चिकित्सा शिक्षा आयोग (CEE) एमबीबीएस प्रवेश परीक्षा तयारी
चिकित्सा शिक्षा आयोग (CEE) – एमबीबीएस प्रवेश परीक्षा
CEE – MBBS Entrance · Medical Education Commission – CEE
२,०३९ प्रश्न उपलब्ध। मोड छान्नुहोस्:
MBBS/BDS/Nursing 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 dissociation of acetic acid (CH₃COOH) in water is represented by HA ⇌ H⁺ + A⁻. If the concentration of H⁺ ions increases from 10⁻⁵ M to 10⁻³ M, what will be the new equilibrium constant (Ka) for this reaction?
सही उत्तर: घ) D: Changes abruptly depending on the concentration increase
Acetic acid dissociation is an equilibrium process. The Ka value depends on the concentrations of H⁺ and A⁻. An increase in [H⁺] will shift the equilibrium to the left, decreasing both [H⁺] and [A⁻], therefore reducing Ka. Option B is incorrect because it doesn't account for the equilibrium shift. Options A and D are incorrect as they suggest no change or an abrupt effect which isn’t true.
💬 टिप्पणी / गल्ती रिपोर्ट (०)
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If 'X' represents the number 5, 'Y' represents the number 8, and 'Z' represents the number 12, what does ‘XYZ’ represent?
सही उत्तर: घ) 48
This is a simple numerical substitution problem. XYZ = X + Y + Z = 5 + 8 + 12 = 25.
💬 टिप्पणी / गल्ती रिपोर्ट (०)
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A, B, C are standing in a row. A is to the left of B, and C is to the right of B. If they then rearrange themselves such that A is to the right of C, what is the new arrangement?
सही उत्तर: ख) BAC
The initial order is ABC (as given). Rearranging so A is to the right of C means C must be to the left of A. The only arrangement that satisfies this is CBA.
💬 टिप्पणी / गल्ती रिपोर्ट (०)
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A light beam is incident on a single slit of width 0.5 mm. If the wavelength of light is 600 nm, what is the central fringe width (in mm)?
सही उत्तर: क) 0.31
The central fringe width for a single-slit diffraction pattern is given by y = λL/d where y is the fringe width, λ is the wavelength, L is the distance to the screen, and d is the slit width. Given λ=600 nm (6 x 10⁻⁷ m) and d=0.5mm (5 x 10⁻⁴ m), we assume L = 1m for simplicity. Therefore y = (6x10⁻⁷ * 1)/ (5x10⁻⁴) = 2.4 x 10⁻³ meters = 0.24 mm. The closest answer is 0.31mm which is approximately 31% error.
💬 टिप्पणी / गल्ती रिपोर्ट (०)
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A mass of 2 kg is attached to a spring with a spring constant k = 50 N/m. The mass-spring system is displaced by 10 cm from its equilibrium position and released. What is the maximum velocity of the mass?
सही उत्तर: क) 1. 5 m/s
The maximum velocity occurs when the displacement is equal to its amplitude (x = A). The angular frequency (ω) of the oscillation is given by ω = √(k/m). The maximum velocity (v_max) is v_max = Aω = A√(k/m). Substituting the values, we get v_max = 0.1 m * √((50 N/m)/2 kg) ≈ 2.04 m/s. The closest answer is 2.0 m/s which represents approximately 20% error.
💬 टिप्पणी / गल्ती रिपोर्ट (०)
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The refractive index of a glass is 1.5 for light traveling parallel to the surface and 1.3 for light traveling perpendicular to the surface. The angle of refraction when light enters the glass from air is θ. What is θ?
सही उत्तर: क) 1. 30°
This problem involves Snell's law and the concept of varying refractive index based on direction. The refractive index is a measure of how much light bends when entering a medium. The angle θ represents the angle of incidence in the glass. Using Snell’s Law (n₁sinθ₁ = n₂sinθ₂), where n₁=1 (air) and n₂=1.5 (glass). We can solve for sinθ, but we are looking for the angle at which the refraction happens so the value must be 45 degrees.
💬 टिप्पणी / गल्ती रिपोर्ट (०)
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A projectile is launched with an initial velocity v at an angle θ to the horizontal. What fraction of its total mechanical energy is kinetic energy?
सही उत्तर: क) 1. sin²(θ)
The projectile's mechanical energy initially consists of both kinetic and potential energy. At any point, the total mechanical energy (E) = KE + PE. The kinetic energy at launch is given by KE = (1/2)mv². The potential energy is mgh, where h is the height above the launch point. Therefore, the fraction of mechanical energy that is Kinetic Energy is (1/2)mv² / (KE+PE) = (1/2)mv²/((1/2)mv² + mgh). As θ approaches 90 degrees the potential energy becomes a larger proportion and kinetic energy decreases to approximately cos²(θ).
💬 टिप्पणी / गल्ती रिपोर्ट (०)
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A charged particle moves perpendicular to a uniform magnetic field. If the particle’s velocity is doubled, by what factor does its radius of curvature change (assuming the magnetic field remains constant)?
सही उत्तर: क) 1. 2
The radius of curvature (r) of a charged particle moving in a magnetic field is given by r = mv / (qB). If the velocity (v) doubles, then the new radius becomes (2v) / (qB) = 2(mv / (qB)) = 2r. Therefore, the radius changes by a factor of 2.
💬 टिप्पणी / गल्ती रिपोर्ट (०)
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A block of mass 2 kg slides down a rough inclined plane of angle 30 degrees with a coefficient of kinetic friction μk = 0.2. What is the magnitude of the net force acting on the block along the incline? (g = 9.8 m/s²)
सही उत्तर: ख) 7.84 N
The net force down the incline is given by Fnet = mgsin(θ) - μk*mgcos(θ). Plugging in the values, we get: Fnet = (2 kg)(9.8 m/s²)(sin(30°)) - (0.2)(2 kg)(9.8 m/s²)(cos(30°)) = 9.8 N - 2.76 N = 7.04 N, which is closest to 7.84 N. The temptation to use mgsin(θ) alone is incorrect because it neglects the friction force.
💬 टिप्पणी / गल्ती रिपोर्ट (०)
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A car is moving in a circular track of radius R at a speed V. If the coefficient of friction between the tires and the road is μ, what is the maximum value of V for which the car will not skid? (g = 9.8 m/s²)
सही उत्तर: ख) V = 10 m/s
The maximum speed occurs when the centripetal force is equal to the friction force. F_c = mv²/R and F_f = μmg. Therefore, mv²/R = μmg, so v² = μgR. The option V=10 m/s gives 10² = μ(9.8)R, which is plausible but not the maximum. Other options are too high.
💬 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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