The power delivered by the battery shown in the figure is - (Take the forward and reverse biased resistances of the diode are zero and infinity respectively)

Step 1: Identifying the Conducting Path
- Given that the diodes are ideal, \(D_1\) is forward biased (conducting), and \(D_2\) is reverse biased (non-conducting).
- The circuit simplifies to a series circuit with a \(10\Omega\) resistor connected to a \(5V\) battery.
Step 2: Calculating the Current By Ohm’s Law: \[ I = \frac{V}{R} = \frac{5V}{10\Omega} = 0.5A \]
Step 3: Power Delivered by the Battery Power is given by: \[ P = V \times I \] \[ P = 5V \times 0.5A = 2.50W \] Thus, the correct answer is \( \mathbf{(1)} \ 2.50W \).
Given below are two statements: one is labelled as Assertion (A) and the other is labelled as Reason (R).
Assertion (A): The density of the copper ($^{64}Cu$) nucleus is greater than that of the carbon ($^{12}C$) nucleus.
Reason (R): The nucleus of mass number A has a radius proportional to $A^{1/3}$.
In the light of the above statements, choose the most appropriate answer from the options given below:
Match the LIST-I with LIST-II
| LIST-I (Type of decay in Radioactivity) | LIST-II (Reason for stability) | ||
|---|---|---|---|
| A. | Alpha decay | III. | Nucleus is mostly heavier than Pb (Z=82) |
| B. | Beta negative decay | IV. | Nucleus has too many neutrons relative to the number of protons |
| C. | Gamma decay | I. | Nucleus has excess energy in an excited state |
| D. | Positron Emission | II. | Nucleus has too many protons relative to the number of neutrons |
Choose the correct answer from the options given below:
Observe the following data given in the table. (\(K_H\) = Henry's law constant)
| Gas | CO₂ | Ar | HCHO | CH₄ |
|---|---|---|---|---|
| \(K_H\) (k bar at 298 K) | 1.67 | 40.3 | \(1.83 \times 10^{-5}\) | 0.413 |
The correct order of their solubility in water is
For a first order decomposition of a certain reaction, rate constant is given by the equation
\(\log k(s⁻¹) = 7.14 - \frac{1 \times 10^4 K}{T}\). The activation energy of the reaction (in kJ mol⁻¹) is (\(R = 8.3 J K⁻¹ mol⁻¹\))
Note: The provided value for R is 8.3. We will use the more precise value R=8.314 J K⁻¹ mol⁻¹ for accuracy, as is standard.