PHYSICS

PHYSICS

Why does the classical electromagnetic theory predict that an atom should collapse?

An accelerating electron in orbit emits radiation, losing energy and spiraling into the nucleus, leading to collapse according to classical theory. 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: 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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PHYSICS

Which of the following best describes the nuclear force?

The nuclear force is a strong, short-range attractive force that binds protons and neutrons in the nucleus, overcoming the Coulomb repulsion between protons. It is much stronger than the Coulomb force and does not depend on the electric charge of the nucleons.

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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PHYSICS

A stone is thrown upwards with a speed of 18 m/s . What is its velocity after 2 s ? (Take g = 10 m/s² )

Given: A stone is thrown upwards with a speed of 18 m/s . What is its velocity after 2 s ? (Take g = 10 m/s² ) These values define the system as per NCERT data. Formula: Use v = v_0 + a t. This is the standard NCERT relation for this phenomenon. Substitution & Calculation: Here, v_0 = 18 m/s, a = -10 m/s², t = 2 s . Substitute: v = 18 - 10 · 2 = 18 - 20 = -2 m/s . The velocity is -2 m/s (downward). 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 shape of the wavefront reflected by a plane mirror when a spherical wave is incident on it?

A spherical wavefront incident on a plane mirror reflects as a spherical wavefront, diverging from the mirror’s virtual focus. 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: Ray Optics and Optical Instruments and Wave Optics, Topic: Refraction, lenses and interference/diffraction.

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PHYSICS

What is the kinetic energy of a 500 kg satellite at 7 R_E from Earth’s nter? ( M_E = 6 × 10²⁴ kg, R_E = 6.4 × 10â

Given: What is the kinetic energy of a 500 kg satellite at 7 R_E from Earth’s nter? ( M_E = 6 × 10²⁴ kg, R_E = 6.4 × 10⁶ m, G = 6.67 × 10⁻¹¹ N m²/kg² ) These values define the system as per NCERT data. Formula: K = G M_E m/2 r. This is the standard NCERT relation for this phenomenon. Substitution & Calculation: r = 7 R_E = 4.48 × 10⁷ m . K = frac6.67 × 10⁻¹¹ × 6 × 10²⁴ × 5002 × 4.48 × 10⁷. K = frac2.001 × 10¹⁷⁸.96 × 10⁷ approx 2.23 × 10⁹ 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: 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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PHYSICS

A 1250 kg car turns on a banked road ( θ = 15°, μ_s = 0.3 ) with radius 45 m . What is the maximum speed without slip

Given: A 1250 kg car turns on a banked road ( θ = 15°, μ_s = 0.3 ) with radius 45 m . What is the maximum speed without slipping? (Take g = 10 m/s², tan 15° approx 0.268 ) These values define the system as per NCERT data. Formula: Maximum speed: v_{max = sqrtrg μ_s + tanθ/1 - μ_s tanθ. This is the standard NCERT relation for this phenomenon. Substitution & Calculation: Numerator: μ_s + tanθ = 0.3 + 0.268 = 0.568 . Denominator: 1 - 0.3 × 0.268 = 1 - 0.0804 = 0.9196 . v_{max² = 45 × 10 × 0.568/0.9196 approx 450 × 0.6175 approx 277.875 . v_{max = sqrt277.875 approx 16.67 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.

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PHYSICS

How much energy is required to move a 100 kg satellite from 4 R_E to 8 R_E from Earth’s nter? ( M_E = 6 × 10²⁴ kg,

Given: How much energy is required to move a 100 kg satellite from 4 R_E to 8 R_E from Earth’s nter? ( M_E = 6 × 10²⁴ kg, R_E = 6.4 × 10⁶ m, G = 6.67 × 10⁻¹¹ N m²/kg² ) These values define the system as per NCERT data. Formula: Δ E = -G M_E m (1/r_2 - 1/r_1). This is the standard NCERT relation for this phenomenon. Substitution & Calculation: r_1 = 2.56 × 10⁷ m, r_2 = 5.12 × 10⁷ m . Δ E = -6.67 × 10⁻¹¹ × 6 × 10²⁴ × 100 (1/5.12 × 10⁷- 1/2.56 × 10⁷) . Δ E = -4.002 × 10¹⁶(-1.953 × 10⁻⁸) approx 7.82 × 10⁸ 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: 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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PHYSICS

A gas occupies 5 L at 2 atm and 27° C . If the pressure is increased to 4 atm and volume reduced to 3 L, what is the fi

Given: A gas occupies 5 L at 2 atm and 27° C . If the pressure is increased to 4 atm and volume reduced to 3 L, what is the final temperature in Celsius? These values define the system as per NCERT data. Formula: Given: V_1 = 5 L, P_1 = 2 atm, T_1 = 27° C = 300 K, P_2 = 4 atm, V_2 = 3 L. This is the standard NCERT relation for this phenomenon. Substitution & Calculation: P_1 V_1/T_1 = P_2 V_2/T_2 . T_2 = T_1 × P_2/P_1 × V_2/V_1 = 300 × 4/2 × 3/5 = 300 × 2 × 0.6 = 360 K . t_C = T_2 - 273.15 = 360 - 273.15 approx 86.85° C approx 87° 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.

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PHYSICS

Which condition must be satisfied for the potential energy in an SHM system to be expressible as a quadratic function of

Potential energy in SHM ( U = 1/2 k x² ) is quadratic when the force is conservative and linear ( F = -kx ), distinguishing SHM from non-linear systems.

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

A car moving at 25 m/s decelerates uniformly to rest over 62.5 m . What is the time taken to stop?

Given: A car moving at 25 m/s decelerates uniformly to rest over 62.5 m . What is the time taken to stop? These values define the system as per NCERT data. Formula: Use v² = v_0² + 2 a x to find a : 0 = (25)² + 2 a (62.5) Rightarrow 0 = 625 + 125 a Rightarrow a = -5 m/s². This is the standard NCERT relation for this phenomenon. Substitution & Calculation: Then, v = v_0 + a t : 0 = 25 - 5 t Rightarrow t = 5 s . The time taken is 5 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.

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PHYSICS

A double convex lens has radii of curvature 30 cm and -30 cm with refractive index 1.6 . What is its focal length?

Given: A double convex lens has radii of curvature 30 cm and -30 cm with refractive index 1.6 . 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 the standard NCERT relation for this phenomenon. Substitution & Calculation: n = 1.6, R_1 = 30 cm, R_2 = -30 cm . 1/f = (1.6 - 1) ( 1/30 - 1/-30 ) = 0.6 ( 1/30 + 1/30 ) = 0.6 × 2/30 = 1.2/30 = 1/25 . f = 25 cm . 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: Ray Optics and Optical Instruments and Wave Optics, Topic: Refraction, lenses and interference/diffraction.

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PHYSICS

A point charge 15 μC is at the origin. What is the electric field magnitude at a point 6 m along the y-axis?

Given: A point charge 15 μC is at the origin. What is the electric field magnitude at a point 6 m along the y-axis? These values define the system as per NCERT data. Formula: E = k |q|/r². This is the standard NCERT relation for this phenomenon. Substitution & Calculation: k = 9 × 10⁹ Nm²/C², q = 15 × 10⁻⁶ C, r = 6 m . E = 9 × 10⁹ × frac15 × 10⁻⁶(6)² = 9 × 10⁹ × frac15 × 10⁻⁶³⁶= 3.75 × 10³ N/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: Electric Charges and Fields and Electrostatic Potential, Topic: Electric field, potential and capacitance concepts.

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