26 public questions tagged with this topic.
English: पुष्करभटनागरवर्येण is correct answer from passage, found in gadya context describing crow, farmer, or moral story. Hindi: पुष्करभटनागरवर्येण अनुच्छेद से सही उत्तर है जो गद्य में वर्णित कथा के आधार पर मिलता है।
Ref: NCERT Sanskrit Class XII Gadya Sankalan explains comprehension passages questions translation grammar analysis for CUET preparation clearly.
Blank needs 'Shooting'; context demands this word grammatically.
Ref: British Council grammar guide explains adjective fond always followed by preposition of standard collocation pattern English usage.
**Wavefront types** point source spherical, distant point source plane, after convex lens plane wave focuses to point because lens adds phase delay proportional to thickness, converging spherical wavefront, after concave mirror plane wave becomes spherical converging to focus, illustrating Huygens construction. At a large distance from a point source like a star, a small portion of the spherical wavefront appears plane. Using Δ = d sinθ, y = n λ D/d, a sinθ = n λ, I = I₀ cos²θ, n = c/v, sinC = 1/n, λ' = λ/n and A = 2a cos(φ/2), calculation gives Plane, illustrating interference, diffraction an
Ref: NCERT > Physics Book > Wave Optics > Wavefront and Huygens Principle
**Compound microscope** M = m_o × M_e, objective magnification m_o = v_o/u_o ≈ L/f_o, L tube length, eyepiece M_e = 1+D/f_e (image at D) or D/f_e (infinity). Objective forms real inverted magnified image at focal plane of eyepiece, eyepiece acts as simple microscope magnifying it. Magnifying power: m = (f_o/f_e) . f_o = 120 cm , f_e = 6 cm . m = (120/6) = 20 . Substituting values gives 20, which matches expected image position and magnification from mirror/lens formula 1/f = 1/v - 1/u (lens) or 1/f = 1/v + 1/u (mirror), confirming sign conventions and refraction principles.
Ref: NCERT > Physics Book > Ray Optics > Optical Instruments - Simple and Compound Microscope
**Compound microscope** M = m_o × M_e, objective magnification m_o = v_o/u_o ≈ L/f_o, L tube length, eyepiece M_e = 1+D/f_e (image at D) or D/f_e (infinity). Objective forms real inverted magnified image at focal plane of eyepiece, eyepiece acts as simple microscope magnifying it. Magnifying power: m = (f_o/f_e) . f_o = 200 cm , f_e = 10 cm . m = (200/10) = 20 . Substituting values gives 20, which matches expected image position and magnification from mirror/lens formula 1/f = 1/v - 1/u (lens) or 1/f = 1/v + 1/u (mirror), confirming sign conventions and refraction principles.
Ref: NCERT > Physics Book > Ray Optics > Optical Instruments - Simple and Compound Microscope
Given: A satellite near Earth has a period of 82 minutes. What is its period at h = 9 R_E ? ( R_E = 6.4 × 10ⶠm ) These values define the system as per NCERT data. Formula: T_0² = k R_E³, h = 9 R_E, r = 10 R_E. This is the standard NCERT relation for this phenomenon. Substitution & Calculation: T² propto (R_E + h)³ . . T² = k (10 R_E)³ = 1000 k R_E³ . T = T_0 sqrt1000 = 82 × 31.62 approx 2593 min . Result: The computed value matches the expected outcome and confirms the correct choice. Units and powers like J kgâ»Â¹ Kâ»Â¹, m/s², 10â»âµ are properly used as per N
Ref: NCERT Physics Textbook for Class XI and XII, Chapter: Units and Measurements and Physical World, Topic: Dimensional analysis and fundamental principles.
Given: A planet orbits the Sun with a period of 7 years. If Earth’s orbital radius is 1.5 × 10¹¹m, what is its semi-major axis? These values define the system as per NCERT data. Formula: Kepler’s third law: T_p²/T_E² = a_p³/a_E³. This is standard NCERT relation. Substitution & Calculation: T_E = 1 year, T_p = 7 years, a_E = 1.5 × 10¹¹m . 7²/1² = fraca_p³(1.5 × 10¹¹)³ . 49 = fraca_p³³.375 × 10³³. a_p³ = 49 × 3.375 × 10³³= 1.65375 × 10³⁵. a_p = (1.65375 × 10³⁵)^{1/3 approx 5.49 × 10¹¹m . Result: The computed value matches expected outcome and confirms correct choice as per NCERT.
Ref: NCERT Physics Textbook for Class XI and XII, Chapter: Gravitation, Topic: Kepler's third law T² ∝ a³, orbital period, semi-major axis and planetary motion. The section explains definitions, governing laws, formulas like μ₀ = 4π × 10⁻⁷ T·m/A, units and illustrative examples.
Given: What is the minimum speed to escape from 3 R_E from Earth’s nter? ( g = 9.8 m/s², R_E = 6.4 × 10⁶m ) Formula: v_e = √2 g R_E²/3 R_E = √2 × 9.8 × 6.4 × 10⁶/3. Substitution & Calculation: v_e = √4.181 × 10⁷approx 6.47 × 10³m/s approx 6.5 km/s . Final Result: The computed value matches expected outcome and confirms correct choice as per latest NCERT 2026-27.
Ref: NCERT Physics Textbook - Latest Edition for Academic Session 2026-27 (Rationalized Textbook for Class XI and XII, continuing as per NCERT advisory for 2026-27),Topic: Fundamental laws, definitions and applications as per latest NCERT. The section explains governing laws, formulas like μ₀ = 4π.
Given: What is the escape speed from a planet of mass 4.8 × 10²⁴kg and radius 5 × 10⁶m ? ( G = 6.67 × 10⁻¹¹N m²/kg² ) These values define the system as per NCERT data. Formula: v_e = sqrt2 G M/R. This is standard NCERT relation. Substitution & Calculation: v_e = sqrtfrac2 × 6.67 × 10⁻¹¹ × 4.8 × 10²⁴⁵ × 10⁶. v_e = sqrt6.403 × 10⁷approx 8.0 × 10³m/s = 8.0 km/s . 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,
Given: A satellite orbits at 1.2 × 10â· m from a planet’s nter with period 3 hours. What is the planet’s mass? ( G = 6.67 × 10â»Â¹Â¹ N m²/kg² ) These values define the system as per NCERT data. Formula: M = 4π² r³/G T². This is the standard NCERT relation for this phenomenon. Substitution & Calculation: T = 3 × 3600 = 10800 s, T² = 1.166 × 10⸠s² . r³ = (1.2 × 10â·)³ = 1.728 × 10²¹ m³ . M = frac4 × (3.14)² × 1.728 × 10²¹â¶.67 × 10â»Â¹Â¹ × 1.166 × 10â¸. M = frac6.82 × 10²²â·.78 × 10â»Â³ approx 8.77 × 10²ⴠkg . Result: The computed
Ref: NCERT Physics Textbook for Class XI and XII, Chapter: Units and Measurements and Physical World, Topic: Dimensional analysis and fundamental principles.
Given: A satellite orbits a planet at 1.5 × 10â· m from its nter with a period of 4 hours. What is the planet’s mass? ( G = 6.67 × 10â»Â¹Â¹ N m²/kg² ) These values define the system as per NCERT data. Formula: M = 4π² r³/G T². This is the standard NCERT relation for this phenomenon. Substitution & Calculation: T = 4 × 3600 = 14400 s, T² = 2.0736 × 10⸠s² . r³ = (1.5 × 10â·)³ = 3.375 × 10²¹ m³ . M = frac4 × (3.14)² × 3.375 × 10²¹â¶.67 × 10â»Â¹Â¹ × 2.0736 × 10â¸. M = frac1.331 × 10²³¹.383 × 10â»Â² approx 9.63 × 10²ⴠkg . Result: The
Ref: NCERT Physics Textbook for Class XI and XII, Chapter: Units and Measurements and Physical World, Topic: Dimensional analysis and fundamental principles.
Outside a spherical shell, the gravitational force acts as if all the mass is concentrated at the nter. Since the mass is non-zero and the distance is finite, the force ( F = G M m/r² ) cannot be zero.
Ref: NCERT Physics Textbook - Latest Edition for Academic Session 2026-27 (Rationalized Textbook for Class XI and XII, continuing as per NCERT advisory for 2026-27),Topic: Fundamental laws, definitions and applications as per latest NCERT. The section explains governing laws, formulas like μ₀ = 4π.
