OM NAMO NCERT SOLUTIONS: 2013

Ex 10.1

Question 1:

Represent graphically a displacement of 40 km, 30° east of north.

Answer:

Here, vector

represents the displacement of 40 km, 30° East of North.

Question 2:

Classify the following measures as scalars and vectors.

(i) 10 kg (ii) 2 metres north-west (iii) 40°

(iv) 40 watt (v) 10^–19 coulomb (vi) 20 m/s²

Answer:

(i) 10 kg is a scalar quantity because it involves only magnitude.

(ii) 2 meters north-west is a vector quantity as it involves both magnitude and direction.

(iii) 40° is a scalar quantity as it involves only magnitude.

(iv) 40 watts is a scalar quantity as it involves only magnitude.

(v) 10^–19 coulomb is a scalar quantity as it involves only magnitude.

(vi) 20 m/s² is a vector quantity as it involves magnitude as well as direction.

Question 3:

Classify the following as scalar and vector quantities.

(i) time period (ii) distance (iii) force

(iv) velocity (v) work done

Answer:

(i) Time period is a scalar quantity as it involves only magnitude.

(ii) Distance is a scalar quantity as it involves only magnitude.

(iii) Force is a vector quantity as it involves both magnitude and direction.

(iv) Velocity is a vector quantity as it involves both magnitude as well as direction.

(v) Work done is a scalar quantity as it involves only magnitude.

Question 4:

In Figure, identify the following vectors.

(i) Coinitial (ii) Equal (iii) Collinear but not equal

Answer:

(i) Vectors

and

are coinitial because they have the same initial point.

(ii) Vectors

and

are equal because they have the same magnitude and direction.

(iii) Vectors

and

are collinear but not equal. This is because although they are parallel, their directions are not the same.

Question 5:

Answer the following as true or false.

(i)

and

are collinear.

(ii) Two collinear vectors are always equal in magnitude.

(iii) Two vectors having same magnitude are collinear.

(iv) Two collinear vectors having the same magnitude are equal.

Answer:

(i) True.

Vectors

and

are parallel to the same line.

(ii) False.

Collinear vectors are those vectors that are parallel to the same line.

(iii) False.

It is not necessary for two vectors having the same magnitude to be parallel to the same line.

(iv) False.

Two vectors are said to be equal if they have the same magnitude and direction, regardless of the positions of their initial points.

Ex10.2

Question 1:

Compute the magnitude of the following vectors:

Answer:

The given vectors are:

Question 2:

Write two different vectors having same magnitude.

Answer:

Hence,

are two different vectors having the same magnitude. The vectors are different because they have different directions.

Question 3:

Write two different vectors having same direction.

Answer:

The direction cosines of

are the same. Hence, the two vectors have the same direction.

Question 4:

Find the values of x and y so that the vectors

are equal

Answer:

The two vectors

will be equal if their corresponding components are equal.

Hence, the required values of x and y are 2 and 3 respectively.

Question 5:

Find the scalar and vector components of the vector with initial point (2, 1) and terminal point (–5, 7).

Answer:

The vector with the initial point P (2, 1) and terminal point Q (–5, 7) can be given by,

Hence, the required scalar components are –7 and 6 while the vector components are

Question 6:

Find the sum of the vectors

Answer:

The given vectors are

Question 7:

Find the unit vector in the direction of the vector

Answer:

The unit vector

in the direction of vector

is given by

Question 1:

For each of the differential equations given below, indicate its order and degree (if defined).

(i)

(ii)

(iii)

Answer:

(i) The differential equation is given as:

The highest order derivative present in the differential equation is

. Thus, its order is two. The highest power raised to

is one. Hence, its degree is one.

(ii) The differential equation is given as:

The highest order derivative present in the differential equation is

. Thus, its order is one. The highest power raised to

is three. Hence, its degree is three.

(iii) The differential equation is given as:

The highest order derivative present in the differential equation is

. Thus, its order is four.

However, the given differential equation is not a polynomial equation. Hence, its degree is not defined.

Question 2:

For each of the exercises given below, verify that the given function (implicit or explicit) is a solution of the corresponding differential equation.

(i)

(ii)

(iii)

(iv)

Answer:

(i)

Differentiating both sides with respect to x, we get:

Again, differentiating both sides with respect to x, we get:

Now, on substituting the values of

and

in the differential equation, we get:

⇒ L.H.S. ≠ R.H.S.

Hence, the given function is not a solution of the corresponding differential equation.

(ii)

Differentiating both sides with respect to x, we get:

Again, differentiating both sides with respect to x, we get:

Now, on substituting the values of

and

in the L.H.S. of the given differential equation, we get:

Hence, the given function is a solution of the corresponding differential equation.

(iii)

Differentiating both sides with respect to x, we get:

Again, differentiating both sides with respect to x, we get:

Substituting the value of

in the L.H.S. of the given differential equation, we get:

Hence, the given function is a solution of the corresponding differential equation.

(iv)

Differentiating both sides with respect to x, we get:

Substituting the value of

in the L.H.S. of the given differential equation, we get:

Hence, the given function is a solution of the corresponding differential equation.

Question 3:

Form the differential equation representing the family of curves given by

wherea is an arbitrary constant.

Answer:

Differentiating with respect to x, we get:

From equation (1), we get:

On substituting this value in equation (3), we get:

Hence, the differential equation of the family of curves is given as

Question 4:

Prove that

is the general solution of differential equation

, where c is a parameter.

Answer:

This is a homogeneous equation. To simplify it, we need to make the substitution as:

Substituting the values of y and

in equation (1), we get:

Integrating both sides, we get:

Substituting the values of I₁ and I₂ in equation (3), we get:

Therefore, equation (2) becomes:

Hence, the given result is proved.

Question 5:

Form the differential equation of the family of circles in the first quadrant which touch the coordinate axes.

Answer:

The equation of a circle in the first quadrant with centre (a, a) and radius (a) which touches the coordinate axes is:

Differentiating equation (1) with respect to x, we get:

Substituting the value of a in equation (1), we get:

Hence, the required differential equation of the family of circles is

Question 6:

Find the general solution of the differential equation

Answer:

Integrating both sides, we get:

Question 7:

Show that the general solution of the differential equation

is given by (x + y + 1) = A (1 – x – y – 2xy), where A is parameter

Answer:

Integrating both sides, we get:

Hence, the given result is proved.

Question 8:

Find the equation of the curve passing through the point

whose differential equation is,

Answer:

The differential equation of the given curve is:

Integrating both sides, we get:

The curve passes through point

On substituting

in equation (1), we get:

Hence, the required equation of the curve is

Question 9:

Find the particular solution of the differential equation

, given that y = 1 when x = 0

Answer:

Integrating both sides, we get:

Substituting these values in equation (1), we get:

Now, y = 1 at x = 0.

Therefore, equation (2) becomes:

Substituting

in equation (2), we get:

This is the required particular solution of the given differential equation.

Question 10:

Solve the differential equation

Answer:

Differentiating it with respect to y, we get:

From equation (1) and equation (2), we get:

Integrating both sides, we get:

Question 11:

Find a particular solution of the differential equation

, given that y = – 1, when x = 0 (Hint: put x – y = t)

Answer:

Substituting the values of x – y and

in equation (1), we get:

Integrating both sides, we get:

Now, y = –1 at x = 0.

Therefore, equation (3) becomes:

log 1 = 0 – 1 + C

⇒ C = 1

Substituting C = 1 in equation (3) we get:

This is the required particular solution of the given differential equation.

Question 12:

Solve the differential equation

Answer:

This equation is a linear differential equation of the form

The general solution of the given differential equation is given by,

Question 13:

Find a particular solution of the differential equation

, given that y= 0 when

Answer:

The given differential equation is:

This equation is a linear differential equation of the form

The general solution of the given differential equation is given by,

Now,

Therefore, equation (1) becomes:

Substituting

in equation (1), we get:

This is the required particular solution of the given differential equation.

Question 14:

Find a particular solution of the differential equation

, given that y = 0 when x = 0

Answer:

Integrating both sides, we get:

Substituting this value in equation (1), we get:

Now, at x = 0 and y = 0, equation (2) becomes:

Substituting C = 1 in equation (2), we get:

This is the required particular solution of the given differential equation.

Question 15:

The population of a village increases continuously at the rate proportional to the number of its inhabitants present at any time. If the population of the village was 20000 in 1999 and 25000 in the year 2004, what will be the population of the village in 2009?

Answer:

Let the population at any instant (t) be y.

It is given that the rate of increase of population is proportional to the number of inhabitants at any instant.