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1.	The water level on a tank is 5m high. There is a hole of `1 cm^(2)` cross-section at the bottom of the tank. Find the initial rate with which water will leak through the hole. (`g= 10ms^(-2)`)A. `10^(-3) m^3s^(-1)`B. `10^(-4)m^3s^(-1)`C. `10m^3s^(-1)`D. `10^(-2)m^3s^(-1)`
Answer» Correct Answer - A Rate of leakage of water from the hole `=Av= A sqrt(2gh)` `=10^(-4) sqrt(2xx10xx5)=10^(-3) m^3 s^(-1)`

2.	At a depth of 1000 m in an ocean (a) what is the absolute pressure? (b) what is the gauge pressure? (c ) Find the force acting on the window of area `20 cm xx 20cm` of a submarine at this depth, the interior of which is maintained at sea-level atmospheric pressure. The density of sea water is `1.03 xx 10^(3) kg m^(-3)`, `g=10ms^(-2)`. Atmospheric pressure = `1.01 xx 10^(5) Pa`.A. 40 atmB. 52 atmC. 32 atmD. 62 atm
Answer» Correct Answer - B Absolute pressure `p=p_a+rhogh` here `p_a=1.01xx10^5Pa` `p=1.03xx10^3kgm^(-3)` `therefore " " p=1.01xx10^5Pa + 51.5 xx 10^5 Pa` `=1.01xx10^5 Pa + 51.5xx10^5 Pa` `=52.5xx10^5 pa = 52` atm

3.	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. `2 sqrt(h(H-h))`B. `4 sqrt(h(H+h))`C. `4 sqrt(h(H-h))`D. `2 sqrt (h(H+h))`
Answer» Correct Answer - A Range , `R=2 sqrt(h_("top") xx h_("bottom")) = 2 sqrt(h(H-h))`

4.	A block is gently placed on a conveyor belt moving horizontal with constant speed After `t = 4s` the velocity of the block becomes equal to velocity of the belt If the coefficient of friction between the block and the belt is `mu = 0.2`, then the velocity of the conveyor belt is .A. `8ms^(-1)`B. `4ms^(-1)`C. `6ms^(-1)`D. `18ms^(-1)`
Answer» Correct Answer - A Due to friction `(a-mug)` , veloctiy of block will become equal to velocity of bell . Relative motion between two will stop. `therefore v=at = mugt = 0.2xx10xx4 = 8ms^(-1)`

5.	The breaking strength of the cable used to pull a body is 40N. A body of mass 8kg is resting on a table of coefficient of friction `mu=0.20` . The maximum acceleration which can be produced by the cable is (take , `g=10ms^(-2)`)A. `6ms^(-2)`B. `3ms^(-2)`C. `8ms^(-2)`D. `9ms^(-2)`
Answer» Correct Answer - B `a_("max") = (T_("max") - mumg)/(m) = (40-0.2xx8xx10)/(8) = 3ms^(-2)`

6.	Water is conveyed through a horizontal tube 8 cm in diameter and 4 kilometer in length at the rate of 20 litre/s Assuming only viscous resistance , calculate the pressure required to maintain the flow . Coefficient of viscosity of water is 0.001 pa sA. 45 cmB. 50 cmC. 59.68cmD. 49.68 cm
Answer» Correct Answer - C Here , 2r=8cm = 0.08 m or r=004m, l=4km=4000m, `V=20litre//s = 20xx10^(-3) m^3 s^(-1), eta=0.001Pa-s, p=?` As `V=(pipr^4)/(8etal) " or " p=(8Vetal)/(pir^4)` `=(8xx(20xx10^(-3))xx0.001xx4000)/((22/7)xx(0.04)^4)` `=7.954xx10^4 Pa` `therefore` Height of mercury column for pressure difference p will be , `h=p/(rhog)=(7.954xx10^4)/((13.6xx10^3)xx9.8)` =0.5968 m = 59.68 cm

7.	Two capillary tubes of the same length but different radii r1 and r2 are fitted in parallel to the bottom of a vessel. The pressure head is P . What should be the radius of a single tube that can replace the two tubes so that the rate of flow is same as beforeA. `r_1+r_2`B. `(r_1r_2)/(r_1+r_2)`C. `(r_1+r_2)/(2)`D. None of these
Answer» Correct Answer - D When the tubes are fitted in parallel, `V=V_1 + V_2` `rArr (pipr^4)/(8etal) = (pr_1^4)/(8etal) + (pipr_2^4)/(8) rArr r^4 = r_1^4 + r_2^4` `therefore r=(r_1^4 + r_2^4)^(1//4)`

8.	The rate of flow of liquid in a tube of radius r, length l, whose ends are maintained at a pressure difference P is `V = (piQPr^(4))/(etal)` where `eta` is coefficient of the viscosity and Q isA. 8B. `1/8`C. 16D. `1/16`
Answer» Correct Answer - B Rate of flow through a tube is given by `V=(pipr^4)/(8etal)`

9.	If the velocity head of a stream of water is equal to 10 cm, then its speed of flow is approximatelyA. `1.0ms^(-1)`B. `1.4ms^(-1)`C. `140 ms^(-1)`D. `10ms^(-1)`
Answer» Correct Answer - B Given , `(v^2)/(2g)=h=10cm` `therefore v=sqrt(2gh) = sqrt(2xx10xx0.1) =sqrt(2) ms^(-1)` `=1.414ms^(-1)`

10.	In a streamline flowA. The speed of a particle always remains sameB. The velocity of a particle always remains sameC. The kinetic energies of all particles arriving at a given point are the sameD. The potential energies of all the particles arriving at a given point are the same
Answer» Correct Answer - C In streamlined flow, velocity of all the particles arriving at a common point reamains same with time (both in magnitude and direction ). So the KE = `1/2mv^2 = a` Constant for all the particles as all fluid particles have identical mass also

11.	Water flows in a streamline manner through a capillary tube of radius a. The pressure difference being P and the rate of flow is Q . If the radius is reduced to `a//2` and the pressure difference is increased to 2P, then find the rate of flow.A. 4QB. QC. `Q/4`D. `Q/8`
Answer» Correct Answer - D As, `V=(pipr^4)/(8etal)` `therefore V prop pr^4 " " (eta " and l are constants")` `therefore V_2/V_1 = (p_2/p_1)(r_2/r_1)^4 = 2xx(1/2)^4 = 1/8` `rArr V_2 = Q/8`

12.	A block of mass 2kg rests on a rough inclined plane making an angle of `30^@` with the horizontal. The coefficient of static friction between the block and the plane is 0.7. The frictional force on the block isA. 9.8 NB. `0.7xx9.8xx sqrt(3)N`C. `9.8xxsqrt(3)N`D. `0.7xx9.8N`
Answer» Correct Answer - A Angle of repose, `theta = tan^(-1)(mu_s)=tan^(-1)(0.7)` or `tan theta_r=0.7` Angle of plane is `theta = 30^@, tan theta= tan30^@ = 0.577` Since `tan theta lt tan theta, theta lt theta` Block will not slide or `f=mgsintheta ne mu mgcostheta` or `f=(2)(9.8)sin 30^@ = 9.8N`

13.	If the terminal speed of a sphere of gold (density `=19.5kg//m^3`) is `0.2m//s` in a viscous liquid (density `=1.5kg//m^3`), find the terminal speed of a sphere of silver (density `=10.5kg//m^3`) of the same size in the same liquidA. `0.4ms^(-1)`B. `0.133ms^(-1)`C. `0.1ms^(-1)`D. `0.2ms^(-1)`
Answer» Correct Answer - C Terminal speed of spherical body in a viscous liquid is given by `v_T=(2r^2(rho-sigma)g)/(9eta)` where , `rho` = density of substance of body and `sigma` = density of liquid from given data `(v_T(Ag))/(V_T(Gold))= (rho_(Ag) - sigma_l)/(rho_(Gold)- sigma_l)` `rArr v_T(Ag) = (10.5-1.5)/(19.5-1.5) xx0.2 = 9/18xx0.2=0.1 ms^(-1)`

14.	Water from a tap emerges vertically downwards an intial speed of 1 m/s . The cross-sectional area of the tap is `10^(-4)m^(2)` . Assume that the pressure is contant throughout the stream of water and that the flow is steady . The cross-sectional area of the steam. `0.15` m below the tap isA. `5.0xx10^(-4)m^2`B. `1.0xx10^(-5)m^2`C. `5.0xx10^(-5)m^2`D. `2.0xx10^(-5)m^2`
Answer» Correct Answer - C Decrease in potential energy = increasing in kinetic energy `therefore rhogh=1/2rho(V_f^2-V_i^2)` or `2(10)(0.15) = V_f^2 - (10)^2` or `V_f=2ms^(-1)` Now , from continuity equation. `A_1V_1A_2V_1 " or " A prop 1/v` Velocity has become two times . Hence , are of cross section will remain half .