इन्जिनियरिङ प्रवेश – Mathematics, Physics, Chemistry, English
इन्जिनियरिङ प्रवेश – Mathematics, Physics, Chemistry, English
Engineering Entrance – Mathematics, Physics, Chemistry & English · ३,८४१ प्रश्न
३,८४१ प्रश्न उपलब्ध। मोड छान्नुहोस्:
इन्जिनियरिङ प्रवेश: प्रश्न र उत्तर
सही लागेको विकल्पमा थिच्नुहोस्। सही भए हरियो, गलत भए रातो देखिन्छ; ३ सेकेन्डपछि सही उत्तर र व्याख्या देखिन्छ।
सही: ० / ०
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A pipe closed at one end has the same fundamental frequency as a pipe of length 60 cm open at both ends. The length of the closed pipe is:
सही उत्तर: ख) 30 cm
v/(4L_c) = v/(2 × 60) gives L_c = 30 cm.
💬 टिप्पणी / गल्ती रिपोर्ट (०)
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A wave is described by y = 0.1 sin(10πt − πx/5) in SI units. Its wavelength is:
सही उत्तर: घ) 10 m
k = π/5 rad/m, so λ = 2π/k = 10 m.
💬 टिप्पणी / गल्ती रिपोर्ट (०)
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For the resonance-tube readings l₁ = 16 cm and l₂ = 50 cm (first and second resonances), the end correction is:
सही उत्तर: क) 1 cm
e = (l₂ − 3l₁)/2 = (50 − 48)/2 = 1 cm.
💬 टिप्पणी / गल्ती रिपोर्ट (०)
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In a resonance-tube experiment with a 512 Hz fork, the first and second resonances occur at air-column lengths of 16 cm and 50 cm. The speed of sound is about:
सही उत्तर: ग) 348 m/s
l₂ − l₁ = λ/2 = 0.34 m, so λ = 0.68 m and v = fλ = 512 × 0.68 ≈ 348 m/s.
💬 टिप्पणी / गल्ती रिपोर्ट (०)
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A 0.5 m string of mass 5 g is stretched between two fixed points with a tension of 100 N. Its fundamental frequency is:
सही उत्तर: घ) 100 Hz
μ = 0.005/0.5 = 0.01 kg/m, v = √(T/μ) = 100 m/s, f = v/(2L) = 100 Hz.
💬 टिप्पणी / गल्ती रिपोर्ट (०)
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For a non-linear polyatomic ideal gas with 6 degrees of freedom (translational + rotational only), γ = Cp/Cv equals:
सही उत्तर: ग) 4/3
Cv = (f/2)R = 3R and Cp = 4R, so γ = 4/3.
💬 टिप्पणी / गल्ती रिपोर्ट (०)
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The entropy change when 1 kg of ice melts at 0 °C (L = 3.36 × 10⁵ J/kg) is about:
सही उत्तर: ख) 1231 J/K
ΔS = Q/T = 3.36 × 10⁵/273 ≈ 1231 J/K.
💬 टिप्पणी / गल्ती रिपोर्ट (०)
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One mole of a monatomic ideal gas expands adiabatically and its temperature falls by 100 K. The work done by the gas is about (R = 8.314 J mol⁻¹ K⁻¹):
सही उत्तर: क) 1247 J
In an adiabatic process W = −ΔU = nCvΔT = 1 × (3/2)(8.314)(100) ≈ 1247 J.
💬 टिप्पणी / गल्ती रिपोर्ट (०)
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A Carnot refrigerator works between 250 K and 300 K. The work needed to remove 1000 J of heat from the cold reservoir is:
सही उत्तर: ग) 200 J
COP = T₂/(T₁ − T₂) = 250/50 = 5, so W = Q₂/COP = 1000/5 = 200 J.
💬 टिप्पणी / गल्ती रिपोर्ट (०)
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A body cools from 70 °C to 50 °C in 5 minutes in a room at 30 °C. Using the average-temperature form of Newton's law of cooling, its temperature after the next 5 minutes is:
सही उत्तर: घ) 40 °C
First interval: 20/5 = k(60 − 30) ⇒ k = 2/15 per min. Next: (50 − T)/5 = (2/15)((50 + T)/2 − 30), which gives T = 40 °C.
💬 टिप्पणी / गल्ती रिपोर्ट (०)
यी अभ्यास प्रश्न हुन्, आधिकारिक प्रश्न होइनन्। पाठ्यक्रम, विज्ञापन र नतिजाका लागि सम्बन्धित निकायको आधिकारिक वेबसाइट हेर्नुहोस्। उत्तरमा गल्ती भेटे प्रश्नमुनिको "टिप्पणी / गल्ती रिपोर्ट" मा लेख्नुहोस्।
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