26.A raindrop of radius 0.3 mm has a terminal velocity in air is 1 m/s. The viscosity of air is 8 x 10-5 poise. The viscous force on it is A) 45.2×10−4dyne B) 101.73×10−5 dyne C) 16.95×10−4 dyne D) 16.95×10−5dyne View Answer Report DiscussAnswer: Option AExplanation:Radius of drop, r = 0.3 mm = 0.03 cm Terminal velocity v = 1 m/s = 100 cm/sec Viscosity of air, η = 8 x 10-5 poise Viscous force, f = 6 πηrv F = 6 x 3.14 × (8 ×10-5) x 0.03 x 100 = 4.52 x10-3 dyne
27.A person swims in a river aiming to reach exactly on the opposite point on the bank of a river His speed of swimming is 0.5 m/s at an angle of 120° with the direction of flow of water. The speed of water is A) 1.0m/s B) 0.5m/s C) 0.25m/s D) 0.43m/s View Answer Report DiscussAnswer: Option CExplanation:Let the speed of water = ¯u Speed of swimmer = ¯v = 0.5m/sec Angle between ¯u and ¯v is 120°. Then sinθ=¯u¯v = u0.5=12 or u=0.25ms-1
28.An electric dipole is kept in a uniform electric field. It experiences A) a force and a torque B) a force, but no torque C) a torque but no net force D) neither a force nor a torque View Answer Report DiscussAnswer: Option CExplanation:In a uniform electric field, net force = 0 but torque ≠ 0
29.The velocity-time graph of a body moving in a straight line is shown in fig. Find the displacement and distance travelled by the body in 10 seconds A) 50 m, 90m B) 5m , 9m C) 9m , 5m D) 90m , 50m View Answer Report DiscussAnswer: Option AExplanation: Total distance covered in 10 s = Area 1 + Area 2 + Area 3 12×6×20+12×2×20+12×2×10 = 90m Total displacement in 10s = Area 1 - Area 2 +Area 3 12×6×20−12×2×20+12×2×10 = 50m
30.The magnetic field at a distance r from a long wire carrying current i is 0.4 tesla. The magnetic field at a distance 2r is A) 0.2tesla B) 0.8tesla C) 0.1tesla D) 1.6 tesla View Answer Report DiscussAnswer: Option AExplanation:Magnetic field due to long wire, B=μ0i2πr or B∝1r When r is doubled, the magnetic field becomes half, i.e., now the magnetic field will be 0.2T.
