भौतिक विज्ञान (Physics) – Physics MCQ
भौतिक विज्ञान (Physics) – CEE प्रवेश
Physics · Medical Entrance (CEE) – Physics, Chemistry, Biology & MAT · ४०७ प्रश्न
४०७ प्रश्न उपलब्ध। मोड छान्नुहोस्:
भौतिक विज्ञान (Physics): 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 block of mass 5 kg slides down a rough inclined plane with an angle of 30 degrees. If the coefficient of friction between the block and the plane is 0.2, what is the acceleration of the block down the incline? (g = 9.8 m/s²)
सही उत्तर: क) 1.43 m/s²
The acceleration can be calculated using Newton's second law: ΣF = ma. The forces acting on the block are gravity and friction. The correct answer is 1.43m/s², accounting for both gravitational force (mgsinθ) and frictional force (μmgcosθ). Option 2.50 m/s² is incorrect.
💬 टिप्पणी / गल्ती रिपोर्ट (०)
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A uniform magnetic field of 0.5 Tesla is applied perpendicular to the plane of motion of a charged particle with charge 2 x 10⁻⁸ C and mass 10⁻³ kg. What force does this magnetic field exert on the particle? (Use F = qvB sinθ, where θ is the angle between v and B)
सही उत्तर: क) 0.5 N
The force exerted on the charged particle is given by the formula F = qvBsin(θ). In this case, q = 2 x 10⁻⁸ C, v = (unknown), B = 0.5 T, and θ is assumed to be 90 degrees since the velocity and magnetic field are perpendicular. The force exerted will be 1N. The incorrect options don't correctly apply this formula.
💬 टिप्पणी / गल्ती रिपोर्ट (०)
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A galvanometer is converted into an ammeter by connecting: (a) A resistor in series (b) A shunt in parallel (c) A resistance in parallel (d) A capacitor in series
सही उत्तर: ख) (b) A shunt in parallel
An ammeter has a low resistance to minimize the effect of its own resistance on the current being measured. To convert a galvanometer into an ammeter, a *shunt* (a resistor with very low resistance) is connected in parallel. This allows most of the current to flow through the shunt, reducing the impact of the galvanometer’s internal resistance. Option (a) uses a series resistor which would increase the total resistance and diminish the current being measured.
💬 टिप्पणी / गल्ती रिपोर्ट (०)
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A monochromatic light of wavelength 500 nm falls on a single slit of width 0.25 mm. What is the value of the diffraction angle (θ) in degrees?
सही उत्तर: क) 1°
The diffraction angle (θ) is determined using the formula θ = λ / d, where λ is the wavelength and d is the slit width. Here, λ = 500 nm = 500 x 10⁻⁹ m and d = 0.25 mm = 0.25 x 10⁻³ m. So, θ = (500 x 10⁻⁹) / (0.25 x 10⁻³) = 2 x 10⁻² radians or approximately 1.14 degrees. The closest answer is 1°.
💬 टिप्पणी / गल्ती रिपोर्ट (०)
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A rectangular metallic plate has dimensions 10 cm x 20 cm. The resistance between two opposite corners is 2 ohms. If the resistance between adjacent corners is 1 ohm, what is the value of the resistivity (ρ) of the metal in ohm-meter (Ωm)?
सही उत्तर: क) 0.05 Ωm
The resistance between opposite corners is calculated using R = ρL/A, where L is the length of the diagonal and A is the area of the plate. The resistance between adjacent corners is R = ρL/W, where W is the width. Solving for ρ in both equations allows us to find that ρ = 0.05 Ωm. The tempting answer of 0.1Ωm is incorrect because it does not account for the geometric relationship between the dimensions and resistances.
💬 टिप्पणी / गल्ती रिपोर्ट (०)
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A satellite is revolving around Earth at a height of 1600 km. Its orbital velocity is 7.5 km/s. What is the gravitational acceleration due to Earth at this altitude? (G = 6.674 × 10⁻¹¹ Nm²/kg², M_Earth = 5.972 × 10²⁴ kg, R_Earth = 6371 km)
सही उत्तर: ग) 11.2 m/s²
The orbital velocity is given by v = √(g * R_Earth), where g is the gravitational acceleration and R_Earth is the radius of Earth (6371 km). Solving for g, we get g = v²/R_Earth = (7.5 km/s)² / (6371 km) = 56.25 / 6371 ≈ 0.0088 m/s². Converting to more familiar units and using the provided values, this approaches close to 10.1m/s².
💬 टिप्पणी / गल्ती रिपोर्ट (०)
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A particle moves with constant acceleration along a straight line. At time t₁ = 2 s, its velocity is 6 m/s and at time t₂ = 5 s, its velocity is 12 m/s. What is the acceleration of the particle? (v = u + at)
सही उत्तर: ख) 4 m/s²
Using the equation v = u + at, we can solve for a. The change in velocity is 12 m/s - 6 m/s = 6 m/s, which occurred over a time interval of t₂ - t₁ = 5 s - 2 s = 3 s. Therefore, acceleration (a) = (Change in Velocity) / (Time Interval) = 6 m/s / 3 s = 2 m/s². This is not one of the options – it’s important to note that this calculation assumes constant acceleration.
💬 टिप्पणी / गल्ती रिपोर्ट (०)
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A mass of 5 kg is attached to a spring with a spring constant k = 100 N/m. The mass is displaced from its equilibrium position by 0.2 meters and released. What will be the angular frequency of the resulting oscillation? (ω = √(k/m))
सही उत्तर: घ) 20 rad/s
The angular frequency (ω) is calculated using the formula ω = √(k/m), where k is the spring constant (100 N/m) and m is the mass (5 kg). Plugging in these values, we get ω = √((100 N/m)/5 kg) = √20 rad²/s = 4.47 rad/s. The correct answer closest to 20 rad/s is a mistake - it’s vital to recognize that the given problem asks for angular frequency.
💬 टिप्पणी / गल्ती रिपोर्ट (०)
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A uniform wire of length 1 meter and cross-sectional area 1 mm² is stretched so that its length becomes 2 meters. Assuming the elastic modulus of the wire is 8 x 10⁹ N/m², what is the stress applied to the wire? (Use Young's Modulus formula: Stress = Elasticity * Strain)
सही उत्तर: ख) 8 x 10^9 N/m²
First, calculate the strain: Strain = (Change in Length) / (Original Length) = 2 m / 1 m = 2. Then, Stress = Young’s Modulus * Strain = (8 x 10⁹ N/m²) * 2 = 16 x 10⁹ N/m². The option ‘4 x 10^9 N/m²’ is too low because it doesn't consider the linear relationship between stress and strain.
💬 टिप्पणी / गल्ती रिपोर्ट (०)
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A vertically thrown ball reaches its maximum height after a time ‘t’. What is the magnitude of its velocity at that instant?
सही उत्तर: घ) 4g
At the maximum height, the ball's vertical velocity becomes zero. Therefore, the velocity at the highest point is g/2 due to the equation v² = u² - 2gh (where u is initial upward velocity and h is the max height attained). The magnitude of the velocity at that instant will be g/2.
💬 टिप्पणी / गल्ती रिपोर्ट (०)
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