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**Q.1. An ideal fluid flows (laminar flow) through a pipe of non-uniform diameter. The maximum and minimum diameters of the pipes are 6.4 cm and 4.8 cm, respectively. The ratio of the minimum and the maximum velocities of fluid in this pipe is (2020)(1) 9/16(2)(3) 3/4(4) 81/256Ans.** (1)

Solution.

Given that, d

Using continuity equation, we have A

(1)

(2)

(3)

(4)

Ans.

Solution.

We know that object float on liquid when its mass is equal to liquid, or force due to weight is equal to buoyancy force.

For minimum density of liquid, sphere float with completely merge in liquid.

So, mass of sphere is equal to mass of liquid displaced, that is m

Given that

So,

Q.3. Two liquids of densities ρ_{1} and ρ_{2}(ρ_{2} = 2ρ_{1}) 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 lower part NO to that at the upper part MN is (Assume that the liquids are not mixing) (2020)**Ans. **(4)

Solution.

Force exert by the liquid on the wall is given by F = ρghA

Given that h_{1} = 5 m,h_{2}= 10 m and ρ_{2} = 2ρ_{1} .

So, force on MN is F_{1 }= ρ_{1}gh_{1}A_{1} ................. (1)

and, force on NO is F_{2} = ρ_{1}gh_{2}A_{1 }+................ (2)

Since, h_{1} = h_{2}

From Eqs. (1) and (2), we get**Q.4. Water flows in a horizontal tube (see figure). The pressure of water changes by 700 N/m ^{2 }between A and B where the area of cross section are 40 cm^{2} and 20 cm^{2}, respectively. Find the rate of flow of water through the tube.(ρ_{water} – 1000 kg/m^{3}) (2020)**(2)

(1) 3020 cm^{3}/s

(2) 2720 cm^{3}/s

(3) 2420 cm^{3}/s

(4) 1810 cm^{3}/s

Ans.

Solution.

Given that

From equation of continuity, we have

2ν

Using Bernoulli's theorem, we have

= 68 cm/s

Therefore, rate of flow is A

(1)

(2)

(3)

(4) r =

Ans.

Solution.

Let the radius of the droplet be r.

Given condition is shown by following figure

Forces acted on the droplet are

(i) Buoyancy force, which act upward on the droplet and given by

(ii) Force due to surface tension, which act upward on the droplet and given by

F = T (2πr)

(iii) weight of the droplet, which act downward on the droplet and given by

At balanced point, we have B + F= W

(1) 0.03 mm

(2) 0.02 mm

(3) 0.04 mm

(4) 0.01 mm

Ans.

Solution.

Given that m = 6 g, L = 60 cm, A = 1mm 2 , v = 90 m/s,Y = 16 ´1011 N/m2 Speed of transverse waves is given by

Where T is tension on wire and m is mass per unit length of wire.

Tension in wire is given by

And μ = m/L

Where Y is young’s modulus, A is area of cross section, ΔL is extension of wire, L is length of wire and m is mass.

So,

= 0.3 x 10^{-3} m = 0.03 mm**Q.7. A body of mass m = 10 kg is attached to one end of a wire of length 0.3 m. The maximum angular speed (in rad/s) with which it can be rotated about its other end in space station is (Breaking stress of wire = 4.8 × 107 N/m ^{2} and area of cross-section of the wire = 10^{−2} cm^{2}) is (2020)**(4)

Ans.

Solution.

Given that m = 10 kg, l =0.3m, P

Tension on a wire is given by T = mlω

Breaking stress is given by

(1) √2 : 1

(2) 1 : 2

(3) 2 : 1

(4) 1 : √2

Ans. (1)

Solution.

Given that

Energy stored in the stretched wire is given by

E = 1/2 x V x (stress) x (strain)

=> ε = E/V = 1/2 x (stress) x (strain) = 1/2 x

So, ε

**Q 9. A rod, of length L at room temperature and uniform area of cross-section A, is made of a metal having coefficient of linear expansion α/°C. It is observed that an external compressive force F, is applied on each of its ends, prevents any change in the length of the rod, when its temperature rises by ΔT K. Young’s modulus, Y, for this metal is (2019)(1) (2) (3) (4) Ans:** (1)

...(1)

The force acts on both the end of rod. So, the length will decrease but temperature will increase.

Put this value in Eq. (1), we get

(1)

(2)

(3)

(4)

Ans:

If the rod is rotating through an angle θ then

where restoring angular acceleration

Time period of S.H.M is

Ω= ωθ (average velocity)

Thus,

Velocity of rotation is

Tension towards the centre is

Moment of inertia is given by

Put these values in Eq. (1), we get

(1) 3.0 mm

(2) 4.0 mm

(3) 5.0 mm

(4) zero

Ans:

ρ is the relative density of material.

Initial length of steel wire, l = 1 mm

Change in length with it loaded mass (mg)Δl = 4 mm Thus, young modulus is

Dividing Eq. (2) by Eq. (1), we get

(1) 6.0 m

(2) 4.8 m

(3) 9.6 m

(4) 2.9 m

Ans:

Volume of water inflow = Volume of water out flow

(1) 5.1 cm

(2) 1.7 cm

(3) 4 cm

(4) 2.9 cm

Ans:

Height of water column is constant.

Flow of water in tank = Area of hole = 1 cm^{2} = 10^{−4} m

Thus, rate of water flow in the tank = rate of water out flow**Q 14. A liquid of density ρ is coming out of a hose pipe of radius a with horizontal speed v and hits a mesh. 50% of the liquid passes through the mesh unaffected. 25% loses all of its momentum and 25% comes back with the same speed. The resultant pressure on the mesh will be (2019)(1) (2) (3) (4) Ans:** (1)

Momentum per second carried by liquid per second = ρAv^{2 }

Force due to 25% which loses all its momentum

Force due to 25% which come back with same speed

Total force

Net pressure,**Q 15. A long cylindrical vessel is half filled with a liquid. When the vessel is rotated about its own vertical axis, the liquid rises up near the wall. If the radius of vessel is 5 cm and its rotational speed is 2 rotations per second, then the difference in the heights between the center and the sides, in cm, will be (2019)(1) 2.0(2) 0.1(3) 0.4(4) 1.2Ans:** (1)

(1)

(2)

(3)

(4)

Ans:

Pressure difference in water droplet is

where R is the radius of drop and T is temperature. Since, volume increases with time at constant rate**Q 17. A solid sphere of radius r made of a soft material of bulk modulus K is surrounded by a liquid in a cylindrical container. A massless piston of area a floats on the surface of the liquid, covering entire cross section of cylindrical container. When a mass m is placed on the surface of the piston to compress the liquid, the fractional decrement in the radius of the sphere, (dr/r) is: (2018)(1) **

**(2) **

**(3) **

**(4) Ans:** (3)

Bulk Modulus,

**(1) **

**(2) **

**(3) **

**(4) Ans: **(3)

h

(R Cos θ + R sinθ)ρ

(ρ

(1) 2.4 cm

(2) 12 cm

(3) 4.8 cm

(4) 6 cm

Ans:

r = 2.4cm**Q 20. A compressive force, F is applied at the two ends of a long thin steel rod. It is heated, simultaneously, such that its temperature increases by ΔT. The net change in its length is zero. Let l be the length of the rod, A its area of cross-section, Y its Young's modulus, and a its coefficient of linear expansion. Then, F is equal to- (2017)**

(1) lA YαΔT

(2) A YαΔT

(4) l^{2} YαΔT**Ans:** (3)**Solution:**

Net change in length = 0

Thermal Exp. = l ∝ Δt**Q 21. Two tubes of radii r _{1} and r_{2}, and lengths l_{1} and l_{2}, respectively, are connected in series and a liquid flows through each of them in stream line conditions. P_{1} and P_{2} are pressure differences across the two tubes. If P_{2} is 4P_{1} and l_{2} is l_{1}/4, then the radius r_{2} will be equal to - (2017)** 4

(1) 4r_{1}

(2) r_{1}

(3) 2r_{1}

(4) r_{1}/2

Ans:

Ans:

Ans:

Using equation of continuity

(2) ΔY/Y gets minimum contribution from the uncertainty in the length.

(3) The figure of merit is the larges for the length of the rod.

(4) ΔY/Y gets its maximum contribution from the uncertainty in strain.

Ans:

(1) (2πμB b)Δa

(2) (πμB b)Δa

(3) (πμB b) a

(4) (4πμB b)Δa

Ans:

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