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The waveform moved from A to B in \(2.5 \times 10^{-3}\) seconds. From the diagram, A to B has \(2\frac{1}{2}\) complete oscillations between it, Therefore, the period of the oscillation is \(\frac{2.5 \times 10^{-3}}{2.5} = 1 \times 10^{-3} s\) \(Frequency = \frac{1}{Period}\) = \(\frac{1}{1 \times 10^{-3}} = 1 \times 10^{3} Hz\)

E = V2t/R t = ER/V2 = 4800 x 1.8 x 103/240 x 240 = 150s

E = 1/2cv2 = 1/2 x 10 x 10-6 x 402 = 1.6 x 10-2J

No expanded working has been attached for this question yet. The verified answer is Option A.

\(R = \frac{R_1 X R_2}{R_1 + R_2}\) = 1 \(E = I^2R = 2^2 x 1 =4W\)

E = Stress/Strain = F/A/e/L E = FL/AE; F = EAe/L

No expanded working has been attached for this question yet. The verified answer is Option B.

The coin will fall back to the hand of the boy (INERTIA OF MOTION). When the coin is thrown vertically up, it moves with the horizontal velocity of the train. It moves along with the train and falls back in the boy's hand. The train in this case, has no acceleration (uniform velocity).

\( u = 0 \, \text{m/s} \) (starts from rest) \( a = 6 \, \text{m/s}^2 \) Distance Covered in \( n \) Seconds The distance covered in \( n \) seconds is given by: \(d_n = u n + \frac{1}{2} a n^2\) Since \( u = 0 \): \(d_n = \frac{1}{2} a n^2 = \frac{1}{2} \cdot 6 \cdot n^2 = 3n^2\) Calculate Distances Distance covered in the first 2 seconds (\( d_2 \)): \(d_2 = 3(2^2) = 3 \cdot 4 = 12 \, \text{m}\) Distance covered in the first 3 seconds (\( d_3 \)): \(d_3 = 3(3^2) = 3 \cdot 9 = 27 \, \text{m}\) Distance Covered in the Third Second The distance covered in the third second is the difference between the distances: \(\text{Distance in the third second} = d_3 - d_2 = 27 \, \text{m} - 12 \, \text{m} = 15 \, \text{m}\)

Range = \(\frac{U^2sin2\theta}{g}\) RangeQ = \(\frac{U^2sin150}{g}\) RangeR = \(\frac{U^2sin30}{g}\) Since, sin30o = sin150o therefore, RQ = RR

power = \(\frac{\text {work done}}{time}\) = \(\frac{800 \times 15}{12.5}\) = 960W

\(\frac{500 \times 1.5}{100.0}\) x 100% = 75%

In an experiment to determine Young's Modulus for a wire, the tensile stress depends on the following factors: Tensile Stress = The tensile stress (\( \sigma \)) is defined as the force (load, \( F \)) applied per unit area (\( A \)): \(\sigma = \frac{F}{A}\) Cross-Sectional Area: For a wire with a circular cross-section, the area (\( A \)) can be expressed in terms of the radius (\( r \)): \(A = \pi r^2\) Combining the Equations: Therefore, the tensile stress can be rewritten as: \(\sigma = \frac{F}{\pi r^2}\) From this, we see that the tensile stress depends on both the load (\( F \)) and the radius (\( r \)) of the wire. Thus, the correct answer is: \(\text{B. Load and the radius of the wire}\)

since the plastic sphere floats on both liquids, it displaces the amount of liquid equal to its mass in each case ∴ Mass of oil displaced = mass of water displaced ρ x 0.4v = 1000 x 0.5v ρ = \(\frac{ 1000 \times 0.5v}{0.4v}\) ρ = 1250k\(gm^{-3}\)

Density = mass/ volume At constant pressure, volume is directly proportional to Temperature Density is inversely proportional to the volume at constant mass At constant pressure, volume = constant x temperature hence, density = mass/ (constant x temperature) which shows density is inversely proportional to temperature