Step 1: Understanding Amplitude Modulation In amplitude modulation (AM), the resultant signal consists of the carrier frequency and sidebands. The sideband frequencies are given by: \[ f_{\text{upper}} = f_c + f_m \] \[ f_{\text{lower}} = f_c - f_m \] where:
- \( f_c = 2000 \) kHz = \(2\) MHz (carrier frequency),
- \( f_m = 5 \) kHz (modulating signal frequency).
Step 2: Calculating Sideband Frequencies \[ f_{\text{upper}} = 2000 + 5 = 2005 \, \text{kHz} \] \[ f_{\text{lower}} = 2000 - 5 = 1995 \, \text{kHz} \] Since the question asks for one possible frequency, the correct answer is: \[ \mathbf{1995 \, \text{kHz}} \] Thus, the correct answer is \( \mathbf{(1)} \ 1995 \, \text{kHz} \).
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.