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6 months ago

Class 11th Sets

12. From the sets given below, select equal sets :

A = { 2, 4, 8, 12}, B = { 1, 2, 3, 4}, C = { 4, 8, 12, 14}, D = { 3, 1, 4, 2}

E = {–1, 1}, F = { 0, a}, G = {1, –1}, H = { 0, 1}

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Payal Gupta

Contributor-Level 10

12. B = D = {1, 2, 3, 4} = {3, 1, 4, 2}

E = G = {–1, 1} = {1, –1}.

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6 months ago

Class 11th Sets

11. Are the following pair of sets equal? Give reasons.

(i) A = {2, 3}, B = {x : x is solution of x²+ 5x + 6 = 0}

(ii) A = { x : x is a letter in the word FOLLOW}

(iii) B = { y : y is a letter in the word WOLF}

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Payal Gupta

Contributor-Level 10

11. (i) A = {2, 3)

B = {x : x is solution of x² + 5x + 6 = 0}

So, x² + 5x + 6 = 0

x² + 2x + 3x + 6 = 0

x (x+2) + 3 (x+2) = 0

(x+2) (x+3) = 0

x = –2, –3

So, B = {–2, –3}

So, A ≠ B.

(ii) A = {x : x is a letter in word FOLLOW}

A = {F, O, L, W}

(iii) B = {x : x is a letter in word WOLF}

B = {W, O, L, F}

So, A = B as the elements are all same.

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6 months ago

Class 11th Sets

10. In the following, state whether A = B or not:

(i) A = { a, b, c, d } B = { d, c, b, a }

(ii) A = { 4, 8, 12, 16 } B = { 8, 4, 16, 18}

(iii) A = {2, 4, 6, 8, 10} B = { x : x is positive even integer and x < 10}

(iv) A = { x : x is a multiple of 10}, B = { 10, 15, 20, 25, 30, . . . }

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Payal Gupta

Contributor-Level 10

10. (i) As A and B has a, b, c and d as elements and are exactly the same, Hence, A = B.

(ii) Hence, 12 $\in$ A but 12 $\notin$ B

Similarly 18 $\in$ B but 18 $\notin$ A.

So, A ≠ B.

(iii) A = {2,4,6,8,10} and B = {2,4,6,8,10}

So, A=B.

(iv) A= {10,20,30,40, …} and B= {10,15,20,25,30, …}

So, A ≠ B.

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New answer posted

6 months ago

Class 11th Sets

9. State whether each of the following set is finite or infinite:

(i) The set of lines which are parallel to the x-axis

(ii) The set of letters in the English alphabet

(iii) The set of numbers which are multiple of 5

(iv) The set of animals living on the earth

(v) The set of circles passing through the origin (0,0)

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Payal Gupta

Contributor-Level 10

9. (i) We can draw infinite number of lines parallel to x-axis.

∴ The set is infinite.

(ii) There are 26 letters in the English alphabet.

∴ The set is finite.

(iii) There are infinite number which are a multiple of 5.

∴ The set is infinite

(iv) The numbers of animals living on earth are finite.

∴ The set is finite.

(v) There can be infinite number of circular passing through origin (0,0).

∴ The set is infinite.

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6 months ago

Class 11th Sets

8. Which of the following sets are finite or infinite

(i) The set of months of a year

(ii) {1, 2, 3, . . .}

(iii) {1, 2, 3, . . ., 99, 100}

(iv) The set of positive integers greater than 100

(v) The set of prime numbers less than 99

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Payal Gupta

Contributor-Level 10

8. (i) The set of months of a year has 12 elements. Here, the set is finite.

(ii) The given set has the natural number as its elements. Hence, the set is infinite.

(iii) The given set has 100 elements i.e., from 1 to 100. Hence, the set is finite.

(iv) There are infinite numbers of positive integers greater than 100. Hence the set is infinite.

(v) The numbers of prime number less than 99 is finite. Hence, the set is finite.

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6 months ago

Class 11th Sets

Class 11 Chapter 1 Set Ex 1.2 NCERT Solutions

7. Which of the following are examples of the null set

(i) Set of odd natural numbers divisible by 2

(ii) Set of even prime numbers

(iii) {x : x is a natural numbers, x < 5 and x > 7}

(iv) {y : y is a point common to any two parallel lines}

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Payal Gupta

Contributor-Level 10

1. (i) There is no odd number that can be divided by 2.

?The given set is a null set.

(ii) An even prime number is 2. Hence, the set will have 2 as element.

? The given set is not a null set.

(iii) As there is no natural number which is both less than 5 and greater than 7.

? The given set is a null set.

(iv) Two parallel lines never meets and hence no common point.

? The given set is a null set.

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New question posted

6 months ago

Class 11th Sets

Class 11 Chapter 1 Set Ex 1.2 NCERT Solutions

Q1. Which of the following are examples of the null set

(i) Set of odd natural numbers divisible by 2

(ii) Set of even prime numbers

(iii) {x : x is a natural numbers, x < 5 and x > 7}

(iv) {y : y is a point common to any two parallel lines}

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New answer posted

6 months ago

Class 11th System of Particles and Rotational Motion

7.23 A man stands on a rotating platform, with his arms stretched horizontally holding a 5 kg weight in each hand. The angular speed of the platform is 30 revolutions per minute. The man then brings his arms close to his body with the distance of each weight from the axis changing from 90cm to 20cm. The moment of inertia of the man together with the platform may be taken to be constant and equal to 7.6 kg m².

(a) What is his new angular speed? (Neglect friction)

(b) Is kinetic energy conserved in the process? If not, from where does the change come about?

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Vishal Baghel

Contributor-Level 10

(a)

Moment of Inertia of the man-platform system = 7.6 kg-m²

Moment of inertia when the man stretches his hands to a distance 90 cm

= 2 x mr² = 2 x 5 x (0.9)2 = 8.1 kg-m²

Initial moment of inertia of the system, $-$ = 7.6 + 8.1 = 15.7 kg-m²

Angular speed $-$ = 30 rev/min

Angular momentum, $-$ = $I_{i}$ = 15.7 x 30 …… (i)

Moment of inertia when the man folds his hands to a distance of 20 cm

= 2 x mmr²= 2 x 5 x (0.2)2 = 0.4 kg-m²

Final moment of inertia, $ω_{i}$ = 7.6 + 0.4 = 8 kg-m²

Final angular speed = $L_{i}$ and final angular of momentum, $I_{i} ω_{i}$ = $I_{f}$ = 8 $ω_{f}$ …… (ii)

From the conservation of a

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(a)

Moment of Inertia of the man-platform system = 7.6 kg-m²

Moment of inertia when the man stretches his hands to a distance 90 cm

= 2 x mr² = 2 x 5 x (0.9)2 = 8.1 kg-m²

Initial moment of inertia of the system, $-$ = 7.6 + 8.1 = 15.7 kg-m²

Angular speed $-$ = 30 rev/min

Angular momentum, $-$ = $I_{i}$ = 15.7 x 30 …… (i)

Moment of inertia when the man folds his hands to a distance of 20 cm

= 2 x mmr²= 2 x 5 x (0.2)2 = 0.4 kg-m²

Final moment of inertia, $ω_{i}$ = 7.6 + 0.4 = 8 kg-m²

Final angular speed = $L_{i}$ and final angular of momentum, $I_{i} ω_{i}$ = $I_{f}$ = 8 $ω_{f}$ …… (ii)

From the conservation of angular momentum, we have $L_{f}$ = $I_{f} ω_{f}$

$ω_{f}$ = $I_{i} ω_{i}$ = 58.88 rev/min

(b) Kinetic energy is not conserved in the given process. In fact, with the decrease in the moment of inertia, kinetic energy increases. The additional kinetic energy comes from the work done by the man to fold his hands towards himself.

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6 months ago

Class 11th System of Particles and Rotational Motion

7.22 As shown in Fig.7.40, the two sides of a step ladder BA and CA are 1.6 m long and hinged at A. A rope DE, 0.5 m is tied half way up. A weight 40 kg is suspended from a point F, 1.2 m from B along the ladder BA. Assuming the floor to be frictionless and neglecting the weight of the ladder, find the tension in the rope and forces exerted by the floor on the ladder. (Take g = 9.8 m/s2)

(Hint: Consider the equilibrium of each side of the ladder separately.)

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Vishal Baghel

Contributor-Level 10

The given situation can be shown as

NB = force exerted on the ladder by the floor point B

T = Tension in the rope

BA = CA = 1.6 m, DE = 0.5 m and BF = 1.2 m

Mass of the weight, m = 40 kg

The perpendicular drawn from point A on the floor BC, this intersects DE at mid-point H.

Δ ABI and ΔACI are congruent. Therefore BI = IC, I is the mid-point of BC. DE is parallel to BC.

BC = 2 x DE = 1 m and AF = BA – BF = 0.4 m…… (i)

D is the mid-point of AB, hence we can write

AD = (1/2) x BA = 0.8 ……. (ii)

Using equation (i) and (ii), we get FE = 0.4 m

Hence, F is the mid-point of AD.

Hence G will

...more

The given situation can be shown as

NB = force exerted on the ladder by the floor point B

T = Tension in the rope

BA = CA = 1.6 m, DE = 0.5 m and BF = 1.2 m

Mass of the weight, m = 40 kg

The perpendicular drawn from point A on the floor BC, this intersects DE at mid-point H.

Δ ABI and ΔACI are congruent. Therefore BI = IC, I is the mid-point of BC. DE is parallel to BC.

BC = 2 x DE = 1 m and AF = BA – BF = 0.4 m…… (i)

D is the mid-point of AB, hence we can write

AD = (1/2) x BA = 0.8 ……. (ii)

Using equation (i) and (ii), we get FE = 0.4 m

Hence, F is the mid-point of AD.

Hence G will also be the midpoint of AH.

ΔAFG and ΔADH are similar

FG = (1/2)DH = (1/2) X 0.25 = 0.125 M

In ΔADH, AH = $\frac{3 A B}{4 g \sin ? θ})$ - $\sqrt \frac{11.46}{19.6}$ ) = $\frac{F G}{D H} = \frac{A F}{A D} = \frac{F G}{D F} = \frac{0.4}{0.8} = \frac{1}{2}$ – $\sqrt ({A D}^{2}$ ) = 0.76 m

For translational equilibrium of the ladder, the upward force should be equal to the downward force, NC + NB = mg = 392 …… (iii)

For rotational equilibrium of the ladder, the net moment about A is

${D H}^{2}$ NB x BI +mg x FG + NC x CI + T x AG – T x AG = 0

$\sqrt {({0.8}^{2}$ NB x 0.5 +40 x g + 0.125 x NC x 0.5 = 0

(NC - NB) x 0.5 = 49

NC - NB = 98 ………. (iv)

Adding equation (iii) and (iv) we get

NC = 245 N, NB = 147 N

For rotational equilibrium of the side AB, considering the moment about A

${0.25}^{2}$ NB x BI + mg x FG + T x AG = 0

$-$ 245 x 0.5 + 40 x 9.8 + T x AG = 0

T = 96.7 N

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6 months ago

Class 11th Sets

6. Match each of the set on the left in the roster form with the same set on the right described in set-builder form:

(i) {1, 2, 3, 6} (a) {x : x is a prime number and a divisor of 6}

(ii) {2, 3} (b) {x : x is an odd natural number less than 10}

(iii) {M,A,T,H,E,I,C,S} (c) {x : x is natural number and divisor of 6}

(iv) {1, 3, 5, 7, 9} &nbs

...more

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Payal Gupta

Contributor-Level 10

6. (i) (c)

(ii) (a)

(iii) (d)

(iv) (b)

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Class 11 Chapter 1 Set Ex 1.2 NCERT Solutions

Class 11 Chapter 1 Set Ex 1.2 NCERT Solutions

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