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Revision Question Bank
Gravitation
1.
Why does the weight of a body vary from poles to equator?
Sol.
The earth being not a perfect sphere, the radius is less at poles than at the equator.
So, g =
2.
GM
varies
R2
Gravitational force acts on all objects in proportion to their masses. Why does a heavy
object not fall faster than a light object?
Sol.
According to the second law of motion,
a
F
m
Gravitational force  m
 g is constant, it does not depend on mass.
3.
Find the expression for the gravitational potential energy of a body of mass ‘m’ at a
height ‘h’.
Sol.
PE = W= Fd= mgh joule [where W=work done].
4.
You buy a bag of sugar of weight w at a place on the equator. You take this to Antarctica.
Would its weight be the same there? If not, would it increase or decrease?
Sol.
No, it increases, since g increases from equator to poles, due to the two reasons
(i) Difference in radius
5.
(ii) Rotation of the earth
If a planet exists with radius double than that of the earth's radius and made of same
material density d, find the new acceleration due to gravity in terms of that on the
surface of the earth.
Sol.
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Acceleration due to gravity on the earth's surface is,
g
GM
R2
or
4
G R 3r
4
g 3 2
 G rR
R
3
g  R
Here density is same, but radius of the planet is double the earth's radius.
g  = 2g (on the earth's surface)
6.
(a) Which is greater, the attraction of the earth for 1 kg of iron or the attraction of 1 kg
iron for the earth? Why?
(b) A boy throws a ball vertically upwards and catches it back in 10s. Calculate (i) the
velocity with which it was thrown up and
(ii) maximum height attained by the ball (take g=10 ms–2)
Sol.
(a) Both have same force as the mass is same.
(b) tT =10 s
(i) Time to thrown up = time to come down At the highest point,
v=0, t=5s
0 = u–gt
 u = gt = 10 × 5 = 50 m/s
(ii) From
v2 =u2+2gh
O2 = u2 – 2gh
u2 50 500
h


 125m
2g
20
20
2

7.
A car falls off a ledge and drops to the ground in 0.5s. Let g=10ms-2, then
(a) What is its speed on striking the ground?
(b) How high is the ledge from the ground?
(c) What is its average speed during 0.5s?
Sol.
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(a) v = 10×0.5 = 5ms–1
(b)Using
v2 =u2 + 2gh,
we get 52 = 02 +2×10×h
h
25
 1.25m
20
(c)Average Speed =
vav=
8.
Total Distance
1.25
Total Time
1.25
= 2.5 ms–1
0.5
Give reasons for the following:
(a) Compare the weight of an object on the surface of earth with its weight on the surface
of moon.
(b) Two objects of different masses are dropped down from top of a tower which one
will reach the ground earlier and why?
Sol.
(a) The design of the ship is in such a way that the volume of liquid displaced can bring
an upthrust equalling the weight of the ship and its contents. So, it floats while an iron
needle sinks (its upthrust is not equal to its weight).
(b) Sharper the knife is, the area is less and so, the pressure exerted for a particular force
is more.
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Chapter Test {Gravitation}
M: Marks: 30
1.
M: Time: 40 Min.
At what height above the surface of the earth, the value of g becomes 64% of its value on
the surface of the earth. Take the radius of the earth = 6400 km.
[3]
Sol.
Let g = acceleration due to gravity at the earth surface.
gh = acceleration due to gravity at height (h)
= 0.64 g
gh 
gR 2E
RE  h
2
where, RE is the radius of earth.
0.64g 
gR 2E
RE  h
2
 0.64  R E  h   R 2E
2
 0.8  R E  h   R E
 0.8h  R E  0.8R E  0.2R E
h
2.
2  6400
 1600km  R E  6400km
8
State universal law of gravitation. Write SI unit of G. The gravitational force between two
objects is 100 N. How should the distance between the objects be changed, so that force
between them becomes 50 N?
[3]
Sol.
According to universal law of gravitation, the force of gravitational attraction between
two objects is directly proportional to the product of their masses and inversely
proportional to the square of the distance between them. If mt and m2 be masses of two
objects separated by a distance r, the gravitational force of attraction between them is
given by
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m1m2
r2
F
or
Gm1m2
r2
where, G is constant and known as the universal constant of gravitation.
SI unit of G is newton (metre)2/(kilogram)2 or Nm2kg–2.
Let for a distance r between two objects of mass m1 and m2, the gravitation force be 100
N.
Then
100=
Gm1m2
r2
F
or
Gm1m2
r2
If the distance be changed to r’, so that force become 50 N, then we have
So, if the distance between two objects is changed to 2 times of its original value then
force between them is reduced from 100 N to son.
3.
To find the height of a bridge over a river, when a stone is dropped freely in the river
from the bridge. The stone takes 2 s to touch the water surface in the river. Calculate the
height of the bridge from the water level (g = 9.8 m/s2).
[2]
Sol.
Initial velocity of the stone, u = 0
Time taken, t = 2s
Acceleration due to gravity, g = 9.8 m/s 2
Height of the bridge, h =?
1
Using equation h = ut + gt2
2
1
2
= 0  2  9.8  2
2
=
1
 9.8  4  19.6m
2
Thus, the height of bridge above the water level is 19.6 m.
4.
An object is thrown vertically upwards and rises to height of 80 m. Find
[3]
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(a) the velocity with which the object was thrown upwards and
(b) the time taken by the object to reach the highest point (g =10 ms–2).
Sol.
As per question, maximum height covered by the
object in vertically upward direction, h = 80m,
acceleration of object, a = acceleration due to
gravity in downward direction = – g = –10 ms –2
and final velocity at highest point v = 0.
(i) Let initial velocity of projection be u, then from the relation v2 –u2 = 2as, we have
(0)2–(u)2 =2×(–10)×80
u2 =1600  h = 600 = 40ms–1
(ii) From the relation, v = u + at, the time taken by the object to reach the
highest point
is
t
5.
v  u 0  40

 4s
a
 10
A small sphere of mass 40 kg is being attracted by another small sphere of mass 80 kg
with a force equal to weight of 1/4 th of a milligram, when their centres are 30 cm apart.
Calculate the value of gravitational constant G in SI units.
[3]
Sol.
Here masses, m1 = 5.2 , m2 = 2.4
Gravitational force, F = 23 × 10–12 and we know that G = 6.67 × 10–11 Nm2 kg–2 If
separation between the particles be d, then
F
Gm1m2
Gm1m2
ord 
2
d
F
6.673  1011  5.2  2.4

 19  approx 
2.3  1012
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What must be the separation between a 5.2 kg particle and a 2.4 kg particle for their
gravitational attraction to have a magnitude of 23 × 10 –12 N.
[3]
Sol.
Here mass of first sphere, m1 =40
Mass of second sphere, m2 = 80 kg
and distance between the centres of two spheres
d=30 cm = 0.3m
Gravitational force of attraction
1
F = weight of th of a milligram
2
1
= weight of ×10–6 kg
4
=
1
×10–4 ×9.8
4
From,
F= G
m1m2
d2
 Gravitational constant,
2
1
 106  9.8   0.3
Fd2
G
4
m1m2
40  80
= 6.89 × 10–11 Nm2 kg –2
7.
A ball thrown up vertically returns to the thrower after 6 s. Find
[3]
(a) velocity with which it was thrown up.
(b) the maximum height it reaches.
(c) its position after 4 s.
Sol.
Let the ball was thrown vertically upwards with an initial velocity u. As ball returns to
6
the thrower after 6 s, it takes = 3 s to reach the highest point 2
2
h where its velocity, v = 0
(a)Using the relation v = u + at, we have
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0 = u + (–9.8)×3
[ Acceleration
a = – g = – 9.8 ms –2 for upward motion]
 u = 9.8 × 3 = 29.4 ms–1
(b)Maximum height reached h can be calculated by using the relation
v2 –u2 =2as
(O)2 – (29.4)2 = 2 ×(–9.8) × h
 h=
29.4  29.4
= 44.1m
2  9.8
(c) Let after 4 s from start, the ball be at a height h' from ground. Out of 4 s, ball takes 3 s
to reach the maximum height and in remaining Is, it falls downward covering a distance,
s,
s=ut +
1
at
2
= 0 ×1 +
1
× (9.8) × (1)2 = 4.9 m
2
 Height of ball from the ground
h' = h–s=44.1–4.9=39.2 m
8.
(a) Distinguish between G and g.
[4]
(b) What is the effect of shape of earth on value of g?
Sol.
(a) The differences between gravitational constant (G) and acceleration due to gravity
(g) are
Gravitational constant
(G)
(i)
Acceleration due to
gravity (g)
It is defined as the
It is defined as the
force of attraction acceleration of an
between two object freely falling
objects of unit
mass each
under the action of
force of gravity,
separated by unit
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distance.
(ii) It is an universal It is a constant at
constant and its • a given place. Its
value is
value changes
6.673 x10-11 from place to
22
place. Mean value
Nm kg
' of g on surface of
earth is 9.8 ms~2.
(b) The earth is not a perfect sphere. The radius of earth increases from poles to the
equator. Hence, as per relation g =
GM
, value of g is greater at the poles and lesser at the
R2
equator.
9.
Two objects of masses m1 and m2, when separated by a distance d, exerts a force F on
each other. What happens when
[4]
(a) value of mass of first is doubled?
(b) masses of both objects are doubled?
(c) masses are brought so closer that distance between them becomes d/2?
(d) the space between the two objects has no air, i.e., it is complete vacuum?
Sol.
We know that initial value of gravitational force between two objects,
F=
Gm1m2
d2
(a) If the mass of first object is doubled, then,
F’  G 2ml  m2
d2
2
Gm1m2
 2F
d2
Hence, the force is doubled.
(b) If the masses of both objects are doubled, then
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F' 
G 2m1 2m2  4Gm1m2

 4F
d2
d2
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Hence, the force is quadrupled.
(c) If distance between the objects is reduced to one
half of its original value, then
F' 
Gm 1m2
 d / 2
2

4Gm1m2
 4F
d2
hence, the force is quadrupled.
(d) There is no effect on force when air medium is removed.
10. Planet Mars has radius one-half of the Earth and mass 1/9th of the Earth. Find the value
of g on the surface of Mars. Given that value of g on the surface of Earth is 9.81 ms–2.
[2]
Sol.
It is given that R =
RE
M
and M = E where, RE and ME are the radius and mass of Earth
2
9
respectively,
Value of g on Earth gE =9.81ms–2 =
Then g =
GM
R 
2

G  ME / 9
2
 RE 
 2 
 

GME
R 2e
4GME
9R 2E
4
4
 g E   9.81 = 4.36 ms–2
3
3
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Improvement in Food Resources
1.
What are macronutrients?
Sol.
The essential elements utilised by plants in large quantities are called macronutrients.
The four essential nutrients that form the macronutrients are
nitrogen, potassium, calcium and sulphur.
2.
Name a chemical fertilizer, which can supply potassium as well as nitrogen.
Sol.
NPK
3.
Jersey and Sahiwal are high milk-yielding breeds of an animal. Name the animal and give
another example of high milk-yielding breed of this animal.
Sol.
Cow, Holstein-Friesian.
4.
Define biotic factors. Name a few biotic factors, which damage the food material during
storage.
Sol.
The living organisms and their products which influence the agriculture and food are
termed as biotic factors.
Examples
(i) Rodents, birds, animals, (ii) Enzymes present in the food material.
5.
Write two advantages of the use of manure over fertilizer.
Sol.
Two advantages of the use of manure over fertilisers are as follows
(i) Use of manures protect our environment from chemicals such as fertilisers.
(ii) Use of manures help in recycling the biological wastes, i.e., animal excreta and plant
wastes, thereby manures help in enriching the soil with various organic matter and
nutrients increasing soil fertility.
6.
What is pasturage? How is it related to honey production?
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Sol.
The flora found around apiary to collect honey and pollen grains is called pasturage. The
value or quality of honey depends upon the pasturage. The adequate quantity of
pasturage or flora determines the quality as well as taste of honey, since the pollen
grains and nectar serve as protein food for bees.
7.
Why has improving crop yields become important these days? List the major groups of
activities for improving crop yield. Which one of these activities is most important and
why?
Sol.
India is a very populous country and its population is still growing. Requirement of food
is also increasing every year to feed this growing population. Additional farming land is
not available in the country to increase production. It is
therefore necessary to increase crop yield to meet the growing demand for food.
The major groups of activities for improving crop yield are
(i) crop variety improvement
(ii) crop production improvement
(iii) crop protection management
Of these, crop variety improvement is most important because the improved variety is
resistant to pests, high yielding and short-lived.
8.
(a) What is the term used for the scientific management of livestock?
(b) What do you mean by the term 'apiary'?
(c)Mention any two desirable traits for which cross-breeding programmes between
Indian and foreign breeds are undertaken in poultry farming.
Sol.
(a) Animal husbandry is the scientific management of animal livestock.
(b) Apiary The commercial production of honeybee farms are established, which are
known as apiary.
(c) Two desirable traits in which cross-breeding between Indian and foreign breeds are
undertaken in poultry farming are
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(i) Number and quality of chicks (ii) Tolerance to high temperature
Chapter Test
Improvement in Food Resources
M: Marks: 30
1.
M: Time: 40 Min.
Name two exotic or foreign and two local breeds of cows selected for long lactation
period. What management practices are common in dairy and poultry farming (Give
any two)?
Sol.
Name two exotic or foreign and two local breeds of cows selected for long lactation
period. What management practices are common in dairy and poultry farming (Give
any two)?
2.
Differentiate between capture fishery and culture fishery. Explain the advantage of
composite fish culture with example. State the problems faced on adopting composite
fish culture and also the solution found to solve it.
Sol.
Differentiate between capture fishery and culture fishery. Explain the advantage of
composite fish culture with example. State the problems faced on adopting composite
fish culture and also the solution found to solve it.
3.
Give technical names for the following with one example
(a) Chemicals used to kill insects.
(b) Chemicals used to kill fungi.
(c) Chemicals used to kill weed.
Give one example of each.
Sol.
By considering following points, we can identify the slide is of striated muscle
(i) Striated muscles show alternate light and dark bands.
(ii) Muscle fibres are unbranched and cylindrical,
(iii) They are multinucleated
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Answer the following questions
(a) Why is white leghorn considered a highly reputed breed of poultry?
(b) What are the main causes of disease in fishes?
(c) How does water pollution affect fish production?
(d) What is the main characteristic of a healthy animal ?
(e) Mention the characteristic of Italian bees.
Sol.
(a) White Leghorn is most popular because of its small size and long legs. This breed
requires less feed for its maintenance.
(b) The main causes of diseases in fishes are viruses and bacteria.
(c) Due to water pollution, oxygen level of waterfalls and pH of water also changes
resulting into death of fishes of all types, thus polluted water adversely affects fish
production.
(d) A healthy animal feeds regularly and has a normal posture.
(e) Characteristics of Italian bees are
(i) They have high honey collection capacity,
(ii) They sting somewhat less.
(iii) They stay in a given bee hive for a long period and breed very well.
5.
How can crop variety improvement methods help farmers facing repeated crop failure?
Describe three factors for which they could do crop improvement, which is the most
common method of obtaining improved variety of crops. Explain briefly.
Sol.
Crop variety improvement basically focuses on finding a crop variety that can provide a
good yield.
Factors for which crop improvement can be done are
(i) Higher Yield Every farmer do agriculture, so that he gets a good yield and he becomes
economically stable. Crops need to be improved, so that they provide good
productivity per acre.
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(ii) Improved Quality Both quantity, quality should also be of almost importance for crop
improvement, e.g., baking quality in wheat, protein quality in pulses, oil quality in oil
seeds should be given consideration, while selecting an improved variety.
(iii) Wider Adaptability Varieties should be developed having wide adaptability, so that
they will stabilise their growth in different environmental conditions.
There are two ways of getting a variety with desirable characteristics
(i) Genetic manipulation Here a gene is introduced that provides the desirable
characters.
(ii) Hybridisation The most common method of obtaining improved variety of crops is
hybridisation. Hybridisation is the cross between genetically dissimilar plants
It can be of three types
(i) Intervarietal Cross between two different varieties.
(ii) Interespecific Cross between two different species of the same genus.
(iii) Intergeneric Cross between two different genera.
6.
Give six useful traits of improved crops.
Sol.
The following two conditions will appear if plant cell is kept in sugar solution
(i) If the sugar solution has higher water potential than the plant cell, water moves into
the plant cell, causing the plant cell to be turgid.
(ii) If the sugar solution has lower water potential then the plant cell, water moves out
from the plant cell, causing the plant cell to lose water and be plasmolysed.
7.
Explain five different factors for which varietal improvement is carried out by the
farmers.
Sol.
(i) Higher Yield To increase the productivity of the crop per acre.
(ii) Improved Quality Quality of crop products varies from crop to crop.
(iii) Biotic and Abiotic Resistances Crop production reduces due to biotic and abiotic
factors. Varieties resistant to these factors can improve crop production.
(iv) Wider Adaptability Developing varieties for wider adaptability will help in
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stabilising the crop production under different environmental conditions.
(v) Change in Maturity Duration The shorter duration of the crop from sowing to
harvesting, the more economical is the variety.
(vi) Desired Agronomic Characters Developing varieties of desired agronomic
characters help to give higher productivity.
8.
(a) How are new varieties of poultry birds with desired traits produced?
(b) Mention any four desirable traits for which new varieties are produced.
(c) List the management practices that are common between dairy and poultry farming.
Sol.
(a) By cross breeding between Indian (Indigenous) and foreign (exotic) poultry birds,
desired traits are produced.
(b)Following are the four Desirable traits for which new varieties are produced
(i) Number and quality of chicks,
(ii) Dwarf broiler parent for commercial chick production.
(iii) Summer adaptation capacity.
(iv) Tolerance to high temperature.
(c)The management practices that common between dairy and poultry farming are
(i) Proper arrangement for housing,
(ii) Proper arrangement for light.
(iii) Proper arrangement of nutrition.
9.
(a) Write a short note on marine fisheries.
(b) Which factors should be taken into consideration for fish culture?
Sol.
(a)India's marine fish area include 7500 km long coastline and deep sea beyond it.
Marine fish are caught using many kinds of fishing nets from fishing boats. The yields are
increased by locating large schools of fish, where large quantities of fishes can be found.
Popular marine fish varieties are pomphret, mackerel, tuna, sardines and Bombay duck.
Large schools are located by using satellites and echo sounders.
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(b)The three important factors to be considered for fish culture are
(i) Topography, i.e., location of pond,
(ii) Water resources and their quality,
(iii) Soil quality.
10. Give any two differences between macronutrients and micronutrients.
Sol.
Differences between macronntrients and micronutrient
S. No. Macronutrient
(i)
Micronutrient
They are utilised by plant
They are utililsed by
in large quantities. plant in small quantities.
(ii)
There are six
microessential
There are seven
microessential nutrients,
nutrients nitrogen, Iron, manganese, boron,
phosphorus, potassium, zinc, copper,
magnesium, sulphur and molybdenum and
calcium.
Chlorine.
11. Answer the following questions
(a) Farmer 'X' planted soyabean + maize + cowpea (lobia) in the same field
simultaneoulsy in a set row pattern. Farmer 'Y' planted cereal crop in one season and
leguminous plant in next season on the same piece of land in preplanned succession.
Name the cropping pattern used by the farmers 'X' and 'Y'.
(b) What are the advantages of different cropping patterns followed by the farmers 'X'
and 'Y'?
(c) Differentiate between mixed cropping and mixed farming.
Sol.
(a) X- Inter cropping, V-Crop rotation.
(b) These cropping systems are beneficial in bisect, pest and weed control, besides
providing nutrients. They reduce the risk and give some insurance against failure of one
of the crops. They give maximum benefit. They ensure maximum
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utilisation of the nutrients supplied.
(c) Mixed cropping is growing of two or more crops simultaneously on the same piece of
land. Mixed farming is a system of farming on a particular farm which includes crop
production, raising of livestock, etc.
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