काठमाडौं विश्वविद्यालय KUUMAT / KU CBT तयारी
काठमाडौं विश्वविद्यालय – KUUMAT / KU CBT
KUUMAT / KU CBT · Kathmandu University (KUUMAT / KU CBT)
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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The minimum energy of a gamma-ray photon that can produce an electron–positron pair is about:
सही उत्तर: घ) 1.02 MeV
It must supply the rest energy of two particles: 2mₑc² = 2 × 0.511 MeV ≈ 1.02 MeV.
💬 टिप्पणी / गल्ती रिपोर्ट (०)
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The antiparticle of the electron is the:
सही उत्तर: क) positron
The positron has the same mass as the electron but charge +e.
💬 टिप्पणी / गल्ती रिपोर्ट (०)
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Which of the following particles is a lepton?
सही उत्तर: ख) electron
Leptons (electron, muon, tau and their neutrinos) do not feel the strong force; protons and neutrons are baryons and pions are mesons.
💬 टिप्पणी / गल्ती रिपोर्ट (०)
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A neutron is composed of the quarks:
सही उत्तर: क) udd
u(+2/3) + d(−1/3) + d(−1/3) = 0, the neutron's charge; the proton is uud.
💬 टिप्पणी / गल्ती रिपोर्ट (०)
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The electric charge of an up quark is:
सही उत्तर: ख) +2e/3
Up quarks carry +2e/3 and down quarks −e/3; a proton (uud) thus has charge +e.
💬 टिप्पणी / गल्ती रिपोर्ट (०)
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Hubble's law states that the recessional velocity v of a distant galaxy and its distance d are related by:
सही उत्तर: ख) v = H₀d
Galaxies recede with speeds proportional to their distances, v = H₀d, which is evidence for an expanding universe.
💬 टिप्पणी / गल्ती रिपोर्ट (०)
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Two long coaxial solenoids of length l have n₁ and n₂ turns per unit length, and the inner one has cross-sectional area A. Their mutual inductance is:
सही उत्तर: ख) μ₀n₁n₂Al
Current I in the outer solenoid gives B = μ₀n₁I; the flux through the n₂l turns of the inner one is μ₀n₁I·A·n₂l, so M = μ₀n₁n₂Al.
💬 टिप्पणी / गल्ती रिपोर्ट (०)
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If the same power is transmitted over a line at a higher voltage, the power loss in the line resistance:
सही उत्तर: ग) decreases in inverse proportion to the square of the voltage
For power P = VI, the current is I = P/V, so the loss I²R = P²R/V² falls as 1/V².
💬 टिप्पणी / गल्ती रिपोर्ट (०)
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In a series LR circuit connected to an AC source, the phase angle φ by which the current lags the voltage is given by:
सही उत्तर: ख) tan φ = ωL/R
The voltage phasors across R and L are at 90°, so tan φ = V_L/V_R = ωL/R.
💬 टिप्पणी / गल्ती रिपोर्ट (०)
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The average value of a sinusoidal alternating current of peak value I₀, taken over one half-cycle, is:
सही उत्तर: घ) 2I₀/π (≈ 0.637 I₀)
Average over a half-cycle = (1/π)∫₀^π I₀ sin θ dθ = 2I₀/π; over a full cycle the average is zero.
💬 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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