चिकित्सा शिक्षा आयोग (CEE) बीएस्सी नर्सिङ र बीएनएस प्रवेश तयारी
चिकित्सा शिक्षा आयोग (CEE) – बीएस्सी नर्सिङ र बीएनएस प्रवेश
B.Sc. Nursing & BNS 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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An electric dipole of moment 2 × 10⁻⁸ C m is placed in a uniform electric field of 5 × 10⁴ N/C with its axis at 30° to the field. The torque on it is:
सही उत्तर: ग) 5 × 10⁻⁴ N m
τ = pE sinθ = 2 × 10⁻⁸ × 5 × 10⁴ × sin 30° = 1 × 10⁻³ × 0.5 = 5 × 10⁻⁴ N m.
💬 टिप्पणी / गल्ती रिपोर्ट (०)
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Sensitive electronic instruments are kept inside closed metal boxes to protect them from external electric fields. This works because:
सही उत्तर: ख) the electric field inside a closed conducting shell (with no charge inside) is zero
Charges on a conductor rearrange so that the field inside a closed conducting cavity is zero (electrostatic shielding); the potential inside is constant, not necessarily zero.
💬 टिप्पणी / गल्ती रिपोर्ट (०)
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Resistors of 2 Ω, 3 Ω and 6 Ω are connected in parallel across a 6 V battery of negligible internal resistance. The total current drawn from the battery is:
सही उत्तर: घ) 6 A
1/R = 1/2 + 1/3 + 1/6 = 1, so R = 1 Ω and I = 6/1 = 6 A.
💬 टिप्पणी / गल्ती रिपोर्ट (०)
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A wire of resistance R is stretched uniformly until its radius becomes half of the original. Its new resistance is:
सही उत्तर: क) 16R
Volume constant: radius halved means area becomes 1/4 and length 4 times; R ∝ L/A becomes 4 × 4 = 16 times.
💬 टिप्पणी / गल्ती रिपोर्ट (०)
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A galvanometer of resistance 50 Ω gives full-scale deflection for a current of 1 mA. To convert it into a voltmeter of range 0–10 V, the resistance needed in series is:
सही उत्तर: घ) 9950 Ω
R = V/I_g − G = 10/10⁻³ − 50 = 10000 − 50 = 9950 Ω.
💬 टिप्पणी / गल्ती रिपोर्ट (०)
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A cell drives a current of 0.9 A through a 2 Ω resistor and 0.3 A through a 7 Ω resistor. The internal resistance of the cell is:
सही उत्तर: घ) 0.5 Ω
E = 0.9(2 + r) = 0.3(7 + r) gives 1.8 + 0.9r = 2.1 + 0.3r, so 0.6r = 0.3 and r = 0.5 Ω (E = 2.25 V).
💬 टिप्पणी / गल्ती रिपोर्ट (०)
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n identical cells, each of emf E, are connected in series, but one of them is connected with its terminals reversed. The net emf of the battery is:
सही उत्तर: ग) (n − 2)E
The reversed cell does not add E but subtracts E, so net emf = (n − 1)E − E = (n − 2)E.
💬 टिप्पणी / गल्ती रिपोर्ट (०)
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A battery of emf 12 V and internal resistance 0.5 Ω has its terminals accidentally joined by a thick copper wire of negligible resistance. The current that flows is:
सही उत्तर: ख) 24 A
On short circuit the only resistance is the internal resistance, so I = E/r = 12/0.5 = 24 A.
💬 टिप्पणी / गल्ती रिपोर्ट (०)
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A 2 μF capacitor charged to 100 V is connected across an uncharged 3 μF capacitor. The energy lost in the process is:
सही उत्तर: ख) 6 mJ
Energy lost = ½ (C₁C₂/(C₁ + C₂)) V² = ½ × 1.2 × 10⁻⁶ × 100² = 6 × 10⁻³ J = 6 mJ (initial 10 mJ, final 4 mJ).
💬 टिप्पणी / गल्ती रिपोर्ट (०)
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The work done in moving a charge of 2 μC between two points having a potential difference of 50 V is:
सही उत्तर: क) 1 × 10⁻⁴ J
W = qV = 2 × 10⁻⁶ × 50 = 1 × 10⁻⁴ J.
💬 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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AI ले गल्ती गर्न सक्छ। महत्वपूर्ण कुरा आधिकारिक स्रोतमा जाँच्नुहोस्।