A \) be a matrix of order \( 3 \times 3 \) such that \( \det(A) = -4 \) and \[ A + I = \begin{bmatrix} 1 & a & 1 \\ 2 & 1 & 0 \\ a & 1 & 2 \end{bmatrix} \] where \( I \) is the identity matrix of order \( 3 \times 3 \).
If \( \det\left( (a + 1) \cdot \text{adj}\left( (a - 1) A \right) \right) \) is \( 2^m 3^n \), \( m, n \in \{ 0, 1, 2, \dots, 20 \} \), then \( m + n \) is equal to:
Rate law for a reaction between $A$ and $B$ is given by $\mathrm{R}=\mathrm{k}[\mathrm{A}]^{\mathrm{n}}[\mathrm{B}]^{\mathrm{m}}$. If concentration of A is doubled and concentration of B is halved from their initial value, the ratio of new rate of reaction to the initial rate of reaction $\left(\frac{\mathrm{r}_{2}}{\mathrm{r}_{1}}\right)$ is
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.