We are given:
- Mass of the block \(m = 2\, kg\)
- Force applied \(F = 10\, N\)
- The surface is smooth, i.e., there is no friction.

Step 1: Use Newton’s Second Law of Motion

According to Newton’s Second Law,
\[ F = m \cdot a \]

Solving for acceleration \(a\):
\[ a = \frac{F}{m} \]
\[ a = \frac{10}{2} = 5 \, m/s^2 \]

Step 2: Final Answer

Thus, the acceleration of the block is \(5 \, m/s^2\).

Answer: The correct answer is option (2) \(5 \, m/s^2\). Quick Tip: When the surface is smooth (frictionless), you can directly apply \( F = ma \) without accounting for frictional forces.

Step 1: Understand the law of reflection
The law of reflection states that: \[ Angle of incidence i = Angle of reflection r \]

Given: \( i = 30^\circ \Rightarrow r = 30^\circ \)

Step 2: Find the angle between the incident and reflected ray

The incident ray and the reflected ray are on opposite sides of the normal, and each makes a \(30^\circ\) angle with the normal. So the angle between them is:
\[ Angle between incident and reflected ray = i + r = 30^\circ + 30^\circ = 60^\circ \]

Answer: The correct answer is option (2) \(60^\circ\). Quick Tip: For a plane mirror, the angle between the incident and reflected ray is always \(2i\), where \(i\) is the angle of incidence.

We are given:
- \(q_1 = +2 \, \muC = 2 \times 10^{-6} \, C\)
- \(q_2 = -2 \, \muC = -2 \times 10^{-6} \, C\)
- Distance \(r = 0.1 \, m\)
- Coulomb’s constant \(k = 9 \times 10^9 \, Nm^2/C^2\)

Step 1: Use Coulomb’s Law
\[ F = \frac{k |q_1 q_2|}{r^2} \]
\[ F = \frac{9 \times 10^9 \cdot (2 \times 10^{-6})^2}{(0.1)^2} \]
\[ F = \frac{9 \times 10^9 \cdot 4 \times 10^{-12}}{0.01} = \frac{36 \times 10^{-3}}{0.01} = 3.6 \, N \]

Step 2: Direction of force

Since the charges are opposite in sign, the force is attractive.

Answer: The electrostatic force between the charges is \(3.6 \, N\). So, the correct answer is option (1). Quick Tip: Always convert microcoulombs (\( \muC \)) to coulombs (\( C \)) before using Coulomb’s Law.

Step 1: Use the relation between energy and frequency

The energy of a photon is given by:
\[ E = h \nu \]

Where:
- \(E = 6.6 \times 10^{-19} \, J\)
- \(h = 6.6 \times 10^{-34} \, Js\)

Step 2: Rearrange to find frequency \(\nu\)
\[ \nu = \frac{E}{h} = \frac{6.6 \times 10^{-19}}{6.6 \times 10^{-34}} = 1 \times 10^{15} \, Hz \]

Answer: The frequency of the photon is \(1 \times 10^{15} \, Hz\). Hence, the correct answer is option (1). Quick Tip: Remember: \(E = h\nu\) is a fundamental relation in quantum physics used to calculate photon frequency from its energy.

Step 1: Use the formula for work done at constant pressure (isobaric process)

Work done by a gas during isobaric expansion:
\[ W = P(V_2 - V_1) \]

Given:
- \(V_1 = V\)
- \(V_2 = 2V\)
- \(P = P\) (constant)
\[ W = P(2V - V) = P \cdot V \]

Step 2: Final Answer

The work done by the gas is \(PV\).

Answer: Hence, the correct answer is option (1) \(PV\). Quick Tip: In an isobaric process, the area under the \(P\)-\(V\) graph (a rectangle) gives the work: \(W = P \Delta V\).

Step 1: Use the formula for number of moles
\[ Number of moles = \frac{Given mass}{Molar mass} \]

Given:
- Mass of \(CO_2 = 44\, g\)
- Molar mass of \(CO_2 = 44\, g/mol\)
\[ Moles = \frac{44}{44} = 1 \, mol \]

Answer: Therefore, the number of moles in \(44\, g\) of \(CO_2\) is 1 mol. So, the correct answer is option (1). Quick Tip: Always remember: \( Moles = \frac{Mass}{Molar Mass} \). Keep units consistent when using this formula.

The elements Na, Mg, Al, and Si all belong to Period 3 of the periodic table.
As we move left to right across a period:

- Atomic number increases
- Effective nuclear charge increases
- Electrons are added to the same shell, so the attraction between the nucleus and outer electrons increases.

Hence, atomic radius decreases across a period.

Order of atomic radius: \[ Na > Mg > Al > Si \]

Answer: Therefore, Si has the smallest atomic radius among the options. Correct answer is option (4). Quick Tip: Across a period in the periodic table, atomic radius decreases due to increasing nuclear charge.

The reaction involves 1-bromobutane (an alkyl halide) with potassium hydroxide (KOH) in ethanol. This is a typical example of an elimination reaction (E2 mechanism), where the hydroxide ion (OH⁻) removes a proton from the carbon adjacent to the carbon that is bonded to the leaving group (Br⁻), leading to the formation of a double bond.

Step 1: Identify the type of reaction

Since the reaction is occurring in ethanol, which is a polar protic solvent, the elimination occurs via the E2 mechanism.

Step 2: Identify the product

- The E2 mechanism will lead to the formation of an alkene.
- The hydrogen is removed from the carbon next to the carbon bearing the bromine (the β-carbon), leading to the formation of But-2-ene.

Answer: The major product of the reaction is But-2-ene, and thus the correct answer is option (2). Quick Tip: In a reaction with KOH in ethanol, the elimination (E2) mechanism typically leads to the formation of an alkene.

For a first-order reaction, the half-life (\(t_{1/2}\)) is given by the formula:
\[ t_{1/2} = \frac{0.693}{k} \]

Given:
- Rate constant \(k = 0.01 \, s^{-1}\)

Step 1: Calculate the half-life
\[ t_{1/2} = \frac{0.693}{0.01} = 69.3 \, s \]

Answer: Therefore, the half-life of the reaction is \(69.3 \, s\), and the correct answer is option (1). Quick Tip: For a first-order reaction, the half-life is independent of the initial concentration and only depends on the rate constant.

A coordination compound is formed when a central metal ion or atom is surrounded by a number of ligands, which are molecules or ions that can donate electron pairs to the metal ion.

Step 1: Review the options
- Option 1: While coordination compounds can have a definite molecular formula, this is not their most distinguishing feature.
- Option 2: Coordination compounds do not necessarily have high melting points or conduct electricity in the solid state. This is more typical of ionic compounds.
- Option 3: The hallmark of a coordination compound is the bonding of a central metal atom/ion with ligands. This option is correct.
- Option 4: Coordination compounds can be soluble in water, especially if they form ions in solution.

Answer: The correct characteristic of a coordination compound is option (3) — it contains a central metal atom/ion bonded to a number of ligands. Quick Tip: Coordination compounds typically involve metal-ligand coordination bonds, and the ligands donate electron pairs to the central metal.

The standard electrode potential of a half-reaction is defined as the potential difference when the half-reaction occurs under standard conditions (1 M concentration, 1 atm pressure, and 25°C).

Step 1: Review the given half-reaction

The given half-reaction is: \[ Ag^+ + e^- \rightarrow Ag(s) \]

The standard electrode potential for this half-reaction is a well-known value in electrochemistry and is given as:
\[ E^\circ = +0.80 \, V \]

Answer: Therefore, the standard electrode potential for the reduction of \(Ag^+\) to \(Ag\) is \(+0.80 \, V\). The correct answer is option (1). Quick Tip: The standard electrode potential for the reduction of \(Ag^+\) to \(Ag\) is \(+0.80 \, V\), which indicates that silver ions are easily reduced to silver metal.