When a long glass capillary tube of radius 0.015 cm is dipped in a liquid , the liquid rises to a height of 15 cm within it …..

Q: When a long glass capillary tube of radius 0.015 cm is dipped in a liquid , the liquid rises to a height of 15 cm within it . If the contact angle between the liquid and glass to close to 0° , the surface tension of the liquid in millinewton m-1 is (ρliquid = 900 kg/m-3 , g = 10 m/s2) (give Answer in closest integer) ….

Click to See Solution :
Ans: (101)

Sol: Here r = 0.015 cm , ρliquid = 900 kg/m-3 , h = 15 cm g = 10 m/s^2 , θ = 0°

Height to which liquid will rise , $\displaystyle h = \frac{2 S cos\theta}{\rho g r }$

$\displaystyle S = \frac{h \rho g r}{2 cos\theta}$

$\displaystyle S = \frac{0.15 \times 900 \times 10 \times 0.015 \times 10^{-2}}{2 cos0^o}$

S = 0.10125 N/m = 101.25 mN/m

 

A liquid is flowing through a horizontal pipe of varying cross section with speed v m/s at a point ….

Q: A liquid is flowing through a horizontal pipe of varying cross section with speed v m/s at a point where the pressure is P pascal . At another point where pressure is P/2 pascal , its speed is V m/s . If the density of the liquid is ρ kg/m3 and flow is streamline , then V is equal to

(a) $\displaystyle \sqrt{\frac{P}{\rho} + v }$

(b) $\displaystyle \sqrt{\frac{2 P}{\rho} + v^2 }$

(c) $\displaystyle \sqrt{\frac{P}{2 \rho} + v^2 }$

(d) $\displaystyle \sqrt{\frac{P}{\rho} + v^2 }$

Click to See Solution :
Ans: (d)
Sol: Applying Bernoulli’s Theorem

$\displaystyle P_1 + \frac{1}{2}\rho v_1^2 = P_2 + \frac{1}{2}\rho v_2^2 $

$\displaystyle P + \frac{1}{2}\rho v^2 = \frac{P}{2} + \frac{1}{2}\rho V^2 $

On solving ,

$\displaystyle V = \sqrt{\frac{P}{\rho} + v^2 }$

 

A cylindrical vessel containing liquid is rotated about its axis so that the liquid rises at its sides as shown in the figure …..

Q: A cylindrical vessel containing liquid is rotated about its axis so that the liquid rises at its sides as shown in the figure . The radius of vessel is 5 cm and the angular speed of rotation is ω rad/s . The difference in the height h (in cm) at the centre of vessel and at the side will be

q & a

(a) $ \displaystyle \frac{2 \omega^2}{25 g}$

(b) $ \displaystyle \frac{5 \omega^2}{2 g}$

(c) $ \displaystyle \frac{25 \omega^2}{2 g}$

(d) $ \displaystyle \frac{2 \omega^2}{25 g}$

Click to See Solution :
Ans: (c)

Sol: q & a

Force acting on a particle A are

(i) Pseudo force m ω2 x

(ii) Weight (mg)

$\displaystyle tan\theta = \frac{m \omega^2 x}{m g}$

$\displaystyle \frac{dy}{dx} = \frac{ \omega^2 x}{ g}$

$\displaystyle \int_{0}^{y}dy = \frac{ \omega^2}{ g} \int_{0}^{x}x dx $

$\displaystyle y = \frac{ \omega^2 x^2}{2 g}$

At x = r

$\displaystyle y = \frac{ \omega^2 r^2}{2 g}$

 

Two liquids of densities ρ1 and ρ2 (ρ2 = 2 ρ1) are filled up behind a square wall of side 10 m as shown in figure ….

Q: Two liquids of densities ρ1 and ρ22 = 2 ρ1) are filled up behind a square wall of side 10 m as shown in figure . Each liquid has a height of 5 m . The ratio of the forces due to these liquids exerted on upper part MN to that at the lower part NO is (Assume that the liquids are not mixing )

qna

(a)1/2

(b) 2/3

(c) 1/4

(d) 1/3

Click to See Solution :
Ans: (c)

Sol: Let ρ1 = ρ , ρ2 = 2ρ

qna

P1 = 0 , P2 = 5 ρ g , P3 = 5 ρ g + 2 × 5 ρ g = 15 ρ g

$\displaystyle F_1 = \frac{P_1 + P_2}{2} A $

$\displaystyle F_2 = \frac{P_2 + P_3}{2} A $

$\displaystyle \frac{F_1}{F_2} = \frac{0 + 5 \rho g}{5 \rho g + 15 \rho g} = \frac{1}{4} $

 

Water flows in a horizontal tube as Shown . The pressure of water changes by 700 N/m^2 between A and B …

Q: Water flows in a horizontal tube as Shown . The pressure of water changes by 700 N/m^2 between A and B where the area of cross section are 40 cm2 and 20 cm2 respectively . Find the rate of flow of water through the tube (density of water = 1000 kg m-3)

q&a

(a) 3020 cm3/s

(b) 2420 cm3/s

(c) 2720 cm3/s

(d) 1810 cm3/s

Click to See Solution :
Ans: (c)

Sol: AA = 40 cm2 , AB = 20 cm2

PA – PB = 700 N/m^2

Applying Bernoulli’s Theorem ,

$\displaystyle P_A + \frac{1}{2}\rho V_A^2 = P_B + \frac{1}{2}\rho V_B^2 $

$\displaystyle P_A – P_B = \frac{1}{2}\rho (V_B^2 – V_A^2 ) $

$\displaystyle 700 = \frac{1}{2}\times 1000 (4 V_A^2 – V_A^2 ) $

$\displaystyle V_A^2 = \frac{7}{5 \times 3} $

VA = 0.68 m/s = 68 cm/s

Rate of flow = AA . VA = 40 × 68 = 2720 cm3/s