Home » HC-Verma Solutions for Class 12 » HC Verma Class XII Physics Chapter 2 – Physics and Mathematics

HC Verma Class XII Physics Chapter 2 – Physics and Mathematics


Exercise : Solution of Questions on page Number : 27


Answer: 1
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: 2
No, it is not possible to obtain zero by adding two vectors of unequal magnitudes.
Example: Let us add two vectors  and  of unequal magnitudes acting in opposite directions. The resultant vector is given by

If two vectors are exactly opposite to each other, then
θ=180°, cos180°=-1R=⇒R=⇒R=(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   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  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°=½ ⇒B sin 60°-C sin 60°=0
Hence, proved.


Answer: 3
A zero vector has physical significance in physics, as the operations on the zero vector gives us a vector.
For any vector assume that

Again, for any real number λ, we have:

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: 4
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 î -î and 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 (î,-î and ) are the same, but their directions are different.
So, the resultant of î and -î is a zero vector.
Now,  (Using the property of zero vector)
∴ The resultant of three unit vectors (î ,-î and ) is a unit vector ().


Exercise : Solution of Questions on page Number : 28


Answer: 5
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: 6
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: 7
No, we cannot add two vectors representing physical quantities of different dimensions. However, we can multiply two vectors representing physical quantities with different dimensions.


Answer: 8
Yes, a vector can have zero components along a line and still have a nonzero magnitude.
Example: Consider a two dimensional vector  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 mangnitude of


Answer: 9
The direction of – is opposite to  So, if vector  and – make the angles εx 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 


Answer: 10
No, the vector sum of the unit vectors and  is not a unit vector, because the magnitude of the resultant of  and  is not one.
Magnitude of the resultant vector is given by
R=
Yes, we can multiply this resultant vector by a scalar number 12 to get a unit vector.


Answer: 11
A vector such that, but A = B are as follows:


Answer: 12
No, we cannot have  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: 13
If =0, then both the vectors are either parallel or antiparallel, i.e., the angle between the vectors is either 0° or 180°.

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


Answer: 14
If =5-4 and=-7·5→+6, then we have =k by putting the value of scalar k as -1.5.
However, we cannot say that because a vector cannot be divided by other vectors, as vector division is not possible.


objective I


Answer: 1
(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 ()directed towards the north be 5 metres. If we slide it parallel to itself, then the direction and magnitude will not change.


Answer: 2
(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: 3
(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: 4
(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: 5
(a) along the west
The vector product  will point towards the west. We can determine this direction using the right hand thumb rule.


Answer: 6
(b) 14.1 cm2
Area of a circle, A = πr2
On putting the values, we get:

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.


objective II


Answer: 1
(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: 2
(b) It is possible to have  
Statements (a), (c) and (d) are incorrect.
Given
Here, the magnitude of the resultant vector may or may not be equal to or less than the magnitudes of  and  or the sum of the magnitudes of both the vectors if the two vectors are in opposite directions.


Answer: 3
(b) C must be less than Ι A-B Ι
Here, we have three vector A, B and C
….(ii)
Subtracting (i) from (ii), we get:

Using the resultant property , we get:

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


Answer: 4
(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: 5
(b) equal to AB
(c) less than AB
(d) equal to zero.
The magnitude of the vector product of two vectors Ι Ιand ΙΙmay be less than or equal to AB, or equal to zero, but cannot be greater than AB.


Exercise : Solution of Questions on page Number : 29


Answer: 1

From the above figure, we have:
Angle between  = 110° − 20° = 90°
ΙΙ=3 m and Ι Ι= 4 m
Magnitude of the resultant vector is given by

Let β be the angle between  and .

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


Answer: 2
Angle between  and  , θ = 60° − 30° = 30°
=10 unitsThe magnitude of the resultant vector is given byR
= =

Let β be the angle between and .

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: 3
First, we will find the components of the vector along the x-axis and y-axis. Then we will find the resultant x and

x-component of  =

Resultant x-component =

Now, y-component of
y-component of
y-component of
Resultant y-component=
Magnitude of the resultant=

Angle made by the resultant vector with the x-axis is given by

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


Answer: 4
Given:
(a) Magnitude of  is given by
(b) Magnitude of  is given by

∴ Magnitude of vector  is given by
(d)
∴ Magnitude of vector  is given by


Answer: 5
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
x-component of

x-component of
= 1 × 0 = 0 m
y-component of
y-component of

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

If it makes an angle α with the positive x-axis, then


Answer: 6
Let the two vectors be and.
Now,
(a) If the resultant vector is 1 unit, then

Squaring both sides, we get:
25+24 cos θ=1⇒24 cos θ=-24⇒cos θ=-1⇒θ=180°
(b) If the resultant vector is 5 units, then

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

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: 7
The displacement of the car is represented by .
=2î+0.5 +4î
6î+0.5
Magnitude of  is given by



Hence, the displacement of the car is 6.02 km along the direction  with positive the x-axis.


Answer: 8

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=
Also, in ∆DCE, we have:
tan Θ=
From (i) and (ii), we get:

(a) In ∆ABC, we have:

(b) In ∆CDE, we have:


(c) In ∆AGE, we have:


Answer: 9
Displacement vector of the mosquito,

(a) Magnitude of displacement=

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


Answer: 10
Given:  is a vector of magnitude 4.5 units due north.
Case (a):

∴ 3 is a vector of magnitude 13.5 units due north.
Case (b):

∴-4 is a vector of magnitude 6 units due south.


Answer: 11
Let the two vectors be = 2 m and =3 m.
Angle between the vectors, θ = 60°
(a) The scalar product of two vectors is given by


(b) The vector product of two vectors is given by


Answer: 12
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.

[As the resultant is zero, the x-component of resultant Rx is zero]



Answer: 13
We have:

Using scalar product, we can find the angle between vectors  and.


Answer: 14
To prove:
Proof: Vector product is given by
is a vector which is perpendicular to the plane containing . This implies that it is also perpendicular to  We know that the dot product of two perpendicular vectors is zero.

Hence, proved.


Answer: 15

The vector product of × can be obtained as follows:


Answer: 16
Given: and   are mutually perpendicular.  is a vector with its direction perpendicular to the plane containing .

∴ The angle between  and  is either 0° or 180°.
i.e., ×()=0
However, the converse is not true. For example, if two of the vectors are parallel, then also, ×=0

So, they need not be mutually perpendicular.


Answer: 17
The particle moves on the straight line XX’ at a uniform speed ν.

In ∆POQ, we have:
OQ = OP sin θ

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 remain constant.
is independent of the position P.


Answer: 18
According to the problem, the net electric and magnetic forces on the particle should be zero.


So, the direction of  × should be opposite to the direction of . Hence,  should be along the positive z-direction.
Again, E = νB sin θ

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

Answer: 19
To prove: ·=· but
Suppose that  is perpendicular to ; is along the west direction.
Also,  is perpendicular to are along the south and north directions, respectively.

 is perpendicular to , so there dot or scalar product is zero.
is perpendicular to  so there dot or scalar product is zero.
·=·but
Hence, proved.


Answer: 20
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:

Here, x = x-coordinate of the point where the slope is to be measured


Answer: 21
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:


Answer: 22
Given: i = i0 e−t/RC

On applying the conditions given in the questions, we get:
(a)At t=0,  
(b)At t=RC ,  
(a) At t=10 RC,


Answer: 23
Electric current in a discharging R-C circuit is given by the below equation:
i = i0 ⋅ e−t/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 e−t/0.3 …(ii)
According to the question, we have:
(a) current at t = 0.3 s

(b) rate of change of current at t = 0.3 s

When t = 0.3 s, we have:
    A/s
(c) approximate current at t = 0.31 s
     A (approx)


Exercise : Solution of Questions on page Number : 30


Answer: 24
The given equation of the curve is y = 3x2 + 6x + 7.

ydx=(3×2+6x+7dx)=1000+300+75-35=1370-235=1135sq.units
The area bounded by the curve and the X-axis with coordinates x1 = 5 and x2 = 10 is given by


Answer: 25
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.


Answer: 26
The given function is y = e−x.
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 ∞.


Answer: 27
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


Answer: 28
According to the question, we have:

Momentum is zero at time, t = 0
Now, dp = [(10 N) + (2 Ns−1)t]dt
On integrating the above equation, we get:


Answer: 29
Changes in a function of y and the independent variable x are related as follows:

Integrating of both sides, we get:
∫dy = ∫x2 dx
where c is a constant
∴ y as a function of x is represented by


Answer: 30
(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: 31
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: 32
(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: 33
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: 34
Average thickness
=2.1733 mm
∴ Rounding off to three significant digits, the average thickness becomes 2.17 mm.


Answer: 35
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.

error: