Linear expansion coefficient α = 2×10⁻⁵/°C. Length increase for 1 m rod heated 50°C: MCQ with Answer and Explanation

Linear expansion coefficient α = 2×10⁻⁵/°C. Length increase for 1 m rod heated 50°C:
A. 1 mm
B. 10 mm
C. 0.1 mm
D. 0.01 mm
Answer: Option A
Solution (By JKSSB Mock Tests)
ΔL = L₀αΔT = 1×2×10⁻⁵×50 = 100×10⁻⁵ = 10⁻³ m = 1 mm. Direct expansion formula application. Memory tip: 'ΔL = LαΔT; convert result to required units'. Tests thermal expansion calculation with unit conversion, frequently appearing in competitive exams.

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The number of significant figures in the measurement 0.0050200 m is:
A. 7
B. 3
C. 5
D. 8

Correct Answer: Option C


Explanation:
According to the rules of significant figures, leading zeros (0.00) are never significant as they only indicate the position of the decimal point. However, trailing zeros after a decimal point are always significant. Therefore, in 0.0050200, the significant digits are 5, 0, 2, 0, and 0, making a total of 5 significant figures.

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Question #2
The product of force and time is called
A. Impulse
B. Energy
C. Power
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Correct Answer: Option A


Explanation:
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Question #3
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. sqrt(K1^2 + K2^2)
B. 2K1 * K2 / (K1 + K2)
C. K1 * K2 / (K1 + K2)
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Correct Answer: Option B


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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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