RD Sharma Class 9 Solutions Chapter 7 Introduction to Euclid’s Geometry Ex 7.3

RD Sharma Class 9 Solutions Chapter 7 Introduction to Euclid’s Geometry Ex 7.3

These Solutions are part of RD Sharma Class 9 Solutions. Here we have given RD Sharma Class 9 Solutions Chapter 7 Introduction to Euclid’s Geometry Ex 7.3

Other Exercises

Question 1.
Draw the graph of each of the following linear equations in two variables.
RD Sharma Class 9 Solutions Chapter 7 Introduction to Euclid’s Geometry Ex 7.3 Q1.1
Solution:
(i)x + y = 4
x = 4 – y
If y = 0, then x = 4
If y = 4, then x = 0
Now plot the points (4, 0) and (0, 4) on the graph and join them ro get the graph of the given equation
RD Sharma Class 9 Solutions Chapter 7 Introduction to Euclid’s Geometry Ex 7.3 Q1.2
(ii)x – y = 2
x = 2 +y
If y = 0, then x = 2 and if y = 1,
Then x = 2 + 1 = 3
RD Sharma Class 9 Solutions Chapter 7 Introduction to Euclid’s Geometry Ex 7.3 Q1.3
Now plot the points (2, 0) and (3, 1) on the graph and join them to get the graph of the equation.
RD Sharma Class 9 Solutions Chapter 7 Introduction to Euclid’s Geometry Ex 7.3 Q1.4
(iii) -x+y = 6 ⇒  y = 6+x
RD Sharma Class 9 Solutions Chapter 7 Introduction to Euclid’s Geometry Ex 7.3 Q1.5
If x = 0, then y = 6 + 0 = 6
If x = -1, then y = 6 – 1 = 5
RD Sharma Class 9 Solutions Chapter 7 Introduction to Euclid’s Geometry Ex 7.3 Q1.6
Now plot the points (0, 6) and (-1, 5) on the graph and join them to get a graph of the line.
(iv) y = 2x
If x = 0, then y =  2 x 0 = 0
If x = 1, then y = 2 x 1 = 2
RD Sharma Class 9 Solutions Chapter 7 Introduction to Euclid’s Geometry Ex 7.3 Q1.7
RD Sharma Class 9 Solutions Chapter 7 Introduction to Euclid’s Geometry Ex 7.3 Q1.8
Now plot the points (0, 0) and (1, 2) on the graph and join them to get the graph of the line.
RD Sharma Class 9 Solutions Chapter 7 Introduction to Euclid’s Geometry Ex 7.3 Q1.9
RD Sharma Class 9 Solutions Chapter 7 Introduction to Euclid’s Geometry Ex 7.3 Q1.10
Now plot the points (5, 0) and (0, 3) on the graph and join them to get the graph of the line.
RD Sharma Class 9 Solutions Chapter 7 Introduction to Euclid’s Geometry Ex 7.3 Q1.11
Now plot the points (4, 0) and (2, -3) on the graph and join them to get the graph of the line.
RD Sharma Class 9 Solutions Chapter 7 Introduction to Euclid’s Geometry Ex 7.3 Q1.12
RD Sharma Class 9 Solutions Chapter 7 Introduction to Euclid’s Geometry Ex 7.3 Q1.13
Now plot the points (-1, 2) and (2, 3) on the graph and join then to get the graph of the line.
RD Sharma Class 9 Solutions Chapter 7 Introduction to Euclid’s Geometry Ex 7.3 Q1.14
RD Sharma Class 9 Solutions Chapter 7 Introduction to Euclid’s Geometry Ex 7.3 Q1.15
Now plot the points (1, 0) and (-1, 1) on the graph and join then to get the graph of the line.

Question 2.
Give the equations of two lines passing through (3, 12). How many more such lines are there, and why ?
Solution:
∵  Points (3, 12) lies on the lines passing through the points
∴ Solutions is x = 3,y- 12
∴  Possible equation can be
x + y = 15
-x+y = 9
4x-y = 0
3x – y + 3 = 0

Question 3.
A three-wheeler scooter charges ₹15 for first kilometer and ₹8 each for every subsequent kilometer. For a distance of x km, an amount of ₹y is paid. Write the linear equation representing the above information.
Solution:
Charges for the first kilometer = ₹15
Charges for next 1 km = ₹8
Distance = x km
and total amount = ₹y
∴ Linear equation will be,
15 + (x- 1) x 8 =y
⇒  15 + 8x – 8 = y
⇒   7 + 8x = y
∴  y = 8x + 7

Question 4.
Plot the points (3, 5) and (-1, 3) on a graph paper and verify that the straight line passing through these points also passes through the point (1, 4).
Solution:
RD Sharma Class 9 Solutions Chapter 7 Introduction to Euclid’s Geometry Ex 7.3 Q4.1
Points (3, 5) and (-1, 3) have been plotted on the graph and joined to get a line. We see that die point (1,4) also lies out.

Question 5.
From the choices given below, choose the equation whose graph is given in figure.
(i) y = x                   
(ii) x + y = 0
(iii) y = 2x                   
(iv) 2 + 3y = 7x
Solution:
RD Sharma Class 9 Solutions Chapter 7 Introduction to Euclid’s Geometry Ex 7.3 Q5.1
From the graph, we see that Points (-1, 1) and (1, -1) be on the graph of the line these will satisfy the equation of the line
∴  -x = y ⇒ x+ y = 0
i.e, required equation
∵ x + y = 0 is the graph of the equation.

Question 6.
From the choices given below, choose the equation whose graph is given in figure.
(i) y = x + 2              
(ii) y = x – 2
(ii) y = -x + 2           
(iv) x + 2y = 6
Solution:
From the graph
Points (-1,3) and (2, 0) lie on the graph of the line
Now there points, by observation, satisfy the equation y= -x+2
∴ Required equation is y = -x + 2 whose graph is given.
RD Sharma Class 9 Solutions Chapter 7 Introduction to Euclid’s Geometry Ex 7.3 Q6.1

Question 7.
If the point (2, -2) lies on the graph of the linear equation 5x + ky =4, find the value of k.
Solution:
∵  Point (2, -2) lies on the graph of the linear equation 5x + ky = 4
∴  It will satisfy it
∴ Now substituting the values of x and y 5 x 2 + k (-2) = 4
⇒ 10 – 2k = 4 ⇒  -2k = 4 – 10 = -6 -6
⇒ k= \(\frac { -6 }{ -2 }\) =3
Hence k = 3

Question 8.
Draw the graph of the equation 2x + 3p = 12. From the graph find the co-ordinates of the point.
(i) whose y -coordinates is 3
(ii) whose x-coordinates is -3
Solution:
RD Sharma Class 9 Solutions Chapter 7 Introduction to Euclid’s Geometry Ex 7.3 Q8.1
RD Sharma Class 9 Solutions Chapter 7 Introduction to Euclid’s Geometry Ex 7.3 Q8.2
Plot the points (6, 0) and (0, 4) on the graph and join them to get the graph if the line.
(i) If y = 3, then draw perpendicular from y = 3 to the line, which get meets it at P then x-coordinate of p will be
∴ coordinates of P are ( \(\frac { 3 }{ 2 }\) ,3)
(ii) If x = -3, draw perpendicular from x = -3 to the line, which meets it Q.
The y coordinates of Q will be y = 6
∴ co-ordinates of Q are (-3, 6)

Question 9.
Draw the graph of each of the equations given below. Also, find the coordinates of the points where the graph cuts the coordinates axes:
(i) 6x – 3y = 12        
(ii) -x + 4y = 8
(iii) 2x + y = 6          
(iv) 3x + 2y + 6 = 0
Solution:
RD Sharma Class 9 Solutions Chapter 7 Introduction to Euclid’s Geometry Ex 7.3 Q9.1
RD Sharma Class 9 Solutions Chapter 7 Introduction to Euclid’s Geometry Ex 7.3 Q9.2
Now plot the points of each equation and join then we get four lines as shown on the graphs.
Equation (i) cuts the axes at (2, 0) and (0, -4)
Equation (ii) cuts the axes at (-8, 0) and (0, 2)
Equation (iii) cuts the axes at (3, 0), (0, 6) and
Equation (iv), cuts the axes at (-2, 0) and (0,-3)
RD Sharma Class 9 Solutions Chapter 7 Introduction to Euclid’s Geometry Ex 7.3 Q9.3

Question 10.
A lending library has a fixed charges for the first three days and an additional charge for each day thereafter. Aarushi paid ₹27 for a book kept for seven days. If fixed charges are ₹x and per day charges are ₹y. Write the linear equation representing the above information.
Solution:
Let fixed charges for first 3 days = ₹x
and additional charges for each day = ₹y
Total period = 7 days
and amount charges = ₹27
∴ x + (7 – 3) x  = 27
⇒  x + 4y = 27
Hence x + 4y = 27

Question 11.
A number is 27 more than the number obtained by reversing its digits. If its unit’s and ten’s digit are x and y respectively, write the linear equation representing the above statement.
Solution:
Let unit’s digit = x
and tens digit = y
∴  Number = x + 10y
By reversing the digits, units digit = y
and ten’s digit = x
∴  number = y + 10x
Now difference of these two numbers = 27 (x + 10y) – (y +10x) = 27
x + 10y – y – 10x = 27
⇒  -9x + 9y – 27 = 0
⇒ x-_y + 3 = 0                   (Dividing by -9)
Hence equation is x – y + 3 = 0

Question 12.
The sum of a two digit number and the number obtained by reversing the order of its digits is 121. If units and ten’s digit of the number are x and y respectively, then write the linear equation representing the above statement.
Solution:
Let unit digit = x
and tens digit = y
∴ Number = x + 10y
By reversing the digits,
units digit = y
and tens digit = x
∴ Number =y+ 10x
Now sum of these two numbers = 121
∴ x + 10y + y + 10x = 121
⇒  1 lx + 11y = 121
⇒  x + y = 11                        (Dividing by 11)
∴  x + y = 11

Question 13.
Draw the graph of the equation 2x + y = 6. Shade the region bounded by the graph and the coordinate axes. Also find the area of the shaded region.
Solution:
2x + y = 6
⇒  y = 6 – 2x
If x = 0, then y = 6- 2 x 0 = 6 – 0 = 6
If x = 2, then y = 6- 2 x 2 = 6- 4 = 2
RD Sharma Class 9 Solutions Chapter 7 Introduction to Euclid’s Geometry Ex 7.3 Q13.1
Now plot the points (0, 6) and (2, 2) on the graph and join them to get a line which intersects x-axis at (3, 0) and y-axis at (0,6)
Now co-ordinates if vertices of the shaded portion are (6, 0) (0, 0) and (3, 0) Now area of the shaded region.
RD Sharma Class 9 Solutions Chapter 7 Introduction to Euclid’s Geometry Ex 7.3 Q13.2

Question 14.
Draw the graph of the equation  \(\frac { x }{ 3 }\) \(\frac { y }{ 4 }\)  = 1 Also find the area of the triangle formed by the line and the co-ordinate axes.
Solution:
RD Sharma Class 9 Solutions Chapter 7 Introduction to Euclid’s Geometry Ex 7.3 Q14.1
Now plot the points (3, 0) and (0, 4) and join them to get a line which interest x-axis at A (3, 0) and y-axis at B (0, 4)
RD Sharma Class 9 Solutions Chapter 7 Introduction to Euclid’s Geometry Ex 7.3 Q14.2

Question 15.
Draw the graph of y = | x |
Solution:
y = | x |
RD Sharma Class 9 Solutions Chapter 7 Introduction to Euclid’s Geometry Ex 7.3 Q15.1
⇒   y = x       [∵ | x |=x]
∴  Now taking z points.

RD Sharma Class 9 Solutions Chapter 7 Introduction to Euclid’s Geometry Ex 7.3 Q15.2
Now plot the points (1, 1) (2, 2) and (3, 3) and join them to get a graph of a line.

Question 16.
Draw the graph of y = | x | + 2
Solution:
y – | x | + 2
⇒  y = x + 2         [| x | = x]
If x = 0, then y = 0 + 2 = 2
If x = 1, then y = 1+2 = 3
If x = 2, then y = 2 + 2 = 4
RD Sharma Class 9 Solutions Chapter 7 Introduction to Euclid’s Geometry Ex 7.3 Q16.1
Now plot the points (0, 2), (1, 3) and (2, 4) on the graph and join them to get a line.
RD Sharma Class 9 Solutions Chapter 7 Introduction to Euclid’s Geometry Ex 7.3 Q16.2

Question 17.
Draw the graphs of the following linear equation on the same graph paper.
2x + 3y = 12, x -y = 1
Find the co-ordinates of the vertices of the triangle formed by the two straight lines and the y-axis. Also find the area of the triangle.
Solution:
RD Sharma Class 9 Solutions Chapter 7 Introduction to Euclid’s Geometry Ex 7.3 Q17.1
Now plot the points (6, 0) (0, 4) on the graph to get a line.
RD Sharma Class 9 Solutions Chapter 7 Introduction to Euclid’s Geometry Ex 7.3 Q17.2
Now plot the points (1, 0) and (2, 1) on the graph to get another line.
Area of the triangle FEB so formed,
= \(\frac { 1 }{ 2 }\) FB x FL = \(\frac { 1 }{ 2 }\) x 5 x 3
= \(\frac { 15 }{ 2 }\)
= 7.5 sq. units
co-ordinates of E, F, B are E (3, 2), (0, -1) and (0, 4)

Question 18.
 Draw the graphs of the linear equations 4x – 3y + 4 = 0 and 4x + 3y – 20 = 0. Find the area bounded by these lines and x-axis.
Solution:
RD Sharma Class 9 Solutions Chapter 7 Introduction to Euclid’s Geometry Ex 7.3 Q18.1
RD Sharma Class 9 Solutions Chapter 7 Introduction to Euclid’s Geometry Ex 7.3 Q18.2
RD Sharma Class 9 Solutions Chapter 7 Introduction to Euclid’s Geometry Ex 7.3 Q18.3
Now plot the points (5, 0) and (2, 4)and join them to get a line we see that the ΔABC is formed by bounding there line with x-axis.

Question 19.
The path of a train A is given by the equation 3x + 4y – 12 = 0 and the path of another train B is given by the equation 6jc + 8y – 48 = 0. Represent this situation graphically.
Solution:
Path of the train A = 3x + 4y – 12 = 0
Path of the train B = 6x + 8y – 48 = 0
Now, 3x + 4y – 12 = 0
RD Sharma Class 9 Solutions Chapter 7 Introduction to Euclid’s Geometry Ex 7.3 Q19.1
RD Sharma Class 9 Solutions Chapter 7 Introduction to Euclid’s Geometry Ex 7.3 Q19.2
Now plot the points (4, 0) and (0, 3) on the graph and join them to get a line, and 6x + 8y – 48 = 0
⇒  6x = 48 – 8y
RD Sharma Class 9 Solutions Chapter 7 Introduction to Euclid’s Geometry Ex 7.3 Q19.3
Now plot the points (0, 6) and (4, 3) on the graph and join them to get another line.

Question 20.
Ravish tells his daughter Aarushi, “Seven years ago, I was seven times as old as you were then. Also, three years from now, I shall be three times as old as you will be”. If present ages of Aarushi and Ravish are x and y years respectively, represent this situation algebraically as well as graphically.
Solution:
Present age of Aarushi = x years
and age of Ravish = y years
7 years ago,
age of Aarushi = x – 7
years and age of Ravish =y-7 years
∴ y- 7 = 7 (X – 7)
⇒  y – 7 = 7x – 49
⇒  7x – y = -7 + 49  = 42
7x – y = 42
⇒  y = 7x – 42
If x = 6, then
y = 7 x 6 – 42 = 42 – 42 = 0,
If x = 7, then
= 7 x 7 – 42 – 49 – 42 = -7
RD Sharma Class 9 Solutions Chapter 7 Introduction to Euclid’s Geometry Ex 7.3 Q20.1
Plot the points (6, 0) (7, -7) on the graph and join them.
After 3 years,
age of Aarushi = x + 3
and age of Ravish = y + 3
⇒  y + 3 = 3(x + 3)
⇒ y + 3 = 3x + 9
⇒ y = 3x+ 9-3
⇒ y = 3x + 6
If x = -2, then y = 3 x (-2) + 6 =6-6=0
If x = 1, then y = 3 x (1) + 6 =3+6=9
RD Sharma Class 9 Solutions Chapter 7 Introduction to Euclid’s Geometry Ex 7.3 Q20.2
Plot the points (1, 9), (-2, 0) on the graph Arundeep’s Mathematics (R.D.) 9th and join them to get another line.
RD Sharma Class 9 Solutions Chapter 7 Introduction to Euclid’s Geometry Ex 7.3 Q20.3

Question 21.
Aarushi was driving a car with uniform speed of 60 km/h. Draw distance-time graph. From the graph, find the distance travelled by Aarushi in.
(i) 2\(\frac { 1 }{ 2 }\) Hours             
(ii) \(\frac { 1 }{ 2 }\) Hour
Solution:
Speed of car = 60 km / h.
RD Sharma Class 9 Solutions Chapter 7 Introduction to Euclid’s Geometry Ex 7.3 Q21.1
Now plot the points (60, 1), (120, 2) are the graph and join then to get the graph of line.
From the graph, we see that
RD Sharma Class 9 Solutions Chapter 7 Introduction to Euclid’s Geometry Ex 7.3 Q21.2
RD Sharma Class 9 Solutions Chapter 7 Introduction to Euclid’s Geometry Ex 7.3 Q21.3

Hope given RD Sharma Class 9 Solutions Chapter 7 Introduction to Euclid’s Geometry Ex 7.3 are helpful to complete your math homework.

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RS Aggarwal Class 9 Solutions Chapter 11 Circle Ex 11C

RS Aggarwal Class 9 Solutions Chapter 11 Circle Ex 11C

These Solutions are part of RS Aggarwal Solutions Class 9. Here we have given RS Aggarwal Solutions Class 9 Chapter 11 Circle Ex 11C.

Other Exercises

Question 1.
Solution:
In cyclic quad. ABCD, ∠ DBC = 60° and ∠BAC = 40°
∴∠ CAD and ∠ CBD are in the same segment of the circle.
∴∠ CAD = ∠ CBD or ∠ DBC
=> ∠ CAD = 60°
∴∠BAD = ∠BAC + ∠CAD
= 40° + 60° = 100°
But in cyclic quad. ABCD,
∠BAD + ∠BCD = 180°
(Sum of opposite angles)
=> 100° + ∠BCD = 180°
=> ∠BCD = 180° – 100°
∴ ∠ BCD = 80°
Hence (i) ∠BCD = 80° and
(ii) ∠CAD = 60° Ans.

Question 2.
Solution:
In the figure, POQ is diameter, PQRS is a cyclic quad, and ∠ PSR =150° In cyclic quad. PQRS.
∠ PSR + ∠PQR = 180°
(Sum of opposite angles)
=> 150° + ∠PQR = 180°
=> ∠PQR = 180°- 150° = 30°
=> ∠PQR =180° – 150° = 30°
Now in ∆ PQR,
∴∠ PRQ = 90° (Angle in a semicircle)
∴∠ RPQ + ∠PQR = 90°
=> ∠RPQ + 30° = 90°
=> ∠RPQ = 90° – 30° = 60° Ans.

Question 3.
Solution:
In cyclic quad. ABCD,
AB || DC and ∠BAD = 100°
∠ ADC = ∠BAD =180°
(co-interior angles)
RS Aggarwal Class 9 Solutions Chapter 11 Circle Ex 11C Q3.1
=> ∠ ADC + 100° = 180°
=> ∠ADC = 180° – 100° = 80°
∴ ABCD is a cyclic quadrilateral.
∴ ∠BAD + ∠BCD = 180°
=> 100° + ∠ BCD = 180°
=> ∠BCD = 180° – 100°
=> ∠BCD = 80°
Similarly ∠ABC + ∠ADC = 180°
=> ∠ABC + 80° = 180°
=> ∠ABC = 180° – 80° = 100°
Hence (i) ∠BCD = 80° (ii) ∠ADC = 80° and (iii) ∠ABC = 100° Ans.

Question 4.
Solution:
O is the centre of the circle and arc ABC subtends an angle of 130° at the centre i.e. ∠AOC = 130°. AB is produced to P
Reflex ∠AOC = 360° – 130° = 230°
Now, arc AC subtends reflex ∠ AOC at the centre and ∠ ABC at the remaining out of the circle.
RS Aggarwal Class 9 Solutions Chapter 11 Circle Ex 11C Q4.1

Question 5.
Solution:
In the figure, ABCD is a cyclic quadrilateral in which BA is produced to F and AE is drawn parallel to CD.
∠ABC = 92° and ∠FAE = 20°
ABCD is a cyclic quadrilateral.
RS Aggarwal Class 9 Solutions Chapter 11 Circle Ex 11C Q5.1
RS Aggarwal Class 9 Solutions Chapter 11 Circle Ex 11C Q5.2

Question 6.
Solution:
In the figure, BD = DC and ∠CBD = 30°
In ∆ BCD,
BD = DC (given)
∠ BCD = ∠ CBD
(Angles opposite to equal sides)
= 30°
But ∠BCD + ∠CBD + ∠BDC = 180° (Angles of a triangle)
=> 30°+ 30°+ ∠BDC = 180°
=> 60°+ ∠BDC = 180°
=> ∠ BDC =180° – 60° = 120°
But ABDC is a cyclic quadrilateral
∠BAC + ∠BDC = 180°
=> ∠BAC + 120°= 180°
=> ∠ BAC = 180° – 120° = 60°
Hence ∠ BAC = 60° Ans.

Question 7.
Solution:
(i) Arc ABC subtends ∠ AOC at the centre , and ∠ ADC at the remaining part of the circle.
∠ AOC = 2 ∠ ADC
RS Aggarwal Class 9 Solutions Chapter 11 Circle Ex 11C Q7.1
RS Aggarwal Class 9 Solutions Chapter 11 Circle Ex 11C Q7.2

Question 8.
Solution:
In the figure, ABC is an equilateral triangle inscribed is a circle
Each angle is of 60°.
∠ BAC = ∠ BDC
(Angles in the same segment)
∠BDC = 60°
BECD is a cyclic quadrilateral.
∠BDC + ∠BEC = 180°
(opposite angles of cyclic quad.)
=> 60°+ ∠BEC = 180°
=> ∠BEC = 180° – 60°= 120°
Hence ∠BDC = 60° and ∠BEC = 120° Ans.

Question 9.
Solution:
ABCD is a cyclic quadrilateral.
∠BCD + ∠BAD = 180°
(opposite angles of a cyclic quad.)
=> 100°+ ∠BAD = 180°
so ∠BAD = 180° – 100° = 80°
Now in ∆ ABD,
∠BAD + ∠ABD + ∠ADB = 180° (Angles of a triangle)
RS Aggarwal Class 9 Solutions Chapter 11 Circle Ex 11C Q9.1
=> 80° + 50° + ∠ADB = 180°
=> 130°+ ∠ADB = 180°
=> ∠ADB = 180° – 130° = 50°
Hence, ∠ADB = 50° Ans.

Question 10.
Solution:
Arc BAD subtends ∠ BOD at the centre and ∠BCD at the remaining part of the circle.
RS Aggarwal Class 9 Solutions Chapter 11 Circle Ex 11C Q10.1

Question 11.
Solution:
In ∆ OAB,
OA = OB (radii of the same circle)
∠OAB = ∠OBA = 50°
and Ext ∠BOD = ∠OAB + ∠OBA
=>x° = 50° + 50° – 100°
ABCD is a cyclic quadrilateral
∠BAD + ∠BCD = 180°
(opposite angles of a cyclic quad.)
=> 50°+ y° = 180°
=> y° = 180° – 50° = 130°
Hence x = 100° and y = 130° Ans.

Question 12.
Solution:
Sides AD and AB of cyclic quadrilateral ABCD are produced to E and F respectively.
∠CBF = 130°, ∠CDE = x.
∠CBF + ∠CBA = 180° (Linear pair)
=> 130°+ ∠CBA = 180°
=> ∠CBA = 180° – 130° = 50°
But Ext. ∠ CDE = Interior opp. ∠ CBA (In cyclic quad. ABCD)
=> x = 50° Ans.

Question 13.
Solution:
In a circle with centre O AB is its diameter and DO || CB is drawn. ∠BCD = 120°
To Find : (i) ∠BAD (ii) ABD
(iii) ∠CBD (iv) ∠ADC
(v) Show that ∆ AOD is an equilateral triangle.
(i) ABCD is a cyclic quadrilateral.
∠BCD + ∠BAD = 180°
120° + ∠BAD = 180°
RS Aggarwal Class 9 Solutions Chapter 11 Circle Ex 11C Q13.1
RS Aggarwal Class 9 Solutions Chapter 11 Circle Ex 11C Q13.2

Question 14.
Solution:
AB = 6cm, BP = 2cm, DP = 2.5cm
Let CD = xcm
Two chords AB and CD
RS Aggarwal Class 9 Solutions Chapter 11 Circle Ex 11C Q14.1

Question 15.
Solution:
O is the centre of the circle
∠ AOD = 140° and ∠CAB = 50°
BD is joined.
(i) ABDC is a cyclic quadrilateral.
RS Aggarwal Class 9 Solutions Chapter 11 Circle Ex 11C Q15.1
RS Aggarwal Class 9 Solutions Chapter 11 Circle Ex 11C Q15.2

Question 16.
Solution:
Given : ABCD is a cyclic quadrilateral whose sides AB and DC are produced to meet each other at E.
To Prove : ∆ EBC ~ ∆ EDA
Proof : In ∆ EBC and ∆ EDA
∠ E = ∠ E (common)
∠ECB = ∠EAD
{Exterior angle of a cyclic quad, is equal to its interior opposite angle}
and ∠ EBC = ∠EDA
∆ EBC ~ ∆ EDA (AAS axiom)
Hence proved

Question 17.
Solution:
Solution Given : In an isosceles ∆ ABC, AB = AC
A circle is drawn x in such a way that it passes through B and C and intersects AB and AC at D and E respectively.
DE is joined.
To Prove : DE || BC
Proof : In ∆ ABC,
AB = AC (given)
∠ B = ∠ C (angles opposite to equal sides)
But ∠ ADE = ∠ C (Ext. angle of a cyclic quad, is equal E to its interior opposite angle)
∠ADE = ∠B
But, these are corresponding angles
DE || BC.
Hence proved.

Question 18.
Solution:
Given : ∆ ABC is an isosceles triangle in which AB = AC.
D and E are midpoints of AB and AC respectively.
DE is joined.
To Prove : D, B, C, E are concyclic.
Proof: D and E are midpoints of sides AB and AC respectively.
DE || BC
In ∆ ABC, AB = AC
∠B = ∠C
But ∠ ADE = ∠ B (alternate angles)
∠ADE =∠C
But ∠ADE is exterior angle of quad. DBCE which is equal to its interior opposite angle C.
DBCE is a cyclic quadrilateral.
Hence D, B, C, E are con cyclic.
Hence proved.

Question 19.
Solution:
Given : ABCD is a cyclic quadrilateral whose perpendicular bisectors l, m, n, p of the side are drawn
RS Aggarwal Class 9 Solutions Chapter 11 Circle Ex 11C Q19.1
To prove : l, m, n and p are concurrent.
Proof : The sides AB, BC, CD and DA are the chords of the circle passing through the vertices’s of quad. A, B, C and D. and perpendicular bisectors of a chord always passes through the centre of the circle.
l,m, n and p which are the perpendicular bisectors of the sides of cyclic quadrilateral will pass through O, the same point Hence, l, m, n and p are concurrent.
Hence proved.

Question 20.
Solution:
Given : ABCD is a rhombus and four circles are drawn on the sides AB, BC, CD and DA as diameters. Diagonal AC and BD intersect each other at O.
RS Aggarwal Class 9 Solutions Chapter 11 Circle Ex 11C Q20.1
RS Aggarwal Class 9 Solutions Chapter 11 Circle Ex 11C Q20.2

Question 21.
Solution:
Given: ABCD is a rectangle whose diagonals AC and BD intersect each other at O.
To prove : O is the centre of the circle passing through A, B, C and D
RS Aggarwal Class 9 Solutions Chapter 11 Circle Ex 11C Q21.1

Question 22.
Solution:
Construction.
(i) Let A, B and C are three points
(ii) With A as centre and BC as radius draw an arc
(iii) With centre C, and radius AB, draw another arc which intersects the first arc at D.
D is the required point.
Join BD and CD, AC and BA and CB
RS Aggarwal Class 9 Solutions Chapter 11 Circle Ex 11C Q22.1
BC = BC (common)
AC = BD (const.)
AB = DC
∴ ∆ ABC ≅ ∆ DBC (SSS axiom)
∴ ∠BAC ≅ ∠BDC (c.p.c.t.)
But these are angles on the same sides of BC
Hence these are angles in the same segment of a circle
A, B, C, D are concyclic Hence D lies on the circle passing througtvA, B and C.
Hence proved.

Question 23.
Solution:
Given : ABCD is a cylic quadrilateral (∠B – ∠D) = 60°
To prove : The small angle of the quad, is 60°
RS Aggarwal Class 9 Solutions Chapter 11 Circle Ex 11C Q23.1
RS Aggarwal Class 9 Solutions Chapter 11 Circle Ex 11C Q23.2

Question 24.
RS Aggarwal Class 9 Solutions Chapter 11 Circle Ex 11C Q24.1
Solution:
Given : ABCD is a quadrilateral in which AD = BC and ∠ ADC = ∠BCD
To prove : A, B, C and D lie on a circle
RS Aggarwal Class 9 Solutions Chapter 11 Circle Ex 11C Q24.2
RS Aggarwal Class 9 Solutions Chapter 11 Circle Ex 11C Q24.3

Question 25.
Solution:
Given : In the figure, two circles intersect each other at D and C
∠BAD = 75°, ∠DCF = x° and ∠DEF = y°
RS Aggarwal Class 9 Solutions Chapter 11 Circle Ex 11C Q25.1
RS Aggarwal Class 9 Solutions Chapter 11 Circle Ex 11C Q25.2

Question 26.
Solution:
Given : ABCD is a cyclic quadrilateral whose diagonals AC and BD intersect at O at right angle.
RS Aggarwal Class 9 Solutions Chapter 11 Circle Ex 11C Q26.1
RS Aggarwal Class 9 Solutions Chapter 11 Circle Ex 11C Q26.2
RS Aggarwal Class 9 Solutions Chapter 11 Circle Ex 11C Q26.3

Question 27.
Solution:
In a circle, two chords AB and CD intersect each other at E when produced.
AD and BC are joined.
RS Aggarwal Class 9 Solutions Chapter 11 Circle Ex 11C Q27.1

Question 28.
Solution:
Given : Two parallel chords AB and CD of a circle BD and AC are joined and produced to meet at E.
RS Aggarwal Class 9 Solutions Chapter 11 Circle Ex 11C Q28.1

Question 29.
Solution:
Given : In a circle with centre O, AB is its diameter. ADE and CBE are lines meeting at E such that ∠BAD = 35° and ∠BED = 25°.
To Find : (i) ∠DBC (ii) ∠DCB (iii) ∠BDC
Solution. Join BD and AC,
RS Aggarwal Class 9 Solutions Chapter 11 Circle Ex 11C Q29.1
RS Aggarwal Class 9 Solutions Chapter 11 Circle Ex 11C Q29.2

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RS Aggarwal Class 9 Solutions Chapter 13 Volume and Surface Area Ex 13C

RS Aggarwal Class 9 Solutions Chapter 13 Volume and Surface Area Ex 13C

These Solutions are part of RS Aggarwal Solutions Class 9. Here we have given RS Aggarwal Class 9 Solutions Chapter 13 Volume and Surface Area Ex 13C.

Other Exercises

Question 1.
Solution:
Radius of base (r) = 35cm
and height (h) = 84cm.
RS Aggarwal Class 9 Solutions Chapter 13 Volume and Surface Area Ex 13C Q1.1
RS Aggarwal Class 9 Solutions Chapter 13 Volume and Surface Area Ex 13C Q1.2

Question 2.
Solution:
Height of cone (h) = 6cm
Slant height (l) = 10cm.
RS Aggarwal Class 9 Solutions Chapter 13 Volume and Surface Area Ex 13C Q2.1

Question 3.
Solution:
Volume of right circular cone = (100 π) cm3
Height (h) = 12cm.
Let r be the radius of the cone
RS Aggarwal Class 9 Solutions Chapter 13 Volume and Surface Area Ex 13C Q3.1

Question 4.
Solution:
Circumference of the base = 44cm
RS Aggarwal Class 9 Solutions Chapter 13 Volume and Surface Area Ex 13C Q4.1
RS Aggarwal Class 9 Solutions Chapter 13 Volume and Surface Area Ex 13C Q4.2

Question 5.
Solution:
Slant height of the cone (l) = 25cm
Curved surface area = 550 cm2
Let r be the radius
πrl = curved surface area
RS Aggarwal Class 9 Solutions Chapter 13 Volume and Surface Area Ex 13C Q5.1

Question 6.
Solution:
Radius.of base (r) = 35cm.
Slant height (l) = 37cm.
We know that
RS Aggarwal Class 9 Solutions Chapter 13 Volume and Surface Area Ex 13C Q6.1

Question 7.
Solution:
Curved surface area = 4070 cm2
Diameter of the base = 70cm
RS Aggarwal Class 9 Solutions Chapter 13 Volume and Surface Area Ex 13C Q7.1

Question 8.
Solution:
Radius of the conical tent = 7m
and height = 24 m.
RS Aggarwal Class 9 Solutions Chapter 13 Volume and Surface Area Ex 13C Q8.1

Question 9.
Solution:
Radius of the first cone (r) = 1.6 cm.
and height (h) = 3.6 cm.
RS Aggarwal Class 9 Solutions Chapter 13 Volume and Surface Area Ex 13C Q9.1

Question 10.
Solution:
Ratio in their heights =1:3
and ratio in their radii = 3:1
Let h1,h2 he their height and r1,r2 be their radii, then
RS Aggarwal Class 9 Solutions Chapter 13 Volume and Surface Area Ex 13C Q10.1
The ratio between their volumes is 3:1
hence proved

Question 11.
Solution:
Diameter of the tent = 105m
RS Aggarwal Class 9 Solutions Chapter 13 Volume and Surface Area Ex 13C Q11.1
RS Aggarwal Class 9 Solutions Chapter 13 Volume and Surface Area Ex 13C Q11.2

Question 12.
Solution:
No. of persons to be s accommodated =11
Area to be required for each person = 4m2
RS Aggarwal Class 9 Solutions Chapter 13 Volume and Surface Area Ex 13C Q12.1

Question 13.
Solution:
Height of the cylindrical bucket (h) = 32cm
Radius (r) = 18cm
Volume of sand filled in it = πr2h
= π x 18 x 18 x 32 cm3
= 10368π cm3
Volume of conical sand = 10368 π cm3
Height of cone = 24 cm
RS Aggarwal Class 9 Solutions Chapter 13 Volume and Surface Area Ex 13C Q13.1

Question 14.
Solution:
Let h be the height and r be the radius of the cylinder and cone.
Curved surface area of cylinder = 2πrh
and curved surface area of cone = πrl
RS Aggarwal Class 9 Solutions Chapter 13 Volume and Surface Area Ex 13C Q14.1
RS Aggarwal Class 9 Solutions Chapter 13 Volume and Surface Area Ex 13C Q14.2

Question 15.
Solution:
Diameter of the pillar = 20cm
Radius (r) = \(\frac { 20 }{ 2 } \) = 10cm
RS Aggarwal Class 9 Solutions Chapter 13 Volume and Surface Area Ex 13C Q15.1
RS Aggarwal Class 9 Solutions Chapter 13 Volume and Surface Area Ex 13C Q15.2

Question 16.
Solution:
Height of the bigger cone (H) = 30cm
By cutting a small cone from it, then volume of smaller cone = \(\frac { 1 }{ 27 } \) of volume of big cone
RS Aggarwal Class 9 Solutions Chapter 13 Volume and Surface Area Ex 13C Q16.1
Let radius and height of the smaller cone be r and h
and radius and height of the bigger cone be R and H.
RS Aggarwal Class 9 Solutions Chapter 13 Volume and Surface Area Ex 13C Q16.2
RS Aggarwal Class 9 Solutions Chapter 13 Volume and Surface Area Ex 13C Q16.3
Hence at the height of 20cm from the base it was cut off. Ans.

Question 17.
Solution:
Height of the cylinder (h) = 10cm.
Radius (r) = 6cm.
Height of the cone = 10cm
RS Aggarwal Class 9 Solutions Chapter 13 Volume and Surface Area Ex 13C Q17.1
RS Aggarwal Class 9 Solutions Chapter 13 Volume and Surface Area Ex 13C Q17.2

Question 18.
Solution:
Diameter of conical vessel = 40cm
Radius (r) = \(\frac { 40 }{ 2 } \) = 20cm
and depth (h) = 24cm.
.’. Volume = \(\frac { 1 }{ 3 } \) πr2h
RS Aggarwal Class 9 Solutions Chapter 13 Volume and Surface Area Ex 13C Q18.1
RS Aggarwal Class 9 Solutions Chapter 13 Volume and Surface Area Ex 13C Q18.2

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Value Based Questions in Science for Class 9 Chapter 2 Is Matter Around Us Pure

Value Based Questions in Science for Class 9 Chapter 2 Is Matter Around Us Pure

These Solutions are part of Value Based Questions in Science for Class 9. Here we have given Value Based Questions in Science for Class 9 Chapter 2 Is Matter Around Us Pure

Question 1.
A student was asked by his teacher to separate an impure sample of sulphur containing sand as the impurity. He tried to purify it with the help of sublimation. But he was not successful. Particles of sulphur could not be separated completely from sand.

  1. Why did not the sublimation process succeed ?
  2. Suggest an alternate method to affect the separation.
  3. What is the value based information associated with this ?

Answer:

  1. Sulphur is not of volatile nature. Upon heating, it melts. Therefore, sublimation process could not succeed.
  2. The student should have dissolved the impure sample in carbon disulphide. It is a liquid in which sulphur completely dissolves while iron does not. From the solution, sulphur can be recovered with the help of crystallization process.
  3. The process of sublimation is useful only if one of the constituents present in the mixture can sublime while the others do not undergo sublimation.

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Question 2:
Amit was asked by his teacher to separate a liquid mixture of acetone and ethyl alcohol. He set up a distillation apparatus and tried to distil the mixture. To his surprise, both the liquids got distilled. Teacher told Amit to repeat the experiment by using a fractionating column in the distillation flask. Amit followed the advice of the teacher and he was able to separate the two liquids.

  1. Why was Amit not successful in separating the liquid mixture earlier ?
  2. Why did teacher ask him to use the fractionating column ?
  3. Which liquid was distilled first ?
  4. As a student of chemistry, what value based information you have gathered ?

Answer:

  1. The difference in boiling point temperatures of acetone (56°C) and ethyl alcohol (78°C) is only 22°C. Therefore, process of simple distillation fails in this case.
  2. Fractionating column is quite effective in this case because it obstructs the distillation of ethyl alcohol which is high boiling and at the same time helps in the distillation of acetone which is low boiling.
  3. Acetone was distilled first since it has comparatively low boiling point.
  4. The process of simple distillation can be used only in case, the liquids present in the mixture differ in their boiling point by 25°C or more.

Question 3.
A housewife got a cut on her finger while working in the Kitchen. She tried to stop the bleeding by applying dettol on it but it was not effective. By chance, her friend was also there. She asked her to rub alum on the cut which she did. The bleeding immediately stopped.

  1. Why was not dettol effective in stopping bleeding ?
  2. Why was alum effective ?
  3. What valuable service was done by the friend to the housewife ?

Answer:

  1. Blood is a colloidal solution and the colloidal particles of RBCs carry charge. Dettol is an organic compound. It could not neutralise the charge on the colloidal particles. Therefore, bleeding did not stop.
  2. Alum is a salt in which charged ions are present. When alum was applied on the cut, the charged ions neutralised the charge on the colloidal particles. In the absence of charge, blood became thick and bleeding stopped. In otherwords, blood got coagulated.
  3. The friend has a proper knowledge of colloidal solutions. She must be a student of chemistry. Timely help by her stopped the bleeding. Otherwise, there would have been a further loss of blood.

Question 4.
Mallika’s mother was suffering from cold and cough. Mallika prepared tea for her mother. She boiled water in a pan, then she added tea leaves, sugar and milk to it. She filtered the tea in a cup and served to her mother.

  1. Explain the values shown by Mallika.
  2. Identify solute, solvent, residue and filtrate in this activity. (CBSE 2013)

Answer:

  1. Mallika used the knowledge of chemistry to provide relief to her mother. Actually she prepared an extract of tea leaves which is helpful in curing cold and cough and gives warmth to the body.
  2. Solute : tea leaves and sugar ; Solvent : water and milk; Filtrate : homogeneous mixture of water, milk, sugar and extract of tea leaves.

Question 5.
Nikhil’s father was suffering from high blood pressure and cardiac problems. Doctor suggested him to take low fat milk. Nikhil churned the milk and separated the butter (fat) from it and then he served that milk to his father. Answer the following questions based on above information :

  1. Name the technique by which Nikhil separated butter from milk.
  2. Write any other application of that technique.
  3. Which values are reflected in Nikhil’s behaviour ?

Answer:

  1. It is called centrifugation carried in a centrifugal machine.
  2. Washing machines that are used for washing dirty clothes are centrifugal machines.
  3. Nikhil had the proper knowledge of chemistry. He used this to separate fat from milk with the help of the centrifugal machine. In this way, he rendered proper service to his father.

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NCERT Exemplar Solutions for Class 9 Science Chapter 2 Is Matter Around Us Pure

NCERT Exemplar Solutions for Class 9 Science Chapter 2 Is Matter Around Us Pure

These Solutions are part of NCERT Exemplar Solutions for Class 9 Science . Here we have given NCERT Exemplar Solutions for Class 9 Science Chapter 2 Is Matter Around Us Pure.

Question 1.
Which of the following statements are true for pure substances ?
(i) Pure substances contain only one kind of particles
(ii) Pure substances may be compounds or mixtures
(iii) Pure substances have the same composition throughout
(iv) Pure substances can be exemplified by all elements other than nickel.
(a) (i) and (ii)            (b) (i) and (iii)
(c) (iii) and (iv)          (d) (ii) and (iii).
Correct Answer:
(b) Both these statements are correct.

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Question 2.
Rusting of an article made up of iron is called
(a) corrosion and it is a physical as well as chemical change
(b) dissolution and it is a physical change
(c) corrosion and it is a chemical change {d) dissolution and it is a chemical change.
Correct Answer:
(c) corrosion and it is a chemical change.

Question 3.
A mixture of sulphur and carbon disulphide is
(a) heterogeneous and shows Tyndall effect
(b) homogeneous and shows Tyndall effect
(c) heterogeneous and does not show Tyndall effect
(d) homogeneous and does not show Tyndall effect.
Correct Answer:
(d) Sulphur dissolves in carbon disulphide to form a solution which is homogeneous in nature. It does not show any Tyndall effect.

Question 4.
Tincture of iodine has antiseptic properties. This solution is made by dissolving
(a) iodine in benzene
(b) iodine in ether
(c) iodine in water
(d) iodine in alcohol.
Correct Answer:
(d) Tincture of iodine is a dilute solution of iodine formed by dissolving iodine in ethyl alcohol also called alcohol.

Question 5.
Which of the following are homogeneous mixtures in nature ?
(i) ice
(ii) wood
(iii) soil
(iv) air.
(a) (i) and (iii)        (b) (ii) and (iv)
(c) (i) and (iv)        (d) (iii) and (iv).
Correct Answer:
(c) Both ice and air are homogeneous mixtures.

Question 6.
Which of the following are physical changes ?
(i) Melting of iron metal
(ii) Rusting of iron
(iii) Bending of an iron rod
(iv) Drawing a wire of iron metal
(a) (i), (ii) and (iii)          (b) (i), (ii) and (iv)
(c) (i), (iii) and (iv)         (d) (ii), (iii) and (iv).
Correct Answer:
(c) All the three are examples of physical changes.

Question 7.
Which of the following are chemical changes ?
(i) Decaying of wood
(ii) Burning of wood
(iii) Growth of wood in a tree
(iv) Hammering of a nail into a piece of wood.
(a) (i) and (ii)             (b) (ii) and (iv)
(c) (iii) and (iv)           (d) (i), (ii) (iiii).
Correct Answer:
(d) All the three are chemical changes.

Question 8.
Two substances, A and B were made to react to form a third substance, A2B according to the following reaction
2A + B —> A2B
Which of the following statements concerning this
reaction are incorrect ?
(i) The product A2B shows the properties of substances A and B
(ii) The product will always have a fixed composition
(iii) The product so formed cannot be classified as a compound
{iv) The product so formed is an element.
(a) (i), (ii) and (iii)           (b) (ii), (iii) and (iv)
(c) (i), (iii) and (iv)          (d) (ii), (iii) and (iv).
Correct Answer:
(c) Except for (ii) all other statements are incorrect.

Question 9.
Two chemical species X and Y combine together to form a product P which contains both X and Y
X + Y —> P
X and Y cannot be broken down into simpler substances by simple chemical reactions. Which of the following concerning the species X, Y and P are correct ?
(i) P is a compound
(ii) X and Y are compounds
(iii) X and Y are elements
(iv) P has a fixed composition
(a) (i), (ii) and {iii)               (b) (i), (ii) and (iv)
(c) (ii), (iii) and (iv)              (d) (i), (iii) and (iv).
Correct Answer:
(d) All these statements are correct.

Short Answer Questions

Question 10.
Suggest separation technique(s) one would need to employ to separate the following mixtures.

  1. Mercury and water
  2. Potassium chloride and ammonium chloride
  3. Common salt, water and sand
  4. Kerosene oil, water and salt.

Answer:

  1. Separation can be done with a separating funnel. Mercury being heavy forms lower layer.
  2. Sublimation can be used. Ammonium chloide sublimes.
  3. Filtration followed by crystallisation techniques can be used here. Sand can be removed as an insoluble residue upon filtration. Upon evaporating the filtrate in a china dish to about one third followed by slow colling crystals of common salt (NaCl) are formed.
  4. Kerosene oil can be separated with a separating funnel. Water and salt can be separated either by distillation (water distils) or by evaporation (water evaporates). Salt is left behind as residue.

Question 11.
Which of the tubes {a) and {b) will be more effective as a condenser in the distillation apparatus ?
NCERT Exemplar Solutions for Class 9 Science Chapter 2 Is Matter Around Us Pure image - 1
Answer:
Condenser
(a) has a number of bends or obstructions. It will provide more opportunity to vapours to condense and act as a better condenser than
(b) which has no such obstructions.

Question 12.
Salt can be recovered from its solution by evaporation. Suggest some other technique for the same ?
Answer:
Crystallisation technique can also be used. For example, both these can be used to separate salts like sodium chloride (common salt) and potasium nitrate (nitre) from their aqueous solutions.

Question 13.
Sea-water can be classified as homogeneous as well as heterogeneous mixture. Comment.
Answer:
The sample of sea water from mid stream is transparent and has certain salts dissolved in water. It is therefore, homogeneous., However, the sample collected from near the seashore contains mud and many suspended particles. It is therefore, heterogeneous in nature.

Question 14.
While diluting a solution of salt in water, a student by mistake added acetone (boiling point 56°C). What technique can be employed to get back acetone ? Justify your choice.
Answer:
Acetone can be recovered by carrying out the distillation in a distillation flask. Its vapours will rise since it is volatile, get condensed and collected in a receiver. Salt being non-volatile in nature remains in the flask.

Question 15.
What would you observe when
(a) a saturated solution of potassium chloride prepared at 60°C is allowed to cool to room temperature.
(b) an aqueous sugar solution is heated to dryness
(c) a mixture of iron filings and sulphur powder is heated strongly.
Answer:
(a) Upon cooling, crystals of potassium chloride would separate.
(b) Initially, entire water slowly gets removed. Upon further heating, sugar gets charred and burning smell will be noticed.
(c) Iron sulphide is formed as a greyish black residue.

Question 16.
Explain why particles of a colloidal solution do not settle down when left undisturbed while in the case of a suspension, they do.
Answer:
In a colloidal solution, the particle size is smaller as compared to particle size in a suspension. The effect of gravity on these particles is less than in a suspension. Therefore, the particles in a suspension settle under the influence of gravity while they remain dispersed in a colloidal solution.

Question 17.
Smoke and fog both are aerosols. In what way are they different ?
Answer:
Air is the dispersion medium in both the cases. However, dispersed phases are not the same. In smoke, carbon particles act as the dispersed phase while water drops represent the dispersed phase in fog.

Question 18.
Classify the following as physical or chemical properties (CBSE 2012)

  1. The composition of a sample of steel is : 98% iron. 1-5% carbon and 0-5% other elements.
  2. Zinc dissolves in hydrochloric acid with the evolution of hydrogen gas.
  3. Metallic sodium is soft enough to be cut with a knife.
  4. Most metal oxides form alkalis on interacting with water.

Answer:

  1. Physical
  2. Chemical
  3. Physical
  4. Chemical.

Question 19.
The teacher instructed three students A’, ‘B’ and ‘C’ respectively to prepare a 50% (mass by volume) solution of sodium hydroxide (NaOH). ‘A’ dissolved 50 g of NaOH in 100 mL of water. ‘B’ dissolved 50 g of NaOH in 100 g of water while ‘C’ dissolved 50 g of NaOH in water to make 100 mL of solution. Which one of them has made the desired solution and why ?
Answer:
The student ‘C’ has prepared the desired solution since it represents mass/volume solution. The student A’ has taken into account volume of the solvent while the student ‘B’ has considered the mass of the solution
NCERT Exemplar Solutions for Class 9 Science Chapter 2 Is Matter Around Us Pure image - 2

Question 20.
Name the process associated with the following :

  1. Dry ice is kept at room temperature and under one atmospheric pressure.
  2. A drop of ink placed on the surface of water contained in a glass spreads throughout the water.
  3. A potassium permanganate crystal is in a beaker and water is poured into the beaker with stirring.
  4. An acetone bottle is left open and the bottle, becomes empty.
  5. Milk is churned to separate cream from it. .
  6. Settling of sand when a mixture of sand and water is left undisturbed for some time.
  7. Fine beam of light entering through a small hole in a dark room, illuminates the particles in its paths.

Answer:

  1. Sublimation
  2. Diffusion/Sedimentation
  3. Dissolution/Dififusion
  4. Evaporation/Diffusion
  5. Centrifugation
  6. Sedimentation
  7. Scattering of light (Tyndall effect).

Question 21.
You are given two samples of water labelled as A and ‘B’. Sample A boils at 100°C and sample ‘B’ boils at 102°C. Which sample of water will not freeze at 0°C ? Comment.
Answer:
The standard boiling point temperature of water is 100°C. Therefore, the sample v/ith boiling point 100°C represents the pure form of water. The sample with boiling point 102° C is impure. Please note that impurities always raise the boiling point temperatures of liquids and lower their freezing point temperature. Therefore, the sample ‘B’ will freeze before 0°C.

Question 22.
What are the favourable qualities given to gold when it is alloyed with copper or silver for the purpose of making ornaments ?
Answer:
Pure gold (24 carats) is very soft and cannot be used for making ornaments. It is alloyed with a small quantity of copper or silver and becomes hard (22 carats). It can be drawn into wires (becomes ductile) or beaten into fine sheets (becomes malleable). It can, therefore, be used for making ornaments.

Question 23.
An element is sonorous and highly ductile. Under which category would you classify this element ? What other characteristics do you expect the element to possess ?
Answer:
The element is a metal. It is expected to possess the important characteristics of the metals.
Semi-Metals or Metalloids
There are a few elements which possess the characteristics of both metals and non-metals. These are actually border-line elements and are known as semi-metals. Semi-metals are also called metalloids. A few common examples are : Silicon, Arsenic, Antimony and Bismuth.

Question 24.
Give an example each for the mixture having the following characteristics. Suggest a suitable method to separate the components of these mixtures

  1. A volatile and a non-volatile component.
  2. Two volatile components with appreciable difference in boiling points.
  3. Two immiscible liquids.
  4. One of the components changes directly from solid to gaseous state.
  5. Two or more coloured constituents soluble in same solvent.

Answer:

  1. A mixture of naphthalene (volatile) and common salt (non-volatile). The separation can be done by sublimation.
  2. A mixture of acetone (boiling point = 56°C) and water (boiling point = 100°C) separation can be done by distillation.
  3. A mixture of kerosene oil and water. Separation can be done in a separating funnel.
  4. Same as in (a).
  5. A mixture of blue/black ink. Separation can be done by chromatography.

Question 25.
Fill in the blanks

  1. A colloid is a …………… mixture and its components can be separated by the technique known as …………… .
  2. Ice, water and water vapours look different and display different …………… properties but they are the …………… same.
  3. A mixture of chloroform and water taken in a separating funnel is mixed and left undisturbed for some time. The upper layer in the separating funnel will be of …………… and the lower layer will be that of …………… .
  4. A mixture of two or more miscible liquids for which the difference in the boiling points is less than 25 K, can be separated by the process called …………… .
  5. When light is passed through water containing a few drops of milk, it shows a bluish tinge. This is due to the …………… of light by milk and the phenomenon is called …………… . This indicates that milk is a …………… solution.

Answer:

  1. heterogenous, centrifugation
  2. physical, chemically
  3. water, chloroform (density of water is less than that of chloroform)
  4. fractional distillation
  5. scattering, Tyndall effect, colloidal (also called emulsion).

Question 26.
Sucrose (sugar) crystals obtained from sugarcane and beetroot are mixed together. Will it be a pure substance or a mixture ? Give reasons for the same.
Answer:
Both are chemically the same and represent a single chemical compound. Flence, they are pure substance.

Question 27.
Give some examples of Tyndall effect observed in your surroundings.
Answer:

  1. Tyndall effect can be observed when sun light is made to enter the room in the morning time through a small slit. Dust particles with zig-zag motion can be seen.
  2. Tyndall effect can be seen when sun light passes through the canopy of a dense forest.

Question 28.
Can we separate a mixture of alcohol and water by the use of a separating funnel ? If not, suggest an alternate method.
Answer:
No, the mixture of alcohol and water cannot be separated by the use of separating funnel because the two are completely miscible with each other. The separation can be done by fractional distillation.
The student should have dissolved the impure sample in carbon disulphide. It is a liquid in which sulphur completely dissolves while iron does not. From the solution, sulphur can be recovered with the help of crystallization process.

Question 29.
On heating, calcium carbonate gets converted into calcium oxide and carbon dioxide.

  1. Is this a physical or a chemical change ?
  2. Can you prepare one acidic and one basic solution by using the products formed in the above process ? If so, write the chemical equation involved.

Answer:

  1. It is chemical change. The chemical equation for the reaction is :
    NCERT Exemplar Solutions for Class 9 Science Chapter 2 Is Matter Around Us Pure image - 3
  2. By adding water to calcium oxide taken in a test tube, a basic solution results. It contains calcium hydroxide
    Calcium oxide + Water —-> Calcium hydroxide
    By passing carbon dioxide through water taken in a tube, carbonic acid is formed. The solution is of acidic nature
    Carbon dioxide + Water —-> Carbonic acid

Note:
All the three chemical equations are word equations. Symbol equations can also be written for these.
For example,
NCERT Exemplar Solutions for Class 9 Science Chapter 2 Is Matter Around Us Pure image - 4
However, you will learn about these in the next class.

Question 30.
Non metals are usually poor conductors of heat and electricity. They are non-lustrous, non-sono- rous, non-malleable and are coloured.

  1. Name a lustrous non-metal.
  2. Name a non-metal which exists as a liquid at room temperature.
  3. The allotropie form of a non-metal is a good conductor of electricity. Name the allotope.
  4. Name a non-metal which is known to form the largest number of compounds.
  5. Name a non-metal other than carbon which shows allotropy.
  6. Name a non-metal which is required for combustion.

Answer:

  1. Graphite (Carbon)
  2. Bromine
  3. Graphite(Carbon)
  4. Carbon
  5. Silicon
  6. Coke (Carbon).

Question 31.
Classify the substances given in the figure into elements and compounds.
NCERT Exemplar Solutions for Class 9 Science Chapter 2 Is Matter Around Us Pure image - 5
Answer:
Elements : Cu, Zn, Hg, Diamond.
Compounds : H2O, CaCO3, O2, F2
Note: Sand, NaCl (aq) and wood are the examples of mixtures.

Question 32.
Which of the following are not compounds ?

  1. Chlorine gas
  2. Potassium chloride
  3. Iron
  4. Iron sulphide
  5. Aluminium
  6. Iodine
  7. Carbon
  8. Carbon monoxide
  9. Sulphur powder.

Answer:
Iron, aluminium, carbon and sulphur powder are not compounds. These are elements.

Long Answer Questions

Question 33.
Fractional distillation is suitable for separation of miscible liquids with a boiling point difference of about 25 K or less. What part of fractional distillation apparatus makes it efficient and possesses an advantage over a simple distillation process. Explain using a diagram.
Answer:
It is the fractionating column which fits into the distillation flask. It is packed with a number of glass beads. A the vapours of both the volatile liquids rise upwards in the distillation flask, the beads obstruct their movement i.e. it slows down. The vapours of higher boiling liquid condense to the liquid state and in doing so they release certain energy’ known as latent heat of condensation. The energy which is released is taken up by the molecules of lower boiling liquid. As a result, they vapourise more quickly and pass through the water condenser and condense as a liquid in the receiver. The higher boiling liquid remains in the distillation flask only. For further details about the fractional distillation
NCERT Exemplar Solutions for Class 9 Science Chapter 2 Is Matter Around Us Pure image - 6

Question 34.
(a) Under which category of mixtures will you classify alloys and why ?
(b) A solution is always a liquid. Comment.
(c) Can a solution be heterogeneous ?
Answer:
(a) Aloys are regarded as homogeneous mixtures since the constituting elements are uniformly mixed in an alloy.
(b) This statement is not true. There may also be solid solutions (solid acts as solvent) and gaseous solutions (gas acts as solvent).
(c) No, a solution is always a homogeneous mixture of two or more substances.

Question 35.
Iron filings and sulphur were mixed together and divided into two parts ‘A’ and ‘B’. Part ‘A’ was heated strongly while Part ‘B’ was not heated. Dilute hydrochloric acid was added to both the Parts and evolution of gas was seen in both the cases. How will you identify the gases evolved ?
Answer:

  1. In part A iron and sulphur will remain as such in the form of a mixture. On adding dilute hydrochlo¬ride acid iron will react to evolve hydrogen gas. The gas can be tested by bringing a burning splinter near its mouth. It will burn with a pop sound.
    NCERT Exemplar Solutions for Class 9 Science Chapter 2 Is Matter Around Us Pure image - 7
  2. In part ‘B’ iron and sulphur will combine on heating to form greyish black solid known as iron sulphide.
    NCERT Exemplar Solutions for Class 9 Science Chapter 2 Is Matter Around Us Pure image - 8
  3. When dilute hydrocloric acid is added to the greyish black mass taken in a tube, a gas with foul smell similar to that of the rotton eggs will evolve. This is hydrogen sulphide (H2S) Cover or Lid
    NCERT Exemplar Solutions for Class 9 Science Chapter 2 Is Matter Around Us Pure image - 9

Question 36.
A child wanted to separate the mixture of dyes constituting a sample of ink. He marked a line by the ink on the filter paper and placed the filter paper in a glass containing water as shown in the fig . The filter paper was removed when the water moved near the top of the filter paper.
NCERT Exemplar Solutions for Class 9 Science Chapter 2 Is Matter Around Us Pure image - 10

  1. What would you expect to see, if the ink contains three different coloured components ?
  2. Name the technique used by the child.
  3. Suggest one more application of this technique

Answer:

  1. Three different bands will be seen on the filter papers.
  2. The technique is known as paper chromatography.
  3. It can be used to separate the coloured pigments from chlorophyll.

Question 37.
A group of students took an old shoe box and covered it with a black paper from all sides. They fixed a source of light (a torch) at one end of the box by making a hole in it and made another hole on the other side to view the light. They placed a milk sample contained in a beaker/tumbler in the box as shown in the figure.
They were amazed to see the milk taken in the tumbler was illuminated. They tried the same activity by taking a salt solution but found that light simply passed through it ?
NCERT Exemplar Solutions for Class 9 Science Chapter 2 Is Matter Around Us Pure image - 11

  1. Explain why the milk sample was illuminated. Name the phenomenon involved.
  2. Same results were not observed with a salt solution. Explain.
  3. Can you suggest two more solutions which would show the same effect as shown by the milk solution ?

Answer:

  1. Milk sample is a colloidal sol also called emulsion. It got illuminated because of the scattering of light on passing through it. The phenomenon is known as Tyndall effect.
  2. The salt solution is a true solution and not a colloidal solution. It will not show any Tyndall effect.
  3. A solution of sulphur in water and a solution of starch in water.

Question 38.
Classify each of the following, as a physical or a chemical change. Give reasons.

  1. Drying of a shirt in the sun.
  2. Rising of hot air over a radiator.
  3. Burning of kerosene in a lantern.
  4. hange in the colour of black tea on adding lemon juice to it.
  5. Churning of milk cream to get butter.

Answer:

  1. It is a physical change. Moisture or water drops present on the shirt will evaporate or change into vapours.
  2. It is a physical change. On cooling, the level of air over the radiator will fall.
  3. It is a chemical change. The hydrocarbons present in kerosene will react chemically with oxygen to form new products.
  4. It is a chemical change. The acid present in lemon juice will react with the constituents (e.g. caffeine) present in black tea.
  5. It is a physical change. However, butter will not change to milk so easily.

Question 39.
During an experiment the students were asked to prepare a 10% (Mass/Mass) solution of sugar in water. Ramesh dissolved 10 g of sugar in 100 g of water while Sarika prepared it by dissolving 10 g of sugar in water to make 100 g of the solution.
(a) Are the two solutions of the same concentration
(b) Compare the mass % of the two solutions.
Answer:
(a) No, the two solutions have different concentrations.
NCERT Exemplar Solutions for Class 9 Science Chapter 2 Is Matter Around Us Pure image - 12

Question 40.
You are provided with a mixture containing sand, iron filings, ammonium chloride and sodium chloride. Describe the procedure you would use to separate these constituents from the mixture ?
Answer:
Separate the constituents from a mixture containing ammonium chloride, sand and iron-filings.
NCERT Exemplar Solutions for Class 9 Science Chapter 2 Is Matter Around Us Pure image - 13
Place the mixture on a paper or petridish. Move a magnet over it. Iron filings get attached to the magnet. Scrap off iron filings from the magnet and collect them separately. Transfer the remaining mixture to a china dish and subject it to the sublimation process. Ammonium chloride forms the sublimate while sand is left as residue in the dish.
NCERT Exemplar Solutions for Class 9 Science Chapter 2 Is Matter Around Us Pure image - 14

Question 41.
Arun has prepared 0301% (by mass) solution of sodium chloride in water. Which of the following correctly represents the composition of the solution ?
(a) 1.00 g of NaCl + 100 g of water
(b) 0.11 g of NaCl + 100 g of water
(c) 0.01 g of NaCl + 99.99 g of water
(d ) 0.10 g of NaCl + 99.90 g of water.
Answer:
By definition 0.01% (by mass) solution means that 0.01 g of sodium chloride is dissolved in 0.99 g (by mass) of water. Therefore (c) represents the correct composition of the solution. This can be further verified as follows :
NCERT Exemplar Solutions for Class 9 Science Chapter 2 Is Matter Around Us Pure image - 15

Question 42.
Calculate the mass of sodium sulphate required to prepare its 20% (mass percent) solution in 100 g of water ?
Answer:
NCERT Exemplar Solutions for Class 9 Science Chapter 2 Is Matter Around Us Pure image - 16

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RS Aggarwal Class 9 Solutions Chapter 9 Quadrilaterals and Parallelograms Ex 9A

RS Aggarwal Class 9 Solutions Chapter 9 Quadrilaterals and Parallelograms Ex 9A

These Solutions are part of RS Aggarwal Solutions Class 9. Here we have given RS Aggarwal Class 9 Solutions Chapter 9 Quadrilaterals and Parallelograms Ex 9A.

Other Exercises

Question 1.
Solution:
We know that sum of angles of a quadrilateral is 360°
Now, sum of three angles = 56° + 115° + 84° = 255°
Fourth angle = 360° – 255° = 105° Ans.

Question 2.
Solution:
Sum of angles of a quadrilateral = 360°
Their ratio = 2 : 4 : 5 : 7
Let first angle = 2x
then second angle = 4x
third angle = 5x
and fourth angle = 7x
∴ 2x + 4x + 5x + 7x = 360°
=> 18x = 360°
=> x = \(\frac { { 360 }^{ o } }{ 18 } \) = 20°
Hence, first angle = 2x = 2 x 20° = 40°
Second angle = 4x = 4 x 20° = 80°
Third angle = 5x = 5 x 20° = 100°
and fourth angle = 7x = 7 x 20° = 140°Ans.

Question 3.
Solution:
In the trapezium ABCD
DC || AB
∴ ∠ A + ∠ D = 180° (Co-intericr angles)
∴ 55°+ ∠D = 180°
∠D = 180° – 55°
∴ ∠D = 125°
Similarly, ∠B + ∠C = 180°
(Co-interior angles)
=> 70° + ∠C = 180°
=> ∠C = 180° – 70°
∠C = 110°
Hence ∠C = 110° and ∠D = 125° Ans.

Question 4.
Solution:
Given : In the figure, ABCD is a square and ∆ EDC is an equilateral triangles on DC. AE and BE are joined.
To Prove : (i) AE = BE
(ii) ∠DAE = 15°
RS Aggarwal Class 9 Solutions Chapter 9 Quadrilaterals and Parallelograms Ex 9A Q4.1
RS Aggarwal Class 9 Solutions Chapter 9 Quadrilaterals and Parallelograms Ex 9A Q4.2

Question 5.
Solution:
Given : In the figure,
BM ⊥ AC, DN ⊥ AC.
BM = DN
RS Aggarwal Class 9 Solutions Chapter 9 Quadrilaterals and Parallelograms Ex 9A Q5.1

Question 6.
Solution:
Given : In quadrilateral ABCD,
AB = AD and BC = DC
AC is joined.
To Prove : (i) AC bisects ∠ A and ∠ C
(ii) BE = DE
(iii) ∠ABC = ∠ADC
Const. Join BD.
Proof : (i) In ∆ ABC and ∆ ADB
RS Aggarwal Class 9 Solutions Chapter 9 Quadrilaterals and Parallelograms Ex 9A Q6.1

Question 7.
Solution:
Given : In square ABCD,
∠ PQR = 90°
PB = QC = DR
RS Aggarwal Class 9 Solutions Chapter 9 Quadrilaterals and Parallelograms Ex 9A Q7.1
RS Aggarwal Class 9 Solutions Chapter 9 Quadrilaterals and Parallelograms Ex 9A Q7.2

Question 8.
Solution:
Given : In quadrilateral ABCD, O is any point inside it. OA, OB, OC and OD are joined.
To Prove : OA + OB + OC + OD > AC + BD
RS Aggarwal Class 9 Solutions Chapter 9 Quadrilaterals and Parallelograms Ex 9A Q8.1
RS Aggarwal Class 9 Solutions Chapter 9 Quadrilaterals and Parallelograms Ex 9A Q8.2

Question 9.
Solution:
Given : In quadrilateral ABCD, AC is its one diagonal.
RS Aggarwal Class 9 Solutions Chapter 9 Quadrilaterals and Parallelograms Ex 9A Q9.1
RS Aggarwal Class 9 Solutions Chapter 9 Quadrilaterals and Parallelograms Ex 9A Q9.2
RS Aggarwal Class 9 Solutions Chapter 9 Quadrilaterals and Parallelograms Ex 9A Q9.3

Question 10.
Solution:
Given : A quadrilateral ABCD
To Prove : ∠A + ∠B + ∠C + ∠D = 360°
Const. Join AC.
RS Aggarwal Class 9 Solutions Chapter 9 Quadrilaterals and Parallelograms Ex 9A Q10.1
RS Aggarwal Class 9 Solutions Chapter 9 Quadrilaterals and Parallelograms Ex 9A Q10.2

 

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RD Sharma Class 9 Solutions Chapter 10 Congruent Triangles Ex 10.1

RD Sharma Class 9 Solutions Chapter 10 Congruent Triangles Ex 10.1

These Solutions are part of RD Sharma Class 9 Solutions. Here we have given RD Sharma Class 9 Solutions Chapter 10 Congruent Triangles Ex 10.1

Other Exercises

Question 1.
Write the complement of each of the following angles:
(i) 20°
(ii) 35°
(iii) 90°
(iv) 77°
(v) 30°
Solution:
We know that two angles are complement to each other if their sum is 90°. Therefore,
(i) Complement of 20° is (90° – 20°) = 70°
(ii) Complement of 35° is (90° – 35°) = 55°
(iii) Complement of 90° is (90° – 90°) = 0°
(iv) Complement of 77° is (90° – 77°) = 13°
(v) Complement of 30° is (90° – 30°) = 60°

Question 2.
Write the supplement of each of the following angles:
(i) 54°
(ii) 132°
(iii) 138°
Solution:
We know that two angles are supplement to each other if their sum if 180°. Therefore,
(i) Supplement of 54° is (180° – 54°) = 126°
(ii) Supplement of 132° is (180° – 132°) = 48°
(iii) Supplement of 138° is (180° – 138°) = 42°

Question 3.
If an angle is 28° less than its complement, find its measure.
Solution:
Let required angle = x, then
Its complement = x + 28°
∴  x + x + 28° = 90° ⇒  2x = 90° – 28° = 62°
∴ x = \(\frac { { 62 }^{ \circ } }{ 2 }\)  = 31°
∴ Required angle = 31°

Question 4.
If an angle is 30° more than one half of its complement, find the measure of the angle.
Solution:
Let the measure of the required angle = x
∴  Its complement =  90° – x
∴  x = \(\frac { 1 }{ 2 }\) (90° – x) + 30°
2x = 90° – x + 60°
⇒ 2x + x = 90° + 60°
⇒  3x = 150°
⇒ x =  \(\frac { { 150 }^{ \circ } }{ 3 }\)  = 50°
∴ Required angle = 50°

Question 5.
Two supplementary angles are in the ratio 4 : 5. Find the angles.
Solution:
Ratio in two supplementary angles = 4 : 5
Let first angle = 4x
Then second angle = 5x
∴  4x + 5x = 180
⇒  9x = 180°
∴ x  = \(\frac { { 180 }^{ \circ } }{ 9 }\) = 20°
∴  First angle = 4x = 4 x 20° = 80°
and second angle = 5x
= 5 x 20° = 100°

Question 6.
Two supplementary angles differ by 48°. Find the angles.
Solution:
Let first angle = x                        ”
Then second angle = x + 48°
∴  x + x + 48° = 180°⇒  2x + 48° = 180°
⇒  2x = 180° – 48° = 132°
x= \(\frac { { 132 }^{ \circ } }{ 2 }\) =66°
∴  First angle = 66°
and second angle = x + 48° = 66° + 48° = 114°
∴ Angles are 66°, 114°

Question 7.
An angle is equal to 8 times its complement. Determine its measure.
Solution:
Let the required angle = x
Then its complement angle = 90° – x
∴ x = 8(90° – x)
⇒ x = 720° – 8x ⇒  x + 8x = 720°
⇒ 9x = 720° ⇒ x =  \(\frac { { 720 }^{ \circ } }{ 9 }\) = 80°
∴  Required angle = 80°

Question 8.
If the angles (2x – 10)° and (x – 5)° are complementary angles, find x.
Solution:
First complementary angle = (2x – 10°) and second = (x – 5)°
∴ 2x – 10° + x – 5° = 90°
⇒ 3x – 15° = 90° ⇒  3x = 90° + 15° = 105°
∴ x = \(\frac { { 105 }^{ \circ } }{ 3 }\) = 35°
∴  First angle = 2x – 10° = 2 x 35° – 10°
= 70° – 10° = 60°
and second angle = x – 5 = 35° – 5 = 30°

Question 9.
If an angle differ from its complement by 10°, find the angle.
Solution:
Let required angle = x°
Then its complement angle = 90° – x°
∴ x – (90° – x) = 10
⇒  x – 90° + x = 10°⇒  2x = 10° + 90° = 100° 100°
⇒ x =  \(\frac { { 100 }^{ \circ } }{ 2 }\) = 50°
∴ Required angle = 50°

Question 10.
If the supplement of an angle is two-third of itself Determine the angle and its supplement.
Solution:
Let required angle = x
Then its supplement angle = 180° – x
∴  (180°-x)= \(\frac { 2 }{ 3 }\)x
540° – 3x = 2x ⇒ 2x + 3x = 540°
⇒ 5x = 540°⇒  x = \(\frac { { 540 }^{ \circ } }{ 5 }\) = 108°
-. Supplement angle = 180° – 108° = 72°

Question 11.
An angle is 14° more than its complementary angle. What is its measure?
Solution:
Let required angle = x
Then its complementary angle = 90° – x
∴  x + 14° = 90° – x
x + x = 90° – 14° ⇒  2x = 76°
⇒ x =  \(\frac { { 76 }^{ \circ } }{ 2 }\) = 38°
∴  Required angle = 38°

Question 12.
The measure of an angle is twice the measure of its supplementary angle. Find its measure.
Solution:
Let the required angle = x
∴  Its supplementary angle = 180° – x
∴  x = 2(180°-x) = 360°-2x
⇒  x + 2x = 360°
⇒ 3x = 360°
⇒  x = \(\frac { { 360 }^{ \circ } }{ 3 }\) = 120°
∴  Required angle = 120°

Question 13.
If the complement of an angle is equal to the supplement of the thrice of it. Find the measure of the angle.
Solution:
Let required angle = x
Then its complement angle = 90° – x
and supplement angle = 180° – x
∴  3(90° – x) = 180° – x
⇒ 270° – 3x = 180° – x
⇒270° – 180° = -x + 3x => 2x = 90°
⇒ x = 45°
∴  Required angle = 45°

Question 14.
If the supplement of an angle is three times its complement, find the angle.
Solution:
Let required angle = x
Then its complement = 90°-  x
and supplement = 180° – x
∴  180°-x = 3(90°-x)
⇒  180° – x = 270° – 3x
⇒  -x + 3x = 270° – 180°
⇒ 2x = 90° ⇒ x = \(\frac { { 90 }^{ \circ } }{ 2 }\) =45°
∴ Required angle = 45°

 

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NCERT Solutions for Class 9 Science Chapter 4 Structure of the Atom

NCERT Solutions for Class 9 Science Chapter 4 Structure of the Atom

These Solutions are part of NCERT Solutions for Class 9 Science. Here we have given NCERT Solutions for Class 9 Science Chapter 4 Structure of the Atom. LearnInsta.com provides you the Free PDF download of NCERT Solutions for Class 9 Science (Chemistry) Chapter 4 – Structure of The Atom solved by Expert Teachers as per NCERT (CBSE) Book guidelines. All Chapter 4 – Structure of The Atom Exercise Questions with Solutions to help you to revise complete Syllabus and Score More marks.

More Resources

Practical Based Questions for Class 9 Science Chemistry

NCERT IN TEXT PROBLEMS

IN TEXT QUESTIONS

Question 1.
What are canal rays ?
Answer:
Canal rays also called anode rays, are seen moving from the anode towards the cathode in the specially designed discharge tube. However, they do not originate from the anode.

Question 2.
If an atom contains one electron and one proton, will it carry any charge or not ?
Answer:
No, it will not carry any charge because an electron has a unit negative charge and a proton has a unit
positive charge. They neutralise each other.

Question 3.
On the basis of Thomson’s model of an atom explain how the atom is neutral as a whole.
Answer:
According to Thomson’s Model of an atom, an atom may be regarded as a positively charged sphere in
which protons are present. The negatively charged electrons are believed to be studded or embedded in the sphere. Since the negative charges due to electrons and positive charges due to protons balance each other, the atom as a whole is neutral.

Question 4.
On the basis of Rutherford’s model of an atom, which sub-atomic particle is present in the nucleus of an atom ?
Answer:
The nucleus of atom is positively charged according to Rutherford’s model of an atom. All the protons in an atom are therefore, present in the nucleus.

Question 5.
Draw a sketch of Bohr’s model of an atom with three shells.
Answer:
NCERT Solutions for Class 9 Science Chapter 4 Structure of the Atom image - 1

Question 6.
What do you think would be the observation if the a-particle scattering experiment is carried out using a foil of a metal other than gold ?
Answer:
If the metal is a heavy one like gold (e.g. silver, platinum etc.) similar results will be obtained. However if the metal is quite light (e.g. sodium, magnesium etc.), it is just possible that the fast moving and massive α-particles (4u mass) may push the nucleus aside and pass through with slight deviations only.

Question 7.
Name the three sub – atomic particles in an atom.
Answer:
The sub-atomic particles in an atom are : protons, electrons and neutrons.

Question 8.
Helium atom has atomic mass of 4u and has two protons in its nucleus. How many neutrons does it have ?
Answer:
Mass number of helium is equal to its atomic mass but has no units.
∴ Mass number (A) of helium = 4
No. of protons in the nucleus = 2 Atomic number (Z) of the element = 2
No. of neutrons (n) = A – Z = 4 – 2 = 2.

Question 9.
Write the distribution of electrons in carbon and sodium atoms.
Answer:
Carbon(Z = 6) : 2 (K-shell) ; 4(L-shell)
Sodium (Z = 11) : 2 (K-shell) ; 10 (L-shell) ; 1 (M-shell)

Question 10.
If the K and L shells of an atom are full, then what would be the number of electrons in the atom ?
(CBSE 2014)
Answer:
Maximum no. of electrons in K-shell = 2
Maximum no. of electrons in L-shell = 8
Total no. of electrons in the atom = 2 + 8 = 10

Question 11.
If the number of electrons in an atom is 8 and the number of protons is also 8, then :
(i) What would be the atomic number of the atom ?
(ii) What is the charge on the atom ?
Answer:
(i) Atomic number (Z) of the atom = No. of protons = No. of electrons = 8
(ii) Charge on the atom = No charge
Actually, when the number of protons and electrons are the same, the atom is neutral.

Question 12.
For the symbols H, D and T, tabulate three sub-atomic particles found in each of them.
Answer:
NCERT Solutions for Class 9 Science Chapter 4 Structure of the Atom image - 2
Question 13.
Write the electronic configuration of any pair of isotopes and isobars.
Answer:
(i) Electronic configuration of pair of isotopes :
NCERT Solutions for Class 9 Science Chapter 4 Structure of the Atom image - 3
(ii) Electronic configuration of pair of isobars :
Isobars have same mass number but different atomic numbers. For example, elements calcium and argon are isobars :
Ca(Z = 20) : 2, 8, 8, 2
Ar (Z = 18) : 2, 8, 8.
Please note that both the elements have mass number (A) equal to 40.

NCERT END EXERCISE

Question 1.
Compare the properties of electrons, protons and neutrons.
Answer:
NCERT Solutions for Class 9 Science Chapter 4 Structure of the Atom image - 4

Question 2.
What are the limitations of J.J. Thomson’s model of an atom ?
Answer:
Limitations of Thomson’s model of an atom
(i) It could not explain the result of the scattering experiment performed by Rutherford.
(ii) It did not have any experimental evidence in its support.

Question 3.
What are the limitations of Rutherford’s model of an atom ?
Answer:
Limitations of Rutherford Model atom

  1. Rutherford model of atom could not explain the stability of the atom.
  2. Rutherford model of atom could not explain as to how the electrons are distributed in the extra nuclear portion in an atom.


Question 4.

Describe Bohr’s model of an atom.
Answer:
In 1913, Neils Bohr gave a theory regarding the distribution of electrons in the extra nuclear space in an atom. The main postulates of the theory are listed :

  1. In the extra nuclear portion of an atom, the electrons revolve in well defined circular paths known as orbits.
  2. These circular orbits are also known as energy levels or energy shells.
  3. These have been designated as K, L, M, N, O, … (or as 1, 2, 3, 4, 5, …) based on the energy present.
  4. The order of the energy of these energy shells is :
    K<L<M<N<0 <…. or 1< 2< 3 < 4<5 <….
  5. While revolving in an orbit, the electron is not in a position to either lose or gain energy. In other words, its energy remains stationary. Therefore, these energy states for the electrons are also known as stationary states.

Question 5.
Compare all the proposed models of an atom given in this chapter.
Answer:
For Thomson’s Model of air atom: 

  1. It could not explain the result of the scattering experiment performed by Rutherford.
  2. It did not have any experimental evidence in its support.

For Rutherford’s Model of an atom:

  1. Rutherford model of atom could not explain the stability of the atom.
  2. Rutherford model of atom could not explain as to how the electrons are distributed in the extra nuclear portion in an atom.

For Bohr’s Model of air atom: 

  1. It gave no idea about the shapes of the molecules formed by the combination of atoms.
  2. According to Bohr’s theory, an electron in an atom follows a well defined circular path called orbit. However, the later studies have revealed that the path of the electron cannot be followed exactly. It is only of probable nature and not exact as stated by Bohr. However, it is not possible to discuss this concept at the present level of the students.

Question 6.
Summarise the rules for writing the distribution of electrons in various shells for first eighteen elements.
Answer:
As the electrons have negligible mass, the entire mass of the atom is that of protons and neutrons present in its nucleus. Since each proton and neutron has mass equal to 4 u, therefore, mass of an atom must be the same as the sum of protons and neutrons present in the nucleus. The latter is known as mass number of an element. It is denoted by the symbol A’. The mass number of an element may be defined as :
The sum of the number of protons and neutrons present in the nucleus of its atom.
Mass number (A) = No. of protons + No. of neutrons

Question 7.
Describe valency by taking the examples of silicon and oxygen.
Answer:
Valency of an element may be defined as the combining capacity of its atom with atoms of other elements in order to acquire 8 electrons (2 in some exceptional cases).
Valency of silicon (Si) : Atomic number of the element is 14. Its electronic distribution is ; K(2), L(8), M(4).
As silicon atom has four valence electrons, it can lose all of them. At the sametime, it can also gain four electrons to have a complete octet. Therefore, the valency of silicon is 4.
Valency of oxygen (O) : Atomic number of the element is 8. Its electronic distribution is ; K(2), L(6)
As oxygen atom has six valence electrons, it needs two more electrons to complete its octet (8-6 = 2). Therefore, valency of oxygen is 2.

Question 8.
Explain with examples
(i) Atomic number
(ii) Mass number
(iii) Isotopes
(iv) Isobars.
Answer:
(i) Atomic number:
Rutherford stated that the number of unit positive charges present in the nucleus of an atom is known as atomic number of the element It is denoted by the symbol Z.
                                        Atomic no. (Z)= No. ofprotons = No. of electrons
(ii) Mass number:
The mass number of an element may be defined as : the sum of the number of protons and neutrons present in the nucleus of its atom.
Mass number (A) = No. of protons + No. of neutrons
(iii) Isotopes:
Isotopes may be defined as the different atoms of the same element having same atomic number but different mass numbers.(iv) Isobars:
isobars may be defined as the atoms belonging to the different elements with same mass numbers but different atomic numbers.

Question 9.
Na+ ion has completely filled K and L shells. Explain. (CBSE 2013)
Answer:
Na+ ion is formed when Na atom loses one electron from the valence shell i.e. M shell. The Na+ ion formed has only two shells which are complete
Na : K(2) L(8) M(1) ;
Na+ : K(2) L(8)

Question 10.
If the element bromine is in the form two isotopes which are 7935Br (49.7%) and 8135Br (30.3%), then calculate the average atomic mass of bromine.
Answer:
NCERT Solutions for Class 9 Science Chapter 4 Structure of the Atom image - 5

Question 11.
The average atomic mass of a sample of element X is 16.2 u. What are the percentages of isotopes 168X and 188X in the sample ? (CBSE 2016)
Answer:
NCERT Solutions for Class 9 Science Chapter 4 Structure of the Atom image - 6

Question 12.
If Z = 3, what would be the valency of the element ? Also name the element.
Answer:
The configuration of the element with Z = 3 is K(2) L(l). It has one electron in the valence shell i.e. L-shell. Its valency is one. The element is lithium (Li).

Question13.
Composition of the nuclei of two atomic species X and Y are given as under
NCERT Solutions for Class 9 Science Chapter 4 Structure of the Atom image - 7
Give the mass numbers of X and Y. What is the relation between the two species ?
Answer:
Mass number of X = protons + neutrons = 6 + 6 =12 Mass number of Y = 6 + 8 = 14
The species X and Y are the isotopes because their atomic numbers are same and their mass numbers are different.

Question 14.
For the following statements, write T for True and F for False.
(a) J.J. Thomson proposed that the nucleus of an atom contains only nucleons.
(b) A neutron is formed by an electron and proton combining together. Therefore, it is neutral.
(c) The mass of an electron is about 1/2000 times that of proton.
(d) An isotope of iodine is used for making tincture of iodine, which is used as medicine.
Answer:
(a) F
(b) F
(c) T
(d) F.

Question 15.
Rutherford’s alpha-particle scattering experiment was responsible for the discovery of
(a) Atomic Nucleus
(b) Electron
(c) Proton
(d) Neutron.
Answer:
(a) is the correct answer.

Question 16.
Isotopes of element have :
(a) the same physical properties
(b) different chemical properties
(c) different number of neutrons
(d) different atomic numbers.
Answer:
(c) Different number of neutrons.

Question 17.
Number of valence electrons in Cl ion are :
(a) 16
(b) 8
(c) 17
(d) 18.
Answer:
(d) CE ion has 18 electrons (17 + 1).

Question 18.
Which one of the following is a correct electronic configuration of sodium ?
(a) 2, 8
(b) 8, 2, 1
(c) 2, 1, 8
(d) 2, 8, 1
Answer:
(d) is the correct answer.

Question 19.
Complete the following table.
NCERT Solutions for Class 9 Science Chapter 4 Structure of the Atom image - 8
Answer:
NCERT Solutions for Class 9 Science Chapter 4 Structure of the Atom image - 9
NCERT Solutions for Class 9 Science Chapter 4 Structure of the Atom image - 10

VERY SHORT ANSWER QUESTIONS

Question 1.
Out of the three sub-atomic particles known, which is called universal particle ?
Answer:
Electron is known as universal particle.

Question 2.
Name a species which has
(i) one proton and one electron
(ii) one proton and two electrons
Answer:
(i) Protium or hydrogen (H)
(ii) Hydride ion (H).

Question 3.
How many electrons are present in the species He2+ ion ? Suggest another name for it.
Answer:
He2+ ion has no electrons. It is also called alpha (α) particle.

Question 4.
How many times is radius of extra nuclear protion more than that of the nucleus of an atom ?
Answer:
Radius of extra nuclear portion is nearly 105 times more as compared to the nucleus.

Question 5.
Is atomic number of an atom always equal to the number of electrons ?
Answer:
No, it is the case when the atom has no charge. In case of cation (positively charged), atomic number is more than the number of electrons and in case of anion (negatively charged), it is less than the number of electrons.

Question 6.
Out of proton and neutron, which is heavier ?
Answer:
Neutron is slightly heavier (1.675 x 10-27 kg) than proton (1.67 x 10-27 kg).

Question 7.
Explain why chlorine has relative atomic mass of about 35.5 u ? (CBSE 2016)
Answer:
Chlorine always exists as two isotopes. These are Cl-35 and Cl-37 isotopes and are present in the ratio of 3:1. The atomic mass of the element is therefore, fractional and is regarded as the relative atomic mass.

Question 8.
Why is proton so named ?
Answer:
Proton (H+) is formed when hydrogen atom also called protium, loses an electron. Therefore, it is called proton.

Question 9.
The cation is represented as M3+. What is the difference between the number of protons and electrons present ?
Answer:
In cation M3+, there are three electrons less as compared to protons.

Question 10.
If K and L shells of an atom are completely filled, what will be its name ?
Answer:
The atom will belong to the noble gas element neon (Ne).

Question 11.
Do isobars have also identical chemical characteristics like isotopes ?
Answer:
No, these are not identical because the isobars have different atomic numbers as well as different electronic configurations. ‘

Question 12.
The electronic configuration of an element is : 2(K), 8(L), 5(M). Predict its valency.
Answer:
The valency of the element is 3. It is calculated as : 8 – 5 = 3.

Question 13.
Write two important postulates of Bohr’s model of an atom. (CBSE 2014)
Answer:
The 2 main postulates of the theory are listed :

  1. In the extra nuclear portion of an atom, the electrons revolve in well defined circular paths known as orbits.
  2. These circular orbits are also known as energy levels or energy shells.

Question 14.
What is the basic difference between the isotopes of an element ?
Answer:
Isotopes of an element differ in the number of neutrons leading to different mass numbers.

Question 15.
The element aluminium is written by the symbol fgAI. Write the number of protons, electrons and neutrons present in it.
Answer:
No. of protons= 13 ; No. of electrons = 13 ; No. of neutrons = 27 – 13 = 14.

Question 16.
What is the number of electrons in the valence shell of chlorine (Z = 17) ?
Answer:
The electronic distribution of the element is : K(2), L(8), M(7). This means that the valence shell of chlorine has 7 electrons.

Question 17.
Which radioisotope is used for the treatment of cancer ?
Answer:
Radioisotope Co-60 is used for the treatment of cancer.

Question 18.
Out of C-12 and C-14 isotopes of carbon, which is of radioactive nature ?
Answer:
C-14 isotope is of radioactive nature.

Question 19.
If an atom has one proton and one electron, state the charge on the atom. Justify your answer. (CBSE 2014)
Answer:
The atoms is neutral. One unit positive charge on proton cancels with one unit negative charge on electron.

Question 20.
The electronic configuration of an element Z is 2, 8, 6. How many electrons does it require to have a stable
configuration ?
Answer:
In order to have a stable configuration, the element Z must have 8 electrons in the valence shell. It needs (8 – 6) = 2 electrons for this purpose.

Question 21.
Will 126C and 146C have different valencies ?
Answer:
Both the species are the isotopes of the same element i.e. the carbon. Since their atomic numbers are same, their electronic configurations as well as valencies will also be the same.

Question 22.
The atom of an element A’ has three electrons in the outermost shell. It loses one of these to the atom of another element ‘B’. What will be the nature and value of charge on the ion which results from ‘A’ ?
Answer:
Since the atom A’ has lost one electron, the ion formed will be positive and will have +1 charge. It can be represented as A+.

Question 23.
The atom of an element has electronic configuration 2, 8, 6. Does it gain two electrons or lose six electrons to have the configuration of the nearest noble gas element ?
Answer:
The atom has six electrons in its valence shell. It is very difficult for it to lose six electrons. However, it can gain two electrons from a neighboring atom of another element. By doing so, it will have the electronic configuration of nearest noble gas element argon (2, 8, 8).

Question 24.
Do the elements X and represent pair of isotopes ?
Answer:
No, they do not because the two elements differ in their atomic numbers. Isotopes have the same atomic
numbers.

Question 25.
An element ‘X’ has mass number 4 and atomic number 2. Find the valency of ‘X’. (CBSE 2012)
Answer:
The electronic configuration : K(2). Since the K-shell is filled, the valency of ‘X’ = zero.

Question 26.
Name the particles which determine the mass of an atom. (CBSE 2014)
Answer:
Protons and neutrons present in the nucleus determine the mass of an atom.

SHORT ANSWER QUESTIONS

Question 27.
What is the number of valence electrons in
(i) Sodium ion (Na+)
(ii) oxide ion (O2-) ? (CBSE 2012)
Answer:
(i) Sodium ion (Na+) :
No. of electrons : (11 – 1) = 10 ; Electronic configuration = 2, 8.
∴ Na+ ion has 8 valence electrons.
(ii) Oxide ion (O2-) :
No. of electrons : (8 + 2) = 10 ; Electronic configuration = 2, 8
∴ O2- ion has 8 valence electrons.
Both Na+ ion and O2- ion are known as isoelectronic species as they have same number of electrons i.e. 10. Their electronic configuration is similar to that of neon (Z = 10) which is a noble gas element.
This means that O2- ion, Na+ ion and Ne atom are isoelectronic species.

Question 28.
Elements A and ‘B’ have atomic numbers 18 and 16 respectively. Which of these two would be more reactive and why ?
Answer:
Electronic configuration of A (Z = 18) ; 2, 8, 8
Electronic configuration of B (Z = 16) ; 2, 8, 6
Element A’ has completely filled outermost shell (M-shell). It would be therefore, least reactive.
Element ‘B’ has incomplete outermost shell (M-shell has 6 electrons) only.
Therefore, element ‘B’ would be more reactive than element ‘A’.

Question 29.
Two atoms A and B have the following composition :
NCERT Solutions for Class 9 Science Chapter 4 Structure of the Atom image - 11
What are their mass numbers ? What is the relation between the species ?
Answer:
Mass number of A =17+18 = 35
Mass number of B = 17 + 20 = 37
The atoms A and B represent a pair of isotopes since thery have the same atomic number (17)

Question 30.
The composition of two atomic particles is given :
NCERT Solutions for Class 9 Science Chapter 4 Structure of the Atom image - 12
(i) What is the mass number of X ?
(ii) What is the mass number of Y
(iii) What is the relation between X and Y ?
(iv) Which element/elements do they represent ? (CBSE 2012)
Answer:
(i) Mass number of X = 8 + 8 =16
(ii) Mass number of Y = 8 + 9 = 17
(iii) The atomic particles represent a pair of isotopes.
(iv) The atomic number of the atomic particles is the same as the number of protons (8). The element with atomic number 8 is oxygen.

Question 31.
State whether the following statements are true or false.
(i) One mole of methane has four H atoms.
(ii) The isotopes of an element differ in the number of electrons.
(iii) The elements are identified by their atomic numbers.
(iv) The mass number of an element has specific units.
(v) The K shell of an element cannot have more than eight electrons.
Answer:
(i) False
(ii) False
(iii) True
(iv) False
(v) False

Question 32.
Name the elements which have the following electronic configuration :
(i) 2, 6
(ii) 2, 7
(iii) 2, 8, 1
(iv) 2, 8, 7
(v) 2, 8
Answer:
(i) oxygen
(ii) fluorine
(iii) sodium
(iv) chlorine
(v) neon.

Question 33.
Give three points of distinction between isotopes and isobars.
Answer:
NCERT Solutions for Class 9 Science Chapter 4 Structure of the Atom image - 13

Question 34.
The details about three atoms X, Y and Z are given. Give information about their atomic number, mass number and valency ?
NCERT Solutions for Class 9 Science Chapter 4 Structure of the Atom image - 14
Answer:
NCERT Solutions for Class 9 Science Chapter 4 Structure of the Atom image - 15

Question 35.
Why do elements helium, neon and argon have zero valencies ?
Answer:
The electronic configuration of three elements are given :
NCERT Solutions for Class 9 Science Chapter 4 Structure of the Atom image - 16
The first element helium (He) has only one shell and has maximum of two electrons in the only shell (K-shell) which it has.
Similarly, the next two elements neon (Ne) and argon (Ar) also have the maximum of eight electrons in their outermost shells L and M respectively.
This means that the atoms of all the three elements donot have any urge to either lose or gain electrons. Therefore, they show zero valencies. These are also called noble gas or inert gas elements as they donot take part in the chemical combination.

Question 36.
Give the electronic distribution in magnesium atom and magnesium ion. How do these electronic configuration differ ?
Answer:
NCERT Solutions for Class 9 Science Chapter 4 Structure of the Atom image - 17
Mg2+ ion has only two shells and its electronic configuration is similar to that of noble gas element neon (Ne). Mg atom has three shells with 2 electrons in the outer shell i.e. M-shell.

LONG ANSWER QUESTIONS

Question 37.
(a) Give the schematic atomic structures of chlorine atom and chloride ion.
(b) Mention two uses of isotopes in the field of medicine.
(c) How are the following pair of elements known as ?
4018Ar,   4020Ca
Answer:
(a)
NCERT Solutions for Class 9 Science Chapter 4 Structure of the Atom image - 18
(b) (i) Radioisotope 1-131 is used to check the working of thyroid gland.
(ii) Radioisotope Co – 60 is used in cancer therapy.
(c) These are known as isobars. Since they have the same mass number but different atomic numbers.

Question 38.
Answer the following in one line or two :
(a) What is the maximum number of electrons that can be accomodated in the outermost energy shell in an atom ?
(b) On the basis of Thomsons model of an atom, explain how an atom is neutral as a whole.
(c) How many neutrons are present in hydrogen atom ?
(d) Do isobars belong to the same element ?
(e) An element has five electrons in the M-shell which is the outermost shell. Write its electronic configuration.
(CBSE 2011)
Answer:
(a) The outermost energy shell in an atom can have a maximum of eight electrons.
(b) According to Thomson’s model of an atoms, all the protons in an atom are present in the positively charged sphere while negatively charged electrons are studded in this sphere. Since the electrons and protons are equal in number each carrying one unit charge, the atom as a whole is electrically neutral.
(c) Hydrogen atom has no neutron.
(d) No, isolars belong to different elements since they differ in their atomic numbers.
(e) The electronic configuration of the element is K(2) L(8) M(5).

Question 39.
Which observations in scattering experiment led Rutherford to make the following conclusions :
(i) Most of the space in an atom is empty.
(ii) Whole mass of an atom is present in its centre.
(iii) Nucleus is positively charged.
Answer:
(i) As most of the alpha particles passed through undeflected, this means that they did not come across any obstruction in their path. Thus, most of the space in an atom is expected to be empty.
(ii)

  1. As a few alpha particles suffered minor deflections and a very few major deflections, this means that these must have met with some obstructions in their path.
  2. This obstruction must be :
    1. Very small : Only a few particles were obstructed by it.Massive : Each alpha particle has 4u mass and is quite heavy.
    2. It could easily pass through a light obstruction by pushing it aside.

(iii) Positively charged : Alpha particles have positive charge. Since they were repelled or deflected back, the obstruction must also carry same charge i.e., positive charge, (similarly charged particles always repel each other


Question 40.

With the help of suitable activities shows that
(i) Cathode rays travel in straight line
(ii) Cathode rays consist of material particles. (CBSE 2012)
Answer:
(i) Cathode rays always travel in straight line. In order to demonstrate this, place an object in the path of the cathode rays that are emitted in a discharge tube. The shadow of the object will immediately appear behind it. This shows that these rays are stopped by the object and thus, travel in straight line.
NCERT Solutions for Class 9 Science Chapter 4 Structure of the Atom image - 19
(ii) Cathode rays consist of negatively charged particles known as electrons which have definite mass. In order to demonstrate this, place a light paddle wheel of mica mounted on an axle in the path of cathode rays. It will start rotating. This shows that the cathode rays or electrons are material particles.
NCERT Solutions for Class 9 Science Chapter 4 Structure of the Atom image - 20

Question 41.
How did Bohr distribute electrons in different shells in an atom ?
Answer:
In 1913, Neils Bohr gave a theory regarding the distribution of electrons in the extra nuclear space in an atom. The main postulates of the theory are listed :

  1. In the extra nuclear portion of an atom, the electrons revolve in well defined circular paths known as orbits.
  2. These circular orbits are also known as energy levels or energy shells.
  3. These have been designated as K, L, M, N, O, … (or as 1, 2, 3, 4, 5, …) based on the energy present.
  4. The order of the energy of these energy shells is :
    K<L<M<N<0 <…. or 1< 2< 3 < 4<5 <….
  5. While revolving in an orbit, the electron is not in a position to either lose or gain energy. In other words, its energy remains stationary. Therefore, these energy states for the electrons are also known as stationary states

Question 42.
The average atomic mass of a sample of an element ‘X’ is 16.2 u. What is the percentage of each isotope 168X and 188X in the sample ?
Answer:
Let the percentage of  168X isotope = x
The percentage of  188X isotope = 100 – x
NCERT Solutions for Class 9 Science Chapter 4 Structure of the Atom image - 21

Question 43.
Give reasons for the following :
(a) Isotopes of an element are chemically similar.
(b) An atom is electrically neutral.
(c) Noble gases show least reactivity
(d) Nucleus of an atom is heavy and and positively charged.
(e) Ions are more stable than atoms.
Answer:
(a) Isotopes of an element have same atomic number as well as electronic distribution. Since the chemical properties of the elements are related to their electronic distribution or configuration, the elements with similar configuration have similar properties. Thus, the isotopes of an element are chemically similar.
(b) In an atom, the number of protons in the nucleus is the equal to the number of electrons in the extra nuclear portion. Since each proton and each electron has the same charge but with opposite magnitude, atom is therefore, electrically neutral.
(c) The atoms of the noble gas elements have complete outermost electron shells. With the exception of helium (He) which has two electrons in its only shell (K-shell), the rest of the members of the family have eight electrons in their outermost or valence shell. This means that atoms of these elements have no urge to take part in chemical combination or they are least reactive in nature.
(d) Positively charged : Alpha particles have positive charge. Since they were repelled or deflected back, the obstruction must also carry same charge i.e., positive charge, (similarly charged particles always repel each other
(e) When an atom changes into ion (cation or anion) the valence shell of the ion has a complete octet, (i.e. it has eight electrons in the outermost or valence shell). In some ions, the valence shell is K-shell and it has only two electrons (e.g. Li+ ion or H ion). The ions are therefore, quite stable in nature.

Question 44.
The following data represents the distribution of electrons, protons and neutrons in atoms of four elements . A, B, C, D.
NCERT Solutions for Class 9 Science Chapter 4 Structure of the Atom image - 22
Answer the following questions :
(a) Describe the electronic distribution in atom of element B.
(b) Is element B a metal or a non-metal ? Why ?
(c) Which two elements form a pair of Isotopes ?
(d) Which two elements form in pair of Isobars ?
Answer:
(a) The electronic distribution of element B is 2, 8, 7 .
(b) The element B is non-metal (Cl). Its atom needs only one electron to have the configuration of nearest noble gas element argon (2, 8, 8).
(c) The elements B and C form pair of isotopes since both of them have same number of electrons 17 and also same atomic number (Z= 17)
(d) The elements A and D represent pair of isobars since both of them have same number (A = 40).

NCERT Solutions for Class 9 Science Chapter 4 Structure of The Atom

Hope given NCERT Solutions for Class 9 Science Chapter 4 are helpful to complete your science homework.

If you have any doubts, please comment below. Learn Insta try to provide online science tutoring for you.

RD Sharma Class 9 Solutions Chapter 7 Introduction to Euclid’s Geometry MCQS

RD Sharma Class 9 Solutions Chapter 7 Introduction to Euclid’s Geometry MCQS

These Solutions are part of RD Sharma Class 9 Solutions. Here we have given RD Sharma Class 9 Solutions Chapter 7 Introduction to Euclid’s Geometry MCQS

Other Exercises

Mark the correct alternative in each of the following:
Question 1.
If (4, 19) is a solution of the equation y = ax + 3, then a =
(a) 3                           
(b) 4
(c) 5            
(d) 6
Solution:
∵  (4, 19) is a solution of equation
y = ax + 3
∴ x = 4, y= 19 will satisfy the equation
∴  19 = a x 4 + 3 = 4a + 3
4a = 19-3 = 16 ⇒ a= \(\frac { 16 }{ 4 }\) = 4
∴  a = 4                                      (b)

Question 2.
If (a, 4) lies on the graph of 3x + y = 10, then the value of a is
(a) 3                           
(b) 1
(c) 2                           
(d) 4
Solution:
∵  (a, 4) is the solution of the equation 3x + y = 10
∴ x = a, y = 4 will satisfy the equation
∴ Substituting the value of x and y in the equation
3 xa + 4= 10 ⇒  3a =10- 4 = 6
⇒  a =  \(\frac { 6 }{ 3 }\) = 2
∴ a = 2                                           (c)

Question 3.
The graph of the linear equation 2x – y= 4 cuts x-axis at
(a) (2, 0)                     
(b) (-2, 0)
(c) (0, -4)                    
(d) (0, 4)
Solution:
∵  graph of the equation,
2x – y = 4 cuts x-axis
∴ y = 0
∴  2x – 0 = 4 ⇒  2x = 4
⇒  x = \(\frac { 4 }{ 2 }\) = 2
∴ The line cuts x-axis at (2, 0)               (a)

Question 4.
How many linear equations are satisfied by x = 2 and y = -3 ?
(a) Only one                
(b)   Two
(c) Three                     
(d)    Infinitely many
Solution:
∵  From a point, infinitely number of lines can pass.
∴  The solution x = 2, y = -3 is the solution of infinitely many linear equations.       (d)

Question 5.
The equation x – 2 = 0 on number line is represented by
(a) aline                      
(b)   a point
(c) infinitely many lines
(d) two lines
Solution:
The equation x – 2 = 0
⇒  x = 2
∴ It is representing by a point on a number line. (b)

Question 6.
x = 2, y = -1 is a solution of the linear equation
(a) x   + 2y  = 0           
(b) x + 2y =  4
(c) 2x + y =  0            
(d) 2x + y =  5
Solution:
x = 2, y = -1
Substituting the values of x and y in the equations one by one, we get (a) x + 2y = 0
⇒ 2 + 2(-1) = 0
⇒ 2 – 2 = 0
⇒ 0 = 0 which is true                             (a)

Question 7.
If (2k – 1, k) is a solution of the equation 10x – 9y = 12, then k =
(a) 1                           
(b) 2
(c) 3                           
(d) 4
Solution:
∵  (2k – 1, k) is a solution of the equation 10x – 9y = 12
Substituting the value of x and y in the equation
10(2k – 1) – 9k = 12
⇒ 20k – 10-9k= 12
⇒  20k – 9k = 12 + 10
⇒  11k = 22
⇒  k =\(\frac { 22 }{ 11 }\)  = 2
∴  k = 2                                                 (b)

Question 8.
The distance between the graph of the equation x = – 3    and x   = 2      is
(a) 1                             
(b) 2
(c) 3                             
(d) 5
Solution:
The distance between the  graphs of the equation
x = -3 and x = 2 will be
2(-3) = 2+ 3 = 5                                     (b) 

Question 9.
The distance   between the graphs of the equations y = -1 and y = 3    is
(a) 2                            
(b) 4
(c) 3                            
(d) 1
Solution:
The distance between the graphs of the equation
y = -1 and y = 3
is 3 – (-1) = 3 + 1 = 4                            (b)

Question 10.
If the graph of the equation 4x + 3y = 12 cuts the co-ordinate axes at A and B, then hypotenuse of right triangle AOB is of length
(a) 4 units
(b) 3 units
(c) 5 units          
(d) none of these
Solution:
Equation is 4x + 3y = 12
If it cuts the x-axis, then y = 0
∴  4x x 3 x 0 = 12
⇒  4x = 12 ⇒  x = \(\frac { 12 }{ 4 }\) = 3
OA = 3 units
∴ The point of intersection of x-axis is (3, 0)
Again if it cuts the y-axis, then x = 0 , Y= 0
∴  4x x 3 x 0 = 12
⇒ 4x = 12 ⇒ x =  \(\frac { 12 }{ 3 }\) = 4
⇒ OB = 4 units
∴ The point of intersection is (0, 4)
∴ In right ΔAOB,
AB2 = AO2 + OB2
= (3)2 + (4)2
= 9 + 16 = 25
= (5)2
∴ AB = 5 units                                        (c)

Hope given RD Sharma Class 9 Solutions Chapter 7 Introduction to Euclid’s Geometry MCQS are helpful to complete your math homework.

If you have any doubts, please comment below. Learn Insta try to provide online math tutoring for you.

RS Aggarwal Class 9 Solutions Chapter 14 Statistics Ex 14B

RS Aggarwal Class 9 Solutions Chapter 14 Statistics Ex 14B

These Solutions are part of RS Aggarwal Solutions Class 9. Here we have given RS Aggarwal Solutions Class 9 Chapter 14 Statistics Ex 14B.

Other Exercises

Question 1.
Solution:
We shall take the game along x-axis and number of students along y-axis.
RS Aggarwal Class 9 Solutions Chapter 14 Statistics Ex 14B Q1.1

Question 2.
Solution:
We shall take the time on x-axis and temperature (in °C) on y-axis.
RS Aggarwal Class 9 Solutions Chapter 14 Statistics Ex 14B Q2.1

Question 3.
Solution:
We shall take name of vehicle on x-axis and velocity (in km/hr) on y-axis
RS Aggarwal Class 9 Solutions Chapter 14 Statistics Ex 14B Q3.1

Question 4.
Solution:
We shall take sports on x-axis and number of students on y-axis.
RS Aggarwal Class 9 Solutions Chapter 14 Statistics Ex 14B Q4.1

Question 5.
Solution:
We shall take years on x-axis and number of students on y-axis.
RS Aggarwal Class 9 Solutions Chapter 14 Statistics Ex 14B Q5.1

Question 6.
Solution:
We shall take years on x-axis and number of students on y-axis.
RS Aggarwal Class 9 Solutions Chapter 14 Statistics Ex 14B Q6.1

Question 7.
Solution:
We shall take years on x-axis and number of students on y-axis.
RS Aggarwal Class 9 Solutions Chapter 14 Statistics Ex 14B Q7.1

Question 8.
Solution:
We shall take years on x-axis and number of students on y-axis.
RS Aggarwal Class 9 Solutions Chapter 14 Statistics Ex 14B Q8.1

Question 9.
Solution:
We shall take years on x-axis and number of students on y-axis.
RS Aggarwal Class 9 Solutions Chapter 14 Statistics Ex 14B Q9.1

Question 10.
Solution:
We shall take years on x-axis and number of students on y-axis.
RS Aggarwal Class 9 Solutions Chapter 14 Statistics Ex 14B Q10.1

Question 11.
Solution:
We shall take week on x-axis and Rate per 10g (in Rs.) on y-axis.
RS Aggarwal Class 9 Solutions Chapter 14 Statistics Ex 14B Q11.1

Question 12.
Solution:
We shall take mode of transport on x-axis and number of students on y-axis.
RS Aggarwal Class 9 Solutions Chapter 14 Statistics Ex 14B Q12.1

Question 13.
Solution:
We see from the graph that
(i) It shows the marks obtained by a student in various subjects.
(ii) The student is very well in mathematics.
(iii) The student is very’ poor, in Hindi.
(iv) Average marks
= \(\frac { 60+35+75+50+60 }{ 5 } \) (Here x = 5)
= \(\frac { 280 }{ 5 } \)
= 56 marks
RS Aggarwal Class 9 Solutions Chapter 14 Statistics Ex 14B Q13.1

Hope given RS Aggarwal Solutions Class 9 Chapter 14 Statistics Ex 14B are helpful to complete your math homework.

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RD Sharma Class 9 Solutions Chapter 6 Factorisation of Polynomials MCQS

RD Sharma Class 9 Solutions Chapter 6 Factorisation of Polynomials MCQS

These Solutions are part of RD Sharma Class 9 Solutions. Here we have given RD Sharma Class 9 Solutions Chapter 6 Factorisation of Polynomials MCQS

Other Exercises

Mark the correct alternative in each of the following:
Question 1.
If x – 2 is a factor of x2 + 3 ax – 2a, then a =
(a) 2
(b) -2
(c) 1                          
(d) -1
Solution:
∴  x – 2 is a factor of
f(x) = x2 + 3 ax – 2a
∴ Remainder = 0
Let x – 2 = 0, then x = 2
Now f(2) = (2)2 + 3a x 2 – 2a
= 4 + 6a – 2a = 4 + 4a
∴ Remainder = 0
∴  4 + 4a = 0 ⇒ 4a = -4
⇒ a = \(\frac { -4 }{ 4 }\) = -1
∴  a= -1                                            (d)

Question 2.
If x3 + 6x2 + 4x + k is exactly divisible by x + 2, then k =
(a) -6                        
(b) -7
(c) -8                        
(d) -10
Solution:
f(x) – x3 + 6x2 + 4x + k is divisible by x + 2
∴ Remainder = 0
Let x + 2 = 0, then x = -2
∴  f(-2) = (-2)3 + 6(-2)2 + 4(-2) + k
= -8 + 24-8 + k = 8 + k
∴  x + 2 is a factor
∴ Remainder = 0                                 .
⇒  8 + k= 0 ⇒ k = -8
k = -8                                                (c)

Question 3.
If x – a is a factor of x3 – 3x2 a + 2a2x + b, then the value of b is
(a) 0                          
(b) 2
(c) 1                          
(d) 3
Solution:
∴  x – a is a factor of x3 – 3x2 a + 2a2x + b
Let f(x) = x3 – 3x2 a + 2a2x+ b
and x – a = 0, then x = a
f(a) = a3 – 3a2.a + 2a2.a + b
= a3 – 3a3 + 2a3 + b = b
∵  x – a is a factor of f(x)
∴   b = 0                                           (a)

Question 4.
If x140 + 2x151 + k is divisible by x + 1, then the value of k is
(a) 1                         
(b) -3
(c) 2                          
(d) -2
Solution:
∴ x + 1 is a factor of f(x) = x140 + 2x151 + k
∴ Remainder will be zero
Let x + 1 = 0, then x = -1
∴  f(-1) = (-1)140 + 2(-1)151 + k
= 1 + 2 x (-1) + k          {∵  140 is even and 151 is odd}
=1-2+k=k-1
∵ Remainder = 0
∴ k – 1=0  ⇒ k=1                                   (a)

Question 5.
If x + 2 is a factor of x2 + mx + 14, then m =
(a) 7                          
(b) 2       
(c) 9                          
(d) 14
Solution:
x + 2 is a factor of(x) = x2 + mx + 14
Let x + 2 = 0, then x = -2
f(-2) = (-2)2 + m{-2) + 14
= 4 – 2m + 14 = 18 – 2m
∴  x + 2 is a factor of f(x)
∴  Remainder = 0
⇒  18 – 2m = 0
2m = 18  ⇒  m = \(\frac { 18 }{ 2 }\) = 9                       (c)

Question 6.
If x – 3 is a factor of x2 – ax – 15, then a =

(a) -2                         
(b) 5
(c) -5                         
(d) 3
Solution:
x – 3 is a factor of(x) = x2 – ax – 15
Let x – 3 = 0, then x = 3
∴ f(3) = (3)2 – a(3) – 15
= 9 -3a- 15
= -6 -3a
∴ x – 3 is a factor
∴  Remainder = 0
-6 – 3a = 0 ⇒  3a = -6
∴   a = \(\frac { -6 }{ 3 }\) = -2            (a)

Question 7.
If x51 + 51 is divided by x + 1, the remainder is
(a) 0                           
(b) 1
(c) 49                         
(d) 50
Solution:
Letf(x) = x51 + 51 is divisible by x + 1
Let x+1=0, then x = -1
∴ f(-1) = (-1)51 + 51 =-1+51   (∵ power 51 is an odd integer)
= 50                                                  (d)

Question 8.
If x+ 1 is a factor of the polynomial 2x2 + kx, then k =
(a) -2                           
(b) -3
(c)  4                           
(d)  2
Solution:
∴ x + 1 is a factor of the polynomial 2x2 + kx
Let x+1=0, then x = -1
Now f(x) = 2x2 + kx
∴ Remainder =f(-1) =  0
= 2(-1)2 + k(-1)
= 2 x 1 + k x (-1) = 2 – k
∴ x + 1 is a factor of f(x)
∴ Remainder = 0
∴ 2 – k = 0 ⇒  k = 2                                 (d)

Question 9.
If x + a is a factor of x4 – a2x2 + 3x – 6a, then a =
(a) 0                           
(b) -1
(c) 1                           
(d) 2
Solution:
x + a is a factor o f(x) = x4– a2x2 + 3x – 6a
Let x + a = 0, then x = -a
Now, f(-a) – (-a)4 -a2(-a)2 + 3 (-a) – 6a
= a4-a4-3a-6a = -9a
∴ x + a is a factor of f(x)
∴Remainder = 0
∴ -9a = 0 ⇒ a = 0                               (a)

Question 10.
The value of k for which x – 1 is a factor of 4x3 + 3x2 – 4x + k, is
(a) 3
(b) 1
(c) -2                          
(d) -3
Solution:
x- 1 is a factor of f(x) = 4x3 + 3x2 – 4x + k
Let x – 1 = 0, then x = 1
f(1) = 4(1 )3 + 3(1)2 – 4 x 1 + k
= 4+3-4+k=3+k
∴ x- 1 is a factor of f(x)
∴ Remainder = 0
∴ 3 + k = 0 ⇒ k = -3                              (d)

Question 11.
If x+2 and x-1 are the factors of x3+ 10x2 + mx + n, then the values of m and n are respectively
(a) 5 and-3                
(b) 17 and-8
(c) 7 and-18              
(d) 23 and -19
Solution:
x+ 2 and x – 1 are the factors of
f(x) = x3 + 10x2 + mx + n
Let x + 2 = 0, then x = -2
∴ f(-2) = (-2)3 + 10(-2)2 + m(-2) + n
= -8 + 40 – 2m + n = 32 – 2m + n
∴  x + 2 is a factor of f(x)
∴ Remainder = 0
∴ 32 – 2m + n = 0 ⇒  2m – n = 32   …(i)
Again x – 1 is a factor of f(x)
Let x-1=0, then x= 1
∴  f(1) = (1)3 + 10(1)2 + m x 1 +n
= 1 + 10+ m + n =m + n+11
∴  x- 1 is a factor of f(x)
∴ m + n+ 11=0 ⇒ m+n =-11       …(ii)
Adding (i) and (ii),
3m = 32 – 11 = 21
⇒ m = \(\frac { 21 }{ 3 }\)  = 7
and n = -11 – m = m = 7, n = -18
∴ m= 7, n = -18                          (c) 

Question 12.
Let f(x) be a polynomial such that  f( \(\frac { -1 }{ 2 }\) )= 0, then a factor of f(x) is
(a) 2x – 1
(b) 2x + b
(c) x- 1
(d) x + 1
Solution:
RD Sharma Class 9 Solutions Chapter 6 Factorisation of Polynomials MCQS Q12.1

Question 13.
When x3 – 2x2 + ax – b is divided by x2 – 2x-3, the remainder is x – 6. The value of a and b are respectively.
(a) -2, -6                    
(b) 2 and -6
(c) -2 and 6               
(d) 2 and 6
Solution:
Let f(x) = x3 – 2x2 + ax – b
and Dividing f(x) by x2 – 2x + 3
Remainder = x – 6
Let p(x) = x3 – 2x2 + ax – b – (x – 6) or x3 – 2x2 + x(a – 1) – b + 6 is divisible by x2 – 2x+ 3 exactly
Now, x2-2x-3 = x2-3x + x- 3
= x(x – 3) + 1(x – 3)
= (x – 3) (x + 1)
∴ x – 3 and x + 1 are the factors of p(x)
Let x – 3 = 0, then x = 3
∴  p(3) = (3)3 – 2(3)2 + (a-1)x3-b + 6
= 27-18 + 3a-3-b + 6
= 33-21+3 a-b
= 12 + 3 a-b
∴ x – 3 is a factor
∴ 12 + 3a-b = 0 ⇒ 3a-b = -12               …(i)
Again let x + 1 = 0, then x = -1
∴p(-1) = (-1)3  –  2(-l)2 + (a – 1) X (-1) – b + 6
= -1-2 + 1 – a – 6 + 6
= 4 – a- b
∴ x + 1 is a factor
4-a-b = 0 ⇒  a + b = 4                  …(ii)
Adding (i) and (ii),
4 a = -12 + 4 = -8 ⇒  a = \(\frac { -8 }{ 4 }\) = -2
From (ii),
and -2 + 6 = 4⇒ 6 = 4 + 2 = 6
∴ a = -2, h = 6                       (b)

Question 14.
One factor of x4 + x2 – 20 is x2 + 5, the other is
(a) x2 – 4
(b) x – 4
(c) x2-5
(d) x + 4
Solution:

RD Sharma Class 9 Solutions Chapter 6 Factorisation of Polynomials MCQS Q14.1

Question 15.
If (x – 1) is a factor of polynomial fix) but not of g(x), then it must be a factor of
(a) f(x) g(x)               
(b) -f(x) + g(x)
(c) f(x) – g(x)             
(d) {f(x) + g(x)}g(x)
Solution:
∴ (x – 1) is a factor of a polynomial f(x)
But not of a polynomial g(x)
∴ (x – 1) will be the factor of the product of f(x) and g(x)          (a)

Question 16.
(x + 1) is a factor of xn + 1 only if
(a) n is an odd integer
(b)  n is an even integer
(c) n is a negative integer
(d) n is a positive integer
Solution:
∴  (x + 1) is a factor of xn+ 1
Let x + 1 = 0, then x = -1
∴ f(x) = xn + 1
and f(-1) = (-1)n + 1
But (-1)n is positive if n is an even integer and negative if n is an odd integer and (-1)n +1=0    {∵ x + 1 is a factor of(x)}
(-1)n must be negative
∴ n is an odd integer                              (a)

Question 17.
If x2 + x + 1 is a factor of the polynomial 3x3 + 8x2 + 8x + 3 + 5k, then the value of k is
(a) 0
(b) \(\frac { 2 }{ 5 }\)
(c) \(\frac { 5 }{ 2 }\)
(d) -1
Solution:
x2 + x + 1 is a factor of
f(x) = 3x3 + 8x2 + 8x + 3 + 5k
Now dividing by x2 + x + 1, we get
RD Sharma Class 9 Solutions Chapter 6 Factorisation of Polynomials MCQS Q17.1

Question 18.
If (3x – 1)7 = a7x7 + a6x6 + a5x5 + … + a1x + a0, then a7 + a6 + a5 + … + a1 + a0 =
(a) 0                          
(b) 1
(c) 128                      
(d) 64
Solution:
f(x) = [3(1) – 1]7 = a7x7 + a6x6 + a5x5 + … + a1x + a0
Let x = 1, then
f(1) = (3x- 1)7 = a7(1)7 + a6(1)6 + a5(1)5 + … + a1 x 1 + a0
⇒  (3 – 1)7 = a7 x 1 + a6 x 1+ a5 x 1 + … + a1x 1 +a0
⇒  (2)7 = a7  + a6 + a5  + … + a1 +a0
∴ a7 + a6 + a5 + … + a1 + a0= 128   (c)

Question 19.
If both x – 2 and x – \(\frac { 1 }{ 2 }\) are factors of px2 + 5x + r, then
(a) p = r                 
(b) p + r = 0
(c) 2p + r = 0         
(d) p + 2r = 0
Solution:
RD Sharma Class 9 Solutions Chapter 6 Factorisation of Polynomials MCQS Q19.1
RD Sharma Class 9 Solutions Chapter 6 Factorisation of Polynomials MCQS Q19.2

Question 20.
If x2 – 1 is a factor of ax4 + bx3 + cx2 + dx + e, then
(a) a + c + e- b + d
(b
)a + b + e = c + d

(c)a + b + c = d+ e
(d
)b + c + d= a + e

Solution:
X2 – 1 is a factor of ax4 + bx3 + cx2 + dx + e
⇒  (x + 1), (x – 1) are the factors of ax4 + bx3 + cx2 + dx + e
Let f(x) = ax4 + bx3 + cx2 + dx + e
and x + 1 = 0 then x = -1
∴  f(-1) = a(-1)4 + b(-1)3 + c(-1)2 + d(-1) + e
= a- b + c- d+ e
∴ x + 1 is a factor of f(x)
∴  a-b + c- d+e = 0
⇒ a + c + e = b + d                 (a)

Hope given RD Sharma Class 9 Solutions Chapter 6 Factorisation of Polynomials MCQS are helpful to complete your math homework.

If you have any doubts, please comment below. Learn Insta try to provide online math tutoring for you.