Title 10. MECHANICAL PROPERTIES OF FLUIDS(MOBILE).pdf ✅

SPARK PCMB | DOWNLOAD APP NOW| +91-920-630-6398 10. MECHANICAL PROPERTIES OF FLUIDS #QID# 40857 (1.) A body floats in a liquid contained in a beaker. If the whole system falls under gravity, them the upthrust on the body due to liquids is (a.) equal to the weight of the body in air (b.) equal to the weight of the body in liquid (c.) zero (d.) equal to the weight of the immersed part of the body ANSWER: a EXPLANATION: (a) Upthrust is independent of all factors of the body such as its mass, size, density etc, except the volume of the body inside the fluid. Fraction of volume immersed in the liquidVin = ( ρ σ ) V ie, it depends upon the densities of the block and liquid. So, there will be no change in it if system moves upward or downward with constant velocity or some acceleration. Therefore, the upthrrust on the body due to liquid is equal tto the weight of the body in air. #QID# 40858 (2.) The working of venturimeter is based on (a.) Torricelli’s law (b.) Pascal’s law (c.) Bernoulli’s theorem (d.) Archimede’s principle ANSWER: c #QID# 40859
SPARK PCMB | DOWNLOAD APP NOW| +91-920-630-6398 (3.) A rain drop of radius 1.5 mm, experiences a drag force F = (2 × 10−5 v) N, while falling through air from a height 2 km, with a velocityv. The terminal velocity of the rain drop will be nearly (use g = 10 ms −2 ) (a.) 200 ms −1 (b.) 80 ms −1 (c.) 7 ms −1 (d.) 3 ms −1 ANSWER: c EXPLANATION: (c) When terminal velocity v is reaching, then F = 2 × 10−5v = 4 3 π r 3ρg 2 × 10−5v = 4 3 × 22 7 × (1.5 × 10−3) 3 × 103 × 10 On solving, v = 7.07ms −1 ≈ 7ms −1 #QID# 40860 (4.) A weightless bag is filled with 5 kg of water and then weighed in water. The reading of spring balance is (a.) 5 kgf (b.) 2. 5 kgf (c.) 1.25 kgf (d.) Zero ANSWER: d EXPLANATION: (d) Since, weight of bag with water is equal to the weight of water displaced, hence reading of spring balance is zero
SPARK PCMB | DOWNLOAD APP NOW| +91-920-630-6398 #QID# 40861 (5.) A rain drop of radius 0.3 mm has a terminal velocity in air = 1 ms −1 . The viscous force on it is (a.) 101.73 × 10−4 dyne (b.) 101.73 × 10−5 dyne (c.) 16.95 × 10−4 dyne (d.) 16.95 × 10−5 dyne ANSWER: a EXPLANATION: (a) F = 6 π η r v = 6 × 3.14 × (18 × 10−5 ) × 0.03 × 100 = 101.73 × 10−4 dyne #QID# 40862 (6.) A rectangular vessel when full of water, takes 10 min to be emptied through an orifice in its bottom. How much time will take to be emptied when half filled with water? (a.) 9 min (b.) 7 min (c.) 5 min (d.) 3 min ANSWER: b EXPLANATION: (b) If A0 in the area orifice at the bottom below the free surface and A that of vessel, then time t taken to be emptied the tank is given as
SPARK PCMB | DOWNLOAD APP NOW| +91-920-630-6398 t = A A0 √ 2H g ∴ t1 t2 = √ H1 H2 ⟹ t t2 = √ H1 H1/2 ⟹ t t2 = √2 ∴ t2 = t √2 = 10 √2 = 5 √2 = 7min #QID# 40863 (7.) A liquid is kept in a cylindrical vessel which is rotated along its axis. The liquid rises at the sides (figure). If the radius of the vessel is 0.05 m and the sped of rotation is 2 rad s −1 , find the difference in the height of the liquid at the centre of the vessel and its sides  (a.) 20 cm (b.) 4 cm (c.) 2 cm (d.) 0.2 cm ANSWER: c EXPLANATION: (c)