A ball is thrown vertically upward with speed 20 m/s. The maximum height reached (g = 10 m/s²) is
A. 40 m
B. 10 m
C. 200 m
D. 20 m
Answer: Option D
Solution (By JKSSB Mock Tests)
Using v² = u² - 2gH, at maximum height v = 0. So 0 = 20² - 2×10×H => 400 = 20H => H = 20 m. Alternative formula: H = u²/(2g) = 400/20 = 20 m. Remember to use consistent units. For quick calculation, u²/(2g) is handy.
Two slabs of equal thickness and cross-sectional area have thermal conductivities K1 and K2. If they are joined in series, the equivalent thermal conductivity of the combination is:
Explanation:
For slabs in series, the rate of heat flow (H) is the same. Total resistance R_eq = R1 + R2. Since thermal resistance R = L / (KA), and total length is 2L: (2L) / (K_eq * A) = (L / K1A) + (L / K2A). Canceling L/A yields 2/K_eq = 1/K1 + 1/K2. Solving gives K_eq = 2K1K2 / (K1 + K2).
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