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Staff Selection Commission Combined Graduate Level Exam

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Question : Most of the Chlorophyceae have one or more storage bodies called ______ located in the chloroplasts.

Option 1: stomata

Option 2: carotenoids

Option 3: pyrenoids

Option 4: chlorophyll

Team Careers360 5th Jan, 2024

Correct Answer: pyrenoids


Solution : The correct answer is pyrenoids.

The majority of chlorophytes have one or more pyrenoids, which are starch-sheathed central proteinaceous bodies that are located all around the chloroplast. A protein-rich body called a pyrenoid is present in green algae, or Chlorophyceae members. They are exclusive

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Question : The length of a side of an equilateral triangle is 8 cm. The area of the region lying between the circumcircle and the incircle of the triangle is __________ ( Use: $\pi = \frac{22}{7}$)

Option 1: $50\frac{1}{7}\ \text{cm}^2$

Option 2: $50\frac{2}{7}\ \text{cm}^2$

Option 3: $75\frac{1}{7}\ \text{cm}^2$

Option 4: $75\frac{2}{7}\ \text{cm}^2$

Team Careers360 17th Jan, 2024

Correct Answer: $50\frac{2}{7}\ \text{cm}^2$


Solution :
Given:
Side of the equilateral triangle, $a$ = 8 cm
Radius of incircle = $\frac{a}{2\sqrt{3}}$ cm
Radius of circumcircle = $\frac{a}{\sqrt{3}}$ cm
So, the required area
= Area of circumcircle – Area of incircle
= $\pi (\frac{a}{\sqrt{3}})^2-\pi (\frac{a}{2\sqrt{3}})^2$
= $\pi (\frac{8}{\sqrt{3}})^2-\pi (\frac{8}{2\sqrt{3}})^2$
= $64\pi(\frac{1}{3}-\frac{1}{12})$

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Question : Direction: Select the missing number from the given alternatives.

  6   18  15
  3   2   5
  4   3   ?
  8  27   9

 

Option 1: 11

Option 2: 6

Option 3: 3

Option 4: 2

Team Careers360 21st Jan, 2024

Correct Answer: 3


Solution : Given: 

  6   18  15
  3   2   5
  4   3   ?
  8  27   9

In the first column, 6 ÷ 3 x 4 = 8

In the second column, 18 ÷ 2 x 3 = 27

In the third column, 15 ÷ 5 x ? =

20 Views

Question : If $\operatorname{cosec} \theta+\cot \theta=p$, then the value of $\frac{p^2-1}{p^2+1}$ is:

Option 1: $\cos \theta$

Option 2: $\sin \theta$

Option 3: $\cot \theta$

Option 4: $\operatorname{cosec} \theta$

Team Careers360 11th Jan, 2024

Correct Answer: $\cos \theta$


Solution : Given, $\operatorname{cosec} \theta+\cot \theta=p$
Squaring both sides, we get,
$\operatorname{cosec}^2 \theta+\cot^2 \theta+2\operatorname{cosec} \theta\cot \theta=p^2$ ----------------------(1)
Adding 1 on both sides of equation (1),
$\operatorname{cosec}^2 \theta+(\cot^2 \theta+1)+2\operatorname{cosec} \theta\cot \theta=p^2+1$
$⇒2\operatorname{cosec}^2 \theta+2\operatorname{cosec} \theta\cot \theta=p^2+1$ ---------------------------(2)
Subtracting 1 from both sides of equation (1),
$(\operatorname{cosec}^2-1) \theta+\cot^2 \theta+2\operatorname{cosec}

13 Views

Question : If 3x + 4y – 2z + 9 = 17, 7x + 2y + 11z + 8 = 23, and 5x + 9y + 6z – 4 = 18, then what is the value of x + y + z – 34?

Option 1: –28

Option 2: –14

Option 3: –31

Option 4: –45

Team Careers360 24th Jan, 2024

Correct Answer: –31


Solution : Given:
3x + 4y – 2z + 9 = 17 .......(i)
7x + 2y + 11z + 8 = 23 .......(ii)
5x + 9y + 6z – 4 = 18 ..........(iii)
By adding equation (i), (ii), and (iii), we get,
15x + 15y + 15z

21 Views

Question : A piece of wire 132 cm long is bent successively in the shapes of an equilateral triangle, a square, and a circle. The area will be largest in the shape of:

Option 1: Circle

Option 2: Equilateral triangle

Option 3: Square

Option 4: Equal in all the shapes

Team Careers360 18th Jan, 2024

Correct Answer: Circle


Solution : The length of the wire is the same in all three cases, so it becomes the perimeter of each shape.
(1). For an equilateral triangle with side length $a$, the perimeter = $3a$
$a = \frac{132}{3} = 44$ cm
The area of an equilateral triangle

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