PHYSICS

PHYSICS

A spring system has m = 1.2 kg, k = 480 N/m, A = 6 cm . What is the potential energy at x = 3 cm ?

Given: A spring system has m = 1.2 kg, k = 480 N/m, A = 6 cm . What is the potential energy at x = 3 cm ? These values define the system as per NCERT data. Formula: Potential energy: U = 1/2 k x². This is standard NCERT relation. Substitution & Calculation: k = 480 N/m, x = 0.03 m . U = 0.5 × 480 × (0.03)² = 0.5 × 480 × 0.0009 = 0.216 J . Result: The computed value matches expected outcome and confirms correct choice as per NCERT.

Ref: NCERT Physics Textbook for Class XI and XII, Chapter: Relevant Physics topic covering fundamental principles,

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PHYSICS

A rectangular loop of area 0.03 m² with 20 turns carries 2 A in a field of 0.7 T at 45° to the plane. What is the torque

Given: A rectangular loop of area 0.03 m² with 20 turns carries 2 A in a field of 0.7 T at 45° to the plane. What is the torque? These values define the system as per NCERT data. Formula: tau = N I A B sin θ, where θ = 45° to plane means sin 45° with normal. This is standard NCERT relation. Substitution & Calculation: tau = 20 × 2 × 0.03 × 0.7 × sin 45° = 0.84 × 0.707 = 0.5939 approx 0.59 N m . Result: The computed value matches expected outcome and confirms correct choice as per NCERT.

Ref: NCERT Physics Textbook for Class XI and XII, Chapter: Relevant Physics topic covering fundamental principles,

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PHYSICS

A dipole p = 7 × 10⁻⁹C m is rotated from θ = 90° to 0° in a field E = 2 × 10⁵N/C . What is the work done?

Given: A dipole p = 7 × 10⁻⁹C m is rotated from θ = 90° to 0° in a field E = 2 × 10⁵N/C . What is the work done? These values define the system as per NCERT data. Formula: Work done: W = p E (cos θ_0 - cos θ_1) = 7 × 10⁻⁹ × 2 × 10⁵ × (cos 90° - cos 0°). This is standard NCERT relation. Substitution & Calculation: W = 7 × 10⁻⁹ × 2 × 10⁵ × (0 - 1) = -1.4 × 10⁻³J . Result: The computed value matches expected outcome and confirms correct choice as per NCERT.

Ref: NCERT Physics Textbook for Class XI and XII, Chapter: Relevant Physics topic covering fundamental principles,

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PHYSICS

A conductor has a resistivity of 1.2 × 10⁻⁷Ω m and α = 4 × 10⁻³°C^{-1 at 20° C . What is its resistivity at 80° C ?

Given: A conductor has a resistivity of 1.2 × 10⁻⁷Ω m and α = 4 × 10⁻³°C^{-1 at 20° C . What is its resistivity at 80° C ? These values define the system as per NCERT data. Formula: Use: rho_t = rho_0 [1 + α (T - T_0)]. This is standard NCERT relation. Substitution & Calculation: Substitute: rho_t = 1.2 × 10⁻⁷[1 + 4 × 10⁻³(80 - 20)] . Calculate: rho_t = 1.2 × 10⁻⁷[1 + 0.24] = 1.2 × 10⁻⁷ × 1.24 = 1.488 × 10⁻⁷Ω m . Result: The computed value matches expected outcome and confirms correct choice as per NCERT.

Ref: NCERT Physics Textbook for Class XI and XII, Chapter: Relevant Physics topic covering fundamental principles,

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PHYSICS

How many joules are equivalent to 350 cal of heat? (1 cal = 4.186 J )

Given: How many joules are equivalent to 350 cal of heat? (1 cal = 4.186 J ) These values define the system as per NCERT data. Formula: Heat in J = Heat in cal × 4.186. This is standard NCERT relation. Substitution & Calculation: 350 × 4.186 = 1465.1 J approx 1465 J . Result: The computed value matches expected outcome and confirms correct choice as per NCERT.

Ref: NCERT Physics Textbook for Class XI and XII, Chapter: Relevant Physics topic covering fundamental principles,

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PHYSICS

A telescope has an objective of focal length 160 cm and an eyepiece of focal length 8 cm . What is its magnifying power?

Given: A telescope has an objective of focal length 160 cm and an eyepiece of focal length 8 cm . What is its magnifying power? These values define the system as per NCERT data. Formula: Magnifying power: m = f_o/f_e. This is standard NCERT relation. Substitution & Calculation: f_o = 160 cm, f_e = 8 cm . m = 160/8 = 20 . Result: The computed value matches expected outcome and confirms correct choice as per NCERT.

Ref: NCERT Physics Textbook for Class XI and XII, Chapter: Relevant Physics topic covering fundamental principles,

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PHYSICS

A gas at 3 atm and 300 K has a volume of 10 litres. If the temperature rises to 900 K at constant pressure, what is the

Given: A gas at 3 atm and 300 K has a volume of 10 litres. If the temperature rises to 900 K at constant pressure, what is the new volume? These values define the system as per NCERT data. Formula: Charles’ law: V_1/T_1 = V_2/T_2. This is standard NCERT relation. Substitution & Calculation: V_1 = 10 litres, T_1 = 300 K, T_2 = 900 K. V_2 = V_1 × T_2/T_1 = 10 × 900/300 = 30 litres. Result: The computed value matches expected outcome and confirms correct choice as per NCERT.

Ref: NCERT Physics Textbook for Class XI and XII, Chapter: Thermodynamics and Kinetic Theory, Topic: Charles' law V/T = constant, volume-temperature relation at constant pressure. The section explains definitions, governing laws, formulas like μ₀ = 4π × 10⁻⁷ T·m/A, units and illustrative examples.

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PHYSICS

A 5 kg mass on a horizontal surface ( μ_k = 0.4 ) is pulled by a 3 kg mass over a pulley with a 10 N force aiding the 5

Given: A 5 kg mass on a horizontal surface ( μ_k = 0.4 ) is pulled by a 3 kg mass over a pulley with a 10 N force aiding the 5 kg mass. What is the acceleration? (Take g = 10 m/s² ) These values define the system as per NCERT data. Formula: For 3 kg : 3g - T = 3a Rightarrow 30 - T = 3a. This is standard NCERT relation. Substitution & Calculation: For 5 kg : T + 10 - f_k = 5a, f_k = 0.4 × 5 × 10 = 20 N . T + 10 - 20 = 5a Rightarrow T - 10 = 5a . Solve: 30 - T = 3a, T - 10 = 5a Rightarrow 30 - (5a + 10) = 3a . 30 - 10 - 5a = 3a Rightarrow 20 = 8a Rightarrow a = 2.5 m/s² . Result: The computed value matches expected outcome and confirms correct choice as per NCERT.

Ref: NCERT Physics Textbook for Class XI and XII, Chapter: Relevant Physics topic covering fundamental principles,

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PHYSICS

A circuit has a 12 V battery with 3 Ω internal resistance and two resistors 6 Ω and 3 Ω in parallel. What is the total c

Given: A circuit has a 12 V battery with 3 Ω internal resistance and two resistors 6 Ω and 3 Ω in parallel. What is the total current? These values define the system as per NCERT data. Formula: Parallel resistance: 1/R_p = 1/6 + 1/3 = 1 + 2/6 = 3/6 = 0.5 Rightarrow R_p = 2 Ω. This is standard NCERT relation. Substitution & Calculation: Total resistance: R_{total = 3 + 2 = 5 Ω . Current: I = fracvarepsilonR_{total = 12/5 = 2.4 A . Result: The computed value matches expected outcome and confirms correct choice as per NCERT.

Ref: NCERT Physics Textbook for Class XI and XII, Chapter: Relevant Physics topic covering fundamental principles,

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PHYSICS

A rod rotates at 25 rad/s in a 0.3 T field. If the length from the axis to the tip is 0.8 m, what is the emf induced?

Given: A rod rotates at 25 rad/s in a 0.3 T field. If the length from the axis to the tip is 0.8 m, what is the emf induced? These values define the system as per NCERT data. Formula: varepsilon = 1/2 B omega R². This is standard NCERT relation. Substitution & Calculation: varepsilon = 1/2 × 0.3 × 25 × (0.8)² = 2.4 V . Result: The computed value matches expected outcome and confirms correct choice as per NCERT.

Ref: NCERT Physics Textbook for Class XI and XII, Chapter: Relevant Physics topic covering fundamental principles,

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PHYSICS

A copper block of mass 0.5 kg at 100° C is dropped into 1 kg of water at 20° C . Find the final temperature. (Specific h

Given: A copper block of mass 0.5 kg at 100° C is dropped into 1 kg of water at 20° C . Find the final temperature. (Specific heat of copper = 386 J kg^{-1 K^{-1, water = 4186 J kg^{-1 K^{-1 ) These values define the system as per NCERT data. Formula: Heat lost by copper = Heat gained by water. This is standard NCERT relation. Substitution & Calculation: m_c s_c (100 - T) = m_w s_w (T - 20) . 0.5 × 386 × (100 - T) = 1 × 4186 × (T - 20) . 19300 - 193 T = 4186 T - 83720 . 19300 + 83720 = 4186 T + 193 T . 103020 = 4379 T Rightarrow T approx 23.53° C . Result: The computed value matches expected outcome and confirms correct choice as per NCERT.

Ref: NCERT Physics Textbook for Class XI and XII, Chapter: Relevant Physics topic covering fundamental principles,

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PHYSICS

A double convex lens of refractive index 1.55 has radii of curvature 30 cm and -30 cm . What is its focal length?

Given: A double convex lens of refractive index 1.55 has radii of curvature 30 cm and -30 cm . What is its focal length? These values define the system as per NCERT data. Formula: Lens maker’s formula: 1/f = (n - 1) ( 1/R_1 - 1/R_2 ). This is standard NCERT relation. Substitution & Calculation: n = 1.55, R_1 = 30 cm, R_2 = -30 cm . 1/f = (1.55 - 1) ( 1/30 - 1/-30 ) = 0.55 ( 1/30 + 1/30 ) = 0.55 × 2/30 = 1.1/30 . f = 30/1.1 approx 27.27 cm . Result: The computed value matches expected outcome and confirms correct choice as per NCERT.

Ref: NCERT Physics Textbook for Class XI and XII, Chapter: Relevant Physics topic covering fundamental principles,

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