Edulodhi

Options: (1) u² / (4g sinθ) (2) u² / (2g cosθ) (3) u² / (2g cosθ) (4) u² / (2g sinθ) Step 1: Acceleration on Incline a = -g sinθ Using equation: v² = u² + 2as At highest point: 0 = u² - 2g sinθ s Step 2: Solve s = u² / (2g sinθ) Final Answer: Correct Option: (4)

Options: $(7.5 \times 10^{-2})$ N $(5.7 \times 10^{-3})$ N $(5.7 \times 10^{-2})$ N $(7.5 \times 10^{-3})$ N Solution: Induced emf: $[ \varepsilon = Blv ]$ Induced current: $[ i = \frac{Blv}{R} ]$ Magnetic force: $[ F = Bil ]$ Substitute: $[ F = \frac{B^2 l^2 v}{R} = \frac{(0.1)^2 \times 1^2 \times 1.5}{2} ]$ $[ F = 7.5 \times 10^{-3}\ \text{N} ]$ ✅ Correct Answer: (4)  

10 kg of ice at –10°C is added to 100 kg of water at 25°C. Assume no heat exchange with surroundings. Given: Specific heat of ice = 2100 J/kg°C Specific heat of water = 4200 J/kg°C Latent heat of fusion of ice = 3.36 × 10^5 J/kg Find the final temperature of the mixture. Options: A) 10°C B) 15°C C) 6.67°C D) 11.6°C Solution Heat required to raise ice from –10°C to 0°C: Q 1 = m c ΔT Q 1 = 10 × 2100 × 10 Q 1 = 210000 J Heat required to melt ice: Q 2 = mL Q 2 = 10 × 3.36 × 10^5 Q 2 = 3.36 × 10^6 J Heat required to raise melted water from 0°C to final temperature T: Q 3 = 10 × 4200 × T Heat lost by hot water cooling from 25°C to T: Q 4 = 100 × 4200 × (25 − T) Apply heat balance: 100 × 4200 (25 − T) = 210000 + 3.36 × 10^6 + 10 × 4200T 420000 (25 − T) = 3570000 + 42000T 10500000 − 420000T = 3570000 + 42000T 6930000 = 462000T T = 15°C Correct Answer: B

Options: 1.08 1.90 1.67 1.18 Solution: Wave speed: $[ v = \sqrt{\frac{T}{\mu}} ]$ $[ v_A = \sqrt{\frac{500}{2\times10^{-4}}} ,\quad v_B = \sqrt{\frac{500}{4\times10^{-4}}} ]$ Time: $[ t = \frac{L}{v} \Rightarrow \frac{t_1}{t_2} = 1.18 ]$ ✅ Correct Answer: (4)

Options: 150 500 100 1000 Solution: Energy of satellite: $[ E = -\frac{GMm}{2r} ]$ Energy required: $[ \Delta E = \frac{GMm}{6R_E} ]$ $[ \alpha = 1000 ]$ ✅ Correct Answer: (4)

Beam expansion ratio: $[ \frac{D_2}{D_1} = \frac{f_2}{f_1} ]$ $[ f_2 = \frac{14}{2} \times 40 = 280\ \text{mm} ]$ ✅ Answer: 280 mm

Circuit is balanced Wheatstone bridge . Current: $[ i = \frac{9}{3} = 3A ]$ Heat: $[ H = i^2Rt = 1080\ \text{J} ]$ ✅ Answer: 1080 J