Gravitation Class IX Chapter 10

Question - Answer

Q.No. 1. State the universal law of gravitation.

Answer:

The statement of universal law of gravitation is as follows.

“Every object in the universe attracts every other object with a force which is directly proportional to the product of their masses and inversely proportional to the square of the distance between them”.

Q.No. 2. Write the formula to find the magnitude of the gravitational force between the earth and object on the surface of the earth.

Answers:

The formula which is used to find the magnitude of the gravitational force between the earth and object on the surface of the earth is given below.

F = GMm/d²

 

Here,

F denotes gravitational force.

M and m denote mass of the earth and object respectively.

d denotes distance between the earth and the object.

G is the universal gravitational constant.

The value of G is 6.673x10^-11 Nm²/kg².

Q.No. 3. What do you mean by free fall?

Answer:

                   When an object falls towards the earth. There is no change in the direction of the motion of the object. But due to the earth’s attraction, there will be a change in the magnitude of the velocity.Therefore, whenever an object falls towards the earth, an acceleration is involved due to the earth’s gravitational force.

                   This acceleration is called acceleration due to the gravitational force of the earth or acceleration due to gravity. It is denoted by g. The SI unit of g is ms^-2.

                   According to universal law of gravitation, we know that
F = GMm/d² ................. (i)
F = mg ................. (ii)

Now, from (i) and (ii)

mg =

Therefore,

g = GM/d²

Q.No. 4. What do you mean by acceleration due to gravity?

Answer:

                   Whenever an object falls towards to the earth, acceleration is involved due to gravitational force. This is called acceleration due to gravity. This is equal to

g = GM/d²

Here,

G is the gravitational constant and equal to 6.673x10^-11 Nm²/kg².

M is the mass of the earth.

Q.No. 5. What are the difference between the mass of an object and its weight?

Answer:

                   The mass of an object is the measure of inertia. The mass of an object is constant. It does not change from place to place. Therefore, the mass of an object remains same whether on the surface of the earth, moon or outer space.

                   The weight of an object is variable. It changes from place to place because it depends on gravitational force. Therefore, weight of an object is not same whether on the surface of the earth, moon and outer space.

Q.No. 6. Why is the weight of an object on the moon 1/6^th its weight of the earth.

Answer:

                   Let the acceleration due to gravity on the surface of the earth is g and mass is m.

The weight of the object on the surface of the earth = w_e = mg

The weight of the object on the surface of the moon = w_m =mg/6

We know that acceleration due to gravity on the surface of the moon is 1/6^th of earth.

Therefore,

w_m =we/6

Therefore, the weight of an object on the moon is 1/6^th its weight of the earth.

Real Numbers

Class X

Chapter 1

NCERT Text Book Solution
Exercise 1.1

Q.No.1. Use Euclid division Algorithm to find out the HCF of the following numbers.

(i) 135 and 225 (ii) 196 and 38220 (iii) 867 and 255

(i)

Solution:

According to Euclid division Algorithm, we know that

From question, it is clear that

a = 225

b = 135

Here, remainder is zero. Therefore, the step is over and divisor is 45.

Hence, HCF = 45 Ans.

(ii)

Solution:

According to Euclid division Algorithm, we know that

From question, it is clear that

a = 38220

b = 196

38220 = 196x195 + 0

Here, remainder is zero. Therefore, the step is over and divisor is 196.

Hence, HCF = 196 Ans.

(iii)

Solution:

According to Euclid division Algorithm, we know that

From question, it is clear that

a = 867

b = 255

867 = 255x3 + 102

255 = 102x2 + 51

102 = 51x2 + 0

Here, remainder is zero. Therefore, the step is over and divisor is 51.

Hence, HCF = 51 Ans.

Q.No. 2. Show that any positive odd integer is of the form 6q + 1, or 6q + 3, or 6q +5, where q is some integer.

Solution:

According to Euclid division Algorithm, we know that

Here,

b = 6

Since

The possible remainders are 0, 1, 2, 3, 4 and 5.

That is a can be

6q, or 6q + 1, or 6q + 2, or 6q + 3, or 6q + 4, or 6q + 5, where q is quotient. However, since a is odd, a cannot be 6q, or 6q + 2, or 6q + 4 (Since they are divisible by 2)

Therefore, any odd integer is of the form 6q + 1, or 6q + 3, or 6q + 5.

Q.No. 3. An army contingent of 616 members is to march behind an army band of 32 members in a parade. The two groups are to march in the same number of columns. What is the maximum number of columns in which they can march?

Solution:

According to Euclid division Algorithm, we know that

Here,

a = 616

b = 32

616 = 32x19 + 8

32 = 8x4 + 0

Here, remainder is zero. Therefore, the step is over and divisor is 8.

Hence, HCF = 8.

Therefore, maximum number of column is 8 Ans.

Q.No. 4. Use Euclid division lemma to show that thee square of any positive integer is either of the form 3m or 3m + 1 for some integer m.

Solution:

According to Euclid division Algorithm, we know that

Here,

a = x

b = 3

Since

The possible remainders are 0, 1, and 2.

That is x can be

3q, or 3q + 1, or 3q + 2 where q is quotient.

Now, squaring these

9q², or 9q² + 6q + 1, or 9q² + 12q + 4

When these are divided by 3

Form of these will be

3x3q², or 3(3q² + 2q) + 1, or 3(3q² + 4q + 1) + 1

Therefore, they can be written in the form

3m or 3m + 1.

Q.No. 5. Use Euclid division lemma to show that the cube of any positive integer is of the form 9m, 9m + 1 or 9m + 8.

Solution:

According to Euclid division Algorithm, we know that

Here,

a = x

b = 3

Since

The possible remainders are 0, 1, 2,3 , 4, 5, 6, 7, 8.

That is x can be

9q, or 9q + 1, or 9q + 2, 9q + 3, 9q + 4, 9q + 5, 9q + 6, 9q + 7 or 9q + 8 where q is quotient.

Now, cubing these

Therefore, cube of these can be written in the form

Cube of 9q = 729q³

Form = 9x81q³ = 9m

Cube of 9q + 1= 729q³ + 243q² + 27q + 1

Form = 9(81xq³ + 27q² + 3q) + 1 = 9m + 1

and so on .....................

9m or 9m + 1.

Projectile Motion

Class XI

Derivation of Projectile Equation

An object that is in flight after being thrown or project is called a projectile.

Consider about an object that is projected with an angle θ from horizon. The velocity in which the object is projected is v.

Velocity along vertical = vsinθ
Velocity along horizon = vcosθ
Time of flight of the projectile = T
Maximum Height of the projectile = H
Horizontal Range of the projecile = R
Time ascent = Time of decent = t
Time of flight = T = Time of ascent + Time of decent = t + t = 2t

Maximum distance covered by projectile along vertical direction = Y

Maximum distance covered by projectile along horizontal direction =X

Therefore,

Y = vsinθ.T - gT²/2  -------------------- (i)

X = vcosθ.T

T = X/vcosθ

Now putting the value of T

Y = vsinθ.X/vcosθ - gX²/2cos²θ

Υ = X.tanθ - gX²/2cos²θ ------------------------------ (ii)

Τhis is known as equation of Trajectory. This represent the equation of parabola.

Hence the path of the projectile is a parabola.

Maximum height of a projectile (H).

Here time of ascent = t = time of flight/2 = T/2 (Time taken by projectile to attains maximum height)

At maximum height the velocity of the projectile = 0

Now from kinematic equation

0 = vsinθ - gt

t = vsinθ /g

Time of ascent = Time of decent = t = vsinθ /g

Time of flight = T = 2vsinθ /g

Therefore, 

H = vsinθ.t - gt²/2

Now putting the value of t.

H = (vsinθ )² /g - g(vsinθ)²/2g²

H =  (vsinθ )² /2g 

This is the equation of maximum height.

Horizontal Range = R = vcosθ.T

Now putting the value of T

R = vcosθ.2vsinθ /g

R = v²sin2θ/g

Important equations derive from projectile are

1. Υ = X.tanθ - gX²/2cos²θ
2.  t = vsinθ /g
3. T = 2vsinθ /g
4. R = v²sin2θ/g 

Integration By Parts

Chapter 7

Class XII

Exercise 7.6

Q.No. 1. x.sinx

Solution:

From question it is clear that

I = ഽx.sinx.dx

Put First function = x
Second function = sinx
Now, applying integration by parts;
I = - x.cosx + ഽcosx.dx
or, - x.cosx + sinx + C Ans.

Q.No. 2. x.sin3x

Solution:

From question it clear that

I = ഽx.sin3x.dx
Put first function = x
Second function = sin3x
Now applying integration by parts
I = - x.cos3x/3 + ഽcos3x.dx/3
or, -x.cos3x/3 + sin3x/9 + C Ans.

Q.No. 3. x².ex

Solution:

From question it is clear that
 I = ഽx².ex.dx

Put first function = ex
Second function =  x²

Now applying integration by parts
I =  x².ex - ഽ2x.ex.dx
Again applying integration by parts
or,x².ex - 2x.ex +  ഽ2ex.dx
or,x².ex - 2x.ex + 2ex + C
or, ex(x² - 2x + 2) + C Ans.

Q.No. 4. x.logx

Solution:

From question it is clear that
I = ഽx.logx.dx
Put first function = logx
Second function = x

Now, applying integration by parts

I = x².logx/2 - ഽx.dx/2
or, x².logx/2 - x²/4 + C Ans.

Q.No. 5. x.log2x

Solution:

From question it is clear that
I = ഽx.log2x.dx

Put first function = log2x
Second function = x

Now applying integration by parts

I = log2x.x²/2 - ഽx².2/4x.dx
or, log2x.x²/2 - x²/4 + C Ans.

Q.No. 6. x².logx

Solution:

From question it is clear that
I = ഽ x².logx.dx

Put first function = logx
Second function = x²

Now applying integration by parts

I = logx. x³/3 - ഽx²dx/3
or, logx.x³/3 - x³/9 + C Ans.

Q.No. 7. xsin⁻¹x

Solution:

From question it is clear that
I =  ഽxsin⁻¹x.dx

Put first function = sin⁻¹x
Second function = x

Now applying integration by parts

I = sin⁻¹x. x²/2 - ഽx².dx/2√1-x²

or,  sin⁻¹x. x²/2+ഽ(1-x²-1).dx/√1-x²
or,  sin⁻¹x. x²/2+ ഽ√1-x².dx/2 -  ഽdx/2√1-x²
or,  sin⁻¹x. x²/2+ √1-x²/4 + sin⁻¹x./4 - sin⁻¹x./2 + C
or, (2x² - 1)sin⁻¹x./4 + √1-x²/4 +C Ans.

Q.No. 8 x.tan⁻¹x

Solution:

From question it is clear that
I = ഽx.tan⁻¹x

Put first function = tan⁻¹x
Second function = x

Now applying integration by parts

I = x².tan⁻¹x/2 - ഽx².dx/2(1 + x²)
or, x².tan⁻¹x/2 -ഽ(1 + x² - 1).dx/2(1 + x²)
or, x².tan⁻¹x/2 -ഽdx/2 + ഽdx/2(1 + x²)
or, x².tan⁻¹x/2 - x/2 + .tan⁻¹x/2 + C Ans.

Rest questions will be solved latter.

Similar Triangles

Exercise 6.4
Class X
NCERT Textbook

Q.No. 1. Let △ ABC ∼△DEF and there areas be, respectively, 64 cm² and 121 cm². If EF = 15.4 cm, find BC.

Solution:
ar(ABC)/ar(DEF) = (BC/EF)²
Now, putting the value of areas and EF
64/121 = (BC/15.4)²
Therefore,
BC = 8x15.4/11
Hence,
BC = 11.2 cm Ans.

Q.No. 2. Diagonals of trapezium ABCD with AB॥DC intersect each other at the point O. If AB = 2CD, find the ratio of the areas of triangles AOB and COD.

Solution:

Consider in triangles AOB and COD.
<OAB = <OCD     (Alternate angles)
<OBA = <ODC      (Alternate angles)
<AOB = <COD      (Vertically opposite angles)
△AOB∼ △COD    (AAA Similarity)

Therefore,
ar(AOB)/ar(COD) = (AB/CD)²
Now, putting the value of AB.
ar(AOB)/ar(COD) = (2CD/CD)²
ar(AOB)/ar(COD) = 4/1
Required ratio = 4:1 Ans.

Q.No.3. In Fig. 6.46 ABC and DBC are two triangles on the same base BC. If AD intersect BC at O, show that
ar(ABC)/ar(DBC) = AO/DO

Solution:

Draw perpendicular AP and DP on BC.
Consider in triangles APO and DQO.
<APO = <DQO (Each 90⁰)
<AOP = <DOQ  (Vertically opposite angles)

△APO ∼ △DQO
Therefore,
ar(APO)/ar(DQO) = (AP/DQ)² = (AO/DO)² = (PO/QO)² -------------- (i)

Again,
ar(ABC) = BCxAP/2
ar(DBC) = BCxDQ/2
ar(ABC)/ar(DBC) = AP/DQ

Hence from (i),
ar(ABC) = ar(DBC) = AO/DO Proved.

Q.No. 4. If the areas of two similar triangles are equal, prove that they are congruent.

Solution:

Consider about two triangles ABC and PQR.
Therefore,
ar(ABC)/ar(PQR) = (AB/PQ)² = (BC/QR)² = (AC/PR)² ------------------ (i)

Again,
ar(ABC) = ar(PQR)       (Given)        ----------------------  (ii)

Now, from (i) and (ii).

AB = PQ
BC = QR
AC = PR

Hence,
△ABC≅△PQR Proved.

Q.No. 5. D, E and F respectively the mid points of the sides AB, BC and CA of triangle ABC. Find the ratio of the areas of △ DEF and △ABC.

Solution:

D and F are the mid points of AB and CA.
Therefore,
AD/DB = AF/FC
So,
DF ॥ BC
and
DF = BC/2 ------------- (i)

Similarly,
DE = AC/2 ------------- (ii)
and
EF = AB/2 ----------- -- (iii)

Now, from (i), (ii) and (iii).
ADEF, BEFD and CFDE are parallelograms.

Now, consider in triangles ABC and DEF.
<A = <E ---------------- (ADEF is a parallelogram)
<B = <F ---------------- (BEFD is a parallelogram)
<C = <D ----------------- ( CFDE is parallelogram)

Hence,
△ABC ∼ △EFD
ar(ABC)/ar(DEF) = (AB/EF)²
Now, putting the value of EF.
ar(ABC)/ar(DEF) = 4:1

Therefore,
ar(DEF)/ar(ABC) = 1:4 Ans.


Q.No. 6. Prove that the ratio of the areas of two similar triangles is equal to the square of the ratio of their corresponding medians.

Solution:

Consider about two similar triangles ABC and PQR.
AX, BY and CZ are the medians of ABC. PK, QL and RM are medians of PQR.

Since,
ABC and PQR are similar triangles.
ar(ABC)/ar(PQR) = (AB/PQ)² = (BC/QR)² = (AC/PR)² ---------------------- (i)

Since,
AX and PK are medians of ABC and PQR.
Therefore,
Triangles ABX and PQK are similar.
ar(ABX)/ar(PQK) = (AB/PQ)² = (AX/PK)² ---------------------- (ii)

Now, from (i) and (ii).

The ratio of the area of two similar triangles is equal to the square of the ratio of their corresponding medians. Proved.

7. Prove that area of an equilateral triangle described on one side of a square is equal to half the area of the equilateral triangle described on one of its diagonals.

Solution:

Let the side of the square be X.
Diagonals of the square = ✔️2 X

Area of equilateral triangle on diagonal = 2✔️3X²/4 ------- (i)
Area of equilateral triangles on side = ✔️3X²/4 ------ (ii)

Now, from (i) and (ii)

Area of an equilateral triangle described on one side of a square is equal to half the area of the equilateral triangle described on one of its diagonals.

Proved

Areas of Similar Triangles

Class X

Theorem

"The ratios of areas of two triangles is equal to the square of the ratio of their corresponding sides".

Solution:-

Given

△ABC ~ △PQR

Therefore,

AB/PQ = BC/QR = CA/RP

To prove that

ar(ABC)/ar(PQR) = (AB/PQ)² = (BC/QR)² = (CA/RP)²

Construction

Draw AM ⫡ BC and PN ⫡ QR

Proof

ar(ABC) = BCxAM/2 ---------------- (i)

ar(PQR) = QRxPN/2 ----------------- (ii)

Now, from (i) and (ii)

ar(ABC)/ar(PQR) = BCxAM/QRxPN ------------ (iii)

Now, consider in △ABM and △PQN

<B = <Q     (△ABC ~ △PQR)

<M = <N    ( Each 90⁰)

Hence,

△ABM ~ △PQN  (AA similarity)

Therefore,

AM/PN = AB/PQ

Hence,

AM/PN = AB/PQ = BC/QR = CA/RP

Now, from (iii)

ar(ABC)/ar(PQR) = (AB/PQ)² = (BC/QR)² = (CA/RP)² Proved.