Class 10

RBSE Solutions for Class 10 Maths Chapter 3 Polynomials Ex 3.5

September 1, 2021 by Fazal Leave a Comment

RBSE Solutions for Class 10 Maths Chapter 3 Polynomials Ex 3.5 is part of RBSE Solutions for Class 10 Maths. Here we have given Rajasthan Board RBSE Class 10 Maths Chapter 3 Polynomials Exercise 3.5.

Board	RBSE
Te×tbook	SIERT, Rajasthan
Class	Class 10
Subject	Maths
Chapter	Chapter 3
Chapter Name	Polynomials
E×ercise	Exercise 3.5
Number of Questions Solved	5
Category	RBSE Solutions

Rajasthan Board RBSE Class 10 Maths Chapter 3 Polynomials Ex 3.5

Question 1.
Find the nature of roots of following (RBSESolutions.com) quadratic equations.
(i) 2x² – 3x + 5 = 0
(ii) 2x² – 4x + 3 = 0
(iii) 2x² + x – 1 = 0
(iv) x² – 4x + 4 = 0
(v) 2x² + 5x + 5 = 0
(vi) 3x² – 2x + $\frac { 1 }{ 3 }$ = 0
Solution
(i) Given equation 2x² – 3x + 5 = 0
Comparing with ax² + bx + c = 0
a = 2, b = – 3 and c = 5
Discriminant (D) = b² – 4ac
= (-3)² – 4 × 2 × 5
= 9 – 40
= -31 < 0
Thus given equation (RBSESolutions.com) has no real roots.
(ii) Given equation
2x² – 4x + 3 =0
Comparing it with ax² + bx + c = 0
a = 2, b = – 4 and c = 3
Discriminant(D) = b² – 4ac
= (-4)² – 4 × 2 × 3
= 16 – 24
= -8 < 0
Thus given equation has no real roots.
(iii) Given equation 2x² + x – 1 = 0
Comparing it with ax² + bx + c = 0
a = 2, b = 1 and c = – 1
Discriminant (D) = b² – 4ac
= (1)² – 4 × 2 × (-1)
= 1 + 8
= 9 > 0
Hence, roots of equation (RBSESolutions.com) are real and distinct
(iv) Given equation x² – 4x + 4 = 0
Comparing it with quadratic equation ax² + bx + c = 0
a = 1, b = -4 and c = 4
Discriminant (D) = b² – 4ac
= (-4)² – 4 × 1 × 4
= 16 – 16
= 0
Hence, roots of equation will be real and equal.
(v) Given equation 2x² + 5x + 5 = 0
Comparing it with quadratic equation ax² + bx + c = 0
a = 2, b = 5 and c = 5
Discriminant (D) = (5)² – 4 × 2 × 5
= 25 – 40
= -15 < 0
Thus, roots of equation are not real.
(vi) Given equation 3x² – 2x + $\frac { 1 }{ 3 }$ = o
Comparing it by general quadratic equation
a = 3, b = -2, c = $\frac { 1 }{ 3 }$
Discriminant (D) = b² – 4ac
= (-2)² – 4 × 3 × $\frac { 1 }{ 3 }$
= 4 – 4
= 0
Hence, root of given equation are real and equal.

The Polynomial Root Calculator can find multiple roots to polynomial equations with just one click. To open the tool, click on it.

Question 2.
Find value of k, for which (RBSESolutions.com) following quadratic equations have real and equal roots.
(i) kx(x – 2) + 6 = 0
(ii) x² – 2(k + 1)x + k² = 0
(iii) 2x² + kx + 3 = 0
(iv) (k + 1)x² – 2(k – 1) x + 1 = 0
(v) (k + 4)x² + (k + 1) x + 1 = 0
(vi) kx² – 5x + k = 0
Solution
(i) Given equation kx(x – 2) + 6 = 0
or kx² – 2kx + 6 = 0
Comparing it with ax² + bx + c = 0
a = k, b = -2k and c = 6
Discriminant D = b² – 4ac
= (-2k)² – 4 × k × 6
= 4k² – 24k
= 4k(k – 6)
For equal roots D = 0
4k(k – 6) = 0
⇒ k = 0 or k – 6 = 0
⇒ k = 0 or k = 6
For equal roots k = 6 because k = 0 is not possible.
(ii) Given (RBSESolutions.com) equation
x² – 2(k + 1)x + k² = o
Comparing it with ax² + bc + c = 0
a = 1, b = -2(k + 1) and c = k²
Discriminant D = b² – 4ac
= {-2(k + 1)}² – 4 × 1 × k²
= 4(k² + 2k + 1) – 4k²
= 4k² + 8k + 4 – 4k²
= 8k + 4
For equal roots D = 0
⇒ 8k + 4 = 0
⇒ 8k = -4
⇒ k = $\frac { -1 }{ 2 }$
Thus k = $\frac { -1 }{ 2 }$
(iii) Given equation
2x² + kr + 3 = 0
Comparing with ax² + bx + c = 0
a = 2, b = k and c = 3
Discriminant D = b² – 4ac
= k² – 4 × 2 × 3
= k² – 24
for equal roots D = 0
⇒ k² – 24 = 0
⇒ k² = 24
⇒ k = ±√24 = ±2√6
for equal roots k = ± 2√6
(iv) Given (RBSESolutions.com) equation (k + 1)x² – 2(k – 1)x + 1 = 0
Comparing it by ax² + bx + c = 0
a = (k + 1), b = -2(k – 1) and c = 1
Discriminant, D = b² – 4ac
= {-2 {k – 1)}² – 4 × (k + 1) × 1
= 4{k² + 1 – 2k) – 4 (k + 1)
= 4k² + 4 – 8k – 4k – 4
= 4k² – 12k
= 4k(k – 3)
for equal roots, D = 0
⇒ 4k(k – 3) = 0
⇒ k(k – 3) =0
⇒ k = 0 or k = 3
for equal roots k = 3, since k = 0
(v) Given equation is (k + 4)x² + (k + 1)x + 1 = 0
Comparing it with ax² + bx + c = 0
a = k + 4, b = k + 1, c = 1
Discriminant (D) = b² – 4ac
= (k + 1 )² – 4 × (k + 4) × 1
= k² + 2k + 1 – 4k – 16
= k² – 2k – 15
= k² – 5k + 3k – 15
= k(k – 5) + 3(k – 5)
= (k – 5) (k + 3)
For equal roots, D = 0
⇒ (k – 5)(k + 3) = 0
⇒ k – 5 = 0 or k + 3 = 0
⇒ k = 5 or k = -3
(vi) Given (RBSESolutions.com) equation kx² – 5x + k = 0
Comparing it with ax² + bx + c = 0
a = k, b = -5. c = k
Discriminant, (D) = b² – 4ac
= (-5)² – 4 × k × k
= 25 – 4k²
For equal roots D = 0

Question 3.
Find value of k for which (RBSESolutions.com) following quadratic equations have real and fractional roots.
(i) kx² + 2x + 1 = 0
(ii) kx² + 6x + 1 = 0
(iii) x² – kx + 9 = 0
Solution
(i) Given roots kx² + 2x + 1 = 0
Comparing it by general quadratic equation
ax² + bx + c = 0
a = k, b = 2, c = 1
Discriminant (D) = b2 – 4ac
= (2)² – 4 × k × 1
= 4 – 4 k
= 4(1 – k)
For real and fractional roots
Discriminant (D) > 0
⇒ 4(1 -k) > 0
⇒ 1 – k > 0
⇒ k < 1
Thus, k will be less than.
(ii) Given equation kx² + 6x + 1 = 0
Comparing it by general (RBSESolutions.com) quadratic equation ax² + bx + c – 0,
a = k, b = 6, c = 1
Discriminant (D) = b² – 4ac
= (6)² – 4 × k × 1
= 36 – 4k
= 4(9 – k)
If discriminant (D) > 0 then roots of equation will be real and fractional.
i.e., D > 0
⇒ 4(9 – k) > 0
⇒ 9 – k > 0
⇒ k > 9
Thus value of k will be greater than 9
(iii) Given equation x² – kx + 9 = 0
Comparing it by general quadratic equation
ax² + bx + c = 0
a = 1, b = -k, c = 9
Discriminant (D) = b² – 4ac
= (-k)² – 4 × 1 × 9
= k² – 36
If D > 0, then roots of (RBSESolutions.com) equation will be real and fractional.
D > 0
⇒ k² – 36 > 0
⇒ (k – 6)(k + 6) > 0
⇒ k – 6 > 0 or k + 6 > 0
⇒ k > 6 or k < -6
Thus, value of k will be greater than 6 or less than -6

Question 4.
Find value of k, for which equation x² + 5kx + 16 = 0 has no real roots.
Solution
Given equation x² + 5kx + 16 = 0
Comparing it by general quadratic equation ax² + bx + c = 0
a = 1, b = 5k, c = 16
Discriminant (D) = b² – 4ac
= (5k)² – 4 × 1 × 16
= 25k² – 64
If Discriminant (D) < 0, then roots will not be real.
i.e. D < 0
⇒ 25k² – 64 < 0
⇒ (5k – 8)(5k + 8) < 0
⇒ 5k – 8 < 0 or 5k + 8 < 0
⇒ k < $\frac { 8 }{ 5 }$ or k > $\frac { -8 }{ 5 }$
Hence, value of k will be smaller than $\frac { 8 }{ 5 }$ or greater than $\frac { -8 }{ 5 }$

Question 5.
If roots of quadratic (RBSESolutions.com) equation (b – c)x² + (c – a)x + (a – b) = 0 are real and equal, then prove that 2b = a + c
Solution
Given equation is (b – c)x² + (c – a)x + (a – b) = 0
Comparing it by general quadratic equation, a = (b – c), b = (c – a) and c = (a – b)
Discriminant (D) = b² – 4ac
= (c – a)² – 4(b – c)(a – b)
= (c² + a² – 2ac) – 4(ab – ac – b² + bc)
= c² + a² – 2ac – 4ab + 4ac + 4b² – 4bc
= c² + a² + 2ac – 4ab – 4bc + 4b²
= (a + c)² – 4 b(a + c) + 4b²
= (a + c)² – 2 × (a + c) × 2b + (2b)²
= [(a + c) – 2b]²
Roots of equation (RBSESolutions.com) are real and equal.
D = 0
⇒ [(a + c) – 2b]² = 0
⇒ (a + c) – 2b = 0
⇒ a + c = 2b
Hence Proved.

We hope the given RBSE Solutions for Class 10 Maths Chapter 3 Polynomials Ex 3.5 will help you. If you have any query regarding Rajasthan Board RBSE Class 10 Maths Chapter 3 Polynomials Exercise 3.5, drop a comment below and we will get back to you at the earliest.

RBSE Solutions for Class 10 Maths Chapter 9 Co-ordinate Geometry Ex 9.2

July 20, 2021 by Fazal Leave a Comment

RBSE Solutions for Class 10 Maths Chapter 9 Co-ordinate Geometry Ex 9.2 is part of RBSE Solutions for Class 10 Maths. Here we have given Rajasthan Board RBSE Class 10 Maths Chapter 9 Co-ordinate Geometry Exercise 9.2.

Rajasthan Board RBSE Class 10 Maths Chapter 9 Co-ordinate Geometry Ex 9.2

Question 1.
Find the co-ordinates of the point which divides the line (RBSESolutions.com) segment joining the points (3, 5) and (7, 9) in the ratio 2 : 3 internally.
Solution :
Let point P(x, y) divides the line segment joining the point A(3, 5) and B(7, 9) internally in the ratio 2 : 3.

Hence, the coordinate of P is $\left( 4\frac { 3 }{ 5 } ,6\frac { 3 }{ 5 } \right)$ .

Question 2.
Find the co-ordinates of the point which (RBSESolutions.com) divides the line segment joining the points (5, -2) and $\left( -1\frac { 1 }{ 2 } ,4 \right)$ in the ratio 7 : 9 externally.
Solution :
Let point P(x, y) divides the line segment joining the points A(5, -2) and B $\left( -1\frac { 1 }{ 2 } ,4 \right)$ externaly in the ratio 7 : 9.
x₁ = 5 y₁ = -2 m₁ = 7
x₂ = -1 $\frac { 1 }{ 2 }$ y₂ = 4 m₂ = 9
By formula of external division

Hence, the coordinate of P is $\left( 27\frac { 3 }{ 4 } ,-23 \right)$

Question 3.
Prove that origin O divides the line joining (RBSESolutions.com) the points 4(1, -3) and B(-3, 9) in the ratio 1 : 3 internally. Find the co-ordinates of the points which externally divides the line.
Solution :
Let the point P(x, y) divides the line segment joining the point A(1, -3) and B(-3, 9) internally in the ratio 1 : 3.

Hence co-ordinate of P(x, y) is (0, 0). So origin O is divides (RBSESolutions.com) the line segment joining the given points internally in the ratio 1 : 3.
Again point P(x, y) divides externally, then by formula of external division

Hence, co-ordinate of point P which divides externally = (3, -9).

Free Distance and Midpoint Calculator calculates the distance and midpoint between two points instantaneously.

Question 4.
Find the mid point of line joining (RBSESolutions.com) the points (22, 20) and (0, 16).
Solution :
Let co-ordinates of mid point is P (x, y) which ¡s line joining the points A(22, 20) and B(0, 16).
x₁ = 22, y₁ = 20, x₂ = 0, y₂ = 16
By formula of mid point

Hence, co-ordinate of mid point P(x, y) = (11, 18)

Question 5.
In which ratio, x-axis divides the line segment (RBSESolutions.com) which joins points (5, 3) and (-3, -2)?
Solution :
On x-axis, the ordinate of each point is zero. So let point P(x, 0) divides the line segment joining the points A(5, 3) and B(-3, -2) internally in the ratio m₁ : m₂

Hence, line segment joining the given points by x-axis divides internally in the ratio 3 : 2.

Question 6.
In which ratio, y-axis, divides the line (RBSESolutions.com) segment which joins points (2, -3) and (5, 6)?
Solution :
On y-axis, coordinate of point x is zero, so let point P(0, y) divides the line segment joining the point A(2, -3) and B(5, 6) externally in the ratio m₁ : m₂.
Then by formula of external division

⇒ 5m₁ – 2m₂ = 0
⇒ 5m₁ = 2m₂
⇒ $\frac { { m }_{ 1 } }{ { m }_{ 2 } } =\frac { 2 }{ 5 }$
⇒ m₁ : m₂ = 2 : 5
Hence, line segment joining these points divides externally in the ratio 2 : 5 by y-axis.

Question 7.
In which ratio, point (11, 15) divides the (RBSESolutions.com) line segment which joins (15, 5) and (9, 20)?
Solution :
Let the point P(1 1, 15) divides the line segment joining the point A(15, 5) and B(9, 20) internally in the ratio λ : 1
Hence x₁ = 15 y₁ = 5 m₁ = λ
x₂ = 9 y₂ = 20 m₂ = 1

⇒ 11(λ + 1) = 9λ + 15
⇒ 11λ + 11 = 9λ + 15
⇒ 11λ – 9λ = 15 – 11
⇒ 2λ = 4
λ = $\frac { 4 }{ 2 }$ = $\frac { 2 }{ 1 }$
∴ λ : 1 = 2 : 1
Hence, required ratio is 2 : 1.

Question 8.
If point P (3, 5) divides line segment which (RBSESolutions.com) joins A(-2, 3) and B(x, y) in the ratio 4 : 7 internally, then find the co-ordinates of B.
Solution :
Let the coordinate of point B is (x, y)

Let the point P(3, 5) divides the line segment joining the points A(-2, 3) and B(x, y) internally in the inthe ratio 4 : 7.
So, x₁ = -2, y₁ = 3 m₁ = 4
x₂ = x y₂ = y m₂ = 7
By section formula

Hence, the co-ordinate of point B is $\left( \frac { 47 }{ 4 } ,\frac { 17 }{ 2 } \right)$

Question 9.
Find the co-ordinates of point which (RBSESolutions.com) trisects the line joining point (11, 9) and (1, 2).
Solution :
Let P(x₁ y₁) and Q(x₂, y₂) are required points which is trisect the line segment joining the points A(11, 9) and B(1, 2) respectively.
Let AP = PQ = BQ = x
PB = x + x = 2x
AQ = x + x = 2x

Hence point P divides line segment AB in the (RBSESolutions.com) ratio 1 : 2 and point Q, divides the line segment Ab in the ratio 2 : 1.
By the section formula for point P,
m₁ = 1, m₂ = 2

∴ Hence the co-ordinate of point P is $\left( \frac { 23 }{ 3 } ,\frac { 20 }{ 3 } \right)$
for point Q,
m₁ = 2, m₂ = 1

So, the co-ordinate of Q = $\left( \frac { 13 }{ 3 } ,\frac { 13 }{ 3 } \right)$
Hence, required coordinate of point P and Q is $\left( \frac { 23 }{ 3 } ,\frac { 20 }{ 3 } \right)$ and $\left( \frac { 13 }{ 3 } ,\frac { 13 }{ 3 } \right)$ respectively.

Question 10.
Find the co-ordinates of point which quarter (RBSESolutions.com) sects the line joining point (-4, 0) and (0, 6).
Solution :
Let C, D and E are required point which quarter sects the line segment joining the points A(-4, 0) and B(0, 6)

Since D is mid point of A and B, C is is midpoint A and D, and E ¡s mid point of D and B, then AC = CD = DE = EB
Now, co-ordinate of mid point D of A and B.

= $\left( \frac { -2 }{ 2 } ,\frac { 9 }{ 2 } \right)$
= $\left( -1,\frac { 9 }{ 2 } \right)$
Hence, co-ordinate of points divided in four equal (RBSESolutions.com) part of AB are $\left( -3,\frac { 3 }{ 2 } \right)$ , (-2, 3) and $\left( -1,\frac { 9 }{ 2 } \right)$ respectively.

Question 11.
Find the ratio in which line 3x + y = 9 divides the line segment which joins points (1, 3) and (2, 7)
Solution :
Let the line 3x + y = 9 divides the line segment joining A(1, 3) and B(2, 7) at point P(x, y) in the ratio λ : 1
x₁ = 1 y ₁ = 3 m₁ = λ
x₂ = 2 y₂ = 7 m₂ = 1

⇒ 6λ + 3 + 7λ + 3 = 9(λ + 1)
⇒ 13λ + 6 = 9λ + 9
⇒ 13λ – 9λ = 9 – 6
⇒ 4λ = 3
⇒ λ = $\frac { 3 }{ 4 }$
λ : 1 = 3 : 4
Hence, required ratio is 3 : 4.

Question 12.
Find the ratio where point (-3, P), divides internally (RBSESolutions.com) the line segment which joins points (-5, -4) and (-2, 3). Also find P.
Solution :

Let the point P(-3, p) divides the line segment joining the points A(-5, -4) and B(-2, 3) internally in the ratio m₁ : m₂.
Hence, x co-ordinate of point P(-3, p)

= -3(m₁ + m₂) = -2m₁ – 5m₂
⇒ -3m₁ – 3m₂ = -2m₁ – 5m₂
⇒ -3m₁ + 2m₁ = -5m₂ + 3m₂
⇒ – m₁ = -2m₂
⇒ $\frac { { m }_{ 1 } }{ { m }_{ 2 } } =\frac { -2 }{ -1 }$ = $\frac { 2 }{ 1 }$
Hence required ratio m₁ : m₂ = 2 : 1
Now, y co-ordinate of point P(-3, p)

Hence, required (RBSESolutions.com) ratio 2 : 1 and p = $\frac { 2 }{ 3 }$

We hope the given RBSE Solutions for Class 10 Maths Chapter 9 Co-ordinate Geometry Ex 9.2 will help you. If you have any query regarding Rajasthan Board RBSE Class 10 Maths Chapter 9 Co-ordinate Geometry Exercise 9.2, drop a comment below and we will get back to you at the earliest.

RBSE Class 10 Science Notes Chapter 11 Work, Energy and Power

July 9, 2021 by Prasanna Leave a Comment

RBSE Class 10 Science Notes Chapter 11 Work, Energy and Power are part of RBSE Solutions for Class 10 Science Notes. Here we have given Rajasthan Board RBSE Class 10 Science Notes Chapter 11 Work, Energy and Power.

Rajasthan Board RBSE Class 10 Science Notes Chapter 11 Work, Energy and Power

Displacement Calculator is really a nice calculator that will help you solve a lot of your problems related to displacement.

Work:
The product of force and displacement (in direction of force) is called work.
Work = Force x Displacement in direction of force
W = F x s

If direction of force is different from direction of displacement then we can find the work done by component of force in direction of displacement.

Let us assume that direction of force makes an angle θ with direction of displacement. Then, component of force in direction of displacement is;
F.cosθ

Work done
W = component of force in direction of displacement x displacement
Or W = F.cosθ x s = F s cosθ = F.S
Work is a scalar quantity, i.e. its value can be positive or negative. If force or component of force is in the direction of displacement then work done is positive. But if force or component of force is in opposite direction of displacement then work done is negative.

Unit of Work:
If force is expressed in Newton and displacement is expressed in meter, then unit of work is Joule.
Work = Newton × meter = Joule
If force is expressed in Dyne and displacement is expressed in centimeter then unit of force is erg.
Work = Dyne×cm = erg 1 Joule = 1 N x 1 m = 10^s Dyne x 10²cm = 10⁷erg

Energy:
The capacity to do work is called energy. Thus, work is the measure of energy and hence unit of energy is same as that of work. Energy too is a scalar quantity. If 1 Joule of work is to be done then required energy for this would also be 1 Joule.

Types of Energy:
Some major forms of energy are as follows:

Mechanical Energy: The energy in an object because of movement or position, or both is called mechanical energy.
Heat Energy: The mobile energy in micro particles due to heat is called heat energy. These micro particles transfer energy from high temperature to low temperature,
Chemical Energy: Energy obtained from chemical reactions is called chemical energy. For example, a battery provides energy because of chemical reaction. Similarly, energy from LPG or coal is obtained because of chemical reaction.
Electrical Energy: Energy produced by electric charge is called electrical energy. Most of the household appliances work because of electrical energy. Gravitational Energy: Energy in an object due to gravitational force is called gravitational energy. Gravitational energy is responsible for falling water in waterfalls.
Nuclear Energy: The energy produced by nuclear fission or fusion is called nuclear energy.

Mechanical Energy:
The sum of kinetic energy and potential energy in an object gives the mechanical energy of that object. Mechanical energy gives an object the capacity to do work.

Kinetic Energy:
The energy in an object due to its motion is called kinetic energy. Let us assume that an object of mass m is moving at uniform velocity μ and a force F is applied on it in the direction of motion then the object is displaced to distance s. Let us assume that the work done on the object increases its speed to ν and acceleration of object becomes. Then work done is equal to change in kinetic energy of object.

If initial velocity of the object is zero, i.e. the object starts from rest and obtains the velocity v then;

It can be said that the kinetic energy of an object of mass m and velocity v is;

Potential Energy :
The energy in an object due to the position of the object is called potential energy. Potential energy gives the capacity to do work to an object. The work done in putting an object in a specific position from normal position, gives the measure of its potential energy.

Potential Energy in Gravitational Field :
When an object is lifted up to a height then some work is done against acceleration due to gravity. The work done on object increases the energy of object. This energy gets stored in the object in the form of potential energy. If an object with mass m is lifted to height h from earth then the minimum force required to do this should be equal to its weight, i.e. w = mg.
Potential energy of object at height h = work done against gravitational force
= Force × Displacement = mgh

Potential Energy of a Simple Pendulum :
When a simple pendulum is displaced from its mean position to a particular side, its centre of gravity is raised. The work done on pendulum in this gets stored as potential energy in pendulum.

The pendulum moves ahead from its mean position due to this kinetic energy. While doing so, its kinetic energy reduces to zero and potential energy increases to maximum when pendulum reaches another extreme (point B). The pendulum swings back to its mean position due to the potential energy.

If mass of pendulum is m and pendulum is suspended by a thread of length / then potential energy of pendulum for displacement x can be obtained as follows:

Similarly, when a spring with spring constant k is displaced to x distance from its mean position (under elastic limits) then potential energy of spring is;

Electrical Energy
The energy in charged particles is called electrical energy. When a particle becomes charged then an electrical field is created around it. This electrical field applies a force on charged particles nearby and makes them move. Thus, transmission of energy takes place through particles in motion.

The electrical field around a positively charged particle repels other positively charged particles but attracts negatively charged particles.
Positively charged particles possess potential energy because of their position. When electrical field applies a force on these particles, they move in a certain direction. For example; we need to apply an external force in order to move a positively charged particle from negatively charged source. This process would increase the potential energy of positively charged particles. When the external force is removed, the positively charged particle would move from area of higher potential energy to area of lower potential energy. Thus, a positively charged particle would naturally move towards negatively charged source. While doing so, potential energy of charged particles changes into kinetic energy. This kinetic energy is obtained in the form of electrical energy.

Different types of electric plants are used for production of electricity. Some of them are as follows.

Thermal Power Plant: In a thermal power plant, heat energy is obtained from chemical energy of coal. The heat energy is utilized to convert highly pure water into steam. The steam is utilized to rotate the turbine and thus electricity is produced by generator which is attached with turbine.

Nuclear Power Plant:
Nuclear fission is carried out in a nuclear power plant. The heat generated from nuclear fission is utilized to convert water into steam. The steam is utilized to rotate the turbine and thus electricity is produced by generator which is attached with turbine.

Hydel Power Plant: In a hydel power plant, potential energy of water is increased by making a dam. When water is released from the dam, potential energy of water changes to kinetic energy. This kinetic energy is utilized to rotate the turbine and thus electricity is produced by generator which is attached with turbine.
Wind Power Plant: In a windmill, kinetic energy of air is utilized to rotate the turbine and thus electricity is produced by generator which is attached with turbine. Wind energy is a renewable energy and hence is environment friendly.
Solar Power Plant: Energy from sun is converged at a point with the help of lens and mirrors and is changed into heat energy. This heat is utilized to produce steam which is utilized to rotate the turbine and thus electricity is produced by generator which is attached with turbine.
Photovoltaic Power Plant: In a photovoltaic power plant, solar panels are installed at open spaces or on rooftops. These panels contain photovoltaic cells. When sunlight falls on these cells, the cells gain photons from light and bring the electrons in excited state. These charged particles provide electricity to circuit in the form of electric current.

Conservation of Energy :
As per law of conservation of energy, the total energy of an isolated system always remains constant. Energy can neither be created nor be destroyed. We can only change the form of energy.
As per law of conservation of mechanical energy, the mechanical energy of a system remains conserved. If kinetic energy of the system is increased then potential energy decreases, and vice- versa is also true.
If change in potential energy and kinetic energy is respectively ΔE_P and ΔEK then;
^ΔE_P^=- Δ^Ek
Or, ΔE_p+ΔE_k = 0
Or, Total mechanical energy E_m = E_p+E_kIn reality, there is some fall in mechanical energy during the whole cycle. This reduction is because of action of some non-conserving forces, like friction, viscosity, etc. These forces convert some of the energy into sound, heat, etc. Thus form of energy of the system changes but total energy of the system remains constant.

Dissipation of Energy:
When energy changes from one form to another then some of the energy is dissipated in the form of heat, light, sound, etc. Dissipation of energy means that energy is converted into some forms which are undesirable and which are not required by us. The total energy remains conserved but because of useless dissipation of energy it is impossible to make a cent per cent efficient system.

Power:
Power is defined as the rate of doing a work. It is a scalar quantity.

Unit of Power:
The unit of power is Watt which is named in honour of James Watt; the inventor of steam engine.

Another unit of power is horse power which is equal to 746 W.

Electrical Power:
The rate of transfer of electrical energy in electric circuit is called electric power. On a commercial scale, the electricity distribution company measures the consumption of electricity in the form of kilo Watt Hour. One kilowatt hour is called one electric unit which is read by electric meter.

This means that when a bulb of 1 KW is used for 1 hour, it will consume 1 unit of electricity

RBSE Solutions for Class 10 Science Chapter 11 Work, Energy and Power

July 7, 2021 by Fazal Leave a Comment

RBSE Solutions for Class 10 Science Chapter 11 Work, Energy and Power are part of RBSE Solutions for Class 10 Science. Here we have given Rajasthan Board RBSE Class 10 Science Solutions Chapter 11 Work, Energy and Power.

Board	RBSE
Textbook	SIERT, Rajasthan
Class	Class 10
Subject	Science
Chapter	Chapter 11
Chapter Name	Work, Energy and Power
Number of Questions Solved	109
Category	RBSE Solutions

Rajasthan Board RBSE Class 10 Science Solutions Chapter 11 Work, Energy and Power

Textbook Questions Solved

I. Multiple Choice Questions

Question 1:
What is the unit of work?
(a) Newton
(b) Joule
(c) Watt
(d) None of these
Answer:
(b) Joule

Question 2:
If angle between force F and displacement s is e then work done is which of the following?
(a) F s sinθ
(b) F s θ
(c) F s cos θ
(d) F s tan θ
Answer:
(c) F s cos θ

Kinetic Energy Calculator is a free online tool that displays the kinetic energy of the object.

Question 3:
An object of mass m is moving with velocity y. What is the value of kinetic energy?
(a) mv
(b) mgv
(c) mv²
(d) 1/2 mv²
Answer:
(d) 1/2 mv²

Question 4:
An object of mass m is at height h. What is its potential energy?
(a) mgh
(b) mg/h
(c) mh/g
(d) ½ mgh²
Answer:
(a) mgh

Question 5:
What is the unit of power?
(a) Newton
(b) Watt
(c) Joule
(d) Newton meter
Answer:
(b) Watt

Question 6:
How much work is done in moving an object of mass 1 kg to a height of 4 m? (g. 10 m/s² )
(a) 1 Joule
(b) 4 Joule
(c) 20 Joule
(d) 40 Joule
Answer:
(d) 40 Joule

Question 7:
What happens to total energy of a free falling object?
(a) Keeps on increasing
(b) Keeps on decreasing
(c) Remains constant
(d) Becomes zero
Answer:
(c) Remains constant

Question 8:
If velocity of an object becomes double then its kinetic energy becomes how much?
(a) One fourth
(b) Half
(c) Double
(d) Four times
Answer:
(d) Four times

Question 9:
What is the commercial unit of electric energy?
(a) Joule
(b) Watt second
(c) Kilowatt hour
(d) Kilowatt per hour
Answer:
(c) Kilowatt hour

Question 10:
If spring constant is k, then what happens to its potential energy on compressing the spring
to distance x within its elastic limit?
(a) kx
(b) 1, kx²
(c) kx²
(d) None of these
Answer:
(b) 1/ kx²

Work, Energy and Power Very Short Answer Type Questions

Question 1:
Define work and write its unit.
Answer:
Force multiplied by displacement is equal to work done. Unit of work is Newton meter.

Question 2:
What is energy? What is the unit of energy?
Answer:
The capacity of doing work is called energy. Unit of energy is Newton meter or Joule.

Question 3:
What do you understand by kinetic energy?
Answer:
Energy in an object due to its velocity is called kinetic energy.

Question 4:
What is potential energy?
Answer:
Energy in an object due to its position is called potential energy.

Question 5:
Write the law of conservation of energy.
Answer:
Energy can neither be created nor be destroyed.

Question 6:
Dissipation of energy generally happens in which forms?
Answer:
Heat, Light and Sound

Question 7:
Is it possible to make a hundred percent efficient system in terms of energy?
Answer:
No

Question 8:
What do you understand by electric energy?
Answer:
Energy in charged particles is called electric energy.

Question 9:
Write the name of any three electrical appliances.
Answer:
Electric iron, ceiling fan, refrigerator

Question 10:
What is power? What is the unit of power?
Answer:
The rate of doing a work is called power. Unit of power is watt.

Question 11:
Which type of light is suitable for reducing electricity consumption in homes?
Answer:
LED and CFL

Question 12:
What should be kept in mind while buying a new electrical appliance?
Answer:
Star Rating

Question 13:
An object is displaced by 10 m when a force of 20 N is applied on it. Find the work done.
Answer:
Here, force = 20 N and displacement = 10 m
Work = Force x Displacement
= 20 N x 10 m = 200 Nm

Question 14:
It takes 1 minute to lift an object of 30 kg mass by 2 m. How much energy is spent in doing this work?
Answer:
Mass = 30 kg, height = 2 m, time = 1 min = 60 second and g = 10 m/s²
Energy spent can be calculated as follows:

Question 15:
If a 60 W bulb is used for 8 hours per day then how many units of electricity is consumed in 30 days?
Answer:
Power of bulb = 60 W, time per day = 8 hour, no. of days = 30
Energy consumed: 60W x 8h x 30d = 14400 W = 14.4 kWh = 14.4 unit

Work, Energy and Power Short Answer Type Questions

Question 1:
What do you understand by work? If direction of displacement is different than the direction of force applied then how will you calculate the work done? Explain with suitable example.
Answer:
Force multiplied by displacement is equal to work done. When direction of displacement is different than the direction of force applied then work done is calculated on the basis of angle between direction of displacement and force applied.
Let us assume that angle between force applied and displacement is θ and displacement is s then

W= F s cos θ
When directions of force applied and displacement are opposite to each other they make an angle of 180°. Work done in this case can be calculated as follows:
W = F s cos θ
Since cos 180° = -1
Hence, W = -Fs
This can be illustrated by a situation when force applied on a moving object brings the object to rest.

Question 2:
An object is moving with velocity u. Its velocity becomes v when force F is applied on it. If the object covers a distance s during this then calculate the increase in kinetic energy of object.
Answer:
According to third equation of motion;

This shows that work done is equal to change in kinetic energy of object.

Question 3:
What is potential energy? If an ideal spring has spring constant k then find the acquired potential energy in compressing the spring by distance x.
Answer:
The energy in an object because of its position is called potential energy.
If k is the spring constant and spring is compressed by distance x then force applied for this can be given as follows:
F = kx
Now, work done in moving the spring by distance x can be given as follows:

Question 4:
An object is moving with velocity v. If mass of the object is m, then how much work needs to be done to bring this object to rest?
Answer:
When an object is moving with a certain velocity, it has some kinetic energy. Same kinetic energy needs to be applied in opposite direction of movement to bring this object to rest. Hence, kinetic energy would give the value of work done.
KE = $\frac { 1 }{ 2 }$ mv²

Question 5:
What do you understand by conservation of mechanical energy?
Answer:
The sum of kinetic energy and potential energy of a system is called mechanical energy. For a given system, the sum of kinetic energy and potential energy remains constant. When kinetic energy increases then potential energy decreases and vice versa is also true. This explains the conservation of mechanical energy of a given system. This can be understood by example of a stone kept at a height. The potential energy of stone is maximum and kinetic energy is zero at that height. When the stone is in free fall, its kinetic energy increases and potential energy decreases. When the stone hits the ground, its kinetic energy becomes maximum and potential energy becomes zero. Throughout this, the mechanical energy remains constant.

Question 6:
When an object is in free fall then its potential energy is decreasing continuously. How is mechanical energy being conserved in this situation?
Answer:
Let us assume that a stone kept at a height. The potential energy of stone is maximum and kinetic energy is zero at that height. When the stone is in free fall, its kinetic energy increases and potential energy decreases. When the stone hits the ground, its kinetic energy becomes maximum and potential energy becomes zero. Throughout this, the mechanical energy remains constant.

Question 7:
How is energy dissipated?
Answer:
During conversion of energy from one form to another, total energy does not get converted into one form of energy rather some of the energy is dissipated in the form of heat, light and sound energy. For example, when a ceiling fan is switched on, electrical energy is converted into kinetic energy. During this change, some of the energy is converted into heat energy which is evident from ceiling fan becoming hot.

Question 8:
Explain dissipation of energy from production of electricity to transmission of electricity to households.
Answer:
Let us take the example of thermal power plant to understand this. First of all, heat energy is utilized to make steam from water. Movement of steam is utilized to turn the turbine to produce electricity. Some of the energy is dissipated while converting water into steam. Some energy is dissipated in the form of heat when turbine moves. During transmission of electricity through wires some energy is dissipated in the form of heat due to heating effect of electric current.

Question 9:
How are work, energy and power related to each other?
Answer:
Capacity to do a work is called energy. Efficiency of doing a work is called power. So, a work cannot be done without energy or without efficiency of doing a work. Thus, it can be said that work, energy and power are related to each other.

Question 10:
What do you understand by electrical energy? How is electricity obtained from thermal power plant?
Answer:
The energy in a charged object is called electric energy. In a thermal power plant, fuel is burnt to convert water into steam. The steam is then channelized to propellers of turbine, The flow of steam facilitates the movement in turbine. While doing so, kinetic energy of turbine is converted into electric energy.

Question 11:
How is electricity produced in hydel power plant?
Answer:
In hydel power plant, water is stored behind a dam. After that, water is released thorough sluice gates. The flow of water facilitated the movement in turbine. While doing so, kinetic energy of turbine is converted into electric energy.

Question 12:
What can be done to reduce dissipation of electric energy?
Answer:
Following steps need to be taken to reduce dissipation of electric energy:

Switch off electrical appliances when not in use.
Use LED or CFL in place of incandescent bulb.
Use electrical appliances with star rating.

Question 13:
What can be done to make air-conditioning more effective in homes?
Answer:
Following steps can be taken to make air-conditioning more effective:

Walls and ceiling should be made heat resistant.
Outer walls should be painted in white or light colour.
Hollow bricks should be used while making the walls.

Question 14:
What is electric power? How is consumption of electricity calculated for households? Explain with suitable example.
Answer:
Efficiency of electric energy to do work is called electric power. If a charge of Q coulomb moves through potential difference of V in time t second, then

Rate of transfer of electric energy through an electric circuit is called electric power. Consumption of electricity is measured in terms of kilowatt hour
1 unit = 1 kWh = 1000Wh
1kWh = 1000W x 60 x 60s = 3600000Ws
= 3.6 x 10⁶ $\frac { J }{ s }$ XS = 3.6 x 10⁶J
For example, if a bulb of 1000 W is used for 1 hour then it will consume 1 unit. If a bulb off 100 W is used for 10 hours then it will consume 1 unit.

Question 15:
Explain conversion of energy when we switch on an electric bulb.
Answer:
When we switch on an electric bulb, electric energy is converted into light energy and heat energy. This happens because of heating effect of electric current. The filament becomes red hot due to heating effect and starts emitting light.

Work, Energy and Power Long Answer Type Questions

Question 1:
What is energy? Prove that work done by an object is equal to difference between kinetic energy in its two states.
Answer:
The capacity of an object to do work is called energy.
According to third equation of motion;

This shows that work done is equal to change in kinetic energy of object.

Question 2:
What is electric energy? How is electricity produced in following:
(a) Hydel electric plant
(b) Wind energy plant
(c) Solar energy plant
Answer:
(a) The energy in charged particles is called electrical energy. Flow of water is used for moving the turbine. Turbine converts kinetic energy into electric energy.
(b) Blades of windmill move due to wind speed. Fan of windmill is attached to a turbine. Turbine converts kinetic energy into electric energy.
(c) Solar energy plant uses two methods to produce electricity. In one method, photovoltaic cells convert heat energy into electric energy. In another method, water is converted into steam by using heat energy. Steam is then utilized to turn the turbines and electricity is produced.

Question 3:
Total energy of an ideal simple pendulum remains conserved. Prove this statement by calculating energy in different stages of simple pendulum.
Answer:
Let us assume that a simple pendulum is suspended from a rigid support. Point A shows the mean position of simple pendulum and points B and C show the maximum displacement from mean position.

When simple pendulum is displaced to B, its potential energy reaches the maximum and its kinetic energy becomes zero.
When simple pendulum comes back to its mean position, its kinetic energy becomes the maximum and potential energy becomes zero.
If m is mass and l is length of simple pendulum and it is displaced by distance x then its potential energy can be given as follows:

In this situation, kinetic energy is zero and hence mechanical energy of simple pendulum:
= $\frac { 1 }{ 2 }$ Kx²
When the simple pendulum is at mean position then potential energy becomes zero and kinetic energy is at maximum.
Here, mechanical energy = Kinetic energy + Potential energy
= KE + 0
= $\frac { 1 }{ 4 }$ Kx²
So, if potential energy changes, then kinetic energy changes accordingly to keep the mechanical energy of the system (simple pendulum) constant.

Question 4:
Explain different types of dissipation of energy during conversion of energy. What can be done to reduce such dissipation?
Answer:
During conversion of energy from one form to another, all of energy does not get converted into one form of energy rather some of the energy is dissipated in the form of heat, light and sound energy. For example; when a ceiling fan is switched on, electrical energy is converted into kinetic energy. During this change, some of the energy is converted into heat energy which is evident from ceiling fan becoming hot.
Following steps can be taken to reduce dissipation of energy:

The machine should be subjected to continuous maintenance and timely repair.
Lubricants should be used in machine.
More efficient machines should be designed, e.g. a car with better mileage.
Electrical appliances of higher star rating should be used.

Question 5:
Prove that the mechanical energy of a free falling object remains constant at every point of motion.
Answer:
Energy can neither be created nor destroyed, it can only be transformed from one form to another. The total energy before and after transformation remains the same.

A body of mass’m’ is raised to height ‘h’. At A its potential energy is maximum and kinetic energy is 0 as it is stationary.
When body is allowed to free fall and it reaches at B, h is decreasing hence potential energy decreases and V is increasing hence kinetic energy is increasing.
When the body is about to reach the ground level, h = 0, ν will be maximum hence kinetic energy > potential energy
∴ Decrease in potential energy = Increase in kinetic energy
This shows the continual transformation of gravitational potential energy into kinetic energy.
Potential energy + Kinetic energy = Constant (Mechanical energy)

Work, Energy and Power Numerical Questions

Question 1:
An electron is moving with velocity 1.2 x 10⁶ m/s. If mass of electron is 9.1 x 10^2-31 kg then find its kinetic energy.
Answer:
Here, velocity = 1.2 x 10 m/s and mass = 9.1 x 10″31 Kinetic energy can be calculated as follows:

Question 2:
A machine moves an object of 40 kg mass to 10 m height. Find the amount of work done.
Answer:
Here mass = 40 kg, h = 10 m and g = 9.8 m/s²
Work done can be calculated as follows:
W = F.s = mgh
W = 40 kg x 9.8 ms^-2 x 10
m = 3920 J = 3.92 kJ

Question 3:
An object of mass 6 kg falls from a height of 5 m. Find the change in its potential energy. (g = 10 m/s2)
Answer:
When the object falls its potential energy becomes zero when it reaches at ground level. So, change in potential energy is equal to potential energy when the object is at full height as in question.
Mass = 6 kg, h = 5 m
PE = mgh
= 6 kg x 10 ms^-2 x 5 ms = 300 J

Question 4:
Spring constant of a spring is 4 x 10³ N/m. How much work is done in compressing this spring by 0.04 m?
Answer:
Work done can be calculated as follows:

Question 5:
When a spring is pulled up to 0.02 m then work done is 0.4 J. Find the spring constant.
Answer:
Spring constant can be calculated as follows:

Question 6:
An engine moves an object of 200 kg mass to a height of 50 m in 10 second. Find the power of the engine, (g = 10 m/s²)
Answer:
Work done can be calculated as follows:

Question 7:
5 electrical appliances are used for 10 hours per day in a home. If two appliances are of 200 W and three appliances are of 400 W then find number of units consumed by these appliances in a day.
Answer:
Total power of 5 electrical appliances
= 2 x 200 W + 3 x 400 W
= 200 W + 1200 W = 1600 W
= 1.6 kW
Units per day = 1.6 x 10 h = 16

Question 8:
An object of 40 kg mass is moving with velocity 2 m/s. A force applied on it increases its velocity to 5 m/s. Find the work done by force.
Answer:
Work done can be calculated as follows:

Question 9:
If an object of mass 50 kg is lifted to height of 3 m then find its potential energy. If this object falls freely from that height then find its kinetic energy when it is at halfway distance. (g = 10 m/s² )
Answer:
Potential energy can be calculated as follows:
PE = mgh
= 50 x 10 x 3 = 1500 J
Since KE is zero at that height hence mechanical energy of object at 3 m height = 1500 J 3
At half the height, i.e. 1.5 m, potential energy can be calculated as follows:
PE = mgh
= 50 x 10 x 1.5 = 750 J
At this stage also the mechanical energy would be 1500 J Hence, KE = Mechanical Energy – Potential Energy = 1500 J – 750 J = 750 J

Question 10:
A block of 8 kg is moving on a frictionless surface with velocity 4 m/s. After compressing a spring, this block comes to rest. What is the compression in spring if spring constant is 2 x 10 ⁴N/m?
Answer:
Kinetic energy of block can be calculated as follows:

The KE will be equal to work done on compressing the spring. Hence, compression can be calculated as follows:

Work, Energy and Power Additional Questions Solved

I. Multiple Choice Questions

Question 1:
The unit of work is joule. The other physical quantity that has same unit is
(a) power
(b) velocity
(c) energy
(d) force
Answer:
(c) energy

Question 2:
The spring will have maximum potential energy when
(a) it is pulled out
(b) it is compressed
(c) both (a) and (b)
(d) neither (a) nor (b)
Answer:
(c) both (a) and (b)

Question 3:
The gravitational potential energy of an object is due to
(a) its mass
(b) its acceleration due to gravity
(c) its height above the earth’s surface
(d) all of the above.
Answer:
(d) all of the above.

Question 4:
A ball is dropped from a height of 10 m.
(a) Its potential energy increases and kinetic energy decreases during the falls.
(b) Its potential energy is equal to the kinetic energy during the fall.
(c) The potential energy decreases and the kinetic energy increases during the fall.
(d) The potential energy is ‘O’ and kinetic energy is maximum while it is falling.
Answer:
(c) The potential energy decreases and the kinetic energy increases during the fall.

Question 5:
If the velocity of a body is doubled its kinetic energy
(a) gets doubled
(b) becomes half
(c) does not change
(d) becomes 4 times
Answer:
(d) becomes 4 times

Question 6:
How much time will be required to perform 520 J of work at the rate of 20 W?
(a) 24 s
(b) 16 s
(c) 20 s
(d) 26 s
Answer:
(d) 26 s

Question 7:
A student carries a bag weighing 5 kg from the ground floor to his class on the first floor that is 2 m high. The work done by the boy is
(a) 1 J
(b) 10 J
(c) 100 J
(d) 1000 J
Answer:
(c) 100 J

Question 8:
The work done is ‘0’ if
(a) The body shows displacement in the opposite direction of the force applied.
(b) The body shows displacement in the same direction as that of the force applied.
(c) The body shows a displacement in perpendicular direction to the force applied.
(d) The body moves obliquely to the direction of the force applied.
Answer:
(c) The body shows a displacement in perpendicular direction to the force applied.

Question 9:
One unit of electrical energy is equal to
(a) 3.6x 10⁵J
(b) 3.6 x 10⁶ J
(c) 36 x 10⁵ J
(d) both (b) and (c)
Answer:
(d) both (b) and (c)

Question 10:
Which method is used to produce electricity in hydro electric power plant?
(a) By boiling the water to produce steam
(b) By ionizing water
(c) By running dynamo by kinetic energy of water
(d) Any of the above
Answer:
(c) By running dynamo by kinetic energy of water

Question 11:
Which method is used to produce electricity in thermal power plant?
(a) By heating chargeable cells
(b) By boiling water to produce steam
(c) By pushing pistons by heat energy
(d) Any of the above
Answer:
(b) By boiling water to produce steam

Question 12:
In which of the following kinetic energy is converted into electrical energy?
(a) Tidal energy
(b) Hydro electric
(c) Wind energy
(d) All of these
Answer:
(d) All of these

Question 13:
Which of the following is the ultimate source of energy for us?
(a) LPG
(b) Nuclear
(c) Solar
(d) CNG
Answer:
(c) Solar

Work, Energy and Power Very Short Answer Type Questions

Question 1:
Define work.
Answer: When a force acts on an object and the object moves, we say that the force has done work on the object.
W = F x s

Question 2:
What is the unit of work done?
Answer:
Joule.

Question 3:
Define 1 Joule of work.
Answer:
When a force of 1 Newton acts on an object and the object moves a distance of 1 metre in the direction of the force, the work done by the force is 1 Joule.

Question 4:
Name two conditions required to do work.
Answer:
Two conditions required to do work are force applied and displacement produced.

Question 5:
What is potential energy?
Answer:
The energy possessed by a body due to its height or position is called potential energy.

Question 6:
What is the formula of work done and give the units of each symbol.
Answer:
Formula → W = Fx s ∴W, work
done → Joule = Newton x Metre; F,
force → Newton; s, displacement → Metre

Question 7:
If the work done is 20 J and displacement is 2 m then find the force applied.
Answer:
W = F x s
20 = F x 2
∴F = 10 N

Question 8:
State whether the given phenomenon is an example of work done: When we push a pebble lying on surface.
Answer:
On pushing a pebble lying on a surface, it moves through a distance so work is done.

Question 9:
Give the formula of kinetic energy and potential energy.
Answer:
K.E. = mν² P.E. = mgh
m = mass m = mass
ν = velocity g = acceleration due to gravity
h = height

Question 10:
Name the energy stored when a rubber band is stretched?
Answer:
On stretching a rubber bajrd, potential energy is stored in it.

Question 11:
How many joules are there in 1 kj?
Answer:
1000 Joules = 1 kj

Question 12:
Define power.
Answer:
Power is defined as the rate of doing work or the rate of transfer of energy.
P = $\frac { W }{ t }$

Question 13:
What is the unit of power? Define it.
Answer:
The unit of power is watt.
1 watt is the power of an agent which does work at the rate of 1 joule per second.

Question 14:
What is the commercial unit of energy? Define it.
Answer:
The commercial unit of energy is kWh [kilowatt-hour]. 1 kWh is the energy used in one hour at the rate of 1000 J/s.

Question 15:
Convert 1 kWh into joules.
Answer:
1 kWh = 1 kW x 1 h
= 1000 W x 3600 s
= 3600000J
= 3.6 x 106J

Question 16:
What is mechanical energy?
Answer:
The sum of kinetic energy and potential energy of an object is called its mechanical energy.

Question 17:
Give one example where work done on an object is zero.
Answer:
A person walking horizontally straight with load on its head is said to be doing no work on the load.
This is because the direction of force acting on the load and the displacement seen is perpendicular.

Question 18:
Give one example where work done on an object is negative.
Answer:
A person walking down the stairs with load on its head, the work done is negative as the force applied on the load is in opposite direction to its displacement.

Question 19:

In the above figure a pendulum is shown oscillating from A to B, B to C and C to A. In its one oscillation state the position. Where its kinetic energy is maximum.
Answer:
A pendulum when raised to certain height and is at B, on releasing it from B it attains motion and hence its potential energy is transformed to kinetic energy.
The maximum kinetic energy is at A when it is in motion.

Question 20:
A man does 60 J of work in 6 seconds. Calculate the power
Answer:

Question 21:
What is a source of energy?
Answer:
A source of energy is one which is capable of providing an adequate amount of useful energy.

Work, Energy and Power Short Answer Type Questions

Question 1:
What is gravitational potential energy?
Answer:
The gravitational potential energy of an object at a point above the ground is defined as the work done in raising the object from the ground to that point against gravity.
G.P.E. = mgh

Question 2:
Differentiate between potential energy and kinetic energy.
Answer:

Potential Energy	Kinetic Energy
1. Energy possessed by a body due to its height.	Energy possessed by a body due to its motion.
2. P.E. = mgh m = mass g = acceleration due to gravity h = height	K.E = 1/2 mν² m = mass v = velocity

Question 3:
Give two situations where energy is supplied but no work is done.
Answer:
(i) A person pushing a heavy rock is using all the energy but if the rock does not move no work is done.
(ii) A person standing with heavy load on his head, energy is used/spent in doing this but no work is done.

Question 4:
Answer:
An object having a capability to do work is said to possess energy.
The object which does the work loses energy and the object on which the work is done gains energy. The unit of both energy and work is Joule.

Question 5:
Give two examples in your daily life where you can see kinetic energy is doing work.
Answer:
(i) Blowing wind
(ii) Rotating wheel

Question 6:
Name the form of energy present in the following conditions/situations:
(a) stretched bow and arrow
(b) coiled spring
(c) a falling coconut
(d) water stored in the dam
Answer:
(a) Potential energy
(b) Potential energy
(c) Kinetic energy
(d) Potential energy

Question 7:
What is energy? Name different forms of energy. Give the unit of energy.
Answer:
Energy is the capacity of a body to do work. Different forms of energy are:
Potential energy, kinetic energy, heat energy, chemical energy, electrical energy,

Question 8:
Two stones A and B of same mass fall from height h₁ and h₂ respectively on sand where h₂ hr Which stone will exert more force on the sand and why? Name the energy present in it.
Answer:
Stone B with height h₂ has more energy, potential energy is present in it, when the stone falls on the sand it exerts more force than stone A due to more energy transformation. When the stone falls all the potential energy present in it, is transformed into kinetic energy.

Question 9:
Derive the formula for kinetic energy.
Answer:
Let an object of mass m, move with uniform velocity u.
Let us displace it by s, due to constant force F, acting on it.
Work done on the object of mass ‘m’ is
W = F x s …(i)
Due to the force, velocity changes to u and the acceleration produced is ‘a’. Relationship between ν, u, a and s can be given by formula ν² – u² = 2as

Question 10:
Derive the formula for potential energy.
Answer:
Consider a body with mass ‘m’ raised through a height ‘h’ from the ground level.
Force required to raise the object = weight of object.

Question 11:
Explain, law of conservation of energy.
Answer:
Law of conservation of energy states
that energy can neither be created nor destroyed but it can be transferred from one form to another. The total energy before and sifter transformation remains the same.
Let the body be at position A from the ground level. It has potential energy PE. When the body falls down and reaches in the middle P.E. = K.E. and when the body is just about to touch the ground its K.E. > P.E. This shows that P.E. is getting transferred into kinetic energy.

Question 12:
Two boys A and B were given a task to carry 20 kg load from ground level to height 10 m. A completed the work in 40 s and B in 60 s. Calculate the power in both the cases. Who has greater power?
Answer:
The height at which the object is raised from the ground is 2.3 m.

Question 13:
An electrical appliance of 100 W is used for 3 h per day. Calculate the units of energy consumed per day and for a month.
Answer:
Power of electrical appliance = 100 W = 0.1 kW
Time used = 3 h
Energy = power x time taken
= 0.1 kW x 3 h
= 0.3 kWh
= 0.3 units (∵ 1 kWh = 1 unit)
The units of energy consumed per day = 0.3 units and for a month = 0.3 x 30
= 9 units.

Question 14:
A body possess potential energy of 460 J whose mass is 20 kg and is raised to a certain height. What is the height when g = 10 m/s²?
Answer:

The height at which the object is raised fromthe ground is 2.3 m.

Question 15:
Explain the working of a hydroelectric power plant to produce electricity.
Answer:
In order to produce hydroelectricity, high rise dams are constructed on the river to obstruct the flow of water and thereby collect water in very large reservoirs. The water from high level in the dam is carried through pipes, to the turbine, at the bottom of the dam. The turbines are connected to the generator. Rotation of the turbines makes the generator produce electricity. Water when stored at a height in the dam has a large amount of potential energy which gets converted into the kinetic energy of flowing water when it is allowed to fall on the turbines it gets converted into mechanical energy of the turbine and finally gets converted into the electric energy by the generator.

Question 16:
What is a solar cell panel? Write two advantages associated with such panels.
Answer:
It is a panel on which many solar cells are installed, which convert Sun’s energy into electrical energy.
Two advantages: Solar cell panels can be setup at remote places and secondly it has no moving parts, require less maintenance and work quite satisfactorily without the use of any focusing device.

Work, Energy and Power Long Answer Type Questions

Question 1:
(a) Define power. What is the S.I. unit of power and commercial unit of power?
(b) A tube light of 80 W is used for 8 hours per day. Calculate the unit of energy consumed in 1 day by the tube light.
Answer:
(a) Power is the rate of doing work.
S.I. unit of power = watt
1 watt = $\frac { 1J }{ 1s }$
Commercial unit of power kWh
1kWh= 3.6 x 106 Joules.
(b) Power of electric tube light = 80 W
=0.08 kW
Time used = 8 hour
Energy = power x time taken = 0.08 x 8
= 0.64 kWh
= 0.64 units.
∴ The energy consumed per day = 0.64 units.

Question 2:
(a) List any 3 situations in your daily life where you can say work has been done.
(b) What work is said to be done to increase the velocity of a car from 15 km/h to 30 km/h, if the mass of the car is 1000 kg?
Answer:
(a) 3 situations where work is done are:
(i) If a pebble lying on a surface is pushed, the pebble moves through a distance. In this situation work is said to be done.
(ii) If a girl pulls a trolley and the trolley moves through a distance. In this situation work is done.
(iii) If a book is lifted to a height, the force is applied and the book rises up, work is said to be done.

Question 3:
Explain the following:
(a) An object increases its potential energy when raised through a height.
(b) Energy is neither created nor destroyed then from where do we get energy.
(c) When we push the wall, the wall does not move and no work is done.
Answer:
(a) An object has some mass ‘M, when it is raised through a certain height, energy is applied on it, this energy gets transformed and is gained by the object hence its potential energy increases.
(b) Energy is present in every object in some or other form. It exist in sun, planet, wind, water etc, the energy gets
transformed and we get, because of its transformation into different form. For e.g., sun energy → plants convert it (due to chemical reaction) into food energy -» we eat green vegetables products of plants, energy enters our body.
(c) Work is said to be done, when force is applied on the body and it moves to certain distance. When a wall is pushed there is no displacement and we say that work done is zero.

Question 4:
State and explain one example where
(i) Kinetic energy is present in a body and is used; and
(ii) Potential energy is present in a body and is used.
Answer:
(i) Kinetic energy is the energy present in a body by virtue of its motion. When water is stored in a dam at particular
height, we say it has maximum potential energy as the water is present at a height. When this water is allowed to fall down from a certain height, it is said to have kinetic energy. When the water falls on a turbine, the paddle wheels rotates to rotate the turbine. Hence the kinetic energy of water is used to push or rotate the turbine which is further used to produce the electricity.
(ii) Potential energy present in a body is due to virtue of its position. In hydel power plant, the kinetic energy of water to rotate the turbine is obtained by the use of potential energy of water stored in dam. When water is stored in dam at a certain height, potential energy is stored in the water which transforms into kinetic energy when it released from that height.

We hope the given RBSE Solutions for Class 10 Science Chapter 11 Work, Energy and Power will help you. If you have any query regarding Rajasthan Board RBSE Class 10 Science Solutions Chapter 11 Work, Energy and Power, drop a comment below and we will get back to you at the earliest.

RBSE Solutions for Class 10 Maths Chapter 1 वैदिक गणित Ex 1.2

December 6, 2019 by Veer Leave a Comment

RBSE Solutions for Class 10 Maths Chapter 1 वैदिक गणित Ex 1.2 is part of RBSE Solutions for Class 10 Maths. Here we have given Rajasthan Board RBSE Class 10 Maths Chapter 1 वैदिक गणित Exercise 1.2.

Board	RBSE
Textbook	SIERT, Rajasthan
Class	Class 10
Subject	Maths
Chapter	Chapter 1
Chapter Name	वैदिक गणित
Exercise	Exercise 1.2
Number of Questions Solved	32
Category	RBSE Solutions

Rajasthan Board RBSE Class 10 Maths Chapter 1 वैदिक गणित Ex 1.2

उपसूत्र यावदूनम तावदूनी द्वारा वर्ग ज्ञात कीजिए

प्रश्न 1.
93
हल:
(93)² = 93 – 07/07²
= 86/49 = 8649 उत्तर
संकेत — आधार = 100
विचलन = 93 – 100 = – 07
सूत्र— (संख्या)² = संख्या + विचलन/(विचलन)²

प्रश्न 2.
(106)
हल:
(106)² = 1(106 + 06)/06²
= 1 x 112/36
= 11236 उत्तर,
संकेत-आधार = 100
उपाधार = 200
उपाधार = 2, विचलन = 6
सूत्र-(संख्या)² = उपाधार(RBSESolutions.com) अंक (संख्या + विचलन)/(विचलन)²

प्रश्न 3.
211
हल:
(211)² = 2(211 + 11)/(11)²
= 2 x 222/121
= 444/21 = 44521 उत्तर
संकेत-
आधार = 100
उपाधार = 200
आधार अंक = 2, विचलन = 11
सूत्र-(संख्या) = उपाधार(RBSESolutions.com) अंक (संख्या + विचलन)/(विचलन)

प्रश्न 4.
405
हल:
(405)² = 4(405 + 05)/(05)²
= 4 x 410/25
= 164025 उत्तर
संकेत-
आधार = 100
उपाधार = 400
उपाधार अंक = 4, विचलन = 05
सूत्र-(संख्या)² – उपाधार(RBSESolutions.com) अंक (संख्या + विचलन)/(विचलन)²

उपसूत्र आनुरूप्येण द्वारा वर्ग ज्ञात कीजिए।

प्रश्न 5.
16
हल:
(16)² =

संकेत-

तीन खण्ड बनाते हैं।
प्रथम में दहाई तथा तृतीय में इकाई के अंक वर्ग लिखते हैं।
द्वितीय खण्ड में इकाई व दहाई का गुणन लिखते हैं तथा उसके नीचे पुनः वही गुणनफल लिखते हैं।
योगफल संख्या का अभीष्ट वर्ग है। मध्य खण्ड व तृतीय खण्ड में 1 – 1 अंक ही लिखते हैं।

प्रश्न 6.
31
हल:
(31)² =

प्रश्न 7.
24
हल:

प्रश्न 8.
56
हल:

सूत्र एकाधिकेन पूर्वेण द्वारा वर्ग ज्ञात करो।

प्रश्न 9.
45
हल:
(45)² = 4 x (4 + 1)/5²
= 4 x 5/25
= 20/25
= 2025 उत्तर
संकेत-

इसमें दहाई के अंक से एक(RBSESolutions.com) अधिक अंक लेकरे गुणा करते हैं।
इकाई के अंक का वर्ग कर इसके साथ लिख देते हैं।

प्रश्न 10.
85
हल:
(85)² = 8 x (8 + 1)/5²
= 72/25
= 7225 उत्तर

प्रश्न 11.
115
हल:
(115)² = 11 x 12/(5)²
= 132/25
= 13225 उत्तर

प्रश्न 12.
125
हल:
(125)² = 12 x 13/5²
= 156/25
= 15625 उत्तर

सूत्र संकलन व्यवकलन द्वारा वर्ग ज्ञात कीजिए।

प्रश्न 13.
23
हल:
(23)² = (23 + 3) (23 – 3) + 3² संकेत-इष्ट संख्या = 3
= 26 x 20 + 9
= 520 + 9 = 529 उत्तर

प्रश्न 14.
38
हल:
(38)² = (38 + 2.) (38 – 2) + 2²
= 40 x 36 + 4
= 1440 + 4
= 1444 उत्तर
संकेत-

इष्ट संख्या = 2
इष्ट संख्या का चयन इस प्रकार किया(RBSESolutions.com) जाता है कि जोड़ने या घटाने पर शून्यान्त संख्या प्राप्त हो।

प्रश्न 15.
(69)
हल:
(69)² = (69+ 1) x (69 – 1) + (1)² संकेत-इष्ट संख्या = 1
= 70 x 68 + 1
= 4760 + 1 = 4761 उत्तर

प्रश्न 16.
89
हल:
(89)² = (89+ 1) (89 – 1) + (1)² संकेत-इष्ट संख्या = 1
= 90 x 88 + 1
= 7920 + 1 = 7921 उत्तर

द्वन्द्व योग द्वारा वर्ग ज्ञात कीजिए।

प्रश्न 17.
362
हल:
362 के पांच अंक समूह बनेंगे
(362)² = 32/3 x 6 x 2/3 x 2 x 2 + 6/6 x 2 x 2/2²
= 9/₃6/₄8/₂4/4
(362)² = 131044 उत्तर
संकेत-

सर्वप्रथम वर्ग ज्ञात करने वाली संख्या के अंक(RBSESolutions.com) समूह बनाते हैं। जैसे-3, 36, 362, 62, 2 इसके समूह.
इनको द्वन्द्व योग के अनुसार क्रम में लिखते हैं।
इकाई अंक की ओर से योग करते हैं। एक खण्ड में एक अंक रखते हैं। योगफल ही अभीष्ट संख्या का वर्ग होगा।

प्रश्न 18.
453
हल:
(453)² = अंक समूह = 4, 45, 453, 53, 3
इन्हें पाँच खण्डों में लिखने पर।
= 4²/4×5 x 2/4×3 x 2 + 5/5 x 3 x 2/3²
= 16/₄0/₄9/₃0/9 योग करने पर
= 205209 उत्तर

प्रश्न 19.
4312
हल:
4312 के सात अंक समूह निम्न प्रकार बनेंगे-
4, 43, 431, 4312, 312, 12, 2
(4312)² = (4)²/4 x 3 x 2/4 x 1 x 2 + 3²/4 x 2 x 2 + 3 x 1 x 2/3 x 2 x 2 + 1²/1 x 2 x 2/2²
= 16/₂4/₁7/₂2/₁3/4/4
(4312)² = 18593344 उत्तर

प्रश्न 20.
2456
हल:
(2456)² इसके सात अंक(RBSESolutions.com) समूह निम्न प्रकार हैं
2, 24, 245, 2456, 456, 56, 6
(2456)² = 22/2×4 x 2/2 x 5 x 2 + 4²/2 x 6 x 2 + 4 x 5
x 2/4 x 6 x 2 + 5²/5 x 6 x 2/6²
= 4/₁6/₃6/₆4/₇3/₆0/₃6
(2456)² = 6031936

सूत्र निखिलम् द्वारा घनफल ज्ञात कीजिए।

प्रश्न 21.
14
हल:
सूत्र–घनफल = संख्या + 2 x विचलन/3(विचलन)²/(विचलन)³
(14)³ = 14 + 2(4)/3 x (4)²/4
= 14 + 8/₄8/₆4
= 2744 उत्तर
संकेत-आधार = 10 तथा विचलन = 14 – 10 = 4

प्रश्न 22.
97
हल:
सूत्र–घनफल = संख्या + 2 x विचलन/3(विचलन)²/(विचलन)³
(97)³ = 97 + 2 (-3)/3 x (- 3)/(- 3)³
(97)³ = 97 – 6/3 x 9/- 27
= 91/26/100 – 27
= 912673 उत्तर
संकेत–यहाँ आधार = 100 तथा विचलन = – 3 है।

प्रश्न 23.
27
हल:
(27)³ = (उपाधार अंक)² (संख्या + 2 x विचलन)(RBSESolutions.com)/उपाधार अंक x 3 x (विचलन)/(विचलन)³
यहाँ उपाधार = 30, उपाधार अंक = 3, विचलन = – 3
अतः (27)³ = (3)² (27 + 2x – 3)/3 x 3 x (-3)²/(- 3)³
= 189/₈1/- 27
= 189/₇8/30 – 27
= 19683 उत्तर

प्रश्न 24.
395
हल:
(395)³ = 4²(395 + (-5) x 2)/4×3 x (- 5)²/(- 5)³
= 16(395 – 10)/300/- 125
= 6160/₂98/200 – 125
= 6160/₂98/75
= 61629875 उत्तर
संकेत–यहाँ आधार = 400, उपाधार(RBSESolutions.com) अंक = 4, विचलन = – 5 मध्य व तृतीय खण्ड में 2 = 2 अंक होंगे।

उपसूत्रं आनुरूप्येण द्वारा घनफल ज्ञात कीजिये।

प्रश्न 25.
16
हल:
16 के घनफल के खण्ड

संकेत

इसके लिए चार खण्ड बनेंगे।
बायीं ओर से प्रथम खण्ड में संख्या के दहाई अंक का घन तथा चतुर्थ खण्ड में संख्या के इकाई अंक का घन है।
दूसरे खण्ड में दहाई अंक का वर्ग x इकाई अंक का वर्ग है।
तीसरे खण्ड में दहाई अंक x इकाई अंक वर्ग है।
दूसरे व तीसरे खण्ड में प्राप्त गुणनफल का दुगुना उन्हीं खण्डों में और जोड़ते हैं।
द्वितीय, तृतीय तथा चतुर्थ खण्ड में एक-एक अंक रहेगा। सबका योगफल ही अभीष्ट घनफल है।

प्रश्न 26.
33
हल:
(33)³

संकेत

इसके लिए चार खण्ड बनेंगे।
बायीं ओर से प्रथम खण्ड में संख्या के दहाई(RBSESolutions.com) अंक का घन तथा चतुर्थ खण्ड में संख्या के इकाई अंक का घन है।
दूसरे खण्ड में दहाई अंक का वर्ग x इकाई अंक का वर्ग है।
तीसरे खण्ड में दहाई अंक x इकाई अंक वर्ग है।
दूसरे व तीसरे खण्ड में प्राप्त गुणनफल का दुगुना उन्हीं खण्डों में और जोड़ते हैं।
द्वितीय, तृतीय तथा चतुर्थ खण्ड में एक-एक अंक रहेगा। सबका योगफल ही अभीष्ट घनफल है।

प्रश्न 27.
41
हल:
(41)³

(41)³ = 68921 उत्तर

प्रश्न 28.
52
हल:
(52)³

सूत्र एकाधिकेन पूर्वेण द्वारा घनफल ज्ञात कीजिए।

प्रश्न 29.
45
हल:

संकेत

आधार = 10, विचलन = 5 x 3 – 10 = 5
गुणन संक्रिया को चार खण्डों में लिखते हैं। बायें से(RBSESolutions.com) प्रथम खण्ड= दहाई अंक का वर्ग x उसको एकाधिक, द्वितीय खण्ड = दहाई अंक का वर्ग x विचलन, तृतीय खण्ड = 3 x दहाई अंक x (इकाई अंक), चतुर्थ खण्ड = (इकाई अंक) जहाँ विचलन = इकाई अंक x 3 – 10

प्रश्न 30.
73
हल:

संकेत–विचलन = 9-10 = – 1, आधार = 10

प्रश्न 31.
24
हल:

संकेत-विचलने = 4 x 3 – 10 = 10 = 2, आधार = 10

प्रश्न 32.
106
हल:

संकेत-विचलन = 6 x 3 – 10 = 8

We hope the RBSE Solutions for Class 10 Maths Chapter 1 वैदिक गणित Ex 1.2 help you. If you have any query regarding Rajasthan Board RBSE Class 10 Maths Chapter 1 वैदिक गणित Exercise 1.2, drop a comment below and we will get back to you at the earliest.

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Rajasthan Board RBSE Class 10 Maths Chapter 1 वैदिक गणित Ex 1.2

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