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PHYSICS [3 Mark Question Answer]

Upthrust and Archimedes principle · ICSE STD 9 (ICSE - English Medium). 21 questions.

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[3 Mark Question Answer]

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21 questions · timed · auto-graded

Question 13 Marks
A block of wood of volume 25 cm floats in water with 20 cm 2 of its volume immersed in water.
Calculate:
(1) density of wood
(2) the weight of block of wood.
Answer
Volume of wooden block $= V =25 cm^2$
Volume of wooden block immersed in water $=20 cm^2$
Volume of water displaced by wooden block=volume of wooden
block immersed in water $=20 cm 2$
Density of water $=\rho_{\text {wetr }}=1 gcm \cdot$
Density of wooden block $=\rho_{\text {wasd }}=$ ?
By law of floatation:
Volume of wooden block x Density of wood=Volume of water displaced $x$ Density of water
$
\begin{array}{l}
25 \times \rho_{\text {wood }}=20 \times 1 \\
\rho_{\text {wood }}=\frac{20}{25}=0.8 g cm^{-3}
\end{array}
$
Weight of wooden block $V \rho_{\text {wood }} g$
$
=25 \times 0.8 \times g=20 gf
$
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Question 23 Marks
A body of mass ‘m’ is floating in a liquid of density ‘p’(1) what is the apparent weight of body?
(2) what is the loss of weight of body?
Answer
Mass of body $=m$
Density of liquid $=\rho$
(1) Apparent weight of body = Weight of body in air - Weight of liquid displaces by body.
When a body floats in the liquid, then weight of the body in a liquid is equal to weight of liquid displaced by the body.
=> Apparent weight of body $=0$
(2) Loss in weight of body is equal to the weight of liquid displaced by the body.
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Question 33 Marks
(a) When a piece of ice floating in water melts, the level of water inside the glass remains same. Explain.
(b) An inflated balloon is placed inside a big glass jar which is connected to an evacuating pump. What will you observe when the evacuating pump starts working? Give a reason for your answer.
Answer
(a) A piece of ice displaces an amount of water equal to its own weight. Volume of
water displaced is equal to volume of submerged part of ice cube. When ice cube melts, its volume decreases and gets occupied in that volume of water which is displaced by it. As a result, level of water inside the glass remains same when piece of ice (ice cube) melts.
(b) When evacuating pump starts working, pressure inside the glass jar reduces. As the pressure inside the balloon is more than pressure outside the balloon inside the glass jar, so balloon with burst.
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Question 43 Marks
A cube of ice whose side is 4.0 cm is allowed to melt. The volume of water formed is found to be $58.24 cm^2$. Find the density of ice.
Answer
Side of ice cube $= l=4 cm$
Volume of ice cube $=V=\beta=(4)^3=64 cm^3$
Volume of water $= V =5824 cm^3$
Density of ice cube $=\rho_1=$ ?
Density of water $=\rho_w=1$ gem $^{-3}$
By law of floatation:
Volume of ice cube $x$ Density of ice=Volume of water $x$ Density of water
$
64 \times \rho_i=58.24 \times 1
$
$
\begin{array}{l}
\rho_i=\frac{58.24}{64} \\
\rho_i=0.91 g cm^{-3}
\end{array}$
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Question 53 Marks
A test tube loaded with lead shots weighs 25 gf and floats upto the mark X in water.When the test tube is made to float in brine solution, it needs 5 gf more of lead shots to float upto level X. Find the relative density of brine solution.
Answer
When test tube floats is water :
Weight of test tube $=25$ gf
By law of floatation
Weight of water displaced = Weight of test tube $=25 gf$
When test tube floats in brine solution, it needs 5 gf more of lead shots to float upto
same level as in water.
Weight of test tube $=25+5=30 gf$
By law of floatation:
Weight of brine solution displaced = Weight of test tube $=30 gf As$, volume of brine
solution displaced = Volume of water displaced
$
\begin{array}{l}
\therefore \text { R.D. of brine solution }=\frac{\text { Weight of brine solution displaced }}{\text { Weight of water displaced }} \\
\quad=\frac{30}{25}=1.2
\end{array}
$
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Question 63 Marks
(a) A body dipped in a liquid experiences an upthrust. State the factors on which the upthrust depends.
(b) While floating, is the weight of body greater than, equal to or less than upthrust?
Answer
(a) Factors on which upthrust depends are:
1. Volume of body immersed in fluid.
Upthrust is maximum when body completely immersed in the fluid.
2. Density of the fluid.
Upthrust a density of fluid
Larger the density of the fluid, large will be the upthrust acting on the body.
(b) When the body floats then weight of the body is equal to the upthrust acting on the body.
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Question 73 Marks
A solid of density $5000 kgm ^2$ weighs 0.5 kgf in air. It is completely immersed in a liquid of density $800 kgm ^{-2}$. Calculate the apparent weight of the solid in liquid.
Answer
Density of solid, $d _{ s }=5000 kg m ^{-3}$
Weight of body in air $=0.5 kgf$
$
mg=0.5 kgf
$
$\therefore m -0.5 kg$
Volume of solid $V =\frac{m}{d_s}=\frac{0.5}{5000}=\frac{1}{10000} m^3$
$\therefore$ Vol. of water displaced $=\frac{1}{10000} m$
Density of water $=800 kg m ^{-3}$
$\therefore$ Mass of water displaced $= V \times D$
$
=\frac{1}{10000} \times 800=\frac{8}{100} kg
$
Wt. of water $=0.08 kg$
(1) Apparent weight of the solid in water $=0.5-0.08$
$=0.42 kgf$
(2) Apparent weight of body in liquid of density $800 kg m ^3$ is zero.
 Density of solid is less than density of liquid i.e. upthrust is more than weight of body.
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Question 83 Marks
(a) A body whose volume is 100 cm weighs 1 kgf in air. Find its weight in water.
(b) Why is it easier to swim in sea water than in river water?
Answer
(a) Volume of body $= V =100 cm^3=10^{-4} m^3$
Weight of body in air = 1 kgf
Density of water $= p _{ w }=1000 kgm { }^3$
We know volume of water displaced = Volume of body
$
=V=10^{-4}  m^3
$
Upthrust = Weight of water displaced by body = Vp.g
$
=10^{-4}  \times 1000 \times g .$
$=10^{-1}  kgf=0.1 kgf
$
Weight of body in water=Weight of body in air - Upthrust
$
=1-0.1=0.90 kgf
$
(b) With smaller portion of man's body submerged in sea water, the wt. of sea water displaced is equal to the total weight of body.
While to displace the same weight of river water, a larger portion of the body will have to be submerged in water. It is easier for man to swim in sea water.
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Question 93 Marks
(a) State the principle of floatation.
(b) The mass of a block made of certain material is 1.35 kg and its volume is $1.5 \times 10$ ${ }^3 m^3$.
1. Find the density of block.
2. Will this block float or sink? Give reasons for your answer.
Answer
(a) PRINCIPLE OF FLOATATION : "When a solid is floating in a fluid, the weight of whole solid acting vertically downward at its CENTRE OF GRAVITY, is equal to the weight of fluid displaced by the IMMERSED part of solid acting upward, at its CENTRE OF BUOYANCY or at the centre of the BULK OF LIQUID displaced."
OR
"The weight of a floating body is equal to the weight of the liquid displaced by its SUBMERGED part."
(b) Mass of block $= m =1.35 kg$
Volume of block $= V =1.5 \times 10^{-3} m^{3}$
$
\text { Density of block }=\frac{m}{V}=\frac{1.35}{1.5 \times 10^{-3}}=900 kgm^{-3}
$
(1)
(2) Density of block ( $900 kgm ^{-3}$ ) is less than density of water ($1000 kgm^{-3}$)
$\therefore$ Block will float in water
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Question 103 Marks
A weather forecasting plastic balloon of volume $15 m^3$ contains hydrogen of density $0.09 kgm - s$. The volume of equipment carried by the balloon is negligible compared to its own volume. The mass of the empty balloon is 7.15 . kg . The balloon is floating in air of density $1.3 kgm - s$.
(1) Calculate the mass of hydrogen in balloon.
(2) Calculate the mass of hydrogen and the balloon.
(3) If the mass of equipment is $x kg$, write down the total mass of hydrogen, the balloon and the equipment,
(4) Calculate the mass of air displaced by balloon.
(5) Using the law of floatation, calculate the mass of equipment.
Answer
Volume of Hydrogen $V =15 m^3$
Density of hydrogen $= d =0.09 kg m ^2$
(1) Mass of hydrogen in balloon $= Vd =15 \times 0.09$
$
=1.35 kg
$
The mass of empty balloon alone $=7.15 kg$
(2) The mass of hydrogen and balloon $=1.35+7.15$
$
=8.50 kg
$
Mass of equipment $=x kg$
(3) Total mass of hydrogen + Balloon + Equipment
$
=(8.50+c) kg
$
Density of air $=1.3 kg m ^2$
(4) Mass of air displaced by balloon $= v \times d =15 \times 1.3$
$
=19.5 kg$
(5) According to law of floatation
Total downward wt. = UPTHRUST
$8.5+x=19.5
$
Mass of equipment $x =11 kg$
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Question 113 Marks
A.metal cube of 5 cm edge and relative density 9 is suspended by a thread so as to be completely immersed in a liquid of relative density 1.2 . Find the tension in the thread.
Answer
Volume of metal cube $=(\text { side })^3=5^3=125 cm^3$
Density of cube $=9 g cm^{-3}$
Weight of cube acting downward $= mg = v \times d$
$
F_1 \downarrow=125 \times 9=1125 gf
$
Density of liquid $d -1.2 g cm ^3$
$\therefore$ Upthrust due to liquid in the upward direction.
$\begin{array}{l}
F_2 {\uparrow}=v d_1=125 \times 1.2=150.0 gf \text { Tension in the string = Net downward force }=F_1-F_2 \\
=1125-150=975 gf=9.75 N
\end{array}
$
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Question 123 Marks
A wooden block floats in water with two third of its volume submerged.
(1) Calculate density of wood.
(2) When the same block is placed in oil, three quarter of its volume is immersed in oil.
Calculate the density of oil.
Answer
(1)
Let vol. of wood $= V$
Vol. of wood submerged $v^{\prime}=\frac{2}{3} V$
$
\begin{array}{l}
\frac{d_{s}}{d_{w}}=\frac{v^{\prime}}{v}-\frac{\frac{2}{3} v}{V}=\frac{2}{3} \\
d_{S}=\frac{2}{3} d_{w}=\frac{2}{3} \times 1000=667 kg m^{-3} \\
\text { but } d_{w}=1000 kg m^{-3} \\
\text { Density of wood } d_{s}=667 kg m^{-3}
\end{array}
$
(2) Now $\frac{d_S}{d_{ L }}=\frac{v^{\prime}}{ V }$
$
\begin{aligned}
& \frac{\frac{2000}{3}}{d_{L}}=\frac{\frac{3}{4} v}{V}=\frac{3}{4} \Rightarrow \frac{2000}{3 d_{L}}=\frac{3}{4} \\
\Rightarrow & d_{L}=\frac{2000 \times 4}{3 \times 3}=\frac{8000}{9}=889 kg m^{-3} \\
\therefore \quad & \text { Density of oil }=889 kg m^{-3}
\end{aligned}$
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Question 133 Marks
A test tube weighing 17 gf , floats in alcohol to the level P . When the test tube is made to float in water to the level P, 3 gf of the lead shots are added in it. find the R.D. of alcohol.
Answer
When tube floats in alcohol:
Weight of test tube $=17 gf$
By law of floatation :
Weight of alcohol displaced by test tube $=$ Weight of test tube $=17 gf$
When tube floats in water :
When test tube is made to float in water to the same level, as in alcohol then 3 g lead stones are added in it.
$\therefore$ Weight of test tube $=17 gf +3 gf =20 gf$
Weight of water displaced by test tube $=20 gf$
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{17}{20}=0.85
\end{array}
$
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Question 143 Marks
An iceberg floats in sea water of density $1.17 g cm { }^3$, such that $2 / 9$ of its volume is above sea water. Find the density of iceberg.
Answer
Density of sea water $=\rho_w=1.17 g cm ^-3$
Density of solid ice berg $=\rho_i=$ ?
$\frac{2}{9}$ th volume of iceberg is above the sea water
$\therefore$ Volume of iceberg inside water $=V_i=\left(1-\frac{2}{9}\right) V$
Where $V =$ Total volume of iceberg
$\Rightarrow V_i=\frac{9-2}{9} v=\frac{7}{9} V
$
$\Rightarrow$ Volume of sea water displaced by immersed partof the iceberg
$
=V_w=\frac{7}{9} V
$
By law of floatation :
Weight of iceberg $=$ Weight of sea water displaced by iceberg
$\begin{aligned}
& V_i \times \rho_i \times g=V_w \times \rho_w \times g \\
& V \times \rho_i=\frac{7}{9} V \times 1.17 \\
\Rightarrow & \rho_i=\frac{7}{9} \times 1.17=0.91 gcm^{-3} \\
\Rightarrow & \text { Density of iceberg }=0.91 g cm^{-3}
\end{aligned}$
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Question 153 Marks
A cylinder made of copper and aluminium floats in mercury of density $13.6 gem ^{-3}$, such that 0.26 th part of it is below mercury. Find the density of solid.
Answer
Density of mercury $=\rho_{Hg}=13.6 g . cm ^2$
Density of solid cylinder $=\rho_{\text {Solid  }}=$ ?
0.26 th part of the cylinder is below mercury
Let $V _{ Solid }=$ Volume of solid cylinder
Volume of mercury displaced by immersed part of the solid cylinder
$
=V_{Hg}=0.26 V_{\text {Solid }}
$
By law of floatation :
Weight of the solid cylinder = Weight of mercury displaced by immersed part of solid cylinder
$
\begin{array}{l}
V_{\text {solid }} \times \rho_{\text {solid }} \times g=V_{Hg} \times \rho_{H_8} \times g \\
V_{\text {solid }} \times \rho_{\text {setid }}=0.26 V_{\text {solid }} \times 13.6 \\
\\
\rho_{\text {solid }}=0.26 \times 13.6 \\
\rho_{\text {solid }}=3.536 gcm^{-3}
\end{array}
$
So, density of solid $=3.536 g cm ^{-3}$
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Question 163 Marks
A solid lighter than water is found to weigh 7.5 gf in air. When tied to a sinker the combination is found to weigh .If the sinker alone weighs 72.5 gf in water, find R.D. of solid.
Answer
Weight of solid in air $=7.5 gf$
Weight of sinker in water $=72.5 gf$
Wt. of sinker in water $+W t$. of solid in air
$
=72.5+7.5=80.0 gf
$
Wt. of solid in water + Wt. of sinker in water $=62.5 gf .$. (2)
Subtract eq. (1) from eq. (2)
Wt. of solid in air - Wt . of solid in water $=80-62.5=17.5 gf$.
$
\begin{array}{l}
\therefore \text { R.D. of solid }=\frac{\text { Weight of solid in air }}{\text { Wt. of solid in air }- \text { Wt. of solid in water }} \\
\quad=\frac{7.5}{17.5}=0.428
\end{array}
$
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Question 173 Marks
A sinker is found to weigh 56.7 gf in water. When the sinker is tied to a cork of weight 6 gf , the combination is found to weigh 40.5 gf in water. Calculate R.D. of cork.
Answer
Weight of sinker in water $=56.7 gf$
Weight of cork $=6 gf$
Wt. of sinker in water + Wt. of cork in air
$
=56.7+6=62.7 gf
$
Wt. of cork in water + Wt. of sinker in water $=40.5 gf . . .(2)$
Subtract eq. (1) from eq. (2)
Wt. of cork in air - Wt. of cork in water $=62.7-40.5=22.2 gf$
$
\begin{array}{l}
\therefore \text { R.D. of cork }=\frac{\text { Weight of cork in air }}{\text { wt. of cork in air }- \text { wt. of cork in water }} \\
\quad=\frac{6}{22.2}=0.27
\end{array}$
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Question 183 Marks
A solid of R.D. 4.2 is found to weigh 0.200 kgf in air. Find its apparent weight in water.
Answer
Relative density of solid = R.D. $=4.2$
Weight of solid in air $=W=0.200 kgf$
$
\text { R.D. }=\frac{\text { Weight of solid in air }}{\text { wt. of solid in air }-w t . \text { of solid in water }}
$
Also, wt. of solid in air - Wt. of solid in water = Upthrust
$
\begin{array}{l}
\text { R.D. }=\frac{\text { Wt. of solid in air }}{\text { Upthrust }} \\
4.2=\frac{0.200}{\text { Upthrust }}
\end{array}
$
$
\text { Upthrust }=\frac{0.200}{4.2}=0.0476 kgf
$
So, apparent wt. of solid in water $=w t$. of solid in air - Upthrust
$
=0.200-0.0476=0.1524 \simeq 0.15 kgf
$
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Question 193 Marks
A solid of R.D. $=2.5$ is found to weigh 0.120 kgf in water. Find the wt. of solid in air.
Answer
Relative density of solid = R.D. $=2.5$
Weight of solid in water $=W^{\prime}=0.120 kgf$
Weight of solid in air $= W =$ ?
$
\begin{array}{l}
\text { R.D. }=\frac{\text { Weight of solid in air }}{\text { wt. of solid in air }-wt . \text { of solid in water }} \\
\text { R.D. }=\frac{W}{W-W^{\prime}}
\end{array}
$
$
2.5=\frac{W}{W-0.120}
$
$
2.5 W-2.5 \times 0.120=W
$
$
2.5 W-0.3=W
$
$
2.5 W-W=0.3
$
$
1.5 W=0.3
$
$
W=\frac{0.3}{1.5}=0.20 kgf
$
So, weight of solid in air $=0.20 kgf$.
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Question 203 Marks
A solid weighs 0.08 kgf in air and 0.065 kgf in water. Find
(1) R.D. of solid
(2) Density of solid in SI system. [Density of water $=1000 kgm^{3}$]
Answer
Weight of solid in air $=0.08 kgf$
Weight of solid in water $=0.065 kgf$
Density of water $=1000 kgm^{3}$
(1) Relative density (RD) of solid
$=$ Weight of solid in air
wt. of solid in air-wt. of solid in water
$
\begin{array}{l}
\quad=\frac{\text { Weight of solid in air }}{\text { wt. of solid in air }- \text { wt. of solid in water }} \\
\text { R.D. }=\frac{0.08}{0.08-0.065} \\
\text { R.D. }=\frac{0.08}{0.015}=5.3333
\end{array}
$
(2)
$\begin{array}{l}
\text { R.D. of solid }=\frac{\text { Density of solid }}{\text { Density of water }} \\
5.3333=\frac{\text { Density of solid }}{1000} \\
\text { Density of solid }=5.3333 \times 1000=5333.3 kgm^{-3}
\end{array}$
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Question 213 Marks
A stone of density $3000 kgm ^3$ is lying submerged in water of density $1000 kgm ^3$. If the mass of stone in air is 150 kg , calculate the force required to lift the stone. [ $g =10 ms^2$ ]
Answer
Density of stone $=\rho=3000 kgm ^3$
Density of water $=\rho^{\prime}=1000 kgm ^3$
Mass of stone $= m =150 kg$
Acceleration due to gravity $= g =10 ms^{-2}$
$
\begin{array}{l}
\text { Volume of stone }=V=\frac{m}{\rho} \\
V=\frac{150}{3000}=\frac{1}{20}=0.05 m^3
\end{array}
$
Actual weight of stone $= mg =150 \times 10=1500 N$
Volume of water displaced = Volume of stone
$
V=0.05 m^3
$
Mass of water displaced $= m ^{\prime}= V \times p ^{\prime}$
$
m^{\prime}=0.05 \times 1000=50 kg
$
Upthrust $=$ m'g $=50 \times 10=500 N$
Force required to lift the stone
= Actual weight of stone - upthrust
$
=1500-500=1000 N
$
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