588 + Interview Questions in FLUID-MECHANICS IN GENERAL AWARENESS Page 4 InterviewSolution

151.	If the container filled with liquid gets accelerated horizontally or vertically, pressure in liquid gets changed. In liquid (`a_(y)`) for calculation of pressure, effective `g` is used. A closed box horizontal base `6 m` by `6 m` and a height `2m` is half filled with liquid. It is given a constant horizontal acceleration `g//2` and vertical downward acceleration `g//2`.The maximum value of water pressure in the box is equal toA. `1.4MPa`B. `0.14MPa`C. `0.104MPa`D. `0.014MPa`
Answer» Correct Answer - B `P=(10^(5)+10^3xx10xx4)N//m^(2)` `=[0.1+0.04]Mpa=0.14MPa`

Discussion

152.	A body of volume V and desnity `rho` is initially submerged in a non-viscous liquid of density `sigma (gt rho)`. If it is rises by itself through a height h in the liquid. Its kinetic energy willA. increase by h V `(sigma-rho)g`B. increase by hV`(rho+sigma)`gC. increase by `(hVrhog)/(sigma)`D. decrease be `(hVrhog)/(sigma)`
Answer» Correct Answer - A Work done by all the force W=change in kinetic energy two forces are acting, weight and upthrust. `therefore" "W=(sigma-rho)g(h/V)`

Discussion

153.	A cylindrical block is floating (partially submerged) in a vessel containing water. Initially, the platform on which the vessel is mounted is at rest. Now the platform along with the vessel is allowed to fall freely under gravity. As a result, the buoyancy force A. becomes zeroB. decreasesC. increasesD. information is insufficient
Answer» Correct Answer - A As the vessel is falling freely, the pressure at all the points in the liqids is same and equal to the atmospheric pressure and hence buoyancy becmes zero.

Discussion

154.	A tank if filled with water upto height H. When a hole is made at a distance h below the level of water. What will be the horizontal range of water jet ?A. `2sqrt(h(H-h))`B. `4sqrt(h(H+h))`C. `4sqrt(h(H-h))`D. `2sqrt(h(H+h))`
Answer» Correct Answer - A Range, `R=2sqrt(h_("top")xxh_("bottom"))=2sqrt(h(H-h))`

Discussion

155.	Equal mass of three liquids are kept in three identical cuylindrical vessels A, B and C. the densities are `rho_A, rho_B, rho_C` with `rho_Altrho_Bltrho_C`. The force on the base will beA. AB. BC. CD. equal in all
Answer» Correct Answer - D `W = mg`

Discussion

156.	Water is used as the manometric liquid in a pitot tube mounted in an aircraft to measure air speed. If the maximum difference in height between the liquid columns is `0.1 m,` what is the maximum air speed that can be measured? Density of air `d_("dir")=1.3kg//m^(3)`
Answer» For gases, velocity is given by `v=sqrt((2hrhog)/d)=sqrt((2xx0.1xx1000xx9.8)/1.3)=38.83m//s`

Discussion

157.	A pitot tube is fixed in a water pipe of diameter `14 cm` and difference of pressure by the gauge is `20 cm` of water column. Calculate the rate of flow of water.
Answer» In Pitot tube velocity of liquids is given by `V=sqrt(2xx1000xx20)=200m//s` `Q=Av=pir^(2)v` `=25/7xx7^(2)xx200=30800cm^(3)//s=30.8 1//s`

Discussion

158.	The cylindrical tube of a spray pump has a cross-section of `8 cm^2`, one end of which has `40` fine holes each of area `10^-8 m^2`. If the liquid flows inside the tube with a speed of `0.15 m min^-1`, the speed with which the liquid is ejected through the holes is.A. `50 ms^-1`B. `5 ms^-1`C. `0.65 ms^-1`D. `0.5 ms^-1`
Answer» Correct Answer - B `a_1 v_1 = a_2 v_2` `8 xx 10^-4 xx (0.15)/(60) = 40 xx 10^-8 v_2` `v_2 = 5 m//s`.

Discussion

159.	A viscous fluid is flowing through a cylindrical tube. The velocity distribution of the fluid is best represented by the diagramA. B. C. D. None of these
Answer» Correct Answer - C

Discussion

160.	There are two identical small holes on the opposite sides of a tank containing a liquid. The tank is open at the top. The difference in height between the two holes is `h`. As the liquid comes out of the two holes. The tank will experience a net horizontal force proportional to. .A. `h^(1//2)`B. `h`C. `h^(3//2)`D. `h^(2)`
Answer» Correct Answer - B

Discussion

161.	A homogeneous solid cylinder of length L(LltH/2), cross-sectional area A/5 is immersed such that it floats with its axis vertical at the liquid-liquid interface with length L/4 in the denser liquid as shown in the figure. The lower density liquid is open to atmosphere having pressure `P_0`. Then density D of solid is given by
Answer» a. For equilibrium of the cylinder, Weight of the cylinder `=` weight of the liquid displaced `implies L(A/5)Dg=(3L)/3(A/5)dg+L/4(A/5)2dg` This gives `D=3/4d+d/2` or `D=5/4d` b. Total pressure at the bottom of cylinder is `P= ("weight of liquids" + "weight of cylinder")/A+P_(0)` `=([Ad(H/2)g+A2d(H/2)g]+A/5(5/4d)Lg)/A+P_(0)` `=((6H+L)/4)dg+P_(0)` Here we are asked to find the total pressure at the bottom of the container and not only the hydrostatic pressue, hence, we need to consider the total weight of liquid and cylinder in the container.

Discussion

162.	A homogeneous solid cylinder of length L(LltH/2), cross-sectional area A/5 is immersed such that it floats with its axis vertical at the liquid-liquid interface with length L/4 in the denser liquid as shown in the figure. The lower density liquid is open to atmosphere having pressure `P_0`. Then density D of solid is given by A. `5/4d`B. `4/5d`C. `d`D. `d/5`
Answer» Correct Answer - A

Discussion

163.	A metal ball immersed in alcohol weighs `w_(1) " and" 0^(@)C " and" w_(2) " at " 59^(@)C`. The coefficient of cubical expansion of the metal is less than that of alcohol. Assuming that the density of the metal is large compared to that of alcohol, it can be shown thatA. `w_(1) gt w_(2)`B. `w_(1)=w_(2)`C. `w_(1) lt w_(2)`D. `w_(1)=(w_(1)//2)`
Answer» Correct Answer - C Upthrust = (volume of metal ball)`xx` (density of liquid)`xx g` with increase in temperature volume of ball will increase and density of liquid will decrease. But coefficient of cubical expansion of liquid is more. Hence second effect is more dominating. Therefore upthrust, at higher temperatures will be less or apparent weight will be more.

Discussion

164.	A piece of steel has a weight `w` in air, `w_(1)` when completely immersed in water and `w_(2)` when completely immersed in an unknown liquid. The relative density (specific gravity) of liquid isA. `(w-w_(1))/(w-w_(2))`B. `(w-w_(2))/(w-w_(1))`C. `(w_(1)-w_(2))/(w-w_(1))`D. `(w_(1)-w_(2))/(w-w_(2))`
Answer» Correct Answer - B Relative density of steel=`("weight in air")/("change in weight in water")` `=(w)/(w-w_(1))` Now change in weight in the given liquid = upthrust in this liquid or `w-w_(2)={(w//g)/((w//w-w_(1))rho_(w))}d_(2)xxrho_(w)xxg` `therefore d_(2)=(w-w_(2))/(w-w_(1))`= relative density of given liquid.

Discussion

165.	A body of weight `w_(1)` when floats in water displaces an amount of water `w_(2)`. Then `w_(1)ltw_(2)`. In this statement true or false?
Answer» Correct Answer - A In floating condition, weight of liquid displaced=weight of solid.

Discussion

166.	A non-viscous liquid of constant density `500 kg//m^(3)` flows in a variable cross-sectional tube. The area of cross section of the tube at two points `P` and `Q` at heights of `3 m` and `6 m` are `2 xx 10^(-3) m^(3)` and `4 xx 10^(-3) m^(3)`, respectively. Find the work done per unit volume by the forces of gravity as the fluid flows from point `P` to `Q`.
Answer» Work done by per unit volume by force of gravity `=-` change in potential energy per unit volume `:. W=/_U=-rhog(h_(2)-h_(1))=-rhot(6-3)` `=-3xx500xx10=-15xx10^(4)J//m^(3)`

Discussion

167.	A non-viscous liquid of constant density `1000kg//m^3` flows in a streamline motion along a tube of variable cross section. The tube is kept inclined in the vertical plane as shown in Figure. The area of cross section of the tube two point P and Q at heights of 2 metres and 5 metres are respectively `4xx10^-3m^2` and `8xx10^-3m^2`. The velocity of the liquid at point P is `1m//s`. Find the work done per unit volume by the pressure and the gravity forces as the fluid flows from point P to Q.
Answer» Correct Answer - `+29625J//m^(3),-30000J//m^(3)` From equation of continuity `A_(P)V_(P)=A_(Q)V_(Q)` `4xx10^(-3)xx1=8xx10^(-3)xxV_(Q)` `V_(Q)=(1)/(2)m//s` From `W_(g)=-DeltaU` `=-mg(h_(2)-h_(1))` work done by gravity per unit volume `=-rhog(h_(2)-h_(1))` `=-1000xx9.8xx(5-2)" "=-2.94xx10^(4)J//m^(3)" "=-29400J//m^(3)` change in kinetic energy per unit volume `=(1)/(2)p(V_(Q)^(2)-v _(P)^(2))` `Delta K=(1)/(2)xx1000xx[((1)/(2))^(2)-1]` `=-375 J` work done by all forces per unit voume `=DeltaK` work done by pressure perunit volume `+` work done by gravity per unit volume `=DeltaK` work done by pressure per unit volume `-29400=-375` work done by pressure per unit voume `=29025J//m^(3)`

Discussion

168.	Water is flowing in streamline motion in the tube shown in fig. pressure is A. more at `A` then that at `B`B. equal to `A` and at `B`C. lesser at `A` then that at `B`D. normal at `A` and `B`
Answer» Correct Answer - A At `A` area is more hence velocity is less, hence more pressure.

Discussion

169.	A horizontal pipeline carries water in a streamline flow. At a point along the tube where the cross sectional area is `10^(-2)(m^2)`, the water velcity is `2m//s` and the pressure is 8000Pa. The pressure of water at another point where cross sectional area is `0.5xx(10)^(-2)(m^2)` isA. `4000Pa`B. `1000Pa`C. `2000Pa`D. `3000Pa`
Answer» Correct Answer - C Area at other point is half. So, speed will be double. Now, `p_(1)+(1)/(2)rhov_(1)^(2)=p_(2)+(1)/(2) rho v_(2)^(2)` `:. p_(2)=p_(1)+(1)/(2) rho v_(1)^(2)-v_(2)^(2)` `=(8000)+(1)/(2)xx1000(4-16)` `=2000Pa`.

Discussion

170.	Water is flowign is streamline motion through a tube with its axis horizontal. Consider two points A and B in the tube at the same horizontal levelA. the pressure at A and B are equal for any shape of the tubeB. the pressure can never be equalC. the pressure are equal if the tube hs a uniform cross-sectionD. the pressure may be equal even if the tube has a non-uniform cross-section
Answer» Correct Answer - B::C::D

Discussion

171.	A manometer connected to a closed tap reads `4.5 xx 10^(5)` pascal. When the tap is opened the reading of the monometer falls is `4 xx 10^(5)` pascal. Then the velocity of flow of water isA. `7 ms^(-1)`B. `8 ms^(-1)`C. `9 ms^(-1)`D. `10 ms^(-1)`
Answer» Correct Answer - D

Discussion

172.	Water falls from a tap, down the streamlineA. Area decreasesB. Area increasesC. Velocity remains sameD. Area remains same
Answer» Correct Answer - A

Discussion

173.	The figure shows the commonly observed decrease in diameter of a water stream as it falls from a tap. The tap has internal diameter `D_(0)` and is connected toa large tank of water. The surface of the water is at a height `b` above the end of the tap. By considering the dynamics of a thin "cylinder" of water in the steam answer the following: (ignore any reistance to the flow and any effects of surface tansion, given `rho_(w) =` density of water) The equation for the water speed `v` as a funcation of the distance `x` below the tap will be:A. `v = sqrt(2gb)`B. `v = [2g (b + x)]^(1//2)`C. `v = sqrt(2gx)`D. `v = [2g (b - x)]^(1//2)`
Answer» Correct Answer - B `v = sqrt(2gh) = sqrt(2g(b + x))`.

Discussion

174.	Water falls from a tap with `A_(0)=4 m^(2)`, `A=1 m^(2)` and h=2 m, then velocity v is A. `2.5 ms^(-1)`B. `6.5 ms^(-1)`C. `4.5 ms^(-1)`D. `1.5 ms^(-1)`
Answer» Correct Answer - B Given, `(v^(2))/(2g)=h=10 cm` `therefore " " v=sqrt(2gh)=sqrt(2xx10xx0.1)=sqrt(2) ms^(-1)`

Discussion

175.	The velocity of efflux of a liquid through an orifice in the bottom of a tank does not depend uponA. density of liquidB. height of the liquid column above orificeC. acceleration due to gravityD. None of the above
Answer» Correct Answer - A Velocity of efflux `=sqrt(2gh)`, which is independent of density of liquid.

Discussion

176.	A cubical block of wood `10cm` on a side floats at the interface between oil and water, as in fig. with its lower face `2 cm` below the interface. The intensity of the oli is `0.6 g cm^(-3)`. The mass of the block is A. 340gB. 680gC. 80gD. 10g
Answer» Correct Answer - B `mg = [100xx2xx1+100xx8xx0.6]g` `:. M = (200+480)g = 680 g`

Discussion

177.	An open rectangular tank 1.5 m wide 2m deep ad 2 m long is half filled with water. It is accelerated horizontally at `3.27m//s^(2)` in the direction of its length determine the depth of water at each end of tank.A. 0.9 mB. 1.2 mC. 1.5 mD. 1.7 m
Answer» Correct Answer - C

Discussion

178.	A cubical block of ice floating in water has to suport metal piece weighing 0.5 kg. What can be the minimum edge of the block so that it does not sink in water/ specific gravity of ice=0.9.
Answer» Correct Answer - A::C Weight of ice block+weight of metal piece =upthrust on 100% volume of ice cube. Let a= side of ice cube, Then, `[(a^3)xx900xxg+0.5g]=(a^3)(1000)(g)` `:. a^(3)=(5xx10^(-3))m^(3)` or, `a~~0.17m` or `17 cm`.

Discussion

179.	Three points `A,B`,and `C` on a steady flow of a non-viscon and inconpressible fluid are observed. The pressure, velocity and height of the point `A, B` and C are (2,3,1),(1,2,2) and (4,1,2) respectively.Density of the fluid is `1kgm^(-3)` and all other parameters are given in SI unit. Then which of the following is correct?(`g=10ms^(-2))`A. Point `A` and `B` lie on the same stream lineB. Point `B` and `C` lie on the same stream lineC. Point `C` and `A` lie on the same stream lineD. None of the above
Answer» Correct Answer - D Sum of all three terms are different at three points A,B, and C.

Discussion

180.	A cubical block of side a and density `rho` slides over a fixed inclined plane with constant velocity v. There is a thin film of viscon fluid of thickness t between the plane and the block. Then the coefficeint of viscosity of the film will be A. `(3rhoag t)/(5v)`B. `(4rhoag t)/(5v)`C. `(rhoag t)/(5v)`D. None of these
Answer» Correct Answer - A Viscon force=`mg sin theta` `:. eta(a^2)v/t=mg sin37^(@)=3/5 mg` `(3)/(5) (a^(3)rho)g rArr :. eta =(3 rhoa g t)/(5v)`

Discussion

181.	The three vessels shown in figure have same base area. Equal volumes of a liquid are poured in the thre vessels. The force on the base will be A. maximum in vessel AB. maximum in vessel BC. maximum in vessel CD. equal in al the vessels
Answer» Correct Answer - C

Discussion

182.	A cylindrical vessel open at the top is `20cm` high and `10 cm` in diameter. A circular hole of cross-sectional area `1cm^(2)` is cut at the centre of the bottom of the vessel. Water flows from a tube above it into the vessel at the rate of `10^(2)cm^(3)//s`. The height of water in the vessel under steady state is (Take `g=10m//s^(2))`.A. `20cm`B. `15cm`C. `10cm`D. `5cm`
Answer» Correct Answer - D In steady state, Volume flow rate entering the vessel =volume flow rate leaving the vessel `:. Q=av=a sqrt(2gh)` or `h=(Q^(2))/(2ga^(2))` `=((10^2)^(2))/((2xx1000)(1)^(2))=5cm`.

Discussion

183.	A jar is filled with two non-mixing liquids 1 and 2 haivng densities `rho_1` and `rho_2` respectively. A solid ball, made of a material of density `rho_3` , is dropped in the jar. It comes to equilibrium in the position shown in the figure. Which of the following is true for `rho_1`, `rho_1` and `rho_3`? A. `rho_1 gt rho_3 gt rho_2`B. `rho_1 gt rho_2 gt rho_3`C. `rho_1 lt rho_3 lt rho_2`D. `rho_1 lt rho_1 lt rho_2`
Answer» Correct Answer - C `W = F_B` `rho_3 Vg = rho_1 (V)/(2) g + rho_2 (V)/(2) g` `rho_3 = (rho_1+ rho_2)/(2), rho_3` lies between `rho_1` and `rho_2` `rho_2 gt rho_1` `rho_1 lt rho_3 lt rho_2`.

Discussion

184.	A fixed container of height `H` with large cross-sectional area `A` is completely filled with water. Two small orifice of cross-sectional area `a` are made, one at the bottom and the other on the vertical side of the container at a distance H/2 from the top of the container find the time taken by the water level to reach a height of H/2 from the bottom of the container.A. `(2A)/(3a)sqrt((2H)/g)`B. `(2A)/(3a)(sqrt2-1)sqrt(H/g)`C. `(3A)/(2a)(sqrt2-1)sqrt(H/g)`D. `(3A)/(2a)sqrt((2H)/g)`
Answer» Correct Answer - A

Discussion

185.	A cylindrical vessel is filled with water upto a height of 1m. The cross-sectional area of the orifice at the bottom is `(1//400)` that of the vessel. (a) What is the tiome required to empty the tank through the orifice at the bottom? (b) What is the time required for the same amount of water to flow out if the water level in tank is maintained always at a height of 1m from orifice?
Answer» Correct Answer - A::B::C (a) Time taken to empty the tank (has been derived in theory) is `t=(2A)/(asqrt(2g))sqrt(H)` Given, `A/a=400` Substituting the value we have, `t=(2xx400)/(sqrt(2xx9.8))sqrt(1)` `=180s=3 min` (b) Rate of flow of water `Q=a sqrt(2gH)=constant` Total volume of water `V=AH` `:.` Time take to empty the tank with constant rate `t=V/Q=(AH)/(a sqrt(2gH))` `(400xx1)/(sqrt(2xx9.8xx1))` `=90s =1.5 min`

Discussion

186.	Water is filled in a cylindrical container to a height of `3m`. The ratio of the cross-sectional area of the orifice and the beaker is `0.1`. The square of the speed of the liquid coming out from the orifice is `(g=10m//s^(2))`. A. `50 m//s`B. `50.5 m//s`C. `51 m//s`D. `52 m//s`
Answer» Correct Answer - A

Discussion

187.	The vessel shown in the figure has a two sections of areas of cross-section `A_1` and `A_2`. A liquid of density `rho` fills both th sections, up to a height `h` in each Neglect atmospheric pressure. Choose the wrong option. .A. The pressure at the base of the vessel is `2 h rho g`.B. The force exerted by the liquid on the base of the vessel is `2 h rho g A_2`C. The weight of the liquid is `= 2h rho g A_2`D. The waals of the vessel at the level `X` exerted a downward force `h rho g(A _2 - A_1)` on the liquid.
Answer» Correct Answer - C The weight of the liquid `lt2 rho gh A_2`.

Discussion

188.	There is a small hole at the bottom of tank filled with water. If total pressure at the bottom is `3 atm(1 atm=10^(5)Nm^(-2))`, then find the velocity of water flowing from hole.
Answer» Correct Answer - B `Delta p=(1)/(2) rho v^(2)` `:. v=sqrt((2Delta p)/(rho))` `=sqrt((2(3 atm-1 atm))/(rho))` `=sqrt((2xx2xx10^(-5))/(10^3))=20m//s`.

Discussion

189.	With what terminal velocity will an air bubble `0.8 mm` in diameter rise in a liquid of viscosity `0.15 N-s//m^(2)` and specific gravity `0.9`? Density of air is `1.293 kg//m^(3)`.
Answer» Correct Answer - A::B::C The terminal velocity of the bubble is given by, `v_(T)=(2)/(9) (r^(2)(rho-sigma)g)/(eta)` Here, `r=0.4xx10^(-3)m, sigma =0.9xx10^(3)kg//m^(3), rho=1.293 kg//m^(3), eta=0.15 N-s//m^(2)` and `g=9.8m//s^(2)` Substituting the values, we have `v_(T)=(2)/(9)xx ((0.4 xx 10^(-3))^(2)(1.293-0.9 xx 10^(3))xx9.8)/(0.15)` `= - 0.0021 m//s` or `v_(T) =- 0.21 cm`

Discussion

190.	Liquid is filled in a container upto a height of `H`. A small hle is made at the bottom of the tank. Time taken to empty from `H` to `(H)/(3)` is `t_(0)`. Find the time taken to empty tank from `(H)/(3)` to zero.
Answer» Correct Answer - A::C `t=(A)/(a) sqrt((2H)/(g))` (to empty the complete tank) Now, `t_(Hrarro)=t_(Hrarr(H)/(3))+tunderset(3)overset(H)rarr0` Given,`(A)/(a) sqrt((2H)/(g))=t_(0)+(A)/(a) sqrt((2H//3)/(g))` from here find, `(A)/(a) sqrt((2H//3)/(g))`.

Discussion

191.	A cylindrical tank has a hole of `1cm^(2)` in its bottom. If the water is allowed to flow into the tank from a tube above it at the rate of `70cm^(3)//sec`, then the maximum height up to which water can rise in the tank isA. 2.5 cmB. 5cmC. 10cmD. 0.25cm
Answer» Correct Answer - A The height of water in tank becomes maximum when the volume of water flowing into the tank per second becomes equal to the volume flowing out per second. Volume of water flowing out per second. `Av=Asqrt(2gh)` ..(i) Volume of water flowing in per second `=70 cm^(3)//s`..(ii) Form (i) and (ii) we get ltbr. e`A sqrt(2gh) = 70 implies 1 xx sqrt(2gh) = 70` `:. h = (4900)/(1960) = 2.5 cm`.

Discussion

192.	A tank has a hole at its bottom. The time needed to empty the tank from level `h_(1)` to `h_(2)` will be proportional toA. `h_(1)-h_(2)`B. `h_(1)+h_(2)`C. `sqrt(h_1)-sqrt(h_2)`D. `sqrt(h_1)+sqrt(h_2)`
Answer» Correct Answer - C Let =(dh)/(dt)` represent the rate of descent of water level. Let `A` and `a` represent the cross -sectional areas of the container and hole repectively Then , `-A(dh)/(dt) = asqrt(2gh)` `dt = -k(dh)/(sqrt(n)) dt` or `int_(0)^(t) dt = -kint_(h_1)^(h_2) (1)/(sqrt(h))dh` or `t=-k\|(h^(1/2+1))/(-1/2+1)\|_(h_(1))^(h_2)` or `t=-2k[sqrt(h_2)-sqrt(h_1)]` or `t prop (sqrt(h_(1) )-sqrt(h_2))`.

Discussion

193.	A large tank filled with water to a height `h` is to be emptied through a small hole at the bottom. The ratio of times taken for the level of water to fall from h to `(h)/(2)` and from `(h)/(2)` to zero isA. `sqrt(2)`B. `1/(sqrt(2))`C. `sqrt(2) - 1`D. `1/(sqrt(2) - 1)`
Answer» Correct Answer - C

Discussion

194.	A large cylindrical tank has a hole of area A at its bottom. Water is oured in the tank by a tube of equal cross sectional area A ejecting water at the speed v.A. the water level in the tank wil keep on rising.B. no water an be stored in the tankC. the water level will rise to a height `v^2/2g` and then stopD. the water level will oscillate.
Answer» Correct Answer - C

Discussion

195.	A large cylindrical tank has a hole of area A at its bottom. Water is oured in the tank by a tube of equal cross sectional area A ejecting water at the speed v.A. The water level in the tank will keep on risingB. No water can be stored in the tankC. The water level will rise to a height `v^2//2 g` and then stop.D. The water level will oscillate.
Answer» Correct Answer - C `Av = A sqrt(2 g h_(max))` `h_(max) = (v^2)/(2 g)`.

Discussion

196.	The pressure of water in a pipe when tap is closed is `5.5xx10^(5) Nm^(-2)`. The velocity with which water comes out on opening the tap isA. `10 ms^(-1)`B. `5 ms^(-1)`C. `20 ms^(-1)`D. `15 ms^(-1)`
Answer» Correct Answer - A Decrese in pressure energy = increase in kinetic energy `or" "Deltarho=1/2rhov^(2)` `therefore" " v=sqrt((2(DeltaP))/rho)=sqrt((2xx0.5xx10^(5))/(10^(3)))=10 ms^(-1)`

Discussion

197.	The level of water in a tank is 5 m high. A hole of area of cross section 1 `cm^(2)` is made at the bottom of the tank. The rate of leakage of water for the hole in `m^(3)s^(-1)` is `(g=10ms^(-2))`A. `10^(-3) m^(3)s^(-1)`B. `10^(-4) m^(3)s^(-1)`C. `10 m^(3)s^(-1)`D. `10^(-2) m^(3)s^(-1)`
Answer» Correct Answer - A Rate of leakage of water of from the hole `=Av=Asqrt(2gh)=10^(-4)sqrt(2xx10xx5)=10^(-3)m^(3)s^(-1)`

Discussion

198.	A raft of wood (density`=600kg//m^(3))` of mass `120 kg` floats in water. How much weight can be put on the raft to make it just sink?A. 120 kgB. 200kgC. 40kgD. 80kg
Answer» Correct Answer - D Volume of raft = `V = (mass)/(density)` `V= 120/600 = 1/5 m^(3)` when the raft is totally immersed in water, it displacess `1/5 m^(3)` water, so the upthrust `=1/5 xx 1000 xx g newton if `x kg` is the extra weight put on the raft to fully immerse it in water, then `(x+120)g = 1/5 xx 1000 xx g or = x + 120 = 200` `x =80 kg`.

Discussion

199.	A metal ball immersed in alcohol weights `W_1` at `0^@C` and `W_2` at `50^@C`. The coefficient of expansion of cubical the metal is less than that of the alcohol. Assuming that the density of the metal is large compared to that of alcohol, it can be shown thatA. `W_1 = W_2`B. `W_1 gt W_2`C. `W_1 lt W_2`D. `none of these
Answer» Correct Answer - C (c ) As the coefficent of cobical expanison of metal is less as compared to the coefficient of cubical expansion of liquid we may neight the expansion of metal ball. So when the ball is immersed in alcohol at `0^@ C` and has weight `W_1` `W_1 = W_0 -Wrho_0g` Where `W_0` = weight of ball in air Similarly, `W_2 = W_0 - V rho_(50)g` Where `rho_g` = density fo alcohol at `0^@C` As `rho_(50) lt rho_0`, `:. W_2 gt W_1 or W_1 lt W_2`

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200.	Water is flowing through a long horizontal tube. Let `P_A and P_B` be the pressures at two points A and B of the tubeA. `P_A` must be equal to `P_B`.B. `P_A` must be greater than `P_B`,C. `P_A` must be smaller than `P_B`,D. `P_A = P_B` only if the cross-sectional area at `A` and `B` are equal.
Answer» Correct Answer - A

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