1. Relative density of silver is 10.8. The density of water is 103 Kgm-3. What is the density
of silver in SI unit?
Ans: Relative density of silver = 10.8 Relative density of water =
103 Kgm-3
Density of silver = Relative density of silver x Relative density
of water = 10.8 x 103 Kgm-3
2. Is potential energy a vector or a scalar quantity?
Ans: Potential energy is a scalar quantity.
3. (i) Give any two examples of longitudinal waves.(ii) What is
the most essential property of a wave motion and why?
Ans: (a) i. Waves produced in air. ii. When a freely suspended
spring is pulled downwards and released, longitudinal waves are produced.
(b) The most essential property of a wave motion is its frequency.
Frequency of a wave is its inherent characteristic and does not change by the
change in temperature, pressure or change in medium.
4. (a) Calculate the work done in lifting 200 kg of water through
a vertical height 6 meter. (Assuming g = 10m/s²) (b) When an object moves on a
circular path, what is the work done?
Ans: 20.(a) Given, mass of water (m) = 200
kg Height (h) = 6 m 1/2
.·. weight of water (mg) = 200 x 10 N .·. work
done = mg x h = 200 x 10 x 6J =12000 J
(b) Work done is zero, because displacement is perpendicular to
direction of force always.
5.(a) Calculate the power of an engine which can lift 200 Kg of
water to store in a tank at a height of 10 m in 4.9 s. Also express in horse
power. (Given = 9.8m/s2) (b) What type of energy is stored in the spring
of a watch? (c) What is the work done by the tension in the string of a sample pendulum?
Ans (a) work done (W) = m x g x h = 200 kg x 9.8 m/s2 x 10 m
Power = Work done /Time=[200 kg x 9.8 m/s2 x 10 m]/ 4.9 s= 4000 W÷746 HP=5.36 HP
(b) Elastic potential energy. (c) Tension acts
perpendicular to the displacement of the simple pendulum hence work done is
zero.
6. The atomic number of chlorine is 17 and mass number is 35.
a. What would be the electronic configuration of a negatively charged
chloride ion, Cl- ?
b. What would be the atomic number and mass number of Cl- ? c. Define valency and calculate the
valency of Cl- .
Ans: a. Chlorine atom (Cl) has atomic number of 17. It contains 17
protons and 17 electrons.
Chloride ion (Cl- )
is formed when Cl gains one electron. So, Cl- has 18 electrons and 17 protons.
Therefore, the electronic configuration of Cl- = 2, 8, 8
b. Atomic number of Cl- = Number of protons = 17 Mass
number of Cl- will be
the same as Cl i.e. 35
c. Valency is defined as the combining capacity of an atom. For a
non metallic element, it is equal to eight minus the number of electrons
present in the outermost shell.
Here, Cl- has 8 electrons in the outermost shell, therefore
valency of Cl- = 8-8 = 0
7. The relative
atomic mass of Boron is 10.8 u. Calculate the percentage of its isotopes 10B5and 11B5, occurring
in nature.
Ans: Let the percentage of 10B5be
y in the sample. The percentage of 11B5will
be (100-y)
Average Mass = (10) ×y / 100 + 11×(100-y)/100 = 10.8 Þ y = 20
The percentage of its isotopes 10B5and 11B5, occurring
in nature. 20% and 80%
8. Give one example where kinetic energy is transferred from one
object to other?
Ans: In the carom, the player provides the kinetic energy to the
striker by pushing it with the finger. If the striker collides with other
coins, it will slow down dramatically and the coins it collided with will gain
speed as the kinetic energy is transferred on to it.
9. Define Avogadro’s constant. Give its value.
Ans: Avogadro constant is the actual number of particles (atoms,
molecules or ions) present in 1 mole of any substance. Its value is 6.022 x 1023
10. (a) Chlorine occurs in nature in two isotopic forms with
masses 35 u and 37 u in the ratio of 3:1. Calculate the average atomic mass of
chlorine atom on the basis of this data. (b) Give any three uses of three
isotopes.
Ans: a) The isotopes of chlorine are in the ratio 3:1. It means
that the two isotopes are (3/4x100%)75% and(1/4x100%) 25% respectively.
Average atomic mass of chlorine =[35x75]÷100 +
[35x25]÷100 =142/4 =35.5u
b) Three uses of isotopes: (i) An isotope of cobalt is used
in the treatment of cancer. (ii) An isotope of iodine is used in the treatment
of goiter. (iii) An isotope of uranium is used as a fuel in unclear reactors.
11. When a person uses deodorant spray, the other person standing
at a distance would hear the sound of spraying first and the fragrance of spray
would reach him later. Why so?
Ans: The sound of spraying deodorant travels through the
vibrations of air layers so it reaches first. But the fragrance of deodorant
reaches the other person through actual movement of air particles, therefore
takes more time.
12. Calculate the number of molecules of sulphur present in 16 g
of solid sulphur.
Ans: Molecular mass of S8 = 8 x 32 = 256 g
The number of molecules of sulphur present in 16 g = [6.022x1023 /256]x16 moles= 3.76 x 1022 molecules
13. Following observations were taken while determining the
relative density of a liquid.
Weight of the solid in air = 0.100 kgf
Weight
of the solid in liquid = 0.080 kgf
Weight of the solid in water = 0.075 kgf
Calculate: (a) the apparent
loss in weight of solid in liquid (b) the apparent loss in weight of
solid in water
Ans: (a) The apparent loss in weight of solid in liquid = weight
in air - weight in liquid = 0.100 - 0.08 = 0.02 kgf (b) Apparent loss in weight
of solid in water = weight in air - weight in water = 0.100 - 0.075= 0.025 kgf
14. Two children are at opposite ends of an aluminium rod. One
strikes the end of the rod with a stone. Find the ratio of times taken by the
sound wave in air and in aluminium to reach the second child.
Ans: Speed of sound through air: 346 m/s Speed of sound through
aluminium : 6420 m/s
The
ratio of times taken by the sound wave in air and in aluminium=346/L
÷6420/L=346/6420
15.
What is the consequence of two sound waves which arrive at the ear in a time
interval shorter than 0.1 s?
Ans: If
the time interval between direct and the reflected sound is less than 0.1 s, it
mixed with the direct sounds and the human ear will be unable to
distinguish between the two.
16. Two
bodies of masses m1 and m2 (m1>m2) have same kinetic energy. They are
stopped by applying same retarding force. Which body will stop first?
Ans: If
time taken to stop the masses m1 and m2 are t1 and t2 respectively, the
retardation produced in each case is, a1=u1/t1
and a2 = u2/t2
Since
F=ma Þ a=F/m Þ t1=u1m1/F and t2 =u2 m2/F Þ t1/t2 =
m1u1/m2u2------------------------(1)
Given
that KE1 = KE2 Þ 1/2 m1(u1)2 =1/2 m2(u2)2 Þ m1(u1)2 = m2(u2)2 Þ [u1/u2]2 = m2/m1----(ii)
From
(i) and (ii), t1/t2 =Öm1/m2
As
given , m1 > m2 Þt1 >
t2 Þ heavier body will take longer to stop than
the lighter body.
17.
Explain the following: (a) Swimmers are provided with an inflated rubber
jacket/tube. Why?
(b) It
is easier to swim in sea water than in river water. Why?
Ans:
(a)An inflated rubber jacket has low weight and large volume Hence, it
displaces large volume of water. As a result, up thrust due to water increases
and the person remains floating and there is no chance of drowning of the
swimmer in such case.
(b) It
is easier to swim in sea water because density of sea water is more due to its
salty nature. Hence, up thrust acting on the swimmer in sea water is more in
case of sea water than in fresh water. So, it is easier to swim in sea water.
18. A
ship sends out ultrasound produced by transmitter that returns from the sea bed
and detected after 3.42 s. If the speed of ultrasound waves through sea water
is 1530 m/s, what is the distance of the sea bed from the ship?
Ans:
Depth of sea bed d = vt/2 =[1530x3.42]÷2=
2616.3 m
19. (a)
What is the amount of work done: (i) By an electron revolving in a
circular orbit of radius r round a nucleus? (ii) By an electron moving with
half the speed of light in empty space free of all forces?
Ans:
(a) ( i) Here a centripetal force provided by electrostatic force of attraction
acts on the electron towards the centre of orbit but motion is along the
tangent to the circular orbit at ecah point. As force and displacement are in
mutually perpendicular directions at each point, the work done is zero.
(ii)
Here, no force of any sort is acting on the electron, so the work done is zero.
20.
(a)An electric pump is used to pump water from an underground sump to the
overhead tank situated 20 m above. It transfers 2000 kg of water to overhead
tank in 15 minutes. Calculate the power of pump.
(b)
What do you mean by instantaneous power of a device?
Ans:
Power of pump =Total work/Total time=[mgh]/t=[2000x 9.8x 20J]÷ 900t=435.5W
(c) The
instantaneous power of a device at a particular instant of time is defined as
the rate of doing work by the device at that very instant.
21. If
the kinetic energy of the body is increased by 300% then determine the
percentage increase in the momentum.
Ans:
KE=1/2mv2 Þ2mE=m2v2 Þp=Ö2mE
Now E1=
E+300%of E=4E
So, New
P= Ö2mx4E= Ö8mE
The
percentage increase in the momentum= [(Ö8mE -Ö2mE )/Ö2mE)]100%
=(Ö4-1)100%=100%
22. A
stone of mass 2 kg is falling from rest from the top of a steep hill. What will
be its kinetic energy after 5 s?
Ans:
Here mass of stone m = 2 kg, intial velocity of stone u = 0 and time = 5 s
.·.
Velocity of stone after 5 s, v = u + gt = 0 + 9.8 x 5 = 49 m/s
.·.
Kinetic energy of stone KE = 1/2mv2 = 1/2 x 2 x( 49)2 = 2401 J
23. (a)
Explain the work done by the person in the following conditions. (i) When he is
standing at a place holding a suitcase in his hand.(ii) When he is moving
holding the suitcase in his hands.
Ans:
(i) When the person is standing at a place holding the suitcase, so there is no
change in the position of man or suitcase. So, displacement (s) = 0
W = F x s = F x 0 = 0 .
(ii)
When the person is moving holding the suitcase in his hand, he applies force in
upward direction and displacement of suitcase is in forward direction that is
perpendicular to the direction of force applied. .
So, q= 90o .·. W = F x s cosq = F x s. cos 90o = 0 Hence work done on the
suitcase is Zero.
24.
Describe the law of conservation of energy by giving two examples
Ans:
According to law of conservation of energy:
“
Energy remains conserve during its transformation from one form to other” or in
other words “during transformation of energy, energy is neither created nor
destroyed”.
Examples-
When we lift a stone to a vertical height h from earth surface, stone gains a
potential energy equal to mgh, but we lose some amount of energy.
When a
person kicks a ball, it gets some velocity or kinetic energy, the amount of
energy gained by ball is equal to amount of energy lost by man.
25.
A compound was found to have the following percentage composition by mass Zn =
22.65%, S = 11.15%, H = 4.88%, O = 61.32%. The relative molecular mass is
287g/mol. Find the molecular formula of the compound, assuming that all the
hydrogen in the compound is present in water of crystallizations.
Solution:
Zn : S:O:H = 22.6565: 11.1532 : 61.3210: 4.88/1 =
0.3485: 0.3484 :3.833: 4.88
To
obtain an integral ratio, we divide by smallest number here 0.3484
0.3485/0.3484: 0.3484/0.3484 : 03.833/0.3484 :
4.88/0.3484 = 1 : 1 : 11 : 14
∴ Empirical
formula is Zn SO11H14
Let
Molecular formula be (Zn SO11H14)n
RMM for the molecular = n [65 + 32 +
(11x16) +14]= 287n
Given,
The relative molecular mass is 287g Þ 287n =
287Þ n = 1 Þ Molecular formula is Zn SO11H14
26. A source of sound produces 20
compressions and 20 rarefactions in 0.2 sec. The distance between a compression
and the next (consecutive) rarefaction is 50 cm. find the wavelength, frequency
and time period of wave.
Ans: Distance between a compression and
the next (consecutive) rarefaction is half a wavelength
l/2=50 Þl=100cm No. of waves =20
time taken to complete these waves is 0.2 sec.
Frequency =
20/0.2=200/2=100Hz Time Period= 1/n =1/100sec =0.01 sec.
27. A flask contains 4.4gm of CO2 calculate (a) How many moles of CO2 gas does it contains
(b) How many molecules of CO2 gas are present it contain? (c) How
many atoms of oxygen and C are present in given sample?
Ans: 1 mole of CO2 =12+2x16=44gm
(a) No. of moles in 44gm of CO2 gas = 1 mole
No. of moles in 4.4gm of CO2 gas = (1/44)x4.4=0.1 mole
(b) No. of molecules in 44gm of CO2 gas= 6.023x1023 molecules
No. of molecules in 4.4gm of CO2 gas= (6.023x1023 ÷ 44)x 4.4 molecules= 6.023x1022
(c) No. of atoms of oxygen (2 atoms in CO2 ) = 6.023x1022 x 2 = 1.204x1023 atoms
No. of atoms of oxygen (1 atoms in CO2 ) = 6.023x1022 x 1 = 6.023x1022
28. A
mass of 4 kg is dropped from a tower of height 45 m. Calculate
(i) the potential energy possessed by the body when it is at the top of the tower (ii) the kinetic energy possessed by the body when it is at a height of 35 m (iii) the kinetic energy just before hitting the ground (iv) the velocity of the body just before hitting the ground.
(i) the potential energy possessed by the body when it is at the top of the tower (ii) the kinetic energy possessed by the body when it is at a height of 35 m (iii) the kinetic energy just before hitting the ground (iv) the velocity of the body just before hitting the ground.
Ans: Mass of the body (m) = 4 kg Height of the
tower (h) = 45 m Acceleration due to gravity = 9.8m/s2
i) Potential energy possessed by the body when it is at a height of 45 m above the ground
i) Potential energy possessed by the body when it is at a height of 45 m above the ground
= mgh = 4x9.8x45=1764J
ii) Kinetic energy of a moving body when it is at a
height of 35 m from the ground is equal to the kinetic energy possessed by the
body after covering 10 m. First we have to calculate the velocity with
which the body covers 10 m. Using the eq. v2-u2=2gS,
and substituting value u = 0m/s S= 45-35=10m We
get V2 =196m2/s2
KE = 1/2 x 4 x196 = 392 Joule
iii) According to the law of conservation of energy the kinetic energy of the body just before hitting the ground is equal to potential energy of the body at a height of 45m above the ground.Kinetic energy of the body just before hitting the ground = 1764 J
iii) According to the law of conservation of energy the kinetic energy of the body just before hitting the ground is equal to potential energy of the body at a height of 45m above the ground.Kinetic energy of the body just before hitting the ground = 1764 J
(IV) 1764 =1/2 mv2 Þ v= Ö882 = 29.69 m/s
Download links
IX Maths Original Paper with OTBA 2014
IX Social Science Sample Paper with OTBA 2014
Download File
Download File
A helpful guide for students as same as the students need thanks for shearing 9th class sample paper
ReplyDeleteIts easy to learn things in form of facts and figures.Thanks for sharing these facts.
ReplyDeleteMolecular Orbital Theory
it was very helpful and easy thank u sir
ReplyDelete