Correct Answer: Rourkela Steel Plant

Solution : The correct option is Rourkela Steel Plant.

Rourkela Steel Plant (RSP) is India's first public-sector integrated steel plant. In the 1960s, it was built with German assistance and had a capacity of one million metric tonnes. The Steel Authority of India operates it.

Correct Answer: 16.5

Solution : Given:
13 Ø 3 ♦ 6 % 9 – 4 Ø 14 = ?

After replacing the symbols with mathematical signs, the equation becomes –
= 13 + 3 – 6 × 9 ÷ 4 + 14
= 13 + 3 – 6 × 2.25

Correct Answer: 25%

Solution : As per the given graph:
The male population in the year 2016 = 40 thousand
The female population in the year 2016 = 72 – 40 = 32 thousand
The difference between the male population and the female population in the year 2016 = 40

Correct Answer: Article 371F

Solution : The correct answer is Article 371F.

The laws governing the constitution of the union of states that make up India as a whole are collected in Part XXI of the Indian Constitution. Articles on transitional, special, and temporary provisions make up this section

Correct Answer: co fu

Solution : Given:
1. time has come ⇒ et im ak
2. come with money ⇒ nd ak co
3. money has power ⇒ fu et nd

By comparing all the three coded sentences, we find that –
In sentences 1 and 2, come and ak

Correct Answer: Estuary

Solution : The correct option isEstuary.

Because of the mixing of freshwater and saltwater, estuaries are the most productive ecosystems in the biosphere. An estuary is a partly contained body of water along the coast where freshwater from rivers and streams meets saltwater from the ocean. Estuaries

Correct Answer: $150\sqrt{15}\text{ cm}^{2}$

Solution :
In $\triangle$ ABD,
AB = 20 cm. AD = 30 cm. BD = 40 cm
Semi perimeter (s) = $\frac{a+b+c}{2}$ = $\frac{20+30+40}{2}$ = 45 cm
Now, the area of $\triangle$ABD
= $\sqrt{s(s - a)(s - b)(s - c)}$
= $\sqrt{45(45 - 20)(45 - 30)(45

Correct Answer: $\pm 7$

Solution : Given: $a^{3} - b^{3} = 117$ and $a - b = 3$
We know, $(a^{3} - b^{3}) = (a - b)(a^{2} + b^{2} + ab)$
$⇒(a^{3} - b^{3}) = (a - b)[(a - b)^{2} + 3ab]$
$⇒117 = 3[3^{2} + 3ab]$
$⇒39 = [3^{2}

Correct Answer: $\frac{8}{\sqrt{113}}$

Solution : Given: $8 \cot A = 7$
⇒ $\cot A = \frac{7}{8}$
We know that $\operatorname{cosec^2 A}-\cot^2A=1$
⇒ $\operatorname{cosec^2 A}- (\frac{7}{8})^2=1$
⇒ $\operatorname{cosec^2 A}=1+\frac{49}{64}$
⇒ $\operatorname{cosec A}=\sqrt{\frac{113}{64}}=\frac{\sqrt{113}}{8}$
⇒ $\sin A=\frac{8}{\sqrt{113}}$
Hence, the correct answer is $\frac{8}{\sqrt{113}}$.

Correct Answer: (3)

Solution : The sentence can be improved using the third option "enrol".

This helps improve the sentence by using the appropriate verb "enrol" to convey the action of getting accepted into a university or registering for courses.

The meanings of the other options are as follows: