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An inverted bell lying at the bottom of a lake 47.6 m deep has 50 cm3 of air trapped in it. The bell is brought to the surface of the lake. The volume of the trapped air will be (atmospheric pressure = 70 cm of Hg and density of Hg = 13.6 g/cm3





No explanation available.

The height of a mercury barometer is 75 cm at sea level and 50 cm at the top of a hill. Ratio of density of mercury to that of air is 104. The height of the hill is





(b) Difference of pressure between sea level and the top of hill
P=(h1-h2)×ρHg×g=(75-50)×10-2×ρHg×g         …(i)
and pressure difference due to h meter of air
P=h×ρair×g                                                          …(ii)
By equating (i) and (ii) we get

h×ρair×g=(75-50)×10-2×ρHg×g

h=25×10-2ρHgρair=25×10-2×104=2500 m

Height of the hill = 2.5 km.

Equal masses of water and a liquid of relative density 2 are mixed together, then the mixture has a density of





(b) If two liquid of equal masses and different densities are mixed together then density of mixture

ρ=2ρ1ρ2ρ1+ρ2=2×1×21+2=43

A body of density d1 is counterpoised by Mg of weights of density d2 in air of density d. Then the true mass of the body is





(d) Let M0=mass of the body in the vacuum.
Apparent weight of the body in air = Apparent weight of standard weights in air
Actual weight – upthrust due to displaced air = Actual weight – upthrust due to displaced air 

M0g-V0dg=Mg-VdgM0g-M0d1dg=Mg-Md2dgM0=M1-dd21-dd1

The value of g at a place decreases by 2%. The barometric height of mercury





(a) h=Pρg  h1g (P and ρ are constant)

If value of g decreased by 2% then h will increase by 2%.

A barometer kept in a stationary elevator reads 76 cm. If the elevator starts accelerating up, the reading will be





(d) h=Pρg  h1g. If lift moves upward with some acceleration then effective g increases. So the value of h decreases i.e. reading will be less than 76 cm.

A barometer tube reads 76 cm of mercury. If the tube is gradually inclined at an angle of 60o with vertical, keeping the open end immersed in the mercury reservoir, the length of the mercury column will be





No explanation available.

A triangular lamina of area A and height h is immersed in a liquid of density ρ in a vertical plane with its base on the surface of the liquid. The thrust on the lamina is





(b) Thrust on lamina = pressure at centroid × Area

                               =hρg3×A=13Aρgh

If two liquids of same masses but densities ρ1 and ρ2 respectively are mixed, then density of mixture is given by





(c)  ρ=Total massTotal volume=2mV1+V2=2mm1ρ1+1ρ2

 ρ=2ρ1ρ2ρ1+ρ2

If two liquids of same volume but different densities ρ1 and ρ2 are mixed, then density of mixture is given by





(a) ρ=Total massTotal volume=m1+m22V=V(ρ1+ρ2)2V=ρ1+ρ22

With rise in temperature, density of a given body changes according to one of the following relations





(b) Since, with increase in temperature, volume of given body increases, while mass remains constant so that density will decrease. 

i.e. ρρ0=m/Vm/V0=V0V=VV0(1+γθ)=(1-γθ) ρ=ρ0(1-γ θ)

An ice berg of density 900 Kg/m3 is floating in water of density 1000 Kg/m3. The percentage of volume of ice-cube outside the water is 





(c) Let the total volume of ice-berg is V and its density is ρ. If this ice-berg floats in water with volume Vin 

inside it then Vinσg=VρgVin=ρσV

or Vout=V-Vin=σ-ρσVVoutV=σ-ρσ=1000-9001000=110 Vout= 10% of V

A hemispherical bowl just floats without sinking in a liquid of density 1.2×103 kg/m3. If outer diameter and the density of the bowl are 1 m and 2×104 kg/m3 respectively, then the inner diameter of the bowl will be 





(c) Weight of the bowl = mg

=Vρg=23πD23-d23ρg

where D = Outer diameter ,

d = Inner diameter

ρ = Density of bowl                               

Weight of the liquid displaced by the bowl

=Vσg=23πD23σg

where σ is the density of the liquid

For the floatation    23πD22σg=23πD23-d23ρg123×1.2×103=123-d232×104

By solving we get d = 0.98 m.

A concrete sphere of radius R has a cavity of radius r which is packed with sawdust. The specific gravities of concrete and sawdust are respectively 2.4 and 0.3 for this sphere to float with its entire volume submerged under water. Ratio of mass of concrete to mass of sawdust will be 





(b) Let specific gravities of concrete and saw dust  are ρ1 and ρ2 respectively.
According to principle of floatation weight of whole sphere = upthrust on the sphere

43π(R3-r3)ρ1g+43πr3ρ2g=43πR3×1×gR3ρ1-r3ρ1 +r3ρ2=R3R3(ρ1-1)=r3(ρ1-ρ2)R3r3=ρ1-ρ2ρ1-1R3-r3r3=ρ1-ρ2-ρ1+1ρ1-1(R3-r3)ρ1r3ρ2=1-ρ2ρ1-1ρ1ρ2Mass of concreteMass of saw dust=1-0.32.4-1×2.40.3=4

A metallic block of density 5 gm cm-3 and having dimensions 5 cm × 5 cm × 5 cm is weighed in water. Its apparent weight will be





(d) Apparent weight 

=V(ρ-σ)g =l×b×h×(5-1)×g=5×5×5×4×g Dyne = 4×5×5×5 gf