पोखरा र पूर्वाञ्चल विश्वविद्यालय PU / PoU BE प्रवेश तयारी
पोखरा र पूर्वाञ्चल विश्वविद्यालय – PU / PoU BE प्रवेश
PU BE Entrance · Pokhara & Purbanchal University BE
३,८४१ प्रश्न उपलब्ध। मोड छान्नुहोस्:
BE/B.Arch 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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In Bohr's model, the radius of the n = 2 orbit of hydrogen compared to the n = 1 orbit is:
सही उत्तर: ख) Four times
rₙ = r₁n², so r₂ = 4r₁. The energy, by contrast, varies as 1/n².
💬 टिप्पणी / गल्ती रिपोर्ट (०)
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Light of photon energy 5 eV falls on a metal of work function 2 eV. The stopping potential is:
सही उत्तर: ग) 3 V
K_max = 5 − 2 = 3 eV, so the stopping potential V₀ = K_max/e = 3 V. 7 V results from adding instead of subtracting the work function.
💬 टिप्पणी / गल्ती रिपोर्ट (०)
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As the frequency of an AC source increases, the reactance of a capacitor:
सही उत्तर: ख) Decreases
X_C = 1/(2πfC), so it decreases with frequency; that is why a capacitor blocks DC (f = 0) but passes high-frequency AC. Inductive reactance increases with frequency.
💬 टिप्पणी / गल्ती रिपोर्ट (०)
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The resonant frequency of a series LCR circuit is:
सही उत्तर: क) 1/(2π√(LC))
At resonance X_L = X_C, i.e. ωL = 1/(ωC), giving ω = 1/√(LC) and f = 1/(2π√(LC)). At resonance the impedance equals R.
💬 टिप्पणी / गल्ती रिपोर्ट (०)
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The peak value of an AC voltage is 311 V. Its rms value is approximately:
सही उत्तर: घ) 220 V
V_rms = V₀/√2 = 311/1.414 ≈ 220 V. 155 V is V₀/2, which is not the rms value.
💬 टिप्पणी / गल्ती रिपोर्ट (०)
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The core of a transformer is laminated in order to reduce:
सही उत्तर: ग) Eddy current losses
Thin insulated laminations increase resistance to eddy currents circulating in the core, reducing heat loss. Copper losses are due to resistance of the windings.
💬 टिप्पणी / गल्ती रिपोर्ट (०)
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A transformer works on the principle of:
सही उत्तर: ख) Mutual induction
An alternating current in the primary produces changing flux that induces emf in the secondary (mutual induction). Hence a transformer does not work with steady DC.
💬 टिप्पणी / गल्ती रिपोर्ट (०)
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The energy stored in an inductor of 2 H carrying a current of 3 A is:
सही उत्तर: क) 9 J
U = ½LI² = ½ × 2 × 9 = 9 J. 18 J comes from omitting the factor ½.
💬 टिप्पणी / गल्ती रिपोर्ट (०)
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The magnetic flux through a coil of 10 turns changes by 0.5 Wb in 0.1 s. The magnitude of the average induced emf is:
सही उत्तर: घ) 50 V
ε = N ΔΦ/Δt = 10 × 0.5 / 0.1 = 50 V. 5 V is the emf per turn, ignoring the number of turns.
💬 टिप्पणी / गल्ती रिपोर्ट (०)
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Above the Curie temperature, a ferromagnetic material becomes:
सही उत्तर: ख) Paramagnetic
Thermal agitation destroys the alignment of domains above the Curie temperature, so the material behaves as paramagnetic.
💬 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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