PHYSICS

PHYSICS

A conductor has a resistivity of 1.0 × 10⁻⁷Ω m and α = 4 × 10⁻³°C^{-1 at 25° C . What is its resistivity at

Given: A conductor has a resistivity of 1.0 × 10⁻⁷Ω m and α = 4 × 10⁻³°C^{-1 at 25° C . What is its resistivity at 85° C ? These values define the system as per NCERT data. Formula: Use: rho_t = rho_0 [1 + α (T - T_0)]. This is the standard NCERT relation for this phenomenon. Substitution & Calculation: Substitute: rho_t = 1.0 × 10⁻⁷[1 + 4 × 10⁻³(85 - 25)] . Calculate: rho_t = 1.0 × 10⁻⁷[1 + 0.24] = 1.0 × 10⁻⁷ × 1.24 = 1.24 × 10⁻⁷Ω 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: Current Electricity, Topic: Resistivity, temperature dependence and circuit analysis.

View Details

PHYSICS

A 0.31 kg stone is whirled in a horizontal circle of radius 1.3 m at 54 rev/min . What is the tension? (Take g = 10 m/sÂ

Given: A 0.31 kg stone is whirled in a horizontal circle of radius 1.3 m at 54 rev/min . What is the tension? (Take g = 10 m/s² ) These values define the system as per NCERT data. Formula: Angular speed: omega = 54 × 2π/60 = 9π/5 rad/s. This is the standard NCERT relation for this phenomenon. Substitution & Calculation: omega² = (9π/5)² = 81π²/25 approx 31.99 . Tension: T = m omega² r = 0.31 × 31.99 × 1.3 . T approx 0.31 × 31.99 × 1.3 approx 12.9 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.

View Details

PHYSICS

A 0.2 kg aluminium block at 120° C is placed in 0.8 kg of water at 20° C in a 0.1 kg copper calorimeter at 20° C . Wh

Given: A 0.2 kg aluminium block at 120° C is placed in 0.8 kg of water at 20° C in a 0.1 kg copper calorimeter at 20° C . What is the final temperature? (Specific heat of aluminium = 900 J kg^{-1 K^{-1, water = 4186 J kg^{-1 K^{-1, copper = 386 J kg^{-1 K^{-1 ) These values define the system as per NCERT data. Formula: Heat lost = Heat gained. This is the standard NCERT relation for this phenomenon. Substitution & Calculation: 0.2 × 900 × (120 - T) = (0.8 × 4186 + 0.1 × 386) × (T - 20) . 21600 - 180 T = (3348.8 + 38.6) × (T - 20) = 3387.4 T - 67748 . 21600 + 67748 = 3387.4 T + 180 T . 89348 = 3567.4 T Rightarrow T approx 25.04° C approx 25° C . 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.

View Details

PHYSICS

Minimum speed to escape from 2 R_E from Earth’s nter is? ( g = 9.8 m/s², R_E = 6.4 × 10⁶ m )

Given: Minimum speed to escape from 2 R_E from Earth’s nter is? ( g = 9.8 m/s², R_E = 6.4 × 10⁶ m ) These values define the system as per NCERT data. Formula: v_e = sqrt2 g R_E²/2 R_E = sqrtg R_E. This is the standard NCERT relation for this phenomenon. Substitution & Calculation: v_e = sqrt9.8 × 6.4 × 10⁶= sqrt6.272 × 10⁷. v_e approx 7.92 × 10³ m/s approx 7.9 km/s . 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.

View Details

PHYSICS

How much heat is required to raise the temperature of 0.2 kg of silver from 30° C to 50° C ? (Specific heat of silver

Given: How much heat is required to raise the temperature of 0.2 kg of silver from 30° C to 50° C ? (Specific heat of silver = 236.1 J kg^{-1 K^{-1 ) These values define the system as per NCERT data. Formula: Δ Q = m s Δ T. This is the standard NCERT relation for this phenomenon. Substitution & Calculation: m = 0.2, s = 236.1, Δ T = 50 - 30 = 20 . Δ Q = 0.2 × 236.1 × 20 = 944.4 J approx 944 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: Units and Measurements and Physical World, Topic: Dimensional analysis and fundamental principles.

View Details

PHYSICS

A projectile is launched at 45 m/s at 30° . What is its speed at maximum height? (Take g = 10 m/s², cos 30° = 0.866 )

Given: A projectile is launched at 45 m/s at 30° . What is its speed at maximum height? (Take g = 10 m/s², cos 30° = 0.866 ) These values define the system as per NCERT data. Formula: At maximum height, speed = v_x = v_0 cos θ_0. This is the standard NCERT relation for this phenomenon. Substitution & Calculation: Given: v_0 = 45 m/s, cos 30° = 0.866 . v_x = 45 × 0.866 approx 38.97 m/s . 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.

View Details

PHYSICS

A sphere of radius 0.025 m moves at 0.08 m/s through blood ( eta = 2.7 × 10⁻³ Pa s ). What is the viscous drag force

Given: A sphere of radius 0.025 m moves at 0.08 m/s through blood ( eta = 2.7 × 10⁻³ Pa s ). What is the viscous drag force? These values define the system as per NCERT data. Formula: Stokes’ law: F = 6 π eta a v. This is the standard NCERT relation for this phenomenon. Substitution & Calculation: eta = 2.7 × 10⁻³ Pa s, a = 0.025 m, v = 0.08 m/s . F = 6 × 3.14 × 2.7 × 10⁻³ × 0.025 × 0.08 = 1.017 × 10⁻⁴ 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: Laws of Motion, Work Energy Power, Gravitation and System of Particles, Topic: Newton's laws, work-energy theorem and rotational dynamics.

View Details

PHYSICS

The number of valence electrons in Si and Ge atoms is:

Si (third orbit) and Ge (fourth orbit) are group IV elements, each with four valence electrons (2s and 2p for Si, 4s and 4p for Ge), forming covalent bonds in their lattice.

Ref: NCERT Physics Textbook for Class XI and XII, Chapter: Dual Nature, Atoms, Nuclei and Electronic Devices, Topic: Atomic structure, nuclear binding and semiconductor physics.

View Details

PHYSICS

A coil of 150 turns and area 0.05 m² is in a field that increases from 0 to 0.06 T in 0.3 s. What is the induced emf?

Given: A coil of 150 turns and area 0.05 m² is in a field that increases from 0 to 0.06 T in 0.3 s. What is the induced emf? These values define the system as per NCERT data. Formula: Δ Phi = B A = 0.06 × 0.05 = 0.003 Wb. This is the standard NCERT relation for this phenomenon. Substitution & Calculation: varepsilon = N Δ Phi/Δ t = 150 × 0.003/0.3 = 150 × 0.01 = 1.5 V . 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.

View Details

PHYSICS

A bar magnet with m = 1.5 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 = 1.5 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 = 1.5 A m², r = 0.5 m, μ_0/4π = 10⁻⁷. B = 10⁻⁷ × 1.5/(0.5)³ = 10⁻⁷ × 1.5/0.125 = 1.2 × 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.

View Details

PHYSICS

What is the effect on the frequency of a simple pendulum if it is taken to a planet where gravity is one-fourth that of

Frequency v = 1/2Ï€ sqrtg/L . If g' = g/4, then v' = 1/2Ï€ sqrtg/4/L = 1/2 v, halving the frequency. This follows from NCERT principle where the relation explains the outcome clearly for students in simple steps.

Ref: NCERT Physics Textbook for Class XI and XII, Chapter: Oscillations and Waves, Topic: SHM, wave motion and superposition.

View Details

PHYSICS

What is the temperature at which the rms speed of helium atoms is 1000 m/s? (Atomic mass of He = 4 u, k_B = 1.38 × 10â

Given: What is the temperature at which the rms speed of helium atoms is 1000 m/s? (Atomic mass of He = 4 u, k_B = 1.38 × 10⁻²³ J K^{-1) These values define the system as per NCERT data. Formula: v_{rms = sqrt3k_B T/m, m = frac4 × 10⁻³⁶.02 × 10²³= 6.64 × 10⁻²⁷ kg. This is the standard NCERT relation for this phenomenon. Substitution & Calculation: 1000² = frac3 × 1.38 × 10⁻²³ × T6.64 × 10⁻²⁷, T = frac10⁶ × 6.64 × 10⁻²⁷⁴.14 × 10⁻²³ approx 1604 K . 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: Dual Nature, Atoms, Nuclei and Electronic Devices, Topic: Atomic structure, nuclear binding and semiconductor physics.

View Details