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 Multiple Choice QuestionsMultiple Choice Questions

111.

Two soap bubbles A & B are kept in a closed chamber where the air is maintained at pressure 8 N/m2 . The radii of bubbles A & B are 2cm and 4cm, respectively. Surface tension of the soap water used to make bubbles is 0.04 N/m. If nA and nB are the number of moles of air in bubbles A and B, respectively, then nBnA = ?[Neglect the effect of gravity]

  • 2

  • 8

  • 9

  • 6


D.

6

PV = nRT

  n = PVRT

n ∝ PV  (as R and T are constant)

⇒  nBnA = PBVBPAVA

For bubble A,

    PA - PB = 4SrA

Given, Pa = 8 N/m2

             rA = 2 nm = 0.02 m

            S = 0.04 N/m-1

           PA = PaSrA

               = 8 + 4 × 0.040.02 = 16 Nm-2

For bubble B,

               PB - Pa = 4SrB            PB = Pa + 4SrB                      = 8 + 4 × 0.040.04 = 12 Nm-2           nBnA = PBVBPAVA = 1216 × rBrA3                      = 1216 × 423                       = 6


112.

A boat strikes an under water rock which punctures a hole 5 cm in diameter in its hull which is 1.5 m below the water line. At what rate in litre per second does water enter.

  • 2.65 Ls-1

  • 8.85 Ls-1

  • 10.65 Ls-1

  • 624 Ls-1


C.

10.65 Ls-1

Here , r = 52 cm = 2.5 cm          h = 1.5 m = 150 cmVelocity of efflux V = 2gh = 2 × 980 × 150                             = 14015 cms-1Rate at which water enters,         Q = aV = πr2V             = 227 × 2.52 × 14015 cm3s-1             = 2750 x 3.873 × 10-3 Ls-1             = 10.65 Ls-1


113.

A small spherical drop fall from rest in viscous liquid. Due to friction, heat is produced. The correct relation between the rate of production of heat and the radius of the spherical drop at terminal velocity will be

  • dHdt  1r5

  • dHdt  r4

  • dHdt  1r4

  • dHdt  r5


D.

dHdt  r5

Viscous force acting on spherical drop

      Fv = 6 πηrv

∴  Terminal velocity v = 29r2 σ - ρη

where, η = coefficient of viscosity of liquid 

  σ = density of material of spherical drop

   ρ = density of liquid

Power imparted by viscous force = Rate of production of heat

     P = dHdt

        = Fv . v

         = 6 πηrv2

          = 6 πηrv . 29 r2 σ - ρ gη2

 ⇒   dHdt ∝ r5


114.

Air of density 1.3 kg/m3 flows horizontally with a speed 106 km/h. A house has a plane roof of area 40 m2. Find the magnitude of aerodynamic lift on the roof.

  • 4.22 × 105 N

  • 2.25 × 104 N

  • 4.22 × 103 N

  • 2.25 × 106 N


B.

2.25 × 104 N

Here ρ = 1.3 kg/m3 , v = 106 km/h = 29.44 m/s and A = 40 m2

Air flows just above the roof and there is no air flows just below the roof inside the room. Therefore, v1 =v and v2 =0. Using Bernoulli's theorem at the points just outside and inside the roof, we have

P1 + ρgh112ρV12 = P2 + ρgh212ρV22

Since,     h1 = h2 = h and v1 = v

and        v2 = 0 , so P2 = P112 ρV2

or   P2 − P112 ρv2

Aerodynamic force exerted on the roof,

               F = (P2 − P1) A

or            F = 12 ρV2 × A = 12 × 1.3 × (29.44)2 × 40

                 = 2.25 × 104 N


115.

A capillary tube of length L and radius r is connected with another capillary tube of the same length but half the radius in series. The rate of steady volume flow of water through first capillary tube under a pressure difference of p is V. The rate of steady volume flow
through the combination will be (the pressure difference across the combination is p)

  • 17 V

  • 1617 V

  • V17

  • 1716 V 


C.

V17

Problems of series and parallel combination of pipes can be solved in the similar manner as is done in case of an electrical circuit. The only difference is potential difference is replaced by Δp and electrical resistance

        R18 η Lπ r4 = R

and    R28 ηLπ r24

               = 16 R

Electric current is replaced by rate of volume flow V'

       p = VR1 = VR                      .....(i)

        p' = p = V' Req

                  = V' (R1 + R2 )  

                   = 17 V'R

      From Eqs. (i) and (ii), we get

         VR = 17 V'R

⇒         V' = V17    


116.

The surface tension and vapour pressure of water at 20°C is 7.28x10-2 Nm-1 and 2.33 x 103 Pa respectively. The radius of the smallest spherical droplet which can form without evaporating is

  • 6.25 x 10-5 m

  • 6.25 x 10-3 m

  • 7.45 x 10-3 m

  • 8.9 x 10-4 m


A.

6.25 x 10-5 m

Given,

          T = 7.28 × 10-2 N/m

          p = 2.33 × 103 Pa

The drop will evaporate if the water pressure is greater than the vapour pressure p.

i.e. p = 2TR, where R = radius of drop

   R = 2Tp = 2 × 7.28 × 10-22.33 × 103 

          = 6.25 × 10-5 m


117.

A cylindrical tank is filled with water to level of 3 m. A hole is opened at height of 52.5 cm from bottom. The ratio of the area of the hole to that of cross-sectional area of the cylinder is 0.1. The square of the speed with which water is coming out from the orifice is (Take g = 10 m s-2)

  • 50 m2 s-2

  • 40 m2 s-2

  • 51.5 m2 s-2

  • 50.5 m2 s-2


A.

50 m2 s-2

Suppose A be the area of cross section of tank, a be the area of hole, ve be the velocity of
efflux, h be the height of liquid above the hole, 

     A

Let v be the speed with which the level decreases in the container. Using equation of
continuity, we get

        ave = Av

⇒       v = aveA 

Using Bernoulli's theorem, we have

      P0 + hρg + 12 ρν2 = P012 ρ ve2

ρ = density of fluid

P = pressure

⇒    hρg  + 12 ρ aveA2  =  12 ρve2

⇒                    ve2 2hg1 - a2A2

                           = 2 ×  3 - 0.525 1 - 0.12

⇒                      ve2 = 50 m2 s-2


118.

By sucking through a straw, a student can reduce the pressure in his lungs to 750 mm of Hg (density 13.6 g/cm3). Using the straw, he can drink water from a glass upto a maximum depth of

  • 10 cm

  • 13.6 cm

  • 75 cm

  • 1.36 cm


B.

13.6 cm

        Pressure of lungs = 750 mm of Hg.

  Atmospheric pressure = 760 mm of Hg.

∴ Difference in pressure = 760-750=10 mm of Hg.

(10 mm of Hg) × ρHg × g = hwater × ρwater × g

                            hwater  = ρHgρwater           = 10 × 10-1 cm × 13.6 g/cm31 g / cm3           = 136 mm           = 13.6 cm


119.

Two non-mixing liquids of densities straight rho and nstraight rho (n >1) are put in a container. The height of each liquid is h. a solid cylinder of length L and density d is put in this container. The cylinder floats with its axis vertical and length pL (p<1) in the denser liquid. The density d is equal to,

  • open curly brackets 2 plus left parenthesis straight n plus 1 right parenthesis straight p close curly brackets rho
  • open curly brackets 2 plus left parenthesis straight n minus 1 right parenthesis straight p close curly brackets rho
  • {1+(n-1)p}straight rho

  • {1+(n+1)p}straight rho


C.

{1+(n-1)p}straight rho

The situation can be depicted as follows:

According to Archimedes principle,

Weight of the cylinder = (upthrust)1 + (upthrust)2

i.e.,

space space space space ALdg space equals space left parenthesis 1 minus straight P right parenthesis LAρg space plus space left parenthesis PLA right parenthesis nρg

rightwards double arrow space straight d space equals space left parenthesis 1 minus straight P right parenthesis straight rho space plus space Pnρ
space space space space space space space space space equals space straight rho space minus space Pρ space plus space nPρ
space space space space space space space space space equals space straight rho space plus space left parenthesis straight n minus 1 right parenthesis Pρ
space space space space space space space space space equals straight rho space left square bracket space 1 plus left parenthesis straight n minus 1 right parenthesis straight rho right square bracket

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120.

A U tube with both ends open to the atmosphere is partially filled with water. Oil, which is immiscible with water, is poured into one side until it stands at a distance of 10 mm above the water level on the other side. Meanwhile, the water rises by 65 mm from its original level (see diagram). The density of the oil is

  • 650 kg m–3

  • 425 kg m–3

  • 800 kg m–3

  • 928 kg m–3


D.

928 kg m–3

hoil ρoil g = hwater ρwater g
140 × ρoil= 130 × ρwater
straight rho subscript oil space equals space 13 over 14 space straight x space 1000 space kg divided by straight m cubed
straight rho subscript oil space equals space 928 space kg space straight m to the power of negative 3 end exponent

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