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PHYSICS [5 Mark Question Answer]

Upthrust and Archimedes principle · ICSE STD 9 (ICSE - English Medium). 19 questions.

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[5 Mark Question Answer]

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Question 14 Marks
A solid body weighs 2.10 N . in air. Its relative density is 8.4 . How much will the body weigh if placed
(1) in water,
(2) in liquid of relative density 1.2 ?
Answer
Weight of solid body in air $=2.10 N$
R.D. of solid $=8.4$
$
\begin{array}{l}
\text { R.D. of solid }=\frac{\text { Density of solid }\left(\rho_{\text {solid }}\right)}{\text { Density of water }} \\
8.4=\frac{\rho_{\text {solid }}}{\rho_{\text {water }}} \\
\rho_{\text {solid }}=8.4 \times \rho_{\text {water }}=8.4 \times 1000=8400 kgm^{-3}
\end{array}
$
(1)
$
\begin{array}{l}
\text { R.D. of solid }=\frac{\text { Weight of solid in air }}{\text { Weight of water displaced by body }} \\
8.4=\frac{2.10}{\text { Weight of water displaced by body }}
\end{array}
$
Weight of water displaced by body $=\frac{2.10}{8.4}=0.25 N$
Weight of body in water = Weight of body in air - Weight of water displaced by body
$
=2.10-0.25=1.85 N$
(2) Upthrust due to water = Weight of water displaced by body
$=0.25 N
$
Upthrust due to liquid $=$ Upthrust due to water $\times$ R.D. of liquid
$
=0.25 \times 1.2=0.30 N
$
Weight of body in liquid = Weight of body in air - Upthrust due to liquid
$
=2.10-0.30=1.80 N$
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Question 24 Marks
A rubber ball floats in water with $2 / 7$ of its volume above the surface of water. Calculate the average relative density of rubber ball.
Answer
Let volume of rubber ball $= V$
Let volume of rubber ball $= V$
Volume of rubber ball above the water surface $=\frac{2}{7} V$
Volume of rubber ball below the water surface $=V-\frac{2}{7} V$
$=\frac{7 V-2 V}{7}=\frac{5}{7} V$
$\Rightarrow$ Volume of water displaced by the immersed part of the rubber
$
\text { ball }=\frac{5}{7} V$
by law of floatation :
Volume of rubber ball $\times$ density of rubber ball
$=$ Volume of water displaced $\times$ density of water
$\begin{array}{c}
V \times \rho=\frac{5}{7} V \times \rho_{w} \\
\text { But, } \rho_{w}=1 g cm^{-3} \\
\Rightarrow V \times \rho=\frac{5}{7} V=1 \\
\rho=\frac{5}{7}=0.71 g cm^{-3}
\end{array}$
$\text { Average relative density of rubber ball }=\frac{\text { Density of rubber ball }}{\text { Density of water }}$
$=\frac{\rho}{\rho_w}=\frac{0.71}{1}=0.71$
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Question 34 Marks
(a) State Archimedes' Principle.
(b) A block of mass 7 kg and volume $0.07 m^3$ floats in a liquid of density $140 kg / m ^3$.
Calculate:
1. Volume of block above the surface of liquid.
2. Density of block.
Answer
(a) ARCHIMEDES' PRINCIPLE: "Whenever a body is immersed in a liquid (fluid), wholly or partially, it loses weight equal to the weight of liquid displaced by it."
(b) Mass of block $=m=7 kg$
Volume of block $= V =0.07 m^3$
Density of liquid $=\rho_1=140 kgm ^{-3}$
Let $V ^{\prime}=$ Volume of block immersed in the liquid
By law of floatation:
Weight of block = Weight of liquid displaced by the immersed pa
$\begin{array}{l}
V^{\prime} \rho g=V^{\prime} \rho_l g \\
V^{\prime}=\frac{\rho}{\rho_l} V
\end{array}$
Density of block $=\rho=\frac{m}{V}=\frac{7}{0.07}=100 kg m ^{-3}$
From (1)
$\begin{array}{l}
V^{\prime}=\frac{100 V}{140}=\frac{5}{7} V=\frac{5}{7} \times 0.7 \\
V^{\prime}=\frac{1}{20}=0.05 m^3
\end{array}$
Volume of block above the surface of liquid $= V = V ^1$
$=0.07-0.05=0.02 m^3$
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Question 44 Marks
A test tube loaded with lead shots, weighs 150 gf and floats upto the mark X in water. The test tube is then made to float in alcohol. It is found that 27 gf of lead shots have to be removed, so as to float it to level X. Find R.D. of alcohol.
Answer
When tube floats in water :
Weight of test tube $=150 gf$
By law of floatation:
Weight of water displaced = Weight of test tube $=150 gf$
When test tube floats in alcohol:
When test tube is made to float in alcohol, then 27 gf of lead shots have to removed, so that it can float upto the same level as in water.
$\therefore$ Weight of test tube in alcohol $=150-27=123 gf$
By law of floatation:
Weight of alcohol displaced by test tube $=$ Weight of test tube in alcohol $=123 gf$
As volume of alcohol displaced = Volume of water displaced
$\therefore$ R.D. of alcohol
$
\begin{array}{l}
=\frac{\text { Weight of alcohol displaced }}{\text { Weight of equal volume of water displaced }} \\
=\frac{123}{150}=0.82
\end{array}
$
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Question 54 Marks
A balloon of volume $120 m^2$ is filled with hot air, of density $38 kg^2$. If the fabric of balloon weighs 12 kg , such that an additional equipment of wt . x is attached to it, calculate the magnitude of Density of cold air is $1.30 kgm =$.
Answer
Volume of balloon $= V =120 m^3$
Density of hot air $=\rho_{\text {hot air }}=0.38 kgm ^3$
Mass of empty balloon $=12 kg$
Weight of the empty balloon $=12 kgf$
Weight of the additional equipment attached with the balloon
=x kgf
Density of cold air $=\rho_{\text {coldair}}=1.30 Kgm^3$
Volume of balloon=Volume of hot air inside the balloon=Volume of cold air displaced by balloon $= V =120 m^3$ Weight of hot air $= V _{\text {hotair}} g$
$=120 \times 0.38 \times g=45.6 kgf
$
Weight of empty balloon + Weight of hot air inside the balloon + Weight of equipment = Downthrust
$
12+45.6+x=\text { Downthrust }
$
Downthrust $=57.6+x$
Upthrust = Weight of cold air diplaced by balloon
$\begin{array}{l}
=V \rho_{\text {coldair }} g \\
=120 \times 1.30 \times g=156 kgf
\end{array}
$
By law of floatation :
Downthrust = Upthrust
$\begin{array}{l}
57.6+x=156 \\
x=156-57.6 \\
x=98.4 kgf
\end{array}
$
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Question 64 Marks
A balloon of volume $800 cm^2$ is filled with hydrogen gas of density $9 \times 10:$ gemr2. If the empty balloon weighs 0.3 gf and density of air is $1.3 \times 10^2 g gem ^2$, calculate the lifting power of balloon.
Answer
$\begin{array}{l}
\text { Volume of balloon }=V=800 cm^3\\
\text { Density of hydrogen gas }=p_H=9 \times 10^{-5} g cm^{-3}\\
\text { Weight of empty balloon }=0.3 gf\\
\text { Density of air }=\rho_s=1.3 \times 10^{-3} g cm^{-3}\\
\text { Weight of hydrogen gas in balloon }=V \rho_{H} g\\
=800 \times 9 \times 10^{-5} \times g\\
=72 \times 10^{-3} gf\\
=0.072 gf\\
\text { Volume of balloon = Volume of hydrogen gas in the balloon = Volume of air displaced }\\
\text { by balloon. }\\
\text { Downthrust }=\text { Wt. of empty balloon }+ \text { wt. of hydrogen gas in balloon }\\
=0.3+0.072=0.372 gf\\
\text { Upthrust }=\text { Wt. of air displaced by balloon }\\
= V \rho_{a} g\\
=800 \times 1.3 \times 10^{-3} \times g\\
=8 \times 1.3 \times 10^{-1} \times g\\
=10.4 \times 10^{-1} \times g=1.04 gf\\
\text { Lifting power of balloon = Upthrust }- \text { Down thrust }\\
=1.04-0.372=0.668 gf
\end{array}$
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Question 74 Marks
A balloon of volume $1000 m^3$ is filled with a mixture of hydrogen and helium of density $0.32 kgm ^{-3}$. If the fabric of balloon weighs 40 kgf and the density of cold air is $1.32 kgm ^{-3}$, find the tension in the tope, which is holding the balloon to ground.
Answer
Volume of balloon $= V =1000 m^3$
Density of mixture of hydrogen and helium $=\rho=0.32 kgm^3$
Density weight of empty balloon $=40 kgf$
Density of cold air = $p ^{\prime}=1.32 kgm { }^3$
Volume of balloon=Volume of mixture of hydrogen and helium gas
= Volume of cold air displaced by balloon
$
=V=1000 m^3
$
Weight of mixture of hydrogen and helium gas in balloon $= V \rho g$
$
=1000 \times 0.32 \times g=320 kgf
$
Down thrust = Weight of empty balloon + Weight of mixture of hydrogen and helium gas
$
=40+320=360 kgf
$
Upthrust $=$ Weight of cold air displaced by balloon $= V \rho$ 'g
$
=1000 \times 1.32 \times g=1320 kgf
$
Tension in the rope = Upthrust - downthrust
$=1320-360=960 kgf$
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Question 84 Marks
What fraction of metal of density $3400 kgm ^{-3}$ will be above the surface of mercury of density 13600 $kgm ^{-3}$, while floating in mercury?
Answer
Density of metal $=\rho_m=3400 kgm ^{-3}$
Density of mercury $= p _{ Hg }=13600 kgm ^{-3}$
Let volume of metal $=x$
and volume of metal inside mercury $=y$
By law of floatation:
Weight of mercury displaced by metal $=$ wt. of metal
$\begin{array}{l}
V_{H_g}=\rho_{H_g} \times g=V_{\text {metal }} \times \rho_{\text {metal }} \times g \\
y \times 13600=x \times 3400
\end{array}
$
$
y=\frac{3400}{13600} x=\frac{1}{4} x
$
Volume of metal inside mercury=Volume of mercury displaced
$
y=\frac{1}{4} \times \text { Volume of metal }
$
Volume of metal above the surface of mercury $=x-y$
$
=x-\frac{1}{4} x=\left(1-\frac{1}{4}\right) x=\frac{4-1}{x} x=\frac{3}{4} x
$
Fraction of metal above surface of mercury $=\frac{3}{4} \times \frac{1}{x}=\frac{3}{4}$
$\Rightarrow \frac{3}{4}$ th part of metal lies above the surface of mercury.
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Question 94 Marks
What fraction of an iceberg of density 910 $kgm^{-3}$ will be above the surface of sea water of density $1170 kgm ^{-3}$ ?
Answer
Let volume of iceberg $= V _i= x$
Volume of iceberg inside sea water $= V _{ w }$
Density of iceberg $=\rho_i=910 kmg ^2$
Density of sea waer $=\rho_{ w }=1170 kgm { }^2$
By law of floatation:
Weight of icebeig = Weight of sea water displaced by the iceberg
$\begin{aligned}
V_i \times \rho_i \times g & =V_{w} \times \rho_{w} \times g \\
x \times 910=V_{w} & \times 1170 \\
V_{w} & =x \times \frac{910}{1170} \\
V_{w} & =\frac{7}{9} x=\frac{7}{9} x
\end{aligned}
$
Volume of iceberg inside sea water $=$ Volume of sea water displaced by iceberg
$
=V_w=\frac{7}{9} x=\frac{7}{9} x
$
Volume of iceberg above the sea water $=x-\frac{7}{9} x$
$
\begin{array}{l}
=\left(1-\frac{7}{9}\right) x=\frac{9-7}{9} x \\
=\frac{2}{9} \times \text { Volume of iceberg }
\end{array}
$
Fraction of iceberg above sea water $=\frac{2}{9} x \times \frac{1}{x}=\frac{2}{9}$ part
$\Rightarrow \frac{2}{9}$ th part of iceberg is above the sea water.
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Question 104 Marks
A hollow metal cylinder of length 10 cm floats in alcohol of density $0.80 g cm ^{-3}$, with 1 cm of its length above it. What length of cylinder will be above copper sulphate solution of density $1.25 g cm ^{-3}$ ?
Answer
Length of hollow metal cylinder $=x=10 cm$
1 cm length of cylinder is above the alcohol
$\therefore$ Length of the cylinder below alcohol $=(10-1)=9 cm$
Density of alcohol $=\rho_{\text {alcohol }}=0.80 g cm ^{-3}$
Density of copper sulphate solution $=\rho_{\text {CuS04 }}=1.25 g cm { }^2$
When block floats in alcohol By the law of floatation:
$\begin{array}{l}
h_{\text {block }} \times \rho_{\text {block }}=h_{\text {alcohol }} \times \rho_{\text {alcohol }} \\
10 \times \rho_{\text {block}}=9 \times 0.80 \\
\rho_{\text {block }}=\frac{9 \times 0.80}{10}=0.72 g cm^{-3}
\end{array}
$
When block is floats in copper sulphate solution :
By law of floatation :
$
\begin{array}{l}
h_{\text {block }} \times \rho_{\text {block }}=h_{\text {CuSO4 }} \times \rho_{\text {CuSO4 }} \\
10 \times 0.72=h_{\text {CuSO4 }} \times 1.25 \\
h_{\text {CuSO4 }}=\frac{7.2}{1.25}=5.76 cm
\end{array}
$
Length of metal block below copper sulphate solution $=5.76 cm$
So, length of metal block above copper sulphate solution
$=10-5.76=4.24 cm$
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Question 114 Marks
A wooden block floats in alcohol with $3 / 8$ of its length above alcohol. If it is made to float in water, what fraction of its length is above water? Density of alcohol is 0.80 g $cm ^{-3}$
Answer
Let length of wooden block $= x$
Let length of wooden block $= x$
$\therefore$ Length of the block above alcohol $=\frac{3}{8} x$
Length of the block below alcohol $=x-\frac{3}{8} x=\frac{8 x-3 x}{8}=\frac{5 x}{8}$
Density of water $=\rho_w=1 gcm ^{-3}$
Density of alcohol $=\rho_{\text {alcohol }}=0.80 gcm ^{-3}$
By the law of floatation:
$\begin{array}{l}
h_{\text {block }} \times \rho_{\text {block }}=h_{\text {alcohol}} \times \rho_{\text {alcohol }} \\
x \times \rho_{\text {block }}=\frac{5 x}{8} \times 0.80 \\
\rho_{\text {block  }}=\frac{5}{8} \times 0.80 \\
\rho_{\text {block }}=0.5 gcm^{-3}
\end{array}
$
When block is floating in water :
By law of floatation :
$
\begin{array}{l}
h_{\text {block }} \times \rho_{\text {block }}=h_{\text {water }} \times \rho_{w} \\
x \times 0.5=h_{\text {water }} \times 1 \\
h_{\text {wate }}=0.5 x \text { or } \frac{1}{2} x
\end{array}
$
So, length of block below water $=\frac{1}{2} x$
Length of block above water $=x-\frac{1}{2} x=\frac{x}{2}$
Fraction of the block above the water $=\frac{x}{2} \times \frac{1}{x}=\frac{1}{2}$ part
$\Rightarrow \frac{1}{2}$ th part of wooden block is above the water.
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Question 124 Marks
A cork cut in the form of a cylinder floats in alcohol of density $0.8 gcm^{-3}$, such that $3 / 4$ of its length is outside alcohol. If the total length of cylinder is 35 cm and area of cross-section 25 $cm ^2$, calculate :
(1) density of cork
(2) wt. of cork
(3) extra force required to submerge it in alcohol
Answer
(1) Density of alcohol $=\rho_{\text {stachol }}=0.8 g cm ^{-3}$
Total length of cork cylinder $=h_{\text {catk }}=35 cm$
Area of cross-section of cork cylinder $= A =25 cm^2$
$\because \quad \frac{3}{7}$ of the length of cork cylinder is outside the alcohol
$\therefore \quad$ Length of cork cylinder inside alcohol $=\left(1-\frac{3}{7}\right) l$
$
\begin{array}{l}
h_{\text {alcohol}}=\frac{7-3}{7} \times 35 \\
h_{\text {alcohol}}=4 \times 5=20 cm
\end{array}
$
Density of water $=\rho_{\text {water}}=1 g cm ^{-3}$
By law of floatation :
$
\begin{array}{l}
h_{\text {cork }} \times \rho_{\text {cork }}=h_{\text {alcohol }} \times \rho_{\text {alcohol}} \\
35 \times \rho_{\text {cork }}=20 \times 0.8 \\
\rho_{\text {cork }}=\frac{20 \times 0.8}{35}=0.457 g cm^{-3}
\end{array}
$
(2) Volume of cork $= V = A h_{\text {cork }}$
$
V=25 \times 35=875 cm^3
$
Mass of cork $= m = V \times \rho_{\text {cork }}$
$m =875 \times 0.457$
$
m=399.9 g \simeq 400 g
$
Wt. of cork $= mg =400 \times g =400 gf$
(3) Total upthrust when cork is completely immersed in water
$
\begin{array}{l}
=V \rho_{\text {alcohol }} \times g \\
=875 \times 0.8 \times g=700 g=700 gf
\end{array}
$
Extra force required to submerge complete the cork in
$
\begin{aligned}
\text { alcohol } & =\text { Upthrust }- \text { down thrust } \\
& =700-400=300 gf
\end{aligned}
$
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Question 134 Marks
A hollow cylinder of copper of length 25 cm and area of cross-section 15 cm , floats in water with $3 / 5$ of its length inside water. Calculate:
(1) apparent density of hollow copper cylinder.
(2) wt. of cylinder.
(3) extra force required to completely submerge it in water.
Answer
(1) Length of hollow cylinder of copper $= h _{ cu }=l=25 m$ Length of hollow cylinder of copper inside water
$\begin{aligned}
& =h_{\text {water }}=\frac{3}{5} l \\
h_{\text {water }} & =\frac{3}{5} \times 25 \\
h_{\text {water }} & =15 cm
\end{aligned}$
Area of cross-section of hollow cylinder of copper $=15 cm^2$
Density of water $=\rho_{\text {water }}=1 g cm ^{-3}$
Apparent density of hollow copper cylinder $=\rho_{ cu }=$ ?
By law of flotation:
$\begin{array}{l}
h_{cu} \times \rho_{cu}=h_{\text {water }} \times \rho_{water} \\
25 \times \rho_{cu}=15 \times 1 \\
\quad \rho_{cu}=\frac{15}{25}=0.6 g cm^{-3}
\end{array}$
(2) Volume of cylinder $= V = A \times h_{ cu }=15 \times 25=375 cm^3$
Mass of hollow cylinder of copper $= V \times \rho_{ cu }$
$\begin{array}{l}
m=A h_{cu} \times \rho_{cu} \\
m=15 \times 25 \times 0.6 \\
m=225 g
\end{array}$
Weight of hollow cylinder of copper $= mg$
$=225 \times g=225 gf$
(3) Total upthrust when hollow copper cylinder is completely immersed in water $= V \rho_{\text {water }} g$
$=375 \times 1 \times g=375 gf$
Extra force requird to submerge complete the cylinder in water $=$ Upthrust - down thrust
$=375-225=150 gf$
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Question 144 Marks
A cube of lead of side 8 cm and R.D. 10.6 is suspended from the hook of a spring balance. Find the reading of spring balance. The cube is now completely immersed in sugar solution of R.D. 1.4. Calculate the new reading of spring balance.
Answer
Length of side of cube $=l =8 cm$
Volume of cube $=\beta=(8)^3=512 cm^3$
$
V=512 cm^{3}$
Relative density of lead cube = R.D. $=10.6$ Relative density of sugar solution = R.D. 1.4
Density of water $=1 g cm ^{-3}$
$\text { R.D. of lead }=\frac{\text { Density of lead }}{\text { Density of water }}
$
$
10.6=\frac{\rho_{\text {lead }}}{1}
$
$
\rho_{\text {lead }}=10.6 \times 1=10.6 gcm^{-3}
$
$
\text { Mass of lead }=m=V \times \rho_{\text {lead }}
$
$
m=512 \times 10.6=5427.2 g
$
Wt . of led cube $= mg$
$=5427.2 \times g=5427.2 gf
$
Volume of sugar solution displaced = Volume of lead cube
$=V=512 cm^3$
$\text { R.D. of sugar solution }=\frac{\text { Density of sugar solution }}{\text { Density of water }}$
$1.4=\frac{\rho_{\text {supar }}}{1}
$
$
\rho_{\text {sugar }}=1.4 \times 1=1.4 g cm^{-3}
$
Upthrust due to sugar solution $= V \times \rho_{\text {sugar }} \times g=512 \times 1.4 \times g$
$
=716.8 g=716.8 gf$
Now reding of spring balance $=$ Actual weight - Upthrust
$
=5427.2-716.8=4710.4 gf$
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Question 154 Marks
An aluminium cube of side 5 cm and RD. 2.7 is suspended by a thread in alcohol of relative density 0.80 . Find the tension in thread.
Answer
Side of an aluminium cube $= l = 5 cm$
Volume of aluminium cube $= V = Z^{3}=(5)^3=125 cm^3$
Relative density of aluminium = R.D. $=2.7$
Relative density of alcohol $=$ R.D. $=0.80$
Density of water $=1 g cm ^{-3}$
$\begin{array}{l}
\text { R.D. of aluminium }=\frac{\text { Density of aluminium }}{\text { Density of water }} \\
2.7=\frac{\text { Density of aluminium }}{1}
\end{array}$
Density of aluminium $=\rho=2.7 g cm ^{-3}$
Mass of aluminium $= V \times \rho$
$
m=125 \times 2.7=337.5 g
$
Wt. of aluminium cube acting downwards $=337.5 gf$
Volume of alcohol displaced = Volume of cube $= V =125 cm{ }^3$
Upthrust due to alcohol $= V \times$ $\rho_{\text {alcohol }}$ $\times$ g
$
\text { Now R.D. of alcohol }=\frac{\rho_{\text {sloatol }}}{\text { Density of water }}
$
$
\begin{array}{l}
0.80=\frac{\rho_{\text {alcothol }}}{1} \\
\rho_{\text {alochol }}=0.80 g cm^{-3}
\end{array}
$
$\Rightarrow$ Upthrust due to alcohol $= V \times \rho_{\text {alcobal }} \times g$
$
=125 \times 0.80 \times g=100 gf
$
So tension in thread $= Wt$. of aluminium cube - Upthrust
$
=337.5-100=237.5 gf$
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Question 164 Marks
A glass cylinder of length $12 \times 10^{-2} m$ and area of crosssection $5 \times 10^{-4} m^2$ has a density of $2500 kgm^{-3}$. It is immersed in a liquid of density $1500 kgm^{-3}$, such that $3 / 8$. of its length is above liquid. Find the apparent weight of glass cylinder in newtons.
Answer
Length of glass cylinder $= I =12 \times 10^{-2} m$
Area of cross-section $= A =5 \times 10^{-4}  m^{2} $
Volume of glass cylinder $= V = A \times I$
$
\begin{array}{l}
V=5 \times 10^{-6} \times 12 \times 10^{-2} \\
V=0.00006 m^2
\end{array}
$
Acceleration due to gravity $= g =9.8 m / s ^2$
Density of glass cylinder $=\rho=2500 kgm { }^3$
Density of liquid $=\rho^{\prime}=1500 kgm ^{-3}$
$\because \frac{3}{8}$ length of glass cylinder is above the liquid
$\therefore$ Length of glass cylinder inside the liquid $=1-\frac{3}{8}=\frac{8-3}{8}=\frac{5}{8}$
$\therefore$ Volume of liquid displaced by glass cylinder $=\frac{5}{8} \times$ Volume of glass cylinder
$
\begin{array}{l}
V^{\prime}=\frac{5}{8}=0.00006 \\
V^{\prime}=\frac{0.0003}{8}=0.0000375 m^3
\end{array}$
Mass of glass cylinder $= m = V \times \rho$
$
m=0.00006 \times 2500 m
$
$
=0.15 kg
$
Weight of glass cylinder $= mg =0.15 \times 10=1.5 N$
Mass of liquid displaced by glass cylinder $= V ^{\prime} \times \rho^{\prime}$
$
m^{\prime}=0.0000375 \times 1500
$
$m ^{\prime}=0.05625 kg$
Upthrust = Weight of liquid displaced by the glass cylinder
$
=m^{\prime} g=0.05625 \times 10=0.5625 N
$
Apparent weight of glass cylinder in liquid = Actual weight of glass cylinder - Upthrust
$
=1.5-0.5625=0.9375 N$
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Question 174 Marks
A solid of density $7600 kgm { }^2$ is found to weigh 0.950 kgf in air. If $4 / 5$ volume of solid is completely immersed in a solution of density $900 kgm ^2$, find the apparent weight of solid in liquid.
Answer
Weight of solid in air $=0.950 kgf$
$\therefore$ Mass of solid in air $= m =0.950 kg$
Density of solid $=\rho=7600 kgm^{3}$
$\begin{array}{l}
\therefore \text { Volume of solid }=V=\frac{m}{\rho} \\
\quad V=\frac{0.950}{7600}=0.000125 m^3
\end{array}$
Density of solution $=\rho^{\prime}=900 kgm ^{-3}$.
$\because \frac{4}{5}$ volume of solid is completely immersed in the given solution.
$\therefore$ Volume of solution displaced $= V =\frac{4}{5} \times$ Volume of solid
$
\begin{array}{l}
V^{\prime}=\frac{4}{5} \times 0.000125 \\
V^{\prime}=0.0001 m^3
\end{array}
$
Mass of solution displaced $= m ^{\prime}= V ^{\prime} \times \rho^{\prime}$
$
\begin{array}{l}
m^{\prime}=0.0001 \times 900 \\
m^{\prime}=0.09 kg
\end{array}
$
Upthrust $=$ Weight of solution displaced
$
\begin{array}{l}
=m^{\prime} g \\
=0.09 \times 10=0.9 N=\frac{0.9}{10} kgf \\
=0.09 kgf
\end{array}
$
Apparent weight of solid in liquid
= Actual weight - Upthrust
$=0.950-0.09=0.860 kgf$
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Question 184 Marks
A solid of area of cross-section $0.004 m^2$ and length 0.60 m is completely immersed in water of density 1000 kgm ${ }^3$. Calculate :
1. Wt of solid in SI system
2. Upthrust acting on the solid in SI system.
3. Apparent weight of solid in water.
4. Apparent weight of solid in brine solution of density $1050 kgm ^3$.
[Take $g =10 N / kg$; Density of solid $=7200 kgm ^3$ ]
Answer
Area of cross-section of solid $= A =0.004 m^2$
Length of the solid $= l =0.60 m$
Density of water $=p^{\prime}=1000 kgm { }^3$
Acceleration due to gravity $= g =10 ms^{-2}$
Density of solid $=p=7200 kgm ^3$
(1) Volume of solid $= V = A \times I$
$
V=0.004 \times 0.60=0.0024 m^3
$
Mass of solid $= m = V \times p$
$m=0.0024 \times 7200=17.28 kg$
Weight of the solid $= mg =17.28 \times 10=172.8 N$
(2) Volume of water displaced = Volume of solid
$=V=0.0024 m^3$
Mass of water displaced $= m ^{\prime}= V \times p ^{\prime}$
$m^{\prime}=0.0024 \times 1000=2.4 kg$
Upthrust = Weight of water displaced
$=m^{\prime} g=2.4 \times 10=24 N$
(3) Apparent weight of solid=Actual weight of solid - upthrust
$=172.8-24=148.8 N$
(4) Density of brine solution $=\rho_b=1050 kgm ^3$
Volume of brine solution displaced $=$ Volume of solid $= V$
$V=0.0024 m^3
$
Mass of brine solution displaced
$\begin{array}{l}
=m_b=V \times \rho_b=0.0024 \times 1050 \\
m_b=2.52 kg
\end{array}
$
Upthrust acting on solid in brine solution = Weight of brine solution displaced $- m _{ b } g$
$
=2.52 \times 10=25.2 N
$
Apparent weight of solid in brine solution
= Actual weight - Upthurst
$=172.8-25.2=147.6 N$
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Question 194 Marks
A solid of density 2700 kgm' and of volume 0.0015 m3 is completely immersed in alcohol of density 800 kgm'.Calculate :
1. Weight of solid in SI system.
2. Upthrust on solid in SI system.
3. Apparent weight of solid in alcohol.
4. Will the apparent weight of solid be less or more, if it is immersed completely in brine solution? Give a reason, [ $\left.g =10 ms^{-2}\right]$
Answer
Density of solid = $\rho=2700$ kgm$^3$
Volume of solid $= V =0.0015 m^3$
Density of alcohol $=\rho^{\prime}=800 kgm ^3$
$ \begin{array}{l}
\text {1. Mass of solid }=m=V \times p \\
\quad m=0.0015 \times 2700=4.05 kg \\
\quad \text {Weight of solid}=m g=4.05 \times 10=40.5 N
\end{array}$
2. Volume of alcohol displaced = Volume of solid
$\quad V=0.0015 m^3$
$\quad$Mass of alcohol displaced $= m ^{\prime}= V \times p ^{\prime}$
$\quad m^{\prime}=0.0015 \times 800=m^{\prime}=1.2 kg
$
$\quad$Upthrust = Weight of alcohol displaced
$\quad =m^{\prime} g=1.2 \times 10=12 N
$
3. Apparent weight of solid in alcohol
$\quad$= Actual weight of solid - Upthrust
$\quad=40.5-12=28.5 N $
4. When a solid is immersed completely in brine solution, then upthrust acts on it in upward direction, as a result, its apparent weight of solid will be less than actual weight of solid will be less than actual weight of solid.
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