A vertical gap 2.2 cm wide of infinite extent contains a fluid of viscosity 2.0 NS / m^2 and specific gravity 0.9.

Question

Comprehension type

Passage A vertical gap 2.2 cm wide of infinite extent contains a fluid of viscosity 2.0 NS / m² and specific gravity 0.9. A metallic plate 1m \(\times\) 1m \(\times\) 0.2cm, which is in the middle of the gap, is to be lifted up with a constant speed 0.15 m/sec through the gap. The weight of the plate is 48N. Assuming pulley is massless and frictionless, string is also massless. (g =10 m/s²)

(i) Buoyant force acting on the plate 

(A) 1800 N 

(B) 900 N 

(C) 180 N 

(D) 18 N 

(ii) Net frictional force exerted by the liquid on the plate 

(A) 30 N 

(B) 60 N 

(C) 15 N 

(D) 120 N 

(iii) Tension in the string 

(A) 90 N 

(B) 108 N 

(C) 240 N 

(D) 120 N 

(iv) For doing so the kinetic friction between the inclined plane and the block should be equals to

(A) \(\frac{\sqrt{3}}{4}\)

(B) \(\frac{\sqrt{3}}{8}\)

(C) \(\frac{1}{\sqrt{3}}\)

(D) \(\frac{1}{2\sqrt{3}}\)

Answer 1

(i) (D) 18 N

(ii) (B) 60 N

(iii) (A) 90 N

(iv) (A) \(\frac {\sqrt 3}4\)

Buoyant force = \(\rho\)gV_imm = \(\rho\)gv

F = (0.9) (10³).(10) (0.2 × 1 × 1 × × 10^-2)

F_B = 18 N.

Now T + F_B = W + F_v.

T = W + F_v – F_B

F_v = 60 N.

T = 48 + 60 – 18

T = 90 N.

Now for this the acceleration of the block should be zero.

mg sin θ = f + T

120 – 90 = f

f = 30 N

µ (mg cos θ) = 30

µ (8√3 . 10 . 1/2) = 30

µ = \(\frac{\sqrt 3}4\).

Answer 2

(i) (D) 18 N

(ii) (B) 60 N

(iii) (A) 90 N

(iv) (A) \(\frac{\sqrt{3}}{4}\)