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: MCQ with Answer and Explanation
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:
A. K1 * K2 / (K1 + K2)
B. 2K1 * K2 / (K1 + K2)
C. sqrt(K1^2 + K2^2)
D. (K1 + K2) / 2
Answer: Option B
Solution (By JKSSB Mock Tests)
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).
Explanation:
Work done is given by W = F * d * cos(theta). In uniform circular motion, the centripetal force is always directed strictly towards the center of the circle, while the instantaneous displacement (velocity vector) is tangential to the circle. The angle (theta) between them is exactly 90 degrees. Since cos(90) = 0, the work done by the centripetal force is always zero.
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