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Since the force is inclined at an angle 60o to the horizontal, the effective component of te force along the horizontal is given by F x cos 60o = 100 x 0.5 = 50N. ∴ work done = F x cos 60 x displacement = 50 x 8.0m = 400.0J

In general, the velocity of sound in air varies directly as the square root of temperature measured in kelvin. That V \( \propto \sqrt{T} \implies V^2 \propto T.\) \(\text{Therefore} \frac{V^2_1}{T_1} = \frac{V^2_2}{T_2}\) \(\text{Thus Let } V_1\) = 4m/s T\(_1\) = 10K \(\text{Therefore } V_2 = 2V_1\) = 8m/s \(\implies \frac{4^2}{10} = \frac{8^2}{T_2}\) T\(_2 = \frac{64 \times 10}{16}\) = 40K T\(_2 = 4T_1 \) Thus, when the velocity of sound in air is doubled, its absolute temperature will be quadrupled.

Let the masses of the two objects be M1 and M2 and their distance apart = r Therefore the force acting F1 = \( \frac{GM1M2}{r^2}\) for a new distance of r/2, we have a force of F2 = \( \frac{GM2M2}{r/2}\) \(\text{Thus } F1 \times r^2 = GM_1M_1 \text{and}\) F2 \(\times (r/2)^2 = GM_1M_1\) \(\text{Since } GM1M1 = GM_1M_1\) \(\implies F_1 \times r^2 = F2 \times (r/2)^2\) \(\text{Therefore } 200 \times r^2 = F2 \times r^2/4\) F2 = \(\frac{4 \times 200 \times r^2}{r^2} = 800N\)

l =l1 + l2 = 3 + 6 = 9H

For Hooke's law F = Ke \( \implies \frac{F}{e} = K\) \(\frac{f_1}{e_1} = \frac{f_2}{e_2}\) \(\text{Let the original length} = t_0 \) \(\text{therefore}\) e\(_1\) = (36 - t_0)\)cm ; e\(_2 = (46 - t_0)\)cm \(f_1\) = 40N, \(f_2\) = 60N \(\frac{40}{36 - t_0} = \frac{60}{45 - t_0}\) \(\implies 40(45 - t_0) = 60(36 - t_0)\) \(\text{therefore } 1800 - 40t_0\) = 2160 - 60\(t_0\) 60\(t_0\) - 40\(t_0\) = 2160 - 1800 20\(t_0\) = 360 \(t_0\) = 18cm

Option D is the graph of stopping potential Vo against the frequency of the incident light I. The straight line graph cuts the intercept I = threshold frequency. As the frequency of the incident light increases above this point(threshold frequency), the stopping potential corresponding to the kinetic energy of the photoelectron also increases.

To calculate the amount of energy released when 0.5 kg of uranium is completely burnt, we use Einstein's mass-energy equivalence formula: \(E = mc^2\) Given: Mass of uranium (\( m \)) = 0.5 kg, Speed of light (\( c \)) = \( 3 \times 10^8 \, \text{m/s} \) Substituting the values into the formula: \(E = 0.5 \, \text{kg} \times (3 \times 10^8 \, \text{m/s})^2\) Calculating \( c^2 \): \((3 \times 10^8)^2 = 9 \times 10^{16} \, \text{m}^2/\text{s}^2\) Now, substituting back: \(E = 0.5 \times 9 \times 10^{16} = 4.5 \times 10^{16} \, \text{J}\) The amount of energy released when 0.5 kg of uranium is burnt completely is \(E = 4.5 \times 10^{16} \, \text{J}\)

To determine the deceleration of the motorcyclist, we can use the kinematic equation that relates initial velocity, final velocity, acceleration, and distance: \(v^2 = u^2 + 2as\) Where: \( v \) = final velocity (0 m/s, since he stops at the traffic light), \( u \) = initial velocity (30 m/s), \( a \) = acceleration (deceleration, which will be negative) \( s \) = distance (50 m) Substituting the values into the equation, we rearrange it to solve for \( a \): \(0 = (30)^2 + 2a(50)\) Calculating \( (30)^2 \): \(0 = 900 + 100a\) Now, rearranging for \( a \): \(100a = -900\) \(a = -\frac{900}{100} = -9 \, \text{m/s}^2\) The deceleration of the motorcyclist is \( \mathbf{9 \, m/s^2} \).

Equimolar solution implies all the aqueous solutions have the same concentration. The pH scale ranges from 0 to 14, with 7 being neutral. Values below 7 indicate acidity, and values above 7 indicate basicity. Among all the aqueous solutions we have in the option, NaOH has the highest value, thus it has the highest pH. NaOH is a strong base, meaning it completely dissociates into ions (Na+ and OH-) in water, resulting in a high concentration of hydroxide ions (OH-) and a high pH.

To get the quantity of electricity for discharge at an electrode for each ions, we multiply the number of electrons by the number of moles, since 1 mole = 1F 4.0 moles of Cl\(^-\) → 1e\(^-\) x 4 = 4F 3.0 moles of Na\(^+\) → 3e\(^-\) x 1 = 3F 2.5 moles of Cu\(^{2+}\) → 2e\(^-\) x 2.5 = 5F 2.0 moles of Al\(^{3+}\) → 3e\(^-\) x 2 = 6F. The Ions that require the largest quantity of electricity for discharge at an electrode is the Al\(^{3+}\).

Phenolphthalein, a common acid-base indicator, remains colourless in acidic solutions but turns pink in basic solutions.

The rusting of iron is a spontaneous reaction, meaning it occurs naturally without external energy input, as iron oxidizes to form rust (iron oxide) when exposed to oxygen and moisture. Although, it is a slow process, but there is an increase in entropy (\DeltaS) as well as the emission of heat which makes it a spontaneous reaction as the resulting change in Gibbs free energy will be negative.i.e \DeltaG = - ve

2H\(_2\) + O\(_2\) → 2H\(_2\)O \DeltaH = -571kJ For Hydrogen, Number of moles = \(\frac{mass}{Molar mass}\) Stoichiometrically, 2 = \(\frac{mass}{2}\) Mass of Hydrogen = 2 x 2 = 4g If 4.0g of Hydrogen burns in O\(_2\) to produce - 571 kJ of energy ∴ 12.0g of Hydrogen will burn in O\(_2\) to produce x kJ of energy x = \(\frac{{12}\times{- 571}}{4}\) = - 1713kJ

When sulphur dioxide (SO₂) reacts with a solution of sodium hydroxide (NaOH), it forms sodium sulphite (Na₂SO₃) and water (H₂O). The balanced chemical equation for this reaction is: SO\(_2\)(g) + 2NaOH(aq) → Na\(_2\)SO\(_3\)(aq) + H\(_2\)O(l). When the gas is passed for a long time, the solution becomes saturated with SO\(_2\)(excess). The initially formed sodium sulfite reacts further with the excess SO2 and water to produce sodium hydrogen sulphite, NaHSO\(_3\)(Sodium bisulphite) Na\(_2\)SO\(_3\) + H\(_2\)O + SO\(_2\) → 2NaHSO\(_3\)

NO\(_{2} ^{-}\) = - 1 N\(O_2\) = - 1 N + (-2 x 2) = - 1 N - 4 = - 1 N = - 1 + 4 = + 3 IUPAC name of NO\(_{2} ^{-}\) is dioxonitrate(III) ion