आईओई इन्जिनियरिङ प्रवेश (त्रिवि) BE / B.Arch प्रवेश परीक्षा तयारी
आईओई इन्जिनियरिङ प्रवेश (त्रिवि) – BE / B.Arch प्रवेश परीक्षा
IOE BE / B.Arch Entrance · IOE Entrance (TU)
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
-
The capacitance of a spherical capacitor with inner radius a and outer radius b (vacuum between) is:
सही उत्तर: क) 4πε₀ab/(b − a)
With charge Q, V = (Q/4πε₀)(1/a − 1/b) = Q(b − a)/(4πε₀ab), so C = 4πε₀ab/(b − a).
💬 टिप्पणी / गल्ती रिपोर्ट (०)
-
An isolated parallel-plate capacitor of plate area A carries charge Q. The force of attraction between its plates is:
सही उत्तर: ख) Q²/(2ε₀A)
Each plate is in the field σ/(2ε₀) of the other, so F = Q × Q/(2ε₀A) = Q²/(2ε₀A).
💬 टिप्पणी / गल्ती रिपोर्ट (०)
-
Two charged conducting spheres of radii r₁ and r₂ are joined by a thin wire. The ratio of their surface charge densities σ₁ : σ₂ becomes:
सही उत्तर: ख) r₂ : r₁
They reach the same potential, so q₁/r₁ = q₂/r₂; then σ = q/(4πr²) ∝ 1/r, giving σ₁ : σ₂ = r₂ : r₁.
💬 टिप्पणी / गल्ती रिपोर्ट (०)
-
For a charged spherical conductor of radius R carrying charge Q, the electric potential at a point inside it is:
सही उत्तर: क) kQ/R, the same as at the surface
The field inside a conductor is zero, so the potential is constant throughout and equals its surface value kQ/R.
💬 टिप्पणी / गल्ती रिपोर्ट (०)
-
Light of wavelength 600 nm in air enters water (n = 4/3). In water its wavelength and frequency are:
सही उत्तर: घ) 450 nm, frequency unchanged
Frequency is fixed by the source; the speed and hence the wavelength fall by the factor n: 600/(4/3) = 450 nm.
💬 टिप्पणी / गल्ती रिपोर्ट (०)
-
A spectral line from a distant galaxy is observed with its wavelength increased by 0.1%. The galaxy is receding at about:
सही उत्तर: घ) 300 km/s
For v ≪ c, Δλ/λ = v/c, so v = 0.001 × 3 × 10⁸ m/s = 3 × 10⁵ m/s = 300 km/s.
💬 टिप्पणी / गल्ती रिपोर्ट (०)
-
In Young's experiment, light of wavelengths 600 nm and 400 nm is used together. The first point beyond the centre where bright fringes of both coincide corresponds to:
सही उत्तर: क) the 2nd bright fringe of 600 nm and the 3rd of 400 nm
Coincidence needs n₁ × 600 = n₂ × 400; the smallest solution is n₁ = 2, n₂ = 3 (path difference 1200 nm).
💬 टिप्पणी / गल्ती रिपोर्ट (०)
-
A thin film of refractive index 1.5 in air is illuminated normally with light of wavelength 600 nm. The minimum thickness for which the reflected light is strongly reinforced is:
सही उत्तर: ग) 100 nm
With a phase reversal at the top surface only, reflection maxima need 2nt = (m + ½)λ; the least thickness is λ/(4n) = 600/6 = 100 nm.
💬 टिप्पणी / गल्ती रिपोर्ट (०)
-
The limit of resolution of the normal human eye is about:
सही उत्तर: ख) 1 minute of arc
Two points can be resolved by the eye if they subtend about 1′ (≈ 3 × 10⁻⁴ rad) or more.
💬 टिप्पणी / गल्ती रिपोर्ट (०)
-
A primary rainbow is seen at an angle of about how much from the antisolar point (the direction opposite the Sun)?
सही उत्तर: घ) 42°
Light refracted, reflected once inside drops and refracted again is concentrated near 42° (red outside, violet inside); the secondary bow is near 51°.
💬 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।
नमस्ते! पाठ्यक्रम, सामान्य ज्ञान, भाषा परीक्षा वा कुनै प्रश्नको उत्तर बुझ्न सोध्नुहोस्।
AI ले गल्ती गर्न सक्छ। महत्वपूर्ण कुरा आधिकारिक स्रोतमा जाँच्नुहोस्।