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Heat loss by hot water = Heat gained by water mc(t - 50) = 2mc(50 - 30) t - 50 = 2(20) t - 50 = 40 t = 40 + 50 \(\therefore\) t = 90^\circC

m\(_s\) = 2 kg; c\(_s\) = 460 Jkg\(^{-1}\)K\(^{-1}\); m\(_w\) = ?; c\(_w\) = 4200 Jkg\(^{-1}\)K\(^{-1}\); \theta\(_1\) = 280^\circC; \theta\(_2\) = 30^\circC; \theta\(_3\) = 70^\circC Heat loss = Heat gained m\(_s\) c\(_s\) (\theta\(_1\) - \theta\(_3\)) = m\(_w\) c\(_w\) (\theta\(_3\) - \theta\(_2\)) 2 \times 460 \times (280 - 70) = m\(_w\) \times 4200 \times (70 - 30) 2 \times 460 \times 210 = m\(_w\) \times 4200 \times 40 193,200 = 168,000 m_w m\(_w\) = \(\frac{193200}{168000}\) ≈ 1.15kg.

The heat required to convert ice at 0^\circC to water at the same temperature is given by the formula: Q = m \times L, where m is the mass of the ice, and L is the specific latent heat of ice. Substituting the given values, we get Q = 20 g \times 336 Jg\(^{-1}\) = 6.72 \times 10\(^3\) J.

\Delta\theta\(_P\) = 2 \times \Delta\theta\(_Q\) m\(_P\) = m\(_Q\) c\(_Q\) = c\(_P\) = ? H\(_P\) = H\(_Q\) (given) i.e. m\(_P\) \times c\(_P\) \times \Delta\theta\(_P\) = m\(_Q\) \times c\(_Q\) \times \Delta\theta\(_Q\) Since \Delta\theta\(_P\) = 2 \times \Delta\theta\(_Q\) and m\(_P\) = m\(_Q\) ∴ m\(_Q\) \times c\(_P\) \times 2 \times \Delta\theta\(_Q\) = m\(_Q\) \times c\(_Q\) \times \Delta\theta\(_Q\) c\(_P\) \times 2 = c\(_Q\) \(\frac{c_Q}{c_P}\) = \(\frac{2}{1}\) \(\therefore\) c\(_Q\): c\(_P\) = 2: 1

The general ordering of the electromagnetic spectrum from shortest to longest wavelength is as follows: 1. Gamma rays 2. X-rays 3. Ultraviolet (UV) rays 4. Visible light 5. Infrared (IR) radiation 6. Microwaves 7. Radio waves Therefore, X-rays are the correct answer.

Standing waves are formed when waves reflect back from a boundary and interfere with the incoming waves of the same frequency and amplitude, creating points that appear to stand still (nodes) and points of maximum oscillation (antinodes).

Water waves are mechanical waves that require a material medium, such as water, to propagate. They involve the vibration of water particles. In contrast, light waves are electromagnetic waves that do not need a medium for propagation; they can travel through a vacuum. For example, sunlight travels through space without the need for any material medium.

Answer: A. 1:4 (Pressure equal: \(\frac{F_1}{(\pi r_1^2)}\) = \(\frac{F_2}{(\pi r_2^2)}\); \(\frac{_r2}{r_1}\) = \(\sqrt{(F_2}{F_1)}\) = \(\sqrt{16}\) = 4.)

Efficiency = \(\frac{\text{MA}{\text{VR}}\) x 100 But V.R = \(\frac{1}{sin\theta}\) = \(\frac{1}{sin 30}\) = 2 M.A = \(\frac{\text{L}}{\text{E}}\) Efficiency = \(\frac{\text{MA}{\text{VR}}\) x 100 0.75 = \(\frac{\text{M.A}}{2}\) 0.75 x 2 = M.A M.A = 1.5 = \(\frac{7.5 x 10}{\text{E}}\) 1.5E = 75 E = \(\frac{75}{1.5}\) = 50N

E = 60%, L = 30N, E = 20N M.A = \(\frac{\text{L}}{\text{E}}\) = \(\frac{30}{20}\) = 1.5

Average speed = \(\frac{\text{total distance}}{\total time taken}}\) d = 1. 0km ⇒ 1000m t = 5minutes ⇒ 5 x 60 = 300 seconds Average speed = \(\frac{1000}{300}\) = 3.3m/s

Uniform speed = 120km/hr, d =?, t = 4minutes ⇒ \(\frac{4}{60}\) = 0.0667hrs Distance = speed x time D = 120 x 0.0667 = 8km.

Answer: B For equilibrium, clockwise moment = anticlockwise moment about the pivot. clockwise distance from pivot: 50cm - 10cm = 40cm anticlockwise distance from pivot: 85cm - 10cm = 75cm Applying the principle of moments W x distance(w) = T x distance(T) 100 x 40 = T x 75 T = \(\frac{ 4000}{75}\) ≈ 53.33N

T x 80 + 15 x 10 = W x 50 80T + 150 = 50W - - -- - - - - -(1) T + 15 = W - - - - - - - - - - - (2) 80T + 150 = 50(T + 15) 80T + 150 = 50T + 750 30T = 750 - 150 30T = 600 T = 20N The closest option is 19.2N

In a vacuum (where air resistance is ignored), the maximum horizontal range is achieved at an angle of 45º. From R = \(\frac{u^2sin2\theta}{\text{g}}\) R\(_{max}\) = \(\frac{u^2sin2(45)}{\text{g}}\) = \(\frac{u^2}{\text{g}}\) \(\therefore\) \(\theta\) = 45º