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Mass of grocer’s box = 2.300 kg
Mass of gold piece I = 20.15g = 0.02015 kg
Mass of gold piece II = 20.17 g = 0.02017 kg
Total mass of the box = 2.3 + 0.02015 + 0.02017 = 2.34032 kg
In addition, the final result should retain as many decimal places as there are in the number with the least decimal places. Hence, the total mass of the box is 2.3 kg.
Difference in masses = 20.17 – 20.15 = 0.02 g
In subtraction, the final result should retain as many decimal places as there are in the number with the least decimal places.

1. Total mass of the box = 2.300 kg + 20.15 g + 20.17 g

= 2.34032 kg = 2.34 kg. 

2. Difference in the masses of the pieces = 20.17 g – 20.15 g = 0.02 g.

(a) Total mass of the box = (2.3 + 0.0217 + 0.0215) kg

= 2.3432 kg.

Since the list number of significant figure is 2, therefore,the total mass of the box = 2.3 kg.

(b) Difference of mass =(2.17 -2.15) = 0.02 g.

Since there are two significant figures so the difference in masses to the correct significant figures is 0.02 g.

The correct answer is (iii) 126

Agriculture sector: The prices of the agricultural crops dropped, so low that many farmers became bankrupt and lost their farms. Livestock was the main source of cash. Com was used to feed Cattle and Pigs. But nothing could help them.

Banking Sector: Confidence in the economy was shattered, Wall street and the banks were no longer seen as reliable. Many refused to put money into stocks. The Federal Reserve did not give aid to banks and small banks collapsed.

Industrial sector: As the country’s economy worsened, local industries affected badly. Production went down. Factories closed. Workers remain unemployed. Almost 15 million people were out of work.

Political Sector: The depression affected politics badly people started disliking Herbert Hoover, the president and his type of laissez faire economics. People voted for Franklin Roosevelt. There was dangerous high U.S debt.

(a) Both A and R are right.

Correct Answer is: (a) Haiti

1. The two main political parties in 

South Africa were:

• (a) The Unionist party which was mainly British and
• (b) The south Africa party which had mainly Afrikaners (or) Boers.

2. Botha was the first Prime Minister who belonged to the South African party ruled in cooperation with the British. 

3. But, the militant section of the South African party formed the National party under Herzog. 

4. In 1912, The African national Congress was formed by Nelson Mandela. But it was banned and he was put into prison for 27 years. 

5. In 1920 elections, the National Party won, with 44 seats.

6. Herzog wanted a twin policy of supremacy of Whites over the Blacks and Afrikaners over British. 7. The South African party was led by smuts, secured 41 seats. 

8. In this moment, the Unionist party and the South African party merged together. Therefore Smuts gained majority over National party. 

9. In 1924 elections, National party won supported by the Labour Movement, (composed of white miners)

10. The Act passed in 1924 prevented blacks from joining trade Unions. Native blacks suffered in Social, Political and Economic spheres. 

11. The Great Economic depression brought sufferings to South Africa, therefore, the South African party and the Nationalist party unite in 1934. This smuts-Herzog alliance lasted till 1939. 

12. Herzog resigned when the parliament decided infavour of second world war. 

13. Smuts continued as prime minister. When Herzog died, many from that party joined the nationalist. 

14. Therefore, in 1948, election Reunified National party won over United party.

(a) Both A and R are correct.

Correct Answer is: (a) Roosevelt

(b) Latin America

The books written at the time of Mahajanapadas were mostly religious. Though they were religious books, they also tell us a lot about the towns and villages and about the rulers and kings of those times. Greece writers wrote some books. Famous books written at Mahajanapada period were ‘Dharmasutras’, ‘Upanishads’, Dighenikaya, Majjhimanikaya and Herodotus History.

Answer is (A) Dasulu

1. F 

2. C 

3. E 

4. A 

5. G

(A) Grihapatis

(A) Vyasa Maharshi

A) 1 – C, 2 – D, 3 – E, 4 – B, 5 – A

(B) Agriculturists

Answer is (C) 3

C) Mahajanapadas

1. F 

2. C 

3. E 

4. A 

5. G

B) 1 – C, 2 – E, 3 – A, 4 – D, 5 – B

(A) Ganga- Indus plain

(B) Battle between Pandavas and Kauravas

(C) Both A & B are correct

Answer is (D) 4

C) Ganga-Yamuna

1. Painted Grey Ware

2. bhaga

3. people of equal status

4. assembly

5. Ganga

6. iron ore

7. Bimbisara

8. Vajji

9. Mahavira

10. Siddhartha

11. Takshashila

12. Gandhara art style

B) Gandhara Art

1. Gangetic valley

2. jana

3. janapada

4. Mahajanapadas

5. Modern Delhi

6. Patna

7. coming and sitting near

8. Grihapatis or Gahapatis

9. dasas or slaves

10. bhartukas

(A) Only A is correct

(A) Janapadas

Answer is (C) Kuru

Answer is (C) 2700

Correct option (a)  π⁴/96

Explanation:

Let   1/1⁴ + 1/3⁴ + 1/5⁴ + ....  ∞ = S

Now  1/1⁴ + 1/2⁴ + 1/3⁴ + ......∞  = π⁴/90

(1/1⁴ + 1/3⁴ + 1/5⁴ + ....) + (1/2⁴ + 1/4⁴ + 1/6⁴ + ...) = π⁴/90

⇒ S + 1/2⁴ (1/1⁴ + 1/2⁴ + 1/3⁴ + ........ ∞)  = π⁴/90

⇒ S + 1/16 .π⁴/90 = π⁴/90

⇒ S = π⁴/90(1 - 1/16) = π⁴/90 x 15/90 = π⁴/96

(a) \(\frac{n}{2}\)(n³ + 4n² + 4n - 1)

nth term (Tn) of the given series 

= (2n – 1) (n + 1)² = (2n – 1) (n² + 2n + 1) 

= 2n³ – n² + 4n² – 2n + 2n – 1 = 2n³ + 3n² – 1

∴ S_n = \(2\displaystyle\sum_{k=1}^{n} k^3\) + \(3\displaystyle\sum_{k=1}^{n} k^2-n\)

= 2 x \(\frac{n^2(n+1)^2}{4}\) + 3 x \(\frac{1}{6}\) x n (n + 1)(2n + 1) - n 

= \(\frac{1}{2}\)[n² (n² + 2n +1) + (n² + n)(2n + 1) -2n]

= \(\frac{1}{2}\)[n⁴ + 2n³ + n³ + 2n³ + 2n² + n² + n - 2n]

= \(\frac{1}{2}\)[n⁴ + 4n³ + 4n² - n] = \(\frac{n}{2}\)(n³ + 4n² + 4n - 1)

Correct option (a)  n²(n + 1)/2

Explanation:

1² + 2.2² + 3² + 2.4² + 5² + ........ + (2k - 1)² + 2(2k)²

⁼ 2k(2k + 1)²/2

Let n = 2k + 1 (odd)

1² + 2.2² + 3² + 2.4² + 5² + ........ + (2k - 1)² + 2(2k)² + (2k + 1)²

= 2k(2k + 1)² /2 + (2k + 1)²

= (2k + 1)² (2k + 2)/2 = n²(n + 1)/2

(c) \(\frac{n(4n^2-1)c^2}{3}\)

Let the sum of first n terms of the A.P, Sn = cn² 

∴ S_{n – 1} = c (n – 1)² = c (n² – 2n + 1) 

∴ nth term of the A.P. = S_n – S_{n – 1} 

= cn² – c (n² – 2n + 1) = c (2n – 1) 

Let \(S_n^2\) be the sum of the squares of these n terms. Then,

\(S_n^2\) = \(\displaystyle\sum_{k=1}^{n} t_k{^2}\)              where t_k = c (2k – 1)

= \(\displaystyle\sum_{k=1}^{n} [c^2(2k-1)^2]\) = \(\displaystyle\sum_{k=1}^{n} [c^2.4k^2-c^2.4k+c^2]\)

= \(4c^2\displaystyle\sum_{k=1}^{n} k^2\) - \(4c^2\displaystyle\sum_{k=1}^{n} k+c^2n\)

= 4c² x \(\frac{n(n+1)(2n+1)}{6}\) - 4c² x \(\frac{n(n+1)}{2}\) + c²n

= \(\frac{c^2n}{6}\)[4(c+1)(2n+1)-12(n+1)+6]

= \(\frac{c^2n}{6}\)[8n² + 12n + 4 - 12n - 12 + 6]

= \(\frac{c^2n}{6}\)[8n² - 2] = \(\frac{n(4n^2-1)c^2}{3}\)

Correct option (b)  2c²/3  n(n - 1)(n - 1)

Explanation:

T_n = S_n –S_n–1 = cn(n–1) –c(n–1)(n–2)

= c(n–1){n–n+2}

= 2c(n–1)

T_n = 4c² (n–1)²

∴ S_n² = 4c²{0² + 1² + 2² +..................+ (n – 1)²}

=  4c² n(n- 1)(2n - 1)/6

=  2c²/3  n(n - 1)(n - 1)

The given nets are of the solid as given below : 

(i) Tetrahedron 

(ii) Triangular prism 

(iii) Cube 

(iv) Cuboid

Radius (r) of each pillar = 28 m 

Height (h) = 4 m 

Curved surface area of each pillar = 2πrh

= 2 x 22/7 x 28 x 4 m² = 704 m² 

Surface area of 24 pillars = 704 x 24 m² = 16,896 m² 

Cost of cleaning = Rs. 8 per m² 

Total cost = Rs. 16,896 x 8 = Rs. 1, 35, 168

Volume of a rectangular solid = 1.92 m³ 

Breadth of a rectangular solid = 1.20 m 

Height of a rectangular solid = 80 cm = 0.8 m 

We know

Length x Breadth x Height = Volume of a rectangular solid (cubical) 

Length x 1.20 x 0.8 = 1.92

Length x 0.96 = 1.92

⇒ Length = 1.92/0.92

⇒ Length = 192/96

⇒ Length = 2 m

Diameter = 28 cm 

Radius = 28/2 cm = 14 cm

Height = 20 cm

Volume = πr²h = 22/7 x 14 x 14 x 20

Volume = 12320 cm³ 

Correct answer is (d) third law

Correct option is (d) All of the above

(c) 1 mole of Na₂CO₃

Find the odd one out  ₆O¹²

(b) 6 protons and 6 neutrons

Atoms of different elements that have same atomic masses but different atomic numbers are called isobars.

e.g., ₁₈Ar⁴⁰ ₂₀Ca⁴⁰