66 public questions tagged with this topic.
Excess pressure in a bubble: ΔP = 4Sr. S = 0.03N/m, r = 6×10−3m. ΔP = 4×0.036×10−3 = 20Pa. As per NCERT, applying relevant law/formula with correct units and sign convention leads to 20 Pa. This satisfies dimensional consistency and physical conditions given, so option B is scientifically correct.
Ref: NCBI Bookshelf, Ideal Gas Law: P1V1/T1 = P2V2/T2.
Pi = Pa+ρgh+2Sr. Pa = 1.01×105Pa, ρgh = 1.2×103×9.8×0.5 = 5880Pa. 2Sr = 2×0.0254×10−3 = 12.5Pa. Pi = 1.01×105+5880+12.5 = 1.06892×105Pa. As per NCERT, applying relevant law/formula with correct units and sign convention leads to 1.069 × 10⁵ Pa. This satisfies dimensional consistency and physical conditions given, so option C is scientifically correct.
Ref: NCBI Bookshelf, Ideal Gas Law: P1V1/T1 = P2V2/T2.
Excess pressure: ΔP = 2Sr. S = 0.4355N/m, r = 4×10−3m. ΔP = 2×0.43554×10−3 = 217.75Pa. As per NCERT, applying relevant law/formula with correct units and sign convention leads to 218 Pa. This satisfies dimensional consistency and physical conditions given, so option B is scientifically correct.
Ref: NCBI Bookshelf, Ideal Gas Law: P1V1/T1 = P2V2/T2.
The capillary rise formula (h = 2Scosθ/ρgr) shows that h is inversely proportional to radius (r) because the pressure difference due to surface tension (2S/r) must balance the weight of the liquid column, which depends on h.
Ref: NCERT Class 11 Physics, Thermal Properties and Gravitation.
Increasing temperature weakens cohesive forces between liquid molecules by increasing their kinetic energy, reducing the contractile force at the surface, thus lowering surface tension, as noted in the chapter. As per NCERT, applying relevant law/formula with correct units and sign convention leads to Cohesive forces weaken. This satisfies dimensional consistency and physical conditions given, so option C is scientifically correct.
Ref: NCERT Class 11 Physics, Thermal Properties and Gravitation.
Pi = Pa+ρgh+2Sr. Pa = 1.01×105Pa, ρgh = 1000×10×0.3 = 3000Pa. 2Sr = 2×0.07273×10−3 = 48.47Pa. Pi = 1.01×105+3000+48.47 = 1.04048×105Pa. As per NCERT, applying relevant law/formula with correct units and sign convention leads to 1.04 × 10⁵ Pa. This satisfies dimensional consistency and physical conditions given, so option C is scientifically correct.
Ref: NCBI Bookshelf, Ideal Gas Law: P1V1/T1 = P2V2/T2.
The angle of contact (θ) determines wetting behavior: an acute angle (strong attraction to the solid) causes spreading (wetting), while an obtuse angle (weak attraction) leads to droplet formation, as explained in the surface tension section.
Ref: NCERT Class 11 Physics, Thermal Properties and Gravitation.
Capillary rise (h = 2Scosθ/ρgr) is inversely proportional to radius, not directly proportional. The incorrect statement is that height increases with tube radius. As per NCERT, applying relevant law/formula with correct units and sign convention leads to Height increases with radius. This satisfies dimensional consistency and physical conditions given, so option C is scientifically correct.
Ref: NCERT Class 11 Physics, Thermal Properties and Gravitation.
Capillary rise: h = 2Scosθρga. S = 0.4355N/m, ρ = 13.6×103kg/m3, g = 10m/s2, a = 0.2×10−3m, cosθ = −0.5. h = 2×0.4355×(−0.5)13.6×103×10×0.2×10−3 = −0.016m = −1.6cm. As per NCERT, applying relevant law/formula with correct units and sign convention leads to -1.6 cm. This satisfies dimensional consistency and physical conditions given, so option B is scientifically correct.
Ref: NCERT Class 11 Physics, Thermal Properties and Gravitation.
S = F2l. F = 2.8×10−2N, l = 0.56m. S = 2.8×10−22×0.56 = 0.025N/m. As per NCERT, applying relevant law/formula with correct units and sign convention leads to 0.025 N/m. This satisfies dimensional consistency and physical conditions given, so option C is scientifically correct.
Ref: NCERT Class 11 Physics, Thermal Properties and Gravitation.
Excess pressure: ΔP = 2Sr. S = 0.3×10−3N/m, r = 2.5×10−3m. ΔP = 2×0.3×10−32.5×10−3 = 0.24Pa. As per NCERT, applying relevant law/formula with correct units and sign convention leads to 0.24 Pa. This satisfies dimensional consistency and physical conditions given, so option B is scientifically correct.
Ref: NCBI Bookshelf, Ideal Gas Law: P1V1/T1 = P2V2/T2.
Pi = Pa+ρgh+2Sr. Pa = 1.01×105Pa, ρgh = 1000×9.8×0.15 = 1470Pa. 2Sr = 2×0.07273.5×10−3 = 41.54Pa. Pi = 1.01×105+1470+41.54 = 1.02471×105Pa. As per NCERT, applying relevant law/formula with correct units and sign convention leads to 1.025 × 10⁵ Pa. This satisfies dimensional consistency and physical conditions given, so option C is scientifically correct.
Ref: NCBI Bookshelf, Ideal Gas Law: P1V1/T1 = P2V2/T2.
