Author name: Prasanna

These NCERT Solutions for Class 6 Maths Chapter 8 Decimals Ex 8.1 Questions and Answers are prepared by our highly skilled subject experts.

NCERT Solutions for Class 6 Maths Chapter 8 Decimals Ex 8.1

Question 1.
Write the following as numbers in the given table.

Answer:

Question 2.
Write the following decimals in the place value table.
(a) 19.4
(b) 0.3
(c) 10.6
(d) 205.9
Answer:

Question 3.
Write each of the following as decimals:
(a) Seven-tenths
(b) Two tens and nine-tenths
(c) Fourteen point six
(d) One hundred and two ones
(e) Six hundred point eight
Answer:
(a) The decimal-form of seven-tenths is 7 = \(\frac{7}{10}\) = 0.7
(b) 2 tens and 9-tenths
= 2 x 10 + \(\frac{9}{10}\) = 20 + 0.9
= 20.9
(c) Fourteen point six = 14.6
(d) One hundred and 2-ones
= 100 + 2 x 1 = 100 + 2 = 102
(e) Six hundred point eight = 600.8

Question 4.
Write each of the following as decimals:
Answer:
(a) \(\frac{5}{10}\) = 0.5
(b) 3 + \(\frac{7}{10}\) = 3 + 0.7 = 3.7
(c) 200 + 60 + 5 + \(\frac{1}{10}\)
= 200 + 60 + 5 + 0.1 = 265.1
(d) 70 + \(\frac{8}{10}\) = 70 + 0.8 = 70.8
(e) \(\frac{88}{10}=\frac{80+8}{10}=\frac{80}{10}+\frac{8}{10}\)
= 8 + \(\frac{8}{10}\) = 8 + 0.8 = 8.8
(f) 4\(\frac{2}{10}\) = 4 + \(\frac{2}{10}\) = 4 + 0.2 = 4.2
(g) \(\frac{3}{2}=\frac{3 \times 5}{2 \times 5}=\frac{15}{10}=\frac{10+5}{10}\)
= \(\frac{10}{10}+\frac{5}{10}\) = 1 + 0.5 = 1.5
(h) \(\frac{2}{5}=\frac{2 \times 5}{5 \times 2}=\frac{4}{10}\) = 0.4
(i) \(\frac{12}{5}=\frac{12 \times 2}{5 \times 2}=\frac{24}{10}=\frac{20+4}{10}\)
= \(\frac{20}{10}+\frac{4}{10}\) = 2 + 0.4 = 2.4
(j) \(3 \frac{3}{5}=3+\frac{3}{5}=3+\frac{3 \times 2}{5 \times 2}\)
= 3 + \(\frac{6}{10}\) = 3 + 0.6 = 3.6
(k) \(4 \frac{1}{2}=4+\frac{1}{2}=4+\frac{1 \times 5}{2 \times 5}\)
= 4 + \(\frac{5}{10}\) = 4 + 0.5 = 4.5

Question 5.
Write the following decimals as fractions. Reduce the fractions to lowest form.
Answer:
(a) 0.6 = \(\frac{6}{10}=\frac{3}{5}\)
(b) 2.5 = \(\frac{25}{10}=\frac{5}{2}\)
(c) 1.0 = \(\frac{10}{10}\) = 1
(d) 3.8 = \(\frac{38}{10}=\frac{19}{5}\)
(e) 13.7 = \(\frac{137}{10}\)
(f) 21.2 = \(\frac{212}{10}=\frac{106}{5}\)
(g) 6.4 = \(\frac{64}{10}=\frac{32}{5}\)

Question 6.
Express the following as cm using decimals.
(a) 2 mm
(b) 30 mm
(c) 116 mm
(d) 4 cm 2 mm
(e) 162 mm
(f) 83 mm
Answer:
(a) ∵ 10 mm = 1 cm
∴ 1 mm = \(\frac { 1 }{ 10 }\) cm
∴ 2 mm = \(\frac { 1 }{ 10 }\) x 2 = 0.2 cm

(b) ∵10 mm = 1 cm
∴ 1 mm = \(\frac { 1 }{ 10 }\) cm
∴ 30 mm = \(\frac { 1 }{ 10 }\) x 20 = 3.0 cm

(c) ∵ 10 mm = 1 cm
∴ 1 mm = \(\frac { 1 }{ 10 }\) cm 10
∴ 116 mm = \(\frac { 1 }{ 10 }\) x 116 = 11.6 cm

(d) ∵ 10 mm = 1 cm
4 cm 2 mm = 4 + \(\frac { 2 }{ 10 }\) cm 10
= 4 + 0.2 = 4.2 cm

(e) ∵ 10 mm = 1 cm
∴ 1 mm = \(\frac { 1 }{ 10 }\) cm 10
∴ 162 mm = \(\frac { 1 }{ 10 }\) x 162 = 16.2 cm 10

(f) ∵ 10 mm = 1 cm
∴ 1 mm = \(\frac { 1 }{ 10 }\) cm 10
∴ 83 mm =\(\frac { 1 }{ 10 }\) x 83 = 8.3 cm 10

Question 7.
Between which two whole numbers on the number line are the given numbers lie? Which of these whole numbers is nearer the number?

(a) 0.8
(b) 5.1
(c) 2.6
(d) 6.4
(e) 9.1
(f) 4.9
Answer:
(a) 0.8 lies between 0 and 1, 0.8 is nearer to 1
(b) 5.1 lies between to 6,5.1 is nearer to 5.
(c) From 2 to 3, 2.6 is nearest to 3.
(d) 2.6 lies between 6 to 7, 6.4 is nearer to 6.
(e) 9.1 lies between 9 to 10,9.1 is nearer to 9.
(f) 4.9 lies between 4 to 5, 4.9 is nearer to 5.

Question 8.
Show the following numbers on the number line:
(a) 0.2
(b) 1.9
(c) 1.1
(d) 2.5
Answer:

Question 9.
Write the decimal number represented by the points A, B, C, D on the given number line.

Answer:
A = 0 + \(\frac{8}{10}\) = 0.8
B = 1 + \(\frac{3}{10}\) = 1.3
C = 2 + \(\frac{2}{10}\) = 2.2
D = 2 + \(\frac{9}{10}\) = 2.9

Question 10.
(a) The length of Ramesh’s notebook is 9 cm and 5 mm. What will be its length in cm?
(b) The length of a young gram plant is 65 mm. Express its length in cm.
Answer:
(a) 9 cm 5 mm = 9 cm + 5 mm
= 9 + \(\frac{5}{10}\) = 9.5 cm
(b) 65 mm = \(\frac{65}{10}\) cm = 6.5 cm

These NCERT Solutions for Class 9 Maths Chapter 10 Circles Ex 10.5 Questions and Answers are prepared by our highly skilled subject experts.

NCERT Solutions for Class 9 Maths Chapter 10 Circles Exercise 10.5

Question 1.
In Fig. 10.36, A, B and C are three points on a circle with centre O such that ∠BOC = 30° and ∠AOB = 60°. If D In a point on the circle other than the arc ABC, find ∠ADC.

Solution:
We have given,
∠BOC = 30° and ∠AOB = 60°
∠AQC = 30° + 60°
∴ ∠AOC = 90° …….(i)
Again, ∠AOC and ∠ADC subtended by same arc AC.
We know that the angle subtended by an arc at the centre is double the angle subtended by it at any point on the remaining part of the circle.
∠AOC = 2∠ADC
⇒ 90° = 2∠ADC (From (i))
⇒ ∠ADC = 45°

Question 2.
A chord of a circle is equal to the radius of the circle. Find the angle subtended by the chord at a point on the minor arc and also at a point on the major arc.

Solution:
In circle whose centre is O. AB is a chord equal to its radius.
∴ OAB is a an equilateral triangle.
We know that each angle of an equilateral triangle is 60°.
∴ AOB = 60°
Now, ∠AOB = 2∠AQB
(The angle subtended by an arc at the centre is double the angle subtanded by it at any point on the remaining part of the circle)
⇒ 60° = 2∠AQB
⇒ ∠AQB = 30°
Again, QAPB is a cyclic quadrilateral.
∴ ∠AQB + ∠APB = 180° (Sum of opposite angles, of a cyclic quadrilateral is 180°)
⇒ 30 + ∠APB = 180°
⇒ ∠APB = 180° – 30° = 150°
Therefore, angle subtended by the chord at a point on the minor arc is 150° and at a point on major arc is 30°.

Question 3.
In Fig. 10.37, ∠PQR = 100°, where P, Q and R are points on a circle with centre O. Find ∠OPR.

Solution:
We have given ∠PQR = 100°.
Since, angle subtended by an arc at the centre is double tire angle subtended by it at any point on the remaining part of the circle.
∴ ∠1 = 2∠PQR
⇒ ∠1 = 2 × 100°
⇒ ∠1 = 200°
Again, ∠1 + ∠2 = 360°
⇒ 200° + ∠2 = 360° (∵ ∠1 = 200°)
⇒ ∠2 = 160°
Now, In ∆POR
OP = OR (Radii of circle)
∴ ∠OPR = ∠ORP (Angle opposite to equal sides of ∆ are equal)
Therefore,
∠OPR + ∠ORP + ∠2 = 180 (Angle sum property of triangle)
⇒ ∠OPR + ∠OPR + 160 = 180
⇒ 2∠OPR = 20
⇒ ∠OPR = 10°

Question 4.
In Fig. 10.38, ∠ABC = 69°, ∠ACB = 31°, find ∠BDC.

Solution:
In ∆ABC,
∠ABC = 69° and ∠ACB = 31°
∴ ∠ABC + ∠ACB + ∠BAC = 180° (Angle sum property of ∆)
⇒ 69°+ 31° + ∠BAC = 180°
⇒ ∠BAC = 80°
We know that angles in the same segment of a circle are equal.
∴ ∠BAC = ∠BDC
∠BDC = 80° (∵ ∠BAC = 80°)

Question 5.
In Fig. 10.39, A, B, C and D are four points on a Circle. AC and BD intersect at a point E such that ∠BEC = 130° and ∠ECD = 20°. Find ∠BAC.

Solution:
We have, ∠BEC = 130° and ∠ECD = 20°
Again, ∠BEC + ∠CED = 180° (linear pair)
⇒ 130° + ∠CED = 180°
⇒ ∠CED = 50°
Now, In triangle CDE
∠CED + ∠ECD + ∠CDE = 180° (Sum of all angles of a ∆)
⇒ 50° + 20° + ∠CDE = 180
⇒ ∠CDE = 110°
We know that angles on the same segments of a circle are equal.
∴ ∠CDB = ∠BAC
or, ∠BAC = 110° (∵ ∠CDB = 110°)

Question 6.
ABCD is a cyclic quadrilateral whose diagonals intersect at a point E. If ∠DBC = 70°, ∠BAC is 30°, find ∠BCD. Further, if AB = BC, find ∠ECD.
Solution:
We know that angles on the same segment of a circle are equal.

∴ ∠DAC = ∠DBC
or, ∠DAC = 70° (∵ ∠DBC = 70°)
Again, ABCD is a cyclic quadrilateral.
∴ ∠DAB + ∠BCD = 180°
(Sum of opposite angles of a cyclic quadrilateral is 180°)
100° + ∠BCD = 180° (∵∠DAB = ∠DAC + ∠BAC)
⇒ ∠BCD = 80°
Now, in triangle ABC,
AB = BC (Given)
∴ ∠BAC = ∠BCA (Angle opposite to equal sides are equal)
or, ∠BCA = 30°
Again, ∠BCD = ∠BCA + ∠ACD
⇒ 80° = 30° + ∠ACD
⇒ ∠ACD = 50°
∴ ∠ECD = 50°

Question 7.
If diagonals of a cyclic quadrilateral are diameter of the circle through the vertices of the quadrilateral, prove that it is a rectangle.

Solution:
Given: ABCD is a cyclic quadrilateral in which diagonal AC and BD are diameter of circle.
To prove that: ABCD is a rectangle.
Proof: In ABCD,
AC = BD (Diameter of circle are equal)
Again, ∠ABC = 90° (Angle in a semicircle is a right angle)
Similarly, ∠BAD = ∠ADC = ∠BCD = 90°
Now, we know that, if a quadrilateral has both diagonals are equal and each angle is 90° then the quadrilateral is a rectangle.
∴ ABCD is a rectangle.

Question 8.
If the non parallel sides of a trapezium are equal, prove that it is cyclic.
Solution:
Given: A trapezium ABCD in which AB || CD and AD = BC.
To prove that: ABCD is a cyclic trapezium.
Construction: Draw DE ⊥ AB and CF ⊥ AB.
Proof: In order to prove that ABCD is a cyclic trapezium, it is sufficient to show that ∠B + ∠D = 180°.

In triangles DEA and CFB, we have
AD = BC (Given)
∠DEA = ∠CFB (Each equal to 90°)
DE = CF (Distance between two parallel lines is always same)
So, by S- A-S congruency condition
∆DEA ≅ ∆CFB
∴ ∠A = ∠B and ∠ADE = ∠BCF (By CPCT)
⇒ 90° + ∠ADE = 90° + ∠BCF (Add 90° both side)
⇒ ∠EDC + ∠ADE = ∠FCD + ∠BCF (∵ ∠EDC = 90° and ∠CD = 90°)
⇒ ∠ADC = ∠BCD
⇒ ∠D = ∠C
Thus, ∠A = ∠B and ∠C = ∠D
∴ ∠A + ∠B + ∠C + ∠D = 360° (Sum of the angles of a equal is 360°)
⇒ 2∠B + 2∠D = 360°
⇒ ∠B + ∠D = 180°
Hence ABCD is a cyclic quadrilateral.

Question 9.
Two circles intersect at two points B and C. Through B, two line segment ABD and PBQ are drawn to intersect the circle at A, D and P, Q respectively (see Fig. 10.40). Prove that ∠ACP = ∠QCD.

Solution:
We know that angles in the same segment of a circle are equal.
∠D = ∠Q and ∠P = ∠A
Again, In ∆PQC,
∠P + ∠Q + ∠PCQ = 180° ……(i)
(Sum of all three angles of the triangle is 180°)
Now, In ∆ADC,
∠A + ∠D + ∠ACD = 180° …….(ii)
(Sum of all three angles of a triangle is 180°)
From equation (i) and (ii)
∠P + ∠Q + ∠PCQ = ∠A + ∠D + ∠ACD
⇒ ∠P + ∠Q + ∠PCQ = ∠P + ∠Q + ∠ACD (∵ ∠A = ∠P and ∠D = ∠Q prove above)
∴ ∠PCQ = ∠ACD
⇒ ∠PCQ – ∠PCD = ∠ACD – ∠PCD (Subtract ∠PCD both side)
⇒ ∠QCD = ∠ACP
⇒ ∠ACP = ∠QCD

Question 10.
If circles are drawn taking two sides of a triangle as diameters, prove that the point of intersection of these circles lie on the third side.
Solution:
Join AD
Since angle in a semicircle is right angle.
Therefore,

∠ADB = 90° and ∠ADC = 90°
⇒ ∠ADB + ∠ADC = 90° + 90°
⇒ ∠ADB + ∠ADC = 180°
⇒ ∠BDC = 180°
Therefore, BDC is a straight line or, D must be lie on BC.

Question 11.
ABC and ADC are two right triangles with common hypotenuse AC. Prove that ∠CAD = ∠CBD.
Solution:
Given: ABC and DBC are two right angle triangles on same base AC.

To prove that: ∠CAD = ∠CBD
Construction: Take AC as a diameter and draw a circle and join BD.
Proof: We know that angle in a semicircle is a right angle.
Therefore, vertex B of ∆ABC and vertex D of ∆ADC must be lie on the circumference of the circle.
Again we know that angles in the same segment of a circle are equal.
∴ ∠CAD = ∠CBD.

Question 12.
Prove that a cyclic parallelogram is a rectangle.
Solution:
Let ABCD be acyclic parallelogram. In order to prove that it is a rectangle, it is sufficient to show that one of the angles of parallelogram ABCD is a right angle.

Now ABCD is a parallelogram.
∠B = ∠D ……(i) [∵ opposite angle of a || gm are equal]
Also, ABCD is a cyclic quadrilateral
∴ ∠B + ∠D = 180° ……(ii) (Sum of opposite angles of cyclic quadrilateral)
⇒ ∠B + ∠B = 180° [∵ ∠B = ∠D from (i)]
⇒ 2∠B = 180°
⇒ ∠B = 90°
Hence ABCD is a rectangle.

These NCERT Solutions for Class 6 Maths Chapter 9 Data Handling Ex 9.1 Questions and Answers are prepared by our highly skilled subject experts.

NCERT Solutions for Class 6 Maths Chapter 9 Data Handling Exercise 9.1

Question 1.
In a Mathematics test, the following marks were obtained by 40 students. Arrange these marks in a table using tally marks.

(a) Find how many students obtained marks equal to or more than 7.
(b) How many students obtained marks below 4?
Answer:
Marks Tally Marks No. of students


(a) Twelve students
(b) Eight students

Question 2.
Following is the choice of sweets of 30 students of Class VI.
Ladoo, Barfi, Ladoo, Jalebi, Ladoo, Rasgul- la, Jalebi, Ladoo, Barfi, Rasgulla, Ladoo, Jalebi, Jalebi, Rasgulla, Ladoo, Rasgulla, Jalebi, Ladoo, Rasgulla, Laddo, Ladoo, Barfi, Rasgulla, Rasgulla, Jalebi, Rasgulla, Ladoo, Rasgulla, Jalebi, Ladoo.
(a) Arrange the names of sweets in a ta¬ble using tally marks.
(b) Which sweet is preferred by most of the students?
Answer:
(a) Table using tally marks:

(b) Ladoo, because 11 students prefer to eat.

Question 3.
Catherine threw a dice 40 times and noted the number appearing each time as shown below:

Make a table and enter the data using tally marks. Find the number that appeared.
(a) The minimum number of times
(b) The maximum number of times
(c) Find those numbers that appear an equal number of times.
Answer:

(a) The minimum number of times = 4
(b) The maximum number of times = 5
(c) 1 and 6 are the numbers that

Question 4.
Following pictograph shows the number of tractors in five villages.

Observe the pictograph and answer the following questions:
(i) Which village has the minimum number of tractors?
(ii) Which village has the maximum number of tractors?
(iii) How many more tractors village C has as compared to village B.
(iv) What is the total number of tractors in all the five villages?
Answer:
(i) Village D
(ii) Village C
(iii) 3 tractor
(iv) 28 tractor

Question 5.
The number of girl students in each class of a co-educational middle school is depicted by the pictograph:

Observe this pictograph and answer the following questions:
(a) Which class has the minimum number of girl students?
(b) Is the number of girls in Class VI less than the number of girls in Class V?
(c) How many girls are there in Class VII?
Answer:
(a) Class VIII
(b) No
(c) 3 × 4 = 12 girls

Question 6.
The sale of electric bulbs on different days of a week is shown below:


Observe the pictograph and answer the following questions:
(a) How many bulbs were sold on Friday?
(b) On which day were the maximum number of bulbs sold?
(c) On which of the days same number of bulbs were sold?
(d) On which of the days minimum number of bulbs were sold?
(e) If one big carton can hold 9 bulbs. How many cartons were needed in the given week?
Answer:
(a) Number of bulbs sold on Friday are 14.
(b) Maximum number of bulbs (18) were sold on Sunday.
(c) Same number of bulbs (8) were sold on Wednesday and Saturday.
(d) Then minimum number of bulbs (8) were sold on Wednesday and Saturday.
(e) The total number of bulbs sold in the given week were 86.
So the number of carton required to hold 86 bulb = \(\frac{86}{9}\) ≈ 10. Hence, 10 cartons required to hold the bulbs.

Question 7.
In a village, six fruit merchants sold the following number of fruit baskets in a particular season:


Observe this pictograph and answer the following question:
(a) Which merchant sold the maximum number of baskets?
(b) How many fruit baskets were sold by Anwar?
(c) The merchants who have sold 600 or more number of baskets are planning to buy a godown for the next season. Can you name them?
Answer:
(a) Martin
(b) 7 × 100 = 700 fruit basket
(c) Anwar, Martin, Ranjit Singh

These NCERT Solutions for Class 8 Science Chapter 12 Friction Questions and Answers are prepared by our highly skilled subject experts to help students while preparing for their exams.

Friction NCERT Solutions for Class 8 Science Chapter 12

Class 8 Science Chapter 12 Friction Textbook Exercise Questions and Answers

Page-155

Question 1.
Fill in the blanks:
a. Friction opposes the ……………. between the surfaces in contact with each other.
b. Friction depends on the ……………. of surfaces.
c. Friction produces …………….
d. Sprinkling of powder on the carrom board ……………. friction.
e. Sliding friction is ……………. than the static friction.
Answer:
a. motion,
b. nature,
c. heat,
d. reduces,
e. less

Question 2.
Four children were asked to arrange forces due to rolling, static and sliding frictions in a ‘decreasing order’.
Their arrangements are given below. Choose the correct arrangement.
a. rolling, static, sliding
b. rolling, sliding, static
c. static, sliding, rolling
d. sliding, static, rolling
Answer:
c. static, sliding, rolling

Question 3.
Alida runs her toy car on dry marble floor, wet marble floor, newspaper and towel spread on the floor. The force of friction acting on the car on different surfaces in ‘increasing order’ will be:
a. wet marble floor, dry marble floor, newspaper and towel.
b. newspaper, towel, dry marble floor, wet marble floor.
c. towel, newspaper, dry marble floor, wet marble floor.
d. wet marble floor, dry marble floor, tow el, newspaper
Answer:
a. wet marble floor, dry marble floor, newspaper and towel.

Question 4.
Suppose your writing desk is tilted a little. A book kept on it starts sliding down. Show the direction of frictional force acting on it.
Answer:
When book slides down the desk, a frictional force acts between the book and the surface of the desk. The direction of the frictional force on the book is opposite to the direction of its motion and acts in an upward direction. It is shown in the diagram below:

Question 5.
You spill a bucket of soapy water on a marble floor accidently. Would it make it easier or more difficult for you to walk on the floor? Why?
Answer:
We are able to walk because of the friction present between our feet and the ground. In order to walk, we push the ground in the backward direction with our feet. The force of friction pushes it in the forward direction and allows us to walk. The force of friction between the ground and feet decreases when there is soapy water spilled on the floor. It acts as a lubricant. Hence, it becomes difficult to walk on the soapy floor.

Question 6.
Explain why sportsmen use shoes with spikes.
Answer:
Shoes with spikes offer more friction because of more irregularities in the surface. This gives better grip to sportsmen while running.

Question 7.
Iqbal has to push a lighter box and Seema has to push a similar heavier box on the same floor. Who will have to apply a larger force and why?
Answer:
Force of friction arises because of interlocking of irregularities on the two surfaces in contact. When a heavy object is placed on the floor, the interlocking of irregularities on the surfaces of box and floor becomes strong. This is because the two surfaces in contact are pressed harder. Hence, more force is required to overcome the interlocking. Thus, to push the heavier box, Seema has to apply a greater force than Iqbal.

Question 8.
Explain why the sliding friction is less than the static friction.
Answer:
When something is static, a greater force is required to break the interlocking between two surfaces. When something is in motion, the surfaces do not get enough time to interlock properly and a smaller force is required to keep the object in motion. Hence, sliding friction is less than static friction.

Question 9.
Give examples to show that friction is both a friend and a foe.
Answer:
Following two examples show that friction is both a friend and a foe:
Friction as a friend:

• We are able to walk because of friction.
• Friction between the tip of the pen and a paper allows us to write.

Friction as a foe:

• Tyres and soles of shoes wear out because of friction.
• Friction between the different parts of machines produces heat. This can damage the machines.

Question 10.
Explain why objects moving in fluids must have special shapes.
Answer:
When a body moves through a fluid, it experiences a force which tries to oppose its motion through the fluid. This opposing force is known as drag. This frictional force depends on the shape of the body. By giving the objects a special shape, called the streamlined shape, the force of friction acting on it can be minimised. Hence, it becomes easier for a body to move through the fluid.

NCERT Extended Learning Activities and Projects

Question 1.
What role does friction play in the sport of your choice? Collect some pictures of that sport in action where friction is either supporting it or opposing it. Display these pictures with proper captions on the bulletin board of your classroom.
Hint:
Supporting friction:
a. Cycling: The friction between the tyres and the road is necessary for the cycle to function.
b. Running: The friction between the athlete’s feet and the ground surface is necessary otherwise the runner would fall over.
c. Swimming: The friction in the water is required for the swimmer to be propelled forward.

Opposing friction:
a. Swimming: The friction makes it more difficult for the swimmer to swim very fast.
b. Bobsleigh: The aim is to go as fast as possible but the friction between the bottom of the sleigh and the ice slows it down.
c. Diving: The friction between the diver and the water causes wastage of kinetic energy and sound energy whereas the aim is to be as streamlined as possible.

Question 2.
Imagine that friction suddenly vanishes. How would the life be affected. List ten such situations.
Hint:
If frictional force were absent, we may face the following consequences:

• We cannot walk.
• We will not get a grip to hold things, and then we cannot eat, write, hold a pen or pencil, etc.
• Moving things cannot be stopped.
• Buildings cannot be constructed.
• We cannot fix a nail in the wall.
• We cannot stand properly without a grip.
• We would keep slipping.
• Nothing will be steady on ground. Things will not be at proper places because of absence of grip.
• Brakes in the vehicles will be useless.
• Finally, life would become miserable in the absence of friction.

Question 3.
Visit a shop which sells sports shoes. Observe the soles of shoes meant for various sports. Describe your observations.
Hint:
Sports shoes have rough sole and spikes and the reason behind this is to give a better grip and traction on the grassy field. Also, if it is a rainy day, players (in football) can make better cuts with the cleats.

Question 4.
A toy to play with: Take an empty match box. Take out its tray. Cut a used refill of a ball pen of the same width as the tray as shown in the figure below. Fix the refill with two pins on the top of the tray as shown in figure. Make two holes on the opposite sides of the tray. Make sure that the holes are large enough to allow a thread to pass through them easily. Take a thread about a metre long and pass it through the holes as shown. Fix beads at the two ends of the thread so that it does not come out. Insert the tray in the outer cover of the matchbox.
Suspend the match box by the thread. Leave the thread loose. The match box will start falling down due to gravity. Tighten the thread now and observe what happens.
Explain your observation. Can you relate it to friction?

Hint:
Do it yourself.

Activity 1

Objective: To show that a rough surface exerts greater frictional force than a smooth surface.
Materials Required: A brick, a string, spring balance, polythene, jute bag.
Procedure:

• Tie a string around a brick.
• Pull the brick by a spring balance. You need to apply some force.
• Note down the reading on the spring balance when the brick just begins to move. It gives you a measure of the force of friction between the surface of the brick and the floor.
• Now wrap a piece of polythene around the brick and repeat the activity and observe.
• Repeat the activity by wrapping a piece of jute bag around the brick and observe again.

Observation: In the first case, as the brick surface in contact with the ground is very rough, the brick experienced greater frictional force. But in the second case, when polythene is wrapped around the brick, the surface becomes smoother. So, in this case, the frictional force is less. The reading of spring balance is more than that of when brick was wrapped with polythene. Jute is also rough and exerts a larger frictional force.

Conclusion: A rough surface exerts a greater frictional force than a smooth surface.

Activity 2

Objective: To study that force of friction depends upon the nature of the two surfaces in contact.
Materials Required: A wooden board, a pencil cell, bricks or books, a piece of cloth and sand.
Procedure:

• Make an inclined plane on a smooth floor, or on a table. (You may use a wooden board supported by bricks or books.)
• Put a mark with a pen at any point A on the inclined plane.
• Now let a pencil cell roll down from this point.
• Observe how far does it move on the plane surface before coming to rest. Note down the distance.
• Now spread a piece of cloth over the table. Make sure that there are no wrinkle in the cloth. Try this activity again.
• Repeat this activity by spreading a thin layer of sand over the table. Maintain the same slope throughout the activity.

Observation: The distance covered is minimum in the case when thin layer of sand is spread over the table. The distance covered is maximum on the plane surface. The distance covered by the pencil cell is different every time because of the difference in the amount of friction provided by the different surfaces.

Conclusion: Friction depends upon the nature of surfaces in contact.

Spring balance: It is a device used for measuring the force acting on an object. It consists of a coiled spring and a pointer moving on a graduated scale. When a force is applied, stretching of spring takes place. The reading on the scale indicated by the pointer gives the magnitude of the force.

Types of Friction:
i. Static Friction: The minimum amount of force that is required to overcome the force of friction is called static friction. In other words, the minimum force that is required to make an object moving is called static friction.
ii. Sliding Friction: Once an object starts moving on a surface, some force is required to keep the object moving. The minimum force that is required to keep a moving object in motion at a constant speed is called sliding friction. Sliding friction is always less than static friction. Due to this, it is easier to push a moving box than to make a static box move.

iii. Rolling Friction: When an object rolls over a surface, the resistance produced to its motion is called rolling friction. Rolling friction is less than both the static and the sliding friction. It is always easy to roll than to slide a body over a surface.

Friction-A Necessary Evil: Friction is a necessary evil because it has both harmful and beneficial effects.
Benefits of friction:

• We can walk easily because ground offers friction. In less friction, walking becomes almost impossible. Therefore, friction is necessary even for a simple task like walking. It is difficult to move on a wet muddy track or wet marble floor. This is because these surfaces offer very little friction to the surface of the feet of a person.
• Writing with pen is possible because paper provides friction to the tip of pen.
• Writing with chalk on the blackboard is also possible because of friction. The blackboard surface rubs off some chalk particles which stick to it.
• Moving objects can be stopped because of friction.
• Friction between tyres of automobiles and road enables them to be started, stopped or change their direction of motion.
• Construction of building, fixing a nail on the wall, all require friction.

Evils of friction:

• Friction produces heat. Hence, when a machine is operated, heat generated due to friction causes much wastage of energy.
• Friction opposes the relative motion between objects.
• It causes wear and tear of materials whether they are shoes, ball bearings, screws, etc.
• Excessive friction may even result in unwanted fires, such as forest fires.

Ways to Increase Friction:
i. Rubber grips are put on bicycle handles, electrical tools and on many other tools. Increased friction helps in giving better grip on these objects.
ii. A surface can be made rough with the help of sand paper in order to increase friction.
iii. Soles of shoes have grooves on them. Grooved soles give more friction to the ground which gives better grip when we walk. Shoes with worn out soles may be slippery.

iv. The tyres of vehicles have treads for better grip over the road. It provides friction to the surface of tyres. When treads are worn out, the tyres need to be replaced with new ones.
v. Holding a bottle cap with a thick towel helps in increasing friction. Increased friction makes it easier to open the bottle cap.

Ways to Reduce Friction:
i. Lubricating oil and grease are used in machines to reduce friction between the moving parts.

ii. Some machines use air cushion for reducing friction because oil is not ideal to be used in such machines.
iii. Graphite powder is used in some machines to reduce friction.
iv. Ball bearing converts sliding friction into rolling friction. You have read that rolling friction is less than sliding friction. Thus, ball bearings help in reducing friction.
v. Using wheels to drag load also covert sliding friction into rolling friction, there reducing its effect to a great extent.

Fluid Friction: When an object moves through a fluid (liquid or gas), the fluid creates friction. Friction created by a fluid is called fluid friction or drag. Force of friction is directly proportional to the density of a fluid. Hence, friction in oil will be greater than in water. Similarly, friction in water will be greater than that in air.

Streamlined Shape: A shape which is wide in middle and tapered at the ends is called a streamlined shape. Such a shape reduces drag or fluid friction. Hence, a streamlined body can easily move through a fluid. Birds and fishes are naturally endowed with streamlined body. Boats and airplanes are also made streamlined so that they can easily move through air or water.

Class 8 Science Chapter 12 Friction Additional Important Questions and Answers

Very Short Answer Type Questions

Question 1.
What is friction?
Answer:
The force which opposes the relative movement between two surfaces in contact is called friction.

Question 2.
Give an example to show that friction produces heat.
Answer:
Rubbing of palms together generates heat and makes us feel warm. This shows that friction produces heat.

Question 3.
Why are truck tyres treaded?
Answer:
Truck tyres are treaded to increase friction so as to provide a better grip on the ground.

Question 4.
Why is sliding replaced by rolling in most of the machines?
Answer:
Because rolling friction is smaller than sliding friction and facilitates smooth running of the machine.

Question 5.
Why do you think rolling friction is less than sliding friction?
Answer:
Rolling friction is less than sliding friction because in rolling, the surface area in contact is less than that in the sliding friction. Moreover, while rolling, the surfaces do not rub against each other properly.

Question 6.
Can we reduce friction to zero by using lubricants?
Answer:
No, it is not possible to eliminate friction entirely. No surface can have zero friction.

Question 7.
How can fluid friction be minimised?
Answer:
Fluid friction can be minimised by making the shape of the body streamlined.

Question 8.
Why is it difficult to walk on ice?
Answer:
It is difficult to walk on ice because ice offers very little friction between the feet and the ice surface.

Question 9.
Why does the force of friction exist?
Answer:
Friction is produced by the interlocking of the irregularities on the two surfaces in contact.

Question 10.
Why do the soles of shoes wear out?
Answer:
The soles of shoes wear out due to the friction between the soles and the ground.

Question 11.
Why are the soles of shoes grooved?
Answer:
Soles of shoes are grooved so as to provide better grip between the shoes and the ground.

Question 12.
Give two examples where rolling friction is utilised.
Answer:
Rolling friction is utilised by using ball bearings between the hub and the axle of a ceiling fan and in the bicycle.

Question 13.
On which factors does the friction depend?
Answer:
Friction depends on the nature and pressure of surfaces in contact.

Question 14.
What is meant by air resistance?
Answer:
The friction between a moving object and the air through which it moves is known as air resistance.

Question 15.
What is a streamlined shape?
Answer:
A shape which is wide in middle and tapered at the ends is called a streamlined shape.

Question 16.
Name the device used for measuring the force acting on an object.
Answer:
Spring balance.

Question 17.
Define viscous drag.
Answer:
The opposition offered to the motion of an object by a fluid is called viscous drag.

Short Answer Type Questions

Question 1.
It is easier to open a bottle cap by gripping it with a towel. Why?
Answer:
Rough surface of towel helps in increasing friction and thus provides a better grip for application of a longer force. The towel, thus, makes it easier to open the bottle cap.

Question 2.
Why is it difficult to drive on a wet road?
Answer:
It is very difficult to drive on a wet road because of less friction. A thin film of water is made over the surface of road. This reduces friction, because of which the driver finds it difficult to control the vehicle while driving on a wet road.

Question 3.
Why is friction said to be a necessary evil?
Answer:
Friction is all pervasive. Even a simple task, like writing on a page, cannot be done without friction. Walking on road will not be possible without friction. But friction causes wear and tear of materials, and wastage of energy. Thus, friction is beneficial as well as harmful. Flence, friction is called a necessary evil.

Question 4.
Boats and aeroplanes are given streamlined shape. Why?
Answer:
A shape which is wide in middle and tapered at the ends is called a streamlined shape. Such a shape reduces drag or fluid friction. Hence, a streamlined body can easily move through a fluid. Birds and fishes are naturally endowed with streamlined body. Boats and airplanes are also made streamlined so that they can easily move through air or water.

Question 5.
Give two situations where it is desirable to increase friction.
Answer:
Two situations where it is desirable to increase friction are:
a. Tyres have designs and patterns with grooves on the surface to increase friction with the road. This prevents slipping of the tyres on the wet road.
b. Spikes are provided in the soles of shoes used by players and athletes to increase friction so that they get a firm grip on the ground and do not slip while running.

Question 6.
Give some advantages of friction.
Answer:
Advantages of friction are as follows:

• Friction between pen and paper enables us to write on the paper.
• It enables us to hold various things.
• Friction between our feet and the ground allows our movements like walking, jogging, etc.
• It holds the screw in a wood.
• It allows dishes to stand on table.
• Friction between the surface of the road and tyres of our vehicles allow the vehicles to move without slipping.

Question 7.
Give some disadvantages of friction.
Answer:
Disadvantages of friction are as follows:

• Friction opposes motion, so some of the energy is wasted in overcoming friction.
• It causes wear and tear of sole of shoes, for example, the sole of our shoes wears out after prolonged use because of friction with the rough road surface.
• The heat product by friction damages rubber parts of machines, and also results in energy loss.
• Friction causes moving objects to stop or slow down.

Question 8.
Why is it easier to tie a jute knot than a silk knot?
Answer:
The coefficient of friction of jute on jute is much more than that of silk on silk. Hence, two strands of a jute rope can ‘grip’ each other much more strongly than the two strands of a silk rope. This makes it easier to tie a knot in a jute rope as compared to that in a silk rope.

Question 9.
What is spring balance?
Answer:
It is a device used for measuring the force acting on an object. It contains a coiled spring. The spring gets stretched when a force is applied on it. Stretching of the spring is measured by a pointer moving on a graduated scale. The reading on the scale gives the magnitude of force.

Question 10.
How does lubrication reduce friction?
Answer:
Friction is due to roughness of surfaces. Any process that makes the contact surfaces smooth will reduce friction. Friction can be decreased by adding lubricants to the surfaces like oil, grease or graphite. The sliding surfaces then have a thin layer of the lubricants between them. The friction is then between the surfaces and the lubricant layer which is much less. For example, graphite is used as a lubricant in machines where the moving parts reach very high temperatures.
While playing carrom, fine talcum powder on the carrom board reduces friction.

Question 11.
Why is it not easy to move an object from its static position?
Answer:
When an object is at rest, it has better hold of the surface on which it is placed. In the static position, the irregularities of the surfaces are interlocked properly, due to which more force is required to overcome friction. Hence, it is not easy to move an object from its static position.

Question 12.
Why is it easier to pull luggage fitted with rollers?
Answer:
When a body rolls over the surface of another body, the friction is reduced. It is always easier to roll than to slide over another body. That’s why it is easier to pull luggage fitted with rollers.

Question 13.
Aman and Rohan are riding their bicycles on the same concrete road. Aman has new tyres on his bicycle while Rohan has tyres that are old and used. Aman suggested Rohan to change the tyres of his bicycle as soon as possible.
a. Which of the two boys is more likely to slip while moving through a patch of the rod which has lubricating oil spilled over it.
b. Why has Aman advised Rohan to change the tyres of his bicycle?
Answer:
a. Rohan.
b. This is because the worn out tyres do not offer friction and chances of skidding of vehicle increase.

Long Answer Type Questions

Question 1.
Discuss the different types of friction.
Answer:
Following are the three types of friction:
a. Static Friction: The minimum amount of force that is required to overcome the force of friction is called static friction. In other words, the minimum force that is required to make an object move is called static friction.

b. Sliding Friction: Once an object starts moving on a surface, some force is required to keep the object moving. The minimum force that is required to keep a moving object in motion at a constant speed is called sliding friction. Sliding friction is always less than static friction. Due to this, it is easier to push a moving box compared to pushing a static box.

c. Rolling Friction: When an object rolls over the surface of another object, the friction created is called rolling friction. Rolling friction is less than sliding friction.

Question 2.
Suggest some methods to increase friction.
Answer:
Some methods to increase friction are as follows:

• Rubber grips are put on bicycle handles, electrical tools and on many other tools. Increased friction helps in giving a better grip on these objects.
• A surface can be made rough with the help of sandpaper, in order to increase friction.
• Soles of shoes are made rough to increase friction. This helps in providing a better grip on the surface while walking.
• Treads of tyres are deliberately made so that a good road grip can be achieved with the help of friction.
• Holding a bottle cap with a thick towel helps in increasing friction. Increased friction makes it easier to open the bottle cap.

Question 3.
Suggest some methods to reduce friction.
Answer:
Some methods to reduce friction are as follows:

• Lubricating oil and grease are used in machines to reduce friction between the moving parts.
• Some machines use air cushions for reducing friction because use of oil is not ideal in such machines.
• Graphite powder is used in some machines to reduce friction.
• Ball bearing converts sliding friction into rolling friction. Rolling friction is less than sliding friction. Thus, ball bearing helps in reducing friction.

Question 4.
Define fluid. Write the factors on which fluid friction depends.
Answer:
The gases and liquids are together called fluids. The friction exerted by fluids on an object is called fluid friction. The fluid friction is also called drag. The factors on which fluid friction depends are as follows:

• The fluid friction of an object depends on its speed with respect to the fluid.
• The fluid friction of an object also depends on the shape of the object.
• The fluid friction also depends on the nature and density of the fluid.

Picture-Based Questions

Question 1.
Observe the following figure and answer the questions.

a. What kind of friction do these things encounter.
b. What type of shape do they have?
c. How does this shape help than.
Answer:
a. Fluid friction.
b. They have a streamlined shape.
c. Streamlined shape reduces fluid friction and makes their movement easy.

Question 2.
Draw a diagram to show the surface irregularities.
Answer:

Question 3.
Observe the following figure and answer the questions.

a. What does the figure depict?
b. Why do soles of shoes wear out?
c. Is it possible to reduce friction up to zero?
Answer:
a. The figure shows grooves in the soles of shoes which increases friction.
b. Because of friction, the soles of shoes wear out.
c. No, it is not possible.

These NCERT Solutions for Class 6 Maths Chapter 7 Fractions Ex 7.5 Questions and Answers are prepared by our highly skilled subject experts.

NCERT Solutions for Class 6 Maths Chapter 7 Fractions Exercise 7.5

Question 1.
Write these fractions appropriately as additions or subtractions:

Answer:

Question 2.
Solve:
Answer:
If the denominator is same, we can add and subtract the numerator

(g) Since 1 and \(\frac { 3 }{ 3 }\) are equivalent fractions.
(h) \(\frac{1}{4}+\frac{0}{4}=\frac{1+0}{4}=\frac{1}{4}\)
∴ \(1-\frac{2}{3}=\frac{2}{3}-\frac{2}{3}=\frac{3-2}{3}=\frac{1}{3}\)

(i) Since 3 = \(\frac { 3 }{ 1 }\), and equivalent fraction of \(\frac { 3 }{ 1 }\) having denominator as 5 is given by

Question 3.
Shubham painted \(\frac { 2 }{ 3 }\) of the wall space in his room. His sister Madhavi helped and painted \(\frac { 1 }{ 3 }\) of the wall space. How much did they paint together?
Answre:
Space painted by them together = Space painted by Shubham + Space painted by Madhavi
= \(\frac{2}{3}+\frac{1}{3}=\frac{3}{3}=1\)
∴ Shubham and Madhavi together painted one complete wall in a room.

Question 4.
Fill in the missing fractions.
Answer:

Question 5.
Javed was given \(\frac { 5 }{ 7 }\) of a basket of oranges. What fraction of oranges was left in the basket?
Answer:
Fraction of oranges given to Javed = \(\frac { 5 }{ 7 }\)
Fraction of oranges left in the basket = \(1-\frac{5}{7}=\frac{2}{7}\)

These NCERT Solutions for Class 6 Maths Chapter 7 Fractions Ex 7.4 Questions and Answers are prepared by our highly skilled subject experts.

NCERT Solutions for Class 6 Maths Chapter 7 Fractions Exercise 7.4

Question 1.
Write shaded portion as fraction. Arrange them in ascending and descending order using correct sign ‘<‘, ‘>’ between the fractions:

(c) Show \(\frac{2}{6}, \frac{4}{6}, \frac{8}{6}, \frac{6}{6}\) on the number line. Put appropriate signs between the fractions given.

Answer:
The fractions represented by the shaded portions are as follows:
(a) \(\frac{3}{8}, \frac{6}{8}, \frac{4}{8}, \frac{1}{8}\)
(b) \(\frac{8}{9}, \frac{4}{9}, \frac{6}{9}, \frac{6}{9}\)
When we compare two fractions
Note:
Like Fractions: If the denominator is same, the fraction with the greater numerator is greater.
Unlike Fractions: If the denominator is not same, we first get their equivalent fractions with a denominator which is the L.C.M of the denominators of both the fractions.
The above fractions in (a) and (b) are like fractions as they have the same denominator.
When comparing fractions with same denominator,
So, arranging the fractions in ascending order, we have
(a) \(\frac{1}{8}<\frac{3}{8}<\frac{4}{8}<\frac{6}{8}\)
(b) \(\frac{3}{9}<\frac{4}{9}<\frac{6}{9}<\frac{8}{9}\)
(c) Let us represent the fractions \(\frac{2}{6}, \frac{4}{6}, \frac{8}{6}, \frac{6}{6}\) on the number line.

Again these comparisons are between the like fractions, so numerator will dictate the sign:

Question 2.
Compare the fractions and put an appropriate sign.

Answer:
(a) Here, we have Tike fractions’ so we compare them by their numerators only.

(b) Here, we have ‘unlike fractions’ with same numerators, so we compare them with their denominators only.

(c) Like fractions, so we compare by their numerators only.

(d) ‘Unlike fractions’ with same nemerators so we compare them by their denominators, only.

Question 3.
Make five more such pairs and put appropriate signs.
Answer:

Question 4.
Look at the figures and write ‘<’ or ‘>’, ‘=’ between the given pairs of fractions.

Make five more such problems and solve them with you friends.
Answer:
(a) <
(b) >
(c) >
(d) =
(e) <

Question 5.
How quickly can you do this? Fill appropriate sign (‘<’ , ‘=’ , ‘>’)

Answer:

Question 6.
The following fractions represent just three different numbers. Separate them into three groups of equivalent fractions, by changing each one to its simplest form.
Answer:
Simplifying the fractions into simplest form.

Totally these are three graps of equivalent fractions 1
\(\frac { 1 }{ 6 }\) = (i) (v) (viii) (x) (xi)
\(\frac { 1 }{ 5 }\) = (ii) (vi) (vii)
\(\frac { 4 }{ 25 }\) = (iii) (iv) (ix) (xii)

Question 7.
Find answers to the following. Write and indicate how you solved them.
(a) Is \(\frac { 5 }{ 9 }\) equal to \(\frac { 4 }{ 5 }\) ?
(b) Is \(\frac { 9 }{ 16 }\) equal to \(\frac { 5 }{ 9 }\) ?
(c) Is \(\frac { 4 }{ 5 }\) equal to \(\frac { 16 }{ 20 }\) ?
(d) Is \(\frac { 1 }{ 15 }\) equal to \(\frac { 4 }{ 30 }\) ?
Answer:
We need to first simplify each fraction. After simplification if they are equal, else it is not equal
(a) Not equal
(b) Not equal
(c) Equal
(d) Not equal

Question 8.
Ila read 25 pages of a book containing 100 pages. Lalita read \(\frac { 2 }{ 5 }\) of the same book. who read less?
Answer:
Total number of pages in the book = 100 pages
Number of pages read by Ila = 25 pages
Fraction of pages read by Ila = \(\frac { 25 }{ 100 }\)
= \(\frac { 1 }{ 4 }\)
Lauta read \(\frac { 2 }{ 5 }\) of the same book
So, basically we need to compare the fraction \(\frac { 1 }{ 4 }\) and \(\frac { 2 }{ 5 }\)
Since, they are unlike fractions, we need to find LCM of 4 and 5 which is 20.
So \(\frac{1}{4}=\frac{5}{20}\)
\(\frac{2}{5}=\frac{8}{20}\)
Now, 8 > 5
So Ila read lesser than Lauta.

Question 9.
Rafiq exercised for \(\frac { 3 }{ 6 }\) of an hour, while Rohit exercised for \(\frac { 3 }{ 4 }\) of an hour. Who exercised for a longer time?
Answer:
Rafiq = \(\frac { 3 }{ 6 }\)
Rohit = \(\frac { 3 }{ 4 }\)
We need to compare the fractions. Since these are unlike fraction, we need to calculate the LCM and 4 and 6. LCM of 4 and 6 is 12.
Rafiq = \(\frac{3}{6}=\frac{6}{12}\)
Rahit = \(\frac{3}{4}=\frac{9}{12}\)
\(\frac{9}{12}>\frac{6}{12}\)
∴ Rohit exercised for a longer time than Rafiq.

Question 10.
In a class A of 25 students, 20 passed with 60% or more marks; in another class B of 30 students, 24 passed with 60% or more marks. In which class was a greater fraction of students getting with 60% or more marks?
Answer:
Total number of students in class A = 25 Number of students passed in first class in class A = 20.
Fraction of students of class A who passed in first class = \(\frac{20}{25}=\frac{4}{5}\)
Total number of students in class B = 30 Number of students passed in first class = 24.
Fraction of students of class B who passed in first class = \(\frac{24}{30}=\frac{4}{5}\)
Both class A and B have the same fraction of students who passed in first class.