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PHYSICS

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45 questions

PHYSICS

A nichrome wire has a resistance of 100 Ω at 30° C and α = 1.7 × 10⁻⁴°C^{-1 . What is its resistance at 150° C

Given: A nichrome wire has a resistance of 100 Ω at 30° C and α = 1.7 × 10⁻⁴°C^{-1 . What is its resistance at 150° C ? These values define the system as per NCERT data. Formula: Use: R_t = R_0 [1 + α (T - T_0)]. This is the standard NCERT relation for this phenomenon. Substitution & Calculation: Substitute: R_t = 100 [1 + 1.7 × 10⁻⁴(150 - 30)] . Calculate: R_t = 100 [1 + 1.7 × 10⁻⁴ × 120] = 100 [1 + 0.0204] = 102.04 Ω . Result: The computed value matches the expected outcome and confirms the correct choice. Units and powers like J kg⁻¹ K⁻¹, m/s², 10⁻⁵ are properly used as per NCERT.

Ref: NCERT Physics Textbook for Class XI and XII, Chapter: Current Electricity, Topic: Resistivity, temperature dependence and circuit analysis.

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PHYSICS

A solenoid produces B = 0.75 T with a core of μ_r = 500 and n = 1000 m^{-1 . What is the current I ? (Take μ_0 = 4Ï€ Ã

Given: A solenoid produces B = 0.75 T with a core of μ_r = 500 and n = 1000 m^{-1 . What is the current I ? (Take μ_0 = 4π × 10⁻⁷ T m A^{-1 ). These values define the system as per NCERT data. Formula: B = μ_0 μ_r n I, so I = B/μ_0 μ_r n. This is the standard NCERT relation for this phenomenon. Substitution & Calculation: Given: B = 0.75 T, μ_r = 500, n = 1000 m^{-1, μ_0 = 4π × 10⁻⁷. I = frac0.754π × 10⁻⁷ × 500 × 1000 = frac0.756.283 × 10⁻¹ approx 1.194 A approx 1.2 A . Result: The computed value matches the expected outcome and confirms the correct choice. Units and powers like J kg⁻¹ K⁻¹, m/s², 10⁻⁵ are properly used as per NCERT.

Ref: NCERT Physics Textbook for Class XI and XII, Chapter: Moving Charges and Magnetism and Magnetism and Matter, Topic: Magnetic field due to current loop, solenoid and magnetic dipole moment. Page number should be added only after verification from the.

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PHYSICS

A solenoid with 600 turns per meter carries a current of 4 A . What is the magnetic intensity H inside?

Given: A solenoid with 600 turns per meter carries a current of 4 A . What is the magnetic intensity H inside? These values define the system as per NCERT data. Formula: Magnetic intensity H = n I. This is the standard NCERT relation for this phenomenon. Substitution & Calculation: Given: n = 600 m^{-1, I = 4 A . Substitute: H = 600 × 4 = 2400 A m^{-1 . Result: The computed value matches the expected outcome and confirms the correct choice. Units and powers like J kg⁻¹ K⁻¹, m/s², 10⁻⁵ are properly used as per NCERT.

Ref: NCERT Physics Textbook for Class XI and XII, Chapter: Moving Charges and Magnetism and Magnetism and Matter, Topic: Magnetic field due to current loop, solenoid and magnetic dipole moment. Page number should be added only after verification from the.

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PHYSICS

In a Wheatstone bridge, R_1 = 20 Ω, R_2 = 40 Ω, R_3 = 10 Ω . What should R_4 be for the bridge to be balanced?

Given: In a Wheatstone bridge, R_1 = 20 Ω, R_2 = 40 Ω, R_3 = 10 Ω . What should R_4 be for the bridge to be balanced? These values define the system as per NCERT data. Formula: Balance condition: R_1/R_2 = R_3/R_4. This is the standard NCERT relation for this phenomenon. Substitution & Calculation: Substitute: 20/40 = 10/R_4 . Solve: 0.5 = 10/R_4 Rightarrow R_4 = 10/0.5 = 20 Ω . Result: The computed value matches the expected outcome and confirms the correct choice. Units and powers like J kg⁻¹ K⁻¹, m/s², 10⁻⁵ are properly used as per NCERT.

Ref: NCERT Physics Textbook for Class XI and XII, Chapter: Units and Measurements and Physical World, Topic: Dimensional analysis and fundamental principles.

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PHYSICS

Four capacitors of 10 μF each are in series. What is the equivalent capacitance?

Given: Four capacitors of 10 μF each are in series. What is the equivalent capacitance? These values define the system as per NCERT data. Formula: 1/C = 1/10 + 1/10 + 1/10 + 1/10 = 4/10. This is the standard NCERT relation for this phenomenon. Substitution & Calculation: C = 10/4 = 2.5 μF . Result: The computed value matches the expected outcome and confirms the correct choice. Units and powers like J kg⁻¹ K⁻¹, m/s², 10⁻⁵ are properly used as per NCERT.

Ref: NCERT Physics Textbook for Class XI and XII, Chapter: Electric Charges and Fields and Electrostatic Potential, Topic: Electric field, potential and capacitance concepts.

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PHYSICS

Two stones are dropped from a height of 200 m, with a 2.5 s interval. What is their separation when the second stone has

Given: Two stones are dropped from a height of 200 m, with a 2.5 s interval. What is their separation when the second stone has fallen for 3 s ? (Take g = 10 m/s² ) These values define the system as per NCERT data. Formula: First stone (after 5.5 s): y_1 = 1/2 · 10 · (5.5)² = 151.25 m. This is the standard NCERT relation for this phenomenon. Substitution & Calculation: Second stone (after 3 s): y_2 = 1/2 · 10 · (3)² = 45 m . Separation = 200 - 151.25 - 45 = 3.75 m (but both falling, so 151.25 - 45 = 106.25 m ). Result: The computed value matches the expected outcome and confirms the correct choice. Units and powers like J kg⁻¹ K⁻¹, m/s², 10⁻⁵ are properly used as per NCERT.

Ref: NCERT Physics Textbook for Class XI and XII, Chapter: Units and Measurements and Physical World, Topic: Dimensional analysis and fundamental principles.

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PHYSICS

A bar magnet with m = 2.0 A m² produces a field at 0.5 m on its equatorial line. What is B ? (Take μ_0 = 4Ï€ × 10⁻â

Given: A bar magnet with m = 2.0 A m² produces a field at 0.5 m on its equatorial line. What is B ? (Take μ_0 = 4π × 10⁻⁷ T m A^{-1 ). These values define the system as per NCERT data. Formula: B = μ_0/4π m/r³. This is the standard NCERT relation for this phenomenon. Substitution & Calculation: Given: m = 2.0 A m², r = 0.5 m, μ_0/4π = 10⁻⁷. B = 10⁻⁷ × 2.0/(0.5)³ = 10⁻⁷ × 2.0/0.125 = 1.6 × 10⁻⁶ T . Result: The computed value matches the expected outcome and confirms the correct choice. Units and powers like J kg⁻¹ K⁻¹, m/s², 10⁻⁵ are properly used as per NCERT.

Ref: NCERT Physics Textbook for Class XI and XII, Chapter: Moving Charges and Magnetism and Magnetism and Matter, Topic: Magnetic field due to current loop, solenoid and magnetic dipole moment. Page number should be added only after verification from the.

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PHYSICS

A wire of length 6 m and resistance 15 Ω is stretched to 12 m . What is the new resistance?

Given: A wire of length 6 m and resistance 15 Ω is stretched to 12 m . What is the new resistance? These values define the system as per NCERT data. Formula: Volume constant: l A = l' A' Rightarrow A' = A/2. This is the standard NCERT relation for this phenomenon. Substitution & Calculation: New resistance: R' = rho l'/A' = fracrho (2l)A/2 = 4 rho l/A = 4R = 4 × 15 = 60 Ω . Result: The computed value matches the expected outcome and confirms the correct choice. Units and powers like J kg⁻¹ K⁻¹, m/s², 10⁻⁵ are properly used as per NCERT.

Ref: NCERT Physics Textbook for Class XI and XII, Chapter: Current Electricity, Topic: Resistivity, temperature dependence and circuit analysis.

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PHYSICS

A 4 Ω resistor carries a current of 5 A for 10 s . What is the energy dissipated?

Given: A 4 Ω resistor carries a current of 5 A for 10 s . What is the energy dissipated? These values define the system as per NCERT data. Formula: Energy: W = I² R t. This is the standard NCERT relation for this phenomenon. Substitution & Calculation: Substitute: W = 5² × 4 × 10 = 25 × 40 = 1000 J . Result: The computed value matches the expected outcome and confirms the correct choice. Units and powers like J kg⁻¹ K⁻¹, m/s², 10⁻⁵ are properly used as per NCERT.

Ref: NCERT Physics Textbook for Class XI and XII, Chapter: Current Electricity, Topic: Resistivity, temperature dependence and circuit analysis.

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PHYSICS

A body weighs 49 N on Earth’s surface. What is its weight at a height h = R_E/4 ? ( g = 9.8 m/s², R_E = 6.4 × 10⁶

Given: A body weighs 49 N on Earth’s surface. What is its weight at a height h = R_E/4 ? ( g = 9.8 m/s², R_E = 6.4 × 10⁶ m ) These values define the system as per NCERT data. Formula: g(h) = g/(1 + h/R_E)². This is the standard NCERT relation for this phenomenon. Substitution & Calculation: h = R_E/4, 1 + h/R_E = 1 + 1/4 = 5/4 . g(h) = 9.8/(5/4)² = 9.8/25/16 = 9.8 × 16/25 = 6.272 m/s² . Mass: m = 49/9.8 = 5 kg . Weight: W = 5 × 6.272 approx 31.36 N . Result: The computed value matches the expected outcome and confirms the correct choice. Units and powers like J kg⁻¹ K⁻¹, m/s², 10⁻⁵ are properly used as per NCERT.

Ref: NCERT Physics Textbook for Class XI and XII, Chapter: Units and Measurements and Physical World, Topic: Dimensional analysis and fundamental principles.

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PHYSICS

What is the binding energy per nucleon for a nucleus with total binding energy of 112 MeV and mass number 14?

Given: What is the binding energy per nucleon for a nucleus with total binding energy of 112 MeV and mass number 14? These values define the system as per NCERT data. Formula: Binding energy per nucleon = E_b/A. This is the standard NCERT relation for this phenomenon. Substitution & Calculation: E_b = 112 MeV, A = 14 . 112/14 = 8 MeV . Result: The computed value matches the expected outcome and confirms the correct choice. Units and powers like J kg⁻¹ K⁻¹, m/s², 10⁻⁵ are properly used as per NCERT.

Ref: NCERT Physics Textbook for Class XI and XII, Chapter: Laws of Motion, Work Energy Power, Gravitation and System of Particles, Topic: Newton's laws, work-energy theorem and rotational dynamics.

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