## Water rises to a height of 10cm in a capillary tube and mercury falls to a depth of 3.42 cm in the same capillary tube. If the density of mercury is 13.6g/cc and the angle of contact of mercury and water are 135° and 0° respectively, the ratio of surface tension of water and mercury is

Q: Water rises to a height of 10cm in a capillary tube and mercury falls to a depth of 3.42 cm in the same capillary tube. If the density of mercury is 13.6g/cc and the angle of contact of mercury and water are 135° and 0° respectively, the ratio of surface tension of water and mercury is
(A) 1 : 0.15
(B) 1 : 3
(C) 1 : 6.5
(D) 1.5 : 1

Solution: h =2Scosθ/ρgr
or S =hρgr/2cosθ
or S ∝ hρ/cosθ
Sw/SHg = h1/h2 × ρ12 × cosθ1/cosθ2
Putting the values, we obtain 1 : 6.5
Hence, (C) is correct.

## A piston of a syringe pushes a liquid with a speed of 1 cm/sec. The radii of the syringe tube and the needle are R = 1cm and r = 0.5 mm respectively. The velocity of the liquid coming out of the needle is

Q: A piston of a syringe pushes a liquid with a speed of 1 cm/sec. The radii of the syringe tube and the needle are R = 1cm and r = 0.5 mm respectively. The velocity of the liquid coming out of the needle is (A) 2 cm/sec
(B) 400 cm/sec
(C) 10 cm/sec
(D) None of these
Solution: According to the equation of continuity,
Av = av’
v’ = (A/a)v
where A = πR² and a = πr2
v’= (R/r)2v = (1cm/0.05cm)(1 cm/sec) = 400 cm/sec.

## A cube of mass m and density D is suspended from the point P by a spring of stiffness k. The system is kept inside a beaker filled with a liquid of density d. The elongation in the spring, assuming D > d is

Q: A cube of mass m and density D is suspended from the point P by a spring of stiffness k. The system is kept inside a beaker filled with a liquid of density d. The elongation in the spring , assuming D > d, is (A) mg/k(1-d/D)
(B) mg/k(1-D/d)
(C) mg/k(1+d/D)
(D) None of these.

Solution: The cube is in equilibrium under the following three forces,
(a) spring force kx, where x = elongation of the spring,
(b) gravitational force w, weight of the cube = mg
(c) buoyant force Fb (or upward thrust) imparted by the liquid on the cube given as Fb = Vdg where
V = volume of the immersed portion of the cube. For complete immersion, V = volume of the cube.
For equilibrium of the cube, kx + Fb = mg
x = (mg-Fb) /k
= (mg-Vdg) /k , Where V = (m/D)
x =mg/k(1-d/D)

## A cube of ice is floating in water. The fraction of its length that lie outside the water is (Sp. Gravity of ice = 0.96)

Q: A cube of ice is floating in water. The fraction of its length that lie outside the water is (Sp. Gravity of ice = 0.96)
(A) 0.04
(B) 0.4
(C) 0.96
(D) None of these.
Solution: Since the ice cube is in equilibrium, the net force acting on it is zero. Hence the buoyant force exerted on it is equal to its weight mg where m is the mass of the ice cube . Fb = Mg …(1)
Let the cube of ice be immersed into water to a depth h.
The volume of the cube immersed = V = L2h
The volume of water displaced = V = L2h
The buoyant force = weight of the water displaced
Fb = (mass of the water displaced)g
Fb = ρwVg = ρwL2hg    …(2)
Equating (1) and (2), we obtain,
ρwL2hg = Mg

ρwL2h = Vice ρice
ρwL2h = L3ρice
h/L = ; ρicew

putting  ρw = 1 gm/cc and ρice = 0.96 gm/cc
we obtain, h/L  = 0.96
The fraction of its length exposed to air = 1 – h/L = 0.04

## The density of ice is x gm/cc. and that of water is y gm/cc. What is the change in volume in cc., when m gm of ice melts

Q: The density of ice is x gm/cc. and that of water is y gm/cc. What is the change in volume in cc., when m gm of ice melts ?
(A) m(y – x)
(B) (y – x)/m
(C) mxy (x – y)
(D) m (1/y – 1/x)

Solution: m = volume × density
Mass remains constant.
Vice × ρice = Vwater  ×ρwater
Vwater = Vice × ρicewater
= m/y
ΔV = Vwater – Vice
= m/y -m/x = m(1/y – 1/x)