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Page No 27:

Question 1:

Answer:

Yes. A vector is defined by its magnitude and direction, so a vector can be changed by changing its magnitude and direction. If we rotate it through an angle, its direction changes and we can say that the vector has changed.

Page No 27:

Question 2:

Yes. A vector is defined by its magnitude and direction, so a vector can be changed by changing its magnitude and direction. If we rotate it through an angle, its direction changes and we can say that the vector has changed.

Answer:

No, it is not possible to obtain zero by adding two vectors of unequal magnitudes.
Example: Let us add two vectors A and B of unequal magnitudes acting in opposite directions. The resultant vector is given by

 R=A2+B2+2ABcosθ

If two vectors are exactly opposite to each other, then
θ=180°, cos180°=-1R=A2+B2-2ABR=A-B2R=A-B or B-A

From the above equation, we can say that the resultant vector is zero (R = 0) when the magnitudes of the vectors A and B are equal (A = B) and both are acting in the opposite directions.

Yes, it is possible to add three vectors of equal magnitudes and get zero.
Lets take three vectors of equal magnitudes A, B and C, given these three vectors make an angle of 120° with each other. Consider the figure below:


Lets examine the components of the three vectors.
Ax=AAy=0Bx=-B cos 60°By=B sin 60°Cx=-C cos 60°Cy=-C sin 60°Here, A=B=CSo, along the x-axis , we have:A-(2A cos 60°)=0, as cos 60°=12 B sin 60°-C sin 60°=0

Hence, proved.

Page No 27:

Question 3:

No, it is not possible to obtain zero by adding two vectors of unequal magnitudes.
Example: Let us add two vectors A and B of unequal magnitudes acting in opposite directions. The resultant vector is given by

 R=A2+B2+2ABcosθ

If two vectors are exactly opposite to each other, then
θ=180°, cos180°=-1R=A2+B2-2ABR=A-B2R=A-B or B-A

From the above equation, we can say that the resultant vector is zero (R = 0) when the magnitudes of the vectors A and B are equal (A = B) and both are acting in the opposite directions.

Yes, it is possible to add three vectors of equal magnitudes and get zero.
Lets take three vectors of equal magnitudes A, B and C, given these three vectors make an angle of 120° with each other. Consider the figure below:


Lets examine the components of the three vectors.
Ax=AAy=0Bx=-B cos 60°By=B sin 60°Cx=-C cos 60°Cy=-C sin 60°Here, A=B=CSo, along the x-axis , we have:A-(2A cos 60°)=0, as cos 60°=12 B sin 60°-C sin 60°=0

Hence, proved.

Answer:

A zero vector has physical significance in physics, as the operations on the zero vector gives us a vector.
For any vector A, assume that
A+0=AA-0=AA×0=0
Again, for any real number λ, we have:
λ0=0
The
significance of a zero vector can be better understood through the following examples:
The displacement vector of a stationary body for a time interval is a zero vector.
Similarly, the velocity vector of the stationary body is a zero vector.
When a ball, thrown upward from the ground, falls to the ground, the displacement vector is a zero vector, which defines the displacement of the ball.


 

Page No 27:

Question 4:

A zero vector has physical significance in physics, as the operations on the zero vector gives us a vector.
For any vector A, assume that
A+0=AA-0=AA×0=0
Again, for any real number λ, we have:
λ0=0
The
significance of a zero vector can be better understood through the following examples:
The displacement vector of a stationary body for a time interval is a zero vector.
Similarly, the velocity vector of the stationary body is a zero vector.
When a ball, thrown upward from the ground, falls to the ground, the displacement vector is a zero vector, which defines the displacement of the ball.


 

Answer:

Yes we can add three unit vectors to get a unit vector.
No, the answer does not change if two unit vectors are along the coordinate axes. Assume three unit vectors i,^-i^ and j^ along the positive x-axis, negative x-axis and positive y-axis, respectively. Consider the figure given below:



The magnitudes of the three unit vectors (i,^-i^ and j^ ) are the same, but their directions are different.
So, the resultant of i^ and -i^ is a zero vector.
Now, j^+0=j^          (Using the property of zero vector)
∴ The resultant of three unit vectors (i,^-i^ and j^)  is a unit vector (j^).



Page No 28:

Question 5:

Yes we can add three unit vectors to get a unit vector.
No, the answer does not change if two unit vectors are along the coordinate axes. Assume three unit vectors i,^-i^ and j^ along the positive x-axis, negative x-axis and positive y-axis, respectively. Consider the figure given below:



The magnitudes of the three unit vectors (i,^-i^ and j^ ) are the same, but their directions are different.
So, the resultant of i^ and -i^ is a zero vector.
Now, j^+0=j^          (Using the property of zero vector)
∴ The resultant of three unit vectors (i,^-i^ and j^)  is a unit vector (j^).

Answer:

Yes, there are physical quantities like electric current and pressure which have magnitudes and directions, but are not considered as vectors because they do not follow vector laws of addition.

Page No 28:

Question 6:

Yes, there are physical quantities like electric current and pressure which have magnitudes and directions, but are not considered as vectors because they do not follow vector laws of addition.

Answer:

Two forces are added using triangle rule, because force is a vector quantity. This statement is more appropriate, because we know that force is a vector quantity and only vectors are added using triangle rule.

Page No 28:

Question 7:

Two forces are added using triangle rule, because force is a vector quantity. This statement is more appropriate, because we know that force is a vector quantity and only vectors are added using triangle rule.

Answer:

No, we cannot add two vectors representing physical quantities of different dimensions. However, we can multiply two vectors representing physical quantities with different dimensions.
Example: Torque, τ=r×F

Page No 28:

Question 8:

No, we cannot add two vectors representing physical quantities of different dimensions. However, we can multiply two vectors representing physical quantities with different dimensions.
Example: Torque, τ=r×F

Answer:

Yes, a vector can have zero components along a line and still have a nonzero magnitude.
Example: Consider a two dimensional vector 2i^+0j^. This vector has zero components along a line lying along the Y-axis and a nonzero component along the X-axis. The magnitude of the vector is also nonzero.
Now, magnitude of 2i^+0j^ = 22+02=2

Page No 28:

Question 9:

Yes, a vector can have zero components along a line and still have a nonzero magnitude.
Example: Consider a two dimensional vector 2i^+0j^. This vector has zero components along a line lying along the Y-axis and a nonzero component along the X-axis. The magnitude of the vector is also nonzero.
Now, magnitude of 2i^+0j^ = 22+02=2

Answer:

The direction of -A is opposite to A. So, if vector A and -A make the angles ε1 and ε2 with the X-axis, respectively, then ε1 is equal to ε2 as shown in the figure:

 
Here, tan ε1 = tan ε2
Because these are alternate angles.
Thus, giving tan ε does not uniquely determine the direction of A.

Page No 28:

Question 10:

The direction of -A is opposite to A. So, if vector A and -A make the angles ε1 and ε2 with the X-axis, respectively, then ε1 is equal to ε2 as shown in the figure:

 
Here, tan ε1 = tan ε2
Because these are alternate angles.
Thus, giving tan ε does not uniquely determine the direction of A.

Answer:

No, the vector sum of the unit vectors i and j is not a unit vector, because the magnitude of the resultant of i and j is not one.

Magnitude of the resultant vector is given by
R = 12+12+cos90°=2

Yes, we can multiply this resultant vector by a scalar number 12 to get a unit vector.

Page No 28:

Question 11:

No, the vector sum of the unit vectors i and j is not a unit vector, because the magnitude of the resultant of i and j is not one.

Magnitude of the resultant vector is given by
R = 12+12+cos90°=2

Yes, we can multiply this resultant vector by a scalar number 12 to get a unit vector.

Answer:

A vector  B such that AB, but A = B are as follows:
(i) B=3i-4j(ii) B=3j+4k(iii) B=3k+4i(iv) B=3j-4k

Page No 28:

Question 12:

A vector  B such that AB, but A = B are as follows:
(i) B=3i-4j(ii) B=3j+4k(iii) B=3k+4i(iv) B=3j-4k

Answer:

No, we cannot have  A×B=A·B with A ≠ 0 and B ≠ 0. This is because the left hand side of the given equation gives a vector quantity, while the right hand side gives a scalar quantity. However, if one of the two vectors is zero, then both the sides will be equal to zero and the relation will be valid.

Page No 28:

Question 13:

No, we cannot have  A×B=A·B with A ≠ 0 and B ≠ 0. This is because the left hand side of the given equation gives a vector quantity, while the right hand side gives a scalar quantity. However, if one of the two vectors is zero, then both the sides will be equal to zero and the relation will be valid.

Answer:

If A×B=0, then both the vectors are either parallel or antiparallel, i.e., the angle between the vectors is either 0° or 180°.
(ABsinθn^=0 sin0°=sin180°=0)

Both the conditions can be satisfied:
(a) A=B, i.e., the two vectors are equal in magnitude and parallel to each other
(b) AB, i.e., the two vectors are unequal in magnitude and parallel or anti parallel to each other

Page No 28:

Question 14:

If A×B=0, then both the vectors are either parallel or antiparallel, i.e., the angle between the vectors is either 0° or 180°.
(ABsinθn^=0 sin0°=sin180°=0)

Both the conditions can be satisfied:
(a) A=B, i.e., the two vectors are equal in magnitude and parallel to each other
(b) AB, i.e., the two vectors are unequal in magnitude and parallel or anti parallel to each other

Answer:

If  A=5i-4j and B=-7·5i+6j, then we have B=kA by putting the value of scalar k as -1.5.
However, we cannot say that BA = k, because a vector cannot be divided by other vectors, as vector division is not possible.

Page No 28:

Question 1:

If  A=5i-4j and B=-7·5i+6j, then we have B=kA by putting the value of scalar k as -1.5.
However, we cannot say that BA = k, because a vector cannot be divided by other vectors, as vector division is not possible.

Answer:

(d) it is slid parallel to itself.

A vector is defined by its magnitude and direction. If we slide it to a parallel position to itself, then none of the given parameters, which define the vector, will change.

Let the magnitude of a displacement vector (A) directed towards the north be 5 metres. If we slide it parallel to itself, then the direction and magnitude will not change.

Page No 28:

Question 2:

(d) it is slid parallel to itself.

A vector is defined by its magnitude and direction. If we slide it to a parallel position to itself, then none of the given parameters, which define the vector, will change.

Let the magnitude of a displacement vector (A) directed towards the north be 5 metres. If we slide it parallel to itself, then the direction and magnitude will not change.

Answer:

(c) 1, 2, 1

1,2 and 1 may represent the magnitudes of three vectors adding to zero.For example one of the vector of length 1 should make an angle of 135 with x axis and the other vector of length 1 makes an angle of 225 with x axis. The third vector of length 2 should lie along x axis.

Page No 28:

Question 3:

(c) 1, 2, 1

1,2 and 1 may represent the magnitudes of three vectors adding to zero.For example one of the vector of length 1 should make an angle of 135 with x axis and the other vector of length 1 makes an angle of 225 with x axis. The third vector of length 2 should lie along x axis.

Answer:

(c) α < β if A > B

The resultant of two vectors is closer to the vector with the greater magnitude.
Thus, α < β if A > B

Page No 28:

Question 4:

(c) α < β if A > B

The resultant of two vectors is closer to the vector with the greater magnitude.
Thus, α < β if A > B

Answer:

(d) None of these.

All the given options are incorrect. The component of a vector may be less than, greater than or equal to its magnitude, depending upon the vector and its components. 

Page No 28:

Question 5:

(d) None of these.

All the given options are incorrect. The component of a vector may be less than, greater than or equal to its magnitude, depending upon the vector and its components. 

Answer:

(a) along the west

The vector product A×B will point towards the west. We can determine this direction using the right hand thumb rule.

Page No 28:

Question 6:

(a) along the west

The vector product A×B will point towards the west. We can determine this direction using the right hand thumb rule.

Answer:

(b) 14.1 cm2

Area of a circle, A = πr2
On putting the values, we get:

A=227×2.12×2.12A=14.1 cm2
 
The rules to determine the number of significant digits says that in the multiplication of two or more numbers, the number of significant digits in the answer should be equal to that of the number with the minimum number of significant digits. Here, 2.12 cm has a minimum of three significant digits. So, the answer must be written in three significant digits.

Page No 28:

Question 1:

(b) 14.1 cm2

Area of a circle, A = πr2
On putting the values, we get:

A=227×2.12×2.12A=14.1 cm2
 
The rules to determine the number of significant digits says that in the multiplication of two or more numbers, the number of significant digits in the answer should be equal to that of the number with the minimum number of significant digits. Here, 2.12 cm has a minimum of three significant digits. So, the answer must be written in three significant digits.

Answer:

(a) the value of a scalar
(c) a vector
(d) the magnitude of a vector

The value of a scalar, a vector and the magnitude of a vector do not depend on a given set of coordinate axes with different orientation. However, components of a vector depend on the orientation of the axes.

Page No 28:

Question 2:

(a) the value of a scalar
(c) a vector
(d) the magnitude of a vector

The value of a scalar, a vector and the magnitude of a vector do not depend on a given set of coordinate axes with different orientation. However, components of a vector depend on the orientation of the axes.

Answer:

(b) It is possible to have  C  <  A  and  C  <  B 

Statements (a), (c) and (d) are incorrect.
Given: C=A+B
Here, the magnitude of the resultant vector may or may not be equal to or less than the magnitudes of A and B or the sum of the magnitudes of both the vectors if the two vectors are in opposite directions.

Page No 28:

Question 3:

(b) It is possible to have  C  <  A  and  C  <  B 

Statements (a), (c) and (d) are incorrect.
Given: C=A+B
Here, the magnitude of the resultant vector may or may not be equal to or less than the magnitudes of A and B or the sum of the magnitudes of both the vectors if the two vectors are in opposite directions.

Answer:

(b) C must be less than  A-B 

Here, we have three vector A, B and C.
A+B2=A2+B2+2A.B       ...(i)A-B2=A2+B2-2A.B         ...(ii)

Subtracting (i) from (ii), we get:
A+B2-A-B2=4A.B

Using the resultant property C=A+B, we get:
C2-A-B2=4A.BC2=A-B2+4A.BC2=A-B2+4ABcos120°

Since cosine is negative in the second quadrant, C must be less than A-B.

 

Page No 28:

Question 4:

(b) C must be less than  A-B 

Here, we have three vector A, B and C.
A+B2=A2+B2+2A.B       ...(i)A-B2=A2+B2-2A.B         ...(ii)

Subtracting (i) from (ii), we get:
A+B2-A-B2=4A.B

Using the resultant property C=A+B, we get:
C2-A-B2=4A.BC2=A-B2+4A.BC2=A-B2+4ABcos120°

Since cosine is negative in the second quadrant, C must be less than A-B.

 

Answer:

(a) is equal to the sum of the x-components of the vectors    
(b) may be smaller than the sum of the magnitudes of the vectors
(d) may be equal to the sum of the magnitudes of the vectors.

The x-component of the resultant of several vectors cannot be greater than the sum of the magnitudes of the vectors.

Page No 28:

Question 5:

(a) is equal to the sum of the x-components of the vectors    
(b) may be smaller than the sum of the magnitudes of the vectors
(d) may be equal to the sum of the magnitudes of the vectors.

The x-component of the resultant of several vectors cannot be greater than the sum of the magnitudes of the vectors.

Answer:

(b) equal to AB
(c) less than AB
(d) equal to zero.

The magnitude of the vector product of two vectors  A  and  B  may be less than or equal to AB, or equal to zero, but cannot be greater than AB.



Page No 29:

Question 1:

(b) equal to AB
(c) less than AB
(d) equal to zero.

The magnitude of the vector product of two vectors  A  and  B  may be less than or equal to AB, or equal to zero, but cannot be greater than AB.

Answer:


From the above figure, we have:
Angle between A and B = 110° − 20° = 90°
A=3 m and B= 4 m
Magnitude of the resultant vector is given by
R=A2+B2+2AB cos θ  =32+42+2×3×4×cos 90°  = 5 m

Let β be the angle between R and A.
β=tan-1 A sin 90°A+B cos 90° =tan-1 4 sin 90°3+4 cos 90° =tan-1 43 =tan-1 1.333 =53°

Now, angle made by the resultant vector with the X-axis = 53° + 20° = 73°
∴ The resultant R  is 5 m and it makes an angle of 73° with the x-axis.

Page No 29:

Question 2:


From the above figure, we have:
Angle between A and B = 110° − 20° = 90°
A=3 m and B= 4 m
Magnitude of the resultant vector is given by
R=A2+B2+2AB cos θ  =32+42+2×3×4×cos 90°  = 5 m

Let β be the angle between R and A.
β=tan-1 A sin 90°A+B cos 90° =tan-1 4 sin 90°3+4 cos 90° =tan-1 43 =tan-1 1.333 =53°

Now, angle made by the resultant vector with the X-axis = 53° + 20° = 73°
∴ The resultant R  is 5 m and it makes an angle of 73° with the x-axis.

Answer:

Angle between A and B, θ = 60° − 30° = 30°
A=B=10 unitsThe magnitude of the resultant vector is given byR=A2+A2+2AAcosθ  =102+102+2×10×10×cos 30°  =200+200 cos 30°  =200+200×32  =19.3 units



Let β be the angle between R and A.
β=tan-1 A sin 30°A+A cos 30°β=tan-1 10 sin 30°10+10 cos 30°β=tan-1 12+3=tan-1 13.372β=tan-1 0.26795=15°

Angle made by the resultant vector with the X-axis = 15° + 30° = 45°

∴ The magnitude of the resultant vector is 17.3 and it makes angle of 45° with the X-axis.

Page No 29:

Question 3:

Angle between A and B, θ = 60° − 30° = 30°
A=B=10 unitsThe magnitude of the resultant vector is given byR=A2+A2+2AAcosθ  =102+102+2×10×10×cos 30°  =200+200 cos 30°  =200+200×32  =19.3 units



Let β be the angle between R and A.
β=tan-1 A sin 30°A+A cos 30°β=tan-1 10 sin 30°10+10 cos 30°β=tan-1 12+3=tan-1 13.372β=tan-1 0.26795=15°

Angle made by the resultant vector with the X-axis = 15° + 30° = 45°

∴ The magnitude of the resultant vector is 17.3 and it makes angle of 45° with the X-axis.

Answer:

First, we will find the components of the vector along the x-axis and y-axis. Then we will find the resultant x and y-components.  
x
-component of A=Acos45°=100 cos 45°=1002 unit
x-component of B = Bcos 135°=-1002
x-component of C = Ccos315° = 100 cos 315°
                            =100 cos 45°=1002
Resultant x-component =1002-1002+1002=1002



Now, y-component of A=100 sin 45°=1002
y-component of B=100 sin 135°=1002
y-component of C = 100 sin 315°=-1002
Resultant y-component=1002+1002-1002=1002

Magnitude of the resultant=10022+10022
                                           =10000=100
Angle made by the resultant vector with the x-axis is given by
tan α=y compx comp       =10021002=1
⇒ α = tan−1 (1) = 45°

∴ The magnitude of the resultant vector is 100 units and it makes an angle of 45° with the x-axis.

Page No 29:

Question 4:

First, we will find the components of the vector along the x-axis and y-axis. Then we will find the resultant x and y-components.  
x
-component of A=Acos45°=100 cos 45°=1002 unit
x-component of B = Bcos 135°=-1002
x-component of C = Ccos315° = 100 cos 315°
                            =100 cos 45°=1002
Resultant x-component =1002-1002+1002=1002



Now, y-component of A=100 sin 45°=1002
y-component of B=100 sin 135°=1002
y-component of C = 100 sin 315°=-1002
Resultant y-component=1002+1002-1002=1002

Magnitude of the resultant=10022+10022
                                           =10000=100
Angle made by the resultant vector with the x-axis is given by
tan α=y compx comp       =10021002=1
⇒ α = tan−1 (1) = 45°

∴ The magnitude of the resultant vector is 100 units and it makes an angle of 45° with the x-axis.

Answer:

Given: a=4i+3j and b=3i+4j

(a) Magnitude of a is given by a=42+32=16+9=5

(b)  Magnitude of b is  given by b=32+42=9+16=5

(c) a+b=(4i^+3j^)+(3i^+4j^)=(7i^+7j^)

∴ Magnitude of vector a+b is given by a+b=49+49=98=72

(d) a-b=4i+3j-3i+4j=i-j

∴ Magnitude of vector a-b is given by a-b=12+-12=2

Page No 29:

Question 5:

Given: a=4i+3j and b=3i+4j

(a) Magnitude of a is given by a=42+32=16+9=5

(b)  Magnitude of b is  given by b=32+42=9+16=5

(c) a+b=(4i^+3j^)+(3i^+4j^)=(7i^+7j^)

∴ Magnitude of vector a+b is given by a+b=49+49=98=72

(d) a-b=4i+3j-3i+4j=i-j

∴ Magnitude of vector a-b is given by a-b=12+-12=2

Answer:

First, let us find the components of the vectors along the x and y-axes. Then we will find the resultant x and y-components.  

x
-component of OA= 2 cos 30°=3 m
x-component of BC =1.5cos120°
                               =-1.52=-7.5 m
x-component of DE = 1cos270°
                               = 1 × 0 = 0 m

y-component of OA = 2 sin 30° = 1
y-component of BC = 1.5 sin 120°
                              =3×1.52=1.3
y-component of DE = 1 sin 270° = −1
x-component of resultant Rx=3-0.75+0=0.98 m

y-component of resultant Ry = 1 + 1.3 − 1 = 1.3 m

Resultant, R=Rx2+Ry2                        = 0.982+1.32                        =0.96+1.69                        =2.65                        =1.6 m

If it makes an angle α with the positive x-axis, then
tan α=y-componentx-component         =1.30.98=1.332
∴ α = tan−1 (1.32)

Page No 29:

Question 6:

First, let us find the components of the vectors along the x and y-axes. Then we will find the resultant x and y-components.  

x
-component of OA= 2 cos 30°=3 m
x-component of BC =1.5cos120°
                               =-1.52=-7.5 m
x-component of DE = 1cos270°
                               = 1 × 0 = 0 m

y-component of OA = 2 sin 30° = 1
y-component of BC = 1.5 sin 120°
                              =3×1.52=1.3
y-component of DE = 1 sin 270° = −1
x-component of resultant Rx=3-0.75+0=0.98 m

y-component of resultant Ry = 1 + 1.3 − 1 = 1.3 m

Resultant, R=Rx2+Ry2                        = 0.982+1.32                        =0.96+1.69                        =2.65                        =1.6 m

If it makes an angle α with the positive x-axis, then
tan α=y-componentx-component         =1.30.98=1.332
∴ α = tan−1 (1.32)

Answer:

Let the two vectors be a and b.
Now, a=3 and b=4

(a) If the resultant vector is 1 unit, then

a2+b2+2.a.b cos θ=132+42+2.3.4 cos θ=1
Squaring both sides, we get:

25+24 cos θ=124 cos θ=-24cos θ=-1θ=180°

Hence, the angle between them is 180°.

(b) If the resultant vector is 5 units, then

a2+b2+2.a.b cos θ=532+42+2.3.4 cos θ=5
Squaring both sides, we get:

    25 + 24 cos θ = 25
⇒ 24 cos θ = 0
⇒ cos θ = 90°

Hence, the angle between them is 90°.

(c) If the resultant vector is 7 units, then

a2+b2+2.a.b cos θ=132+42+2.3.4 cos θ=7
Squaring both sides, we get:

    25 + 24 cos θ = 49,
⇒ 24 cos θ = 24
⇒ cos θ = 1
⇒ θ = cos−1 1 = 0°

Hence, the angle between them is 0°.

Page No 29:

Question 7:

Let the two vectors be a and b.
Now, a=3 and b=4

(a) If the resultant vector is 1 unit, then

a2+b2+2.a.b cos θ=132+42+2.3.4 cos θ=1
Squaring both sides, we get:

25+24 cos θ=124 cos θ=-24cos θ=-1θ=180°

Hence, the angle between them is 180°.

(b) If the resultant vector is 5 units, then

a2+b2+2.a.b cos θ=532+42+2.3.4 cos θ=5
Squaring both sides, we get:

    25 + 24 cos θ = 25
⇒ 24 cos θ = 0
⇒ cos θ = 90°

Hence, the angle between them is 90°.

(c) If the resultant vector is 7 units, then

a2+b2+2.a.b cos θ=132+42+2.3.4 cos θ=7
Squaring both sides, we get:

    25 + 24 cos θ = 49,
⇒ 24 cos θ = 24
⇒ cos θ = 1
⇒ θ = cos−1 1 = 0°

Hence, the angle between them is 0°.

Answer:

The displacement of the car is represented by AD.
AD=2i^+0.5j^+4i^      =6i^+0.5j^

Magnitude of AD is given by

AD=AE2+DE2      =62+0.52      =36+0.25=6.02 km



Now, tan θ=DEAE=112
θ=tan-1 112
Hence, the displacement of the car is 6.02 km along the direction tan-1 112 with positive the x-axis.

Page No 29:

Question 8:

The displacement of the car is represented by AD.
AD=2i^+0.5j^+4i^      =6i^+0.5j^

Magnitude of AD is given by

AD=AE2+DE2      =62+0.52      =36+0.25=6.02 km



Now, tan θ=DEAE=112
θ=tan-1 112
Hence, the displacement of the car is 6.02 km along the direction tan-1 112 with positive the x-axis.

Answer:



Consider that the queen is initially at point A as shown in the figure.
Let AB be x ft.
So, DE = (2 - x) ft
In ∆ABC, we have:
tan θ=x2   ...(i)
Also, in ∆DCE, we have:
tan θ=2-x4  ...(ii)
From (i) and (ii), we get:
x2=2-x422-x=4x4-2x=4x6x=4x=23ft

(a) In ∆ABC, we have:
AC=AB2+BC2
    =232+22=49+4=409=2310 ft

(b) In ∆CDE, we have:
DE=2-23=6-23=43 ft
CD = 4 ft
       CE=CD2+DE2=42+432=4310 ft

(c) In ∆AGE, we have:
AE=AG2+GE2
    =22+22=8+22 ft

Page No 29:

Question 9:



Consider that the queen is initially at point A as shown in the figure.
Let AB be x ft.
So, DE = (2 - x) ft
In ∆ABC, we have:
tan θ=x2   ...(i)
Also, in ∆DCE, we have:
tan θ=2-x4  ...(ii)
From (i) and (ii), we get:
x2=2-x422-x=4x4-2x=4x6x=4x=23ft

(a) In ∆ABC, we have:
AC=AB2+BC2
    =232+22=49+4=409=2310 ft

(b) In ∆CDE, we have:
DE=2-23=6-23=43 ft
CD = 4 ft
       CE=CD2+DE2=42+432=4310 ft

(c) In ∆AGE, we have:
AE=AG2+GE2
    =22+22=8+22 ft

Answer:

Displacement vector of the mosquito, r=7i^+4i^+3k^




(a) Magnitude of displacement=72+42+32
                                                 =74 ft

(b) The components of the displacement vector are 7 ft, 4 ft and 3 ft along the X, Y and Z-axes, respectively.

Page No 29:

Question 10:

Displacement vector of the mosquito, r=7i^+4i^+3k^




(a) Magnitude of displacement=72+42+32
                                                 =74 ft

(b) The components of the displacement vector are 7 ft, 4 ft and 3 ft along the X, Y and Z-axes, respectively.

Answer:

Given: a is a vector of magnitude 4.5 units due north.

Case (a):
3a=3×4.5=13.5
3a is a vector of magnitude 13.5 units due north.

Case (b):
-4a=4×1.5=6 units
-4a is a vector of magnitude 6 units due south.

Page No 29:

Question 11:

Given: a is a vector of magnitude 4.5 units due north.

Case (a):
3a=3×4.5=13.5
3a is a vector of magnitude 13.5 units due north.

Case (b):
-4a=4×1.5=6 units
-4a is a vector of magnitude 6 units due south.

Answer:

Let the two vectors be a= 2 m and b=3 m.
Angle between the vectors, θ = 60°

(a) The scalar product of two vectors is given by a.b=a.b cosθ°.

∴ a.b=a.b cos 60°
          =2×3×12=3 m2

(b) The vector product of two vectors is given by a×b=a a sinθ°.
∴ a×b=a a sin 60°
             =2×3×32=33 m2

Page No 29:

Question 12:

Let the two vectors be a= 2 m and b=3 m.
Angle between the vectors, θ = 60°

(a) The scalar product of two vectors is given by a.b=a.b cosθ°.

∴ a.b=a.b cos 60°
          =2×3×12=3 m2

(b) The vector product of two vectors is given by a×b=a a sinθ°.
∴ a×b=a a sin 60°
             =2×3×32=33 m2

Answer:

According to the polygon law of vector addition, the resultant of these six vectors is zero.
Here, a = b = c = d = e = f (magnitudes), as it is a regular hexagon. A regular polygon has all sides equal to each other.
So, Rx=A cos 0 + A cos π3+A cos 2π3+A cos 3π3+ A cos 4π3+A cos 5π3=0
[As the resultant is zero, the x-component of resultant Rx is zero]
 cos 0 + cos π3+cos 2π3+cos3π3+cos 4π3+cos 5π5=0



Note: Similarly, it can be proven that sin 0+sin π3+sin 2π3+sin 3π3+sin 4π3+sin 5π3=0.

Page No 29:

Question 13:

According to the polygon law of vector addition, the resultant of these six vectors is zero.
Here, a = b = c = d = e = f (magnitudes), as it is a regular hexagon. A regular polygon has all sides equal to each other.
So, Rx=A cos 0 + A cos π3+A cos 2π3+A cos 3π3+ A cos 4π3+A cos 5π3=0
[As the resultant is zero, the x-component of resultant Rx is zero]
 cos 0 + cos π3+cos 2π3+cos3π3+cos 4π3+cos 5π5=0



Note: Similarly, it can be proven that sin 0+sin π3+sin 2π3+sin 3π3+sin 4π3+sin 5π3=0.

Answer:

We have:
a=2i+3j+4kb=3i+4j+5k

Using scalar product, we can find the angle between vectors a and b.
i.e., a.b=ab cos θ
So, θ=cos-1a.bab

=cos-12×3+3×4+4×522+32+42 32+42+52=cos-13829 50=cos-1381450

∴ The required angle is cos-1381450.

Page No 29:

Question 14:

We have:
a=2i+3j+4kb=3i+4j+5k

Using scalar product, we can find the angle between vectors a and b.
i.e., a.b=ab cos θ
So, θ=cos-1a.bab

=cos-12×3+3×4+4×522+32+42 32+42+52=cos-13829 50=cos-1381450

∴ The required angle is cos-1381450.

Answer:

To prove: A.A×B=0

Proof: Vector product is given by A×B= AB sin n^.

 AB sin n^ is a vector which is perpendicular to the plane containing A and B. This implies that it is also perpendicular to A. We know that the dot product of two perpendicular vectors is zero.
A.A×B=0

Hence, proved.

Page No 29:

Question 15:

To prove: A.A×B=0

Proof: Vector product is given by A×B= AB sin n^.

 AB sin n^ is a vector which is perpendicular to the plane containing A and B. This implies that it is also perpendicular to A. We know that the dot product of two perpendicular vectors is zero.
A.A×B=0

Hence, proved.

Answer:

Given: A=2i^+3j^+4k^ and B=4i^+3j^+2k^
The vector product of A×B can be obtained as follows:
A×B=i^j^k^234432         =i^6-12-j^4-16+k^6-12         =-6i^+12j^-6k^

Page No 29:

Question 16:

Given: A=2i^+3j^+4k^ and B=4i^+3j^+2k^
The vector product of A×B can be obtained as follows:
A×B=i^j^k^234432         =i^6-12-j^4-16+k^6-12         =-6i^+12j^-6k^

Answer:

Given: A, B and C are mutually perpendicular. A×B is a vector with its direction perpendicular to the plane containing A and B.



∴ The angle between C and A×B is either 0° or 180°.

i.e., C×A×B=0

However, the converse is not true. For example, if two of the vectors are parallel, then also, C×A×B=0


So, they need not be mutually perpendicular.

Page No 29:

Question 17:

Given: A, B and C are mutually perpendicular. A×B is a vector with its direction perpendicular to the plane containing A and B.



∴ The angle between C and A×B is either 0° or 180°.

i.e., C×A×B=0

However, the converse is not true. For example, if two of the vectors are parallel, then also, C×A×B=0


So, they need not be mutually perpendicular.

Answer:

The particle moves on the straight line XX' at a uniform speed ν.



In POQ, we have:
OQ = OP sin θ

OP×ν=OP ν sin θ n^            =ν OP sin θ n^            =νOQ n^
This product is always equal to the perpendicular distance from point O. Also, the direction of this product remains constant. 
So, irrespective of the the position of the particle, the magnitude and direction of OP×ν remain constant.
OP ×ν is independent of the position P.

Page No 29:

Question 18:

The particle moves on the straight line XX' at a uniform speed ν.



In POQ, we have:
OQ = OP sin θ

OP×ν=OP ν sin θ n^            =ν OP sin θ n^            =νOQ n^
This product is always equal to the perpendicular distance from point O. Also, the direction of this product remains constant. 
So, irrespective of the the position of the particle, the magnitude and direction of OP×ν remain constant.
OP ×ν is independent of the position P.

Answer:

According to the problem, the net electric and magnetic forces on the particle should be zero. 



i.e., F=qE+qν×B=0
E=-ν ×B
So, the direction of ν ×B should be opposite to the direction of E. Hence, ν should be along the positive z-direction.

Again, E = νB sin θ
ν=EB sin θ
For ν to be minimum, θ=90° and, thus, νmin=EB
So, the particle must be projected at a minimum speed of EB along the z-axis.

Page No 29:

Question 19:

According to the problem, the net electric and magnetic forces on the particle should be zero. 



i.e., F=qE+qν×B=0
E=-ν ×B
So, the direction of ν ×B should be opposite to the direction of E. Hence, ν should be along the positive z-direction.

Again, E = νB sin θ
ν=EB sin θ
For ν to be minimum, θ=90° and, thus, νmin=EB
So, the particle must be projected at a minimum speed of EB along the z-axis.

Answer:

To prove: A·B=C·B, but A C
Suppose that A is perpendicular to B; B is along the west direction.
Also, B is perpendicular to C; A and C are along the south and north directions, respectively.


A is perpendicular to B, so there dot or scalar product is zero.
i.e., A·B=ABcosθ=ABcos90°=0
B is perpendicular to C, so there dot or scalar product is zero.
i.e., C·B=CBcosθ=CBcos90°=0

A·B=B·C, but A C

Hence, proved.

Page No 29:

Question 20:

To prove: A·B=C·B, but A C
Suppose that A is perpendicular to B; B is along the west direction.
Also, B is perpendicular to C; A and C are along the south and north directions, respectively.


A is perpendicular to B, so there dot or scalar product is zero.
i.e., A·B=ABcosθ=ABcos90°=0
B is perpendicular to C, so there dot or scalar product is zero.
i.e., C·B=CBcosθ=CBcos90°=0

A·B=B·C, but A C

Hence, proved.

Answer:

Note: Students should draw the graph y = 2x2 on a graph paper for results.

To find a slope at any point, draw a tangent at the point and extend the line to meet the x-axis. Then find tan θ as shown in the figure.



The above can be checked as follows:
Slope = tan θ=dydx         =ddx2x2=4x
Here, x = x-coordinate of the point where the slope is to be measured

 

 

Page No 29:

Question 21:

Note: Students should draw the graph y = 2x2 on a graph paper for results.

To find a slope at any point, draw a tangent at the point and extend the line to meet the x-axis. Then find tan θ as shown in the figure.



The above can be checked as follows:
Slope = tan θ=dydx         =ddx2x2=4x
Here, x = x-coordinate of the point where the slope is to be measured

 

 

Answer:

y = sin x   ...(i)

       

Now, consider a small increment ∆x in x.
Then y + ∆y = sin (x + ∆x)   ...(ii)
Here, ∆y is the small change in y.

Subtracting (ii) from (i), we get:
∆y = sin (x + ∆x) − sin x
     =sin π3+π100-sin π3
       = 0.0157

Page No 29:

Question 22:

y = sin x   ...(i)

       

Now, consider a small increment ∆x in x.
Then y + ∆y = sin (x + ∆x)   ...(ii)
Here, ∆y is the small change in y.

Subtracting (ii) from (i), we get:
∆y = sin (x + ∆x) − sin x
     =sin π3+π100-sin π3
       = 0.0157

Answer:

Given: i = i0et/RC
∴ Rate of change of current =didt
                                           =i0 -1RC e-t/RC
                                          =-i0RC×e-t/RC
On applying the conditions given in the questions, we get:
(a) At t=0, didt=-i0RC×e0=-i0RC
(b) At  t=RC, didt=-i0RC×e-1=-i0RCe
(c) At t=10 RC, dtdi=-i0RC×e10=-i0RCe10

Page No 29:

Question 23:

Given: i = i0et/RC
∴ Rate of change of current =didt
                                           =i0 -1RC e-t/RC
                                          =-i0RC×e-t/RC
On applying the conditions given in the questions, we get:
(a) At t=0, didt=-i0RC×e0=-i0RC
(b) At  t=RC, didt=-i0RC×e-1=-i0RCe
(c) At t=10 RC, dtdi=-i0RC×e10=-i0RCe10

Answer:

Electric current in a discharging R-C circuit is given by the below equation:
i
= i0et/RC   ...(i)
Here, i0 = 2.00 A
R = 6 × 105 Ω
C = 0.0500 × 10−6 F
    = 5 × 10−7 F

On substituting the values of R, C and i0 in equation (i), we get:
i = 2.0 et/0.3   ...(ii)

According to the question, we have:

(a) current at t = 0.3 s
i=2 ×e-1=2e A

(b) rate of change of current at t = 0.3 s
didt=-i0RC·e-t/RC
When t = 0.3 s, we have:

didt=20.30·e-0.30.3=-203e A/s

(c) approximate current at t = 0.31 s
i=2e-0.30.3 =5.83e A approx.



Page No 30:

Question 24:

Electric current in a discharging R-C circuit is given by the below equation:
i
= i0et/RC   ...(i)
Here, i0 = 2.00 A
R = 6 × 105 Ω
C = 0.0500 × 10−6 F
    = 5 × 10−7 F

On substituting the values of R, C and i0 in equation (i), we get:
i = 2.0 et/0.3   ...(ii)

According to the question, we have:

(a) current at t = 0.3 s
i=2 ×e-1=2e A

(b) rate of change of current at t = 0.3 s
didt=-i0RC·e-t/RC
When t = 0.3 s, we have:

didt=20.30·e-0.30.3=-203e A/s

(c) approximate current at t = 0.31 s
i=2e-0.30.3 =5.83e A approx.

Answer:

The given equation of the curve is y = 3x2 + 6x + 7.
          
The area bounded by the curve and the X-axis with coordinates x1 = 5 and x2 = 10 is given by
x1x2 y dx=510 3x2+6x+7dx=3x33+6x22+7x510=1000-125+300-75+70-35=1370-235=1135 sq. units

Page No 30:

Question 25:

The given equation of the curve is y = 3x2 + 6x + 7.
          
The area bounded by the curve and the X-axis with coordinates x1 = 5 and x2 = 10 is given by
x1x2 y dx=510 3x2+6x+7dx=3x33+6x22+7x510=1000-125+300-75+70-35=1370-235=1135 sq. units

Answer:

The given equation of the curve is y = sin x.
     
The required area can found by integrating y w.r.t x within the proper limits.
Area=x1x2 y dx =0π sin x dx       =-cos x0π       =-cos π --cos 0       =1+1=2 sq. unit

Page No 30:

Question 26:

The given equation of the curve is y = sin x.
     
The required area can found by integrating y w.r.t x within the proper limits.
Area=x1x2 y dx =0π sin x dx       =-cos x0π       =-cos π --cos 0       =1+1=2 sq. unit

Answer:

The given function is y = ex.
When x = 0, y = e−0 = 1

When x increases, the value of y decrease. Also, only when x = ∞, y = 0

So, the required area can be determined by integrating the function from 0 to ∞.
Area=0e-x dx           =-e-x0           =-0-1=1 sq. unit

Page No 30:

Question 27:

The given function is y = ex.
When x = 0, y = e−0 = 1

When x increases, the value of y decrease. Also, only when x = ∞, y = 0

So, the required area can be determined by integrating the function from 0 to ∞.
Area=0e-x dx           =-e-x0           =-0-1=1 sq. unit

Answer:

ρ = mass/length = a + bx

So, the SI unit of ρ is kg/m.

(a)
SI unit of a = kg/m
SI unit of b = kg/m2
(From the principle of homogeneity of dimensions)

(b) Let us consider a small element of length dx at a distance x from the origin as shown in the figure given below:


dm = mass of the element
      = ρdx
      = (a + dx) dx
∴ Mass of the rod = ∫ dm
=0L a+bx dx=ax+bx220L=aL+bL22

Page No 30:

Question 28:

ρ = mass/length = a + bx

So, the SI unit of ρ is kg/m.

(a)
SI unit of a = kg/m
SI unit of b = kg/m2
(From the principle of homogeneity of dimensions)

(b) Let us consider a small element of length dx at a distance x from the origin as shown in the figure given below:


dm = mass of the element
      = ρdx
      = (a + dx) dx
∴ Mass of the rod = ∫ dm
=0L a+bx dx=ax+bx220L=aL+bL22

Answer:

According to the question, we have:
dpdt=10 N+2 N/st
Momentum is zero at time, t = 0
Now, dp = [(10 N) + (2 Ns−1)t]dt

On integrating the above equation, we get:

p=010 dp=010 10 dt+010 2t dtp=10t+2t22010p=10×10+100-0 p=100+100=200 kg m/s

Page No 30:

Question 29:

According to the question, we have:
dpdt=10 N+2 N/st
Momentum is zero at time, t = 0
Now, dp = [(10 N) + (2 Ns−1)t]dt

On integrating the above equation, we get:

p=010 dp=010 10 dt+010 2t dtp=10t+2t22010p=10×10+100-0 p=100+100=200 kg m/s

Answer:

Changes in a function of y and the independent variable x are related as follows:
dydx=x2dy=x2 dx
Integrating of both sides, we get:
dy = ∫x2dx
y=x33+c, where c is a constant
y as a function of x is represented by y=x33+c.

Page No 30:

Question 30:

Changes in a function of y and the independent variable x are related as follows:
dydx=x2dy=x2 dx
Integrating of both sides, we get:
dy = ∫x2dx
y=x33+c, where c is a constant
y as a function of x is represented by y=x33+c.

Answer:

(a) 1001
Number of significant digits = 4
(b) 100.1
Number of significant digits = 4
(c) 100.10
Number of significant digits = 5
(d) 0.001001
Number of significant digits = 4

Page No 30:

Question 31:

(a) 1001
Number of significant digits = 4
(b) 100.1
Number of significant digits = 4
(c) 100.10
Number of significant digits = 5
(d) 0.001001
Number of significant digits = 4

Answer:

The metre scale is graduated at every millimetre.
i.e., 1 m = 1000 mm
The minimum number of significant digits may be one (e.g., for measurements like 4 mm and 6 mm) and the maximum number of significant digits may be 4 (e.g., for measurements like 1000 mm). Hence, the number of significant digits may be 1, 2, 3 or 4.

Page No 30:

Question 32:

The metre scale is graduated at every millimetre.
i.e., 1 m = 1000 mm
The minimum number of significant digits may be one (e.g., for measurements like 4 mm and 6 mm) and the maximum number of significant digits may be 4 (e.g., for measurements like 1000 mm). Hence, the number of significant digits may be 1, 2, 3 or 4.

Answer:

(a) In 3472, 7 comes after the digit 4. Its value is greater than 5. So, the next two digits are neglected and 4 is increased by one.
∴ The value becomes 3500.
(b) 84
(c) 2.6
(d) 29

Page No 30:

Question 33:

(a) In 3472, 7 comes after the digit 4. Its value is greater than 5. So, the next two digits are neglected and 4 is increased by one.
∴ The value becomes 3500.
(b) 84
(c) 2.6
(d) 29

Answer:

Length of the cylinder, l = 4.54 cm
Radius of the cylinder, r = 1.75 cm
Volume of the cylinder, V = πr2l
                                   = (π) × (4.54) × (1.75)2



The minimum number of significant digits in a particular term is three. Therefore, the result should have three significant digits, while the other digits should be rounded off.

∴ Volume, V = πr2l
= (3.14) × (1.75) × (1.75) × (4.54)
= 43.6577 cm3
Since the volume is to be rounded off to 3 significant digits, we have:
V = 43.7 cm3

Page No 30:

Question 34:

Length of the cylinder, l = 4.54 cm
Radius of the cylinder, r = 1.75 cm
Volume of the cylinder, V = πr2l
                                   = (π) × (4.54) × (1.75)2



The minimum number of significant digits in a particular term is three. Therefore, the result should have three significant digits, while the other digits should be rounded off.

∴ Volume, V = πr2l
= (3.14) × (1.75) × (1.75) × (4.54)
= 43.6577 cm3
Since the volume is to be rounded off to 3 significant digits, we have:
V = 43.7 cm3

Answer:


Average thickness=2.17+2.17+2.183                            =2.1733 mm

∴ Rounding off to three significant digits, the average thickness becomes 2.17 mm.

Page No 30:

Question 35:


Average thickness=2.17+2.17+2.183                            =2.1733 mm

∴ Rounding off to three significant digits, the average thickness becomes 2.17 mm.

Answer:

Consider the figure shown below:



Actual effective length = (90.0 + 2.13) cm
However, in the measurement 90.0 cm, the number of significant digits is only two.
So, the effective length should contain only two significant digits.
i.e., effective length = 90.0 + 2.13 = 92.1 cm.



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