Explore topic-wise InterviewSolutions in Current Affairs.

1. Think ahead: Study the steps of the investigation, so you know what to expect. 

2. Be neat: Keep your work area clean. 

3. Oops!: If you should spill or break something or get cut, tell your teacher right away. 

4. Watch your eyes: Wear safety goggles anytime you are directed to do so. 

5. Yuck!: Never eat or drink anything during a scientific activity. 

6. Protect yourself from shocks: Be especially careful if an electric appliance is used. Be sure that electric cords are in a safe place where you can’t trip over them. Don’t ever pull a plug out an outlet by pulling on the cord. 

7. Keep it clean: Always clean up when you have finished.

Answer is (C) Problem

(D) Karl Popper

(C) Experiment

Predict: To announce or tell before head, forecast, prophesy, foretell.

Scientists observe nature and its laws. Scientists follow a specific way for their innovations. The way they follow is called Scientific Method.

Questioning is the primary or fundamental step in scientific thinking.

Answer is (B) Chemist

Hypothesis: A tentative explanation for a phenomenon used as a basis for further investigation.

a) A geologist examines the distribution of fossils and makes observations to find patterns in natural phenomena.

b) A chemist observes the rate of a chemical reaction at a variety of temperatures. 

c) A biologist observes the reaction of a particular tissue to various stimulants. 

d) An ecologist observes the territorial behaviours of different animals and birds. 

e) A climatologist collects data from weather balloons and makes observations basing on it.

To find out solutions for the problems in scientific way we need to follow a sequential order. So we go through the following.

1. Identifying problem : 

Let us identify any problems from your surroundings. 

Ex: The bulb did not lit in the room. 

2. Making hypothesis: 

List out different solutions which your think for the identifying problem. 

Ex: De filament, fuse failure, switch problem, wire problem. 

3. Collecting information: 

To solve the identifying problem collect material, apparatus, Information, persons. Ex: Collect material like tester,

screwdriver, wooden scale, wires, insulation tape, table and blade. 

4. Data analysis: 

Arrange the collected data or information to conduct experiment. 

5. Experimentation: 

To prove selecting hypothesis conduct experiment. 

Ex: Observe filament of the bulb. 

6. Result analysis: 

Analyzing the results to find out the solution for the problem based on the results you need to select another hypothesis to prove. 

Ex: Filament of the bulb is good in condition so we need to observe fuse. 

7. Generalisation: 

Based on the experiment and its results explain the solution for the problem. 

Ex: Fuse is damaged so the bulb not glow, so we need to replace the fuse.

1. Making a statement about an expected outcome is called Hypothesis. 

2. Variables are factors that can affect the outcome of the investigation.

Scientific Methods:

1. Scientists solve a problem or answer a question by using organized ways. They are called “Scientific Methods”. 

2. The following are the steps involved in a scientific work. 

a) Step -1 : Ask questions. 

b) Step – 2 : Form hypothesis. 

c) Step – 3 : Plan experiments. 

d) Step – 4 : Conduct experiments. 

e) Step – 5 : Draw conclusions.

1. Measuring, interpreting data, using number sense are few methods used by scientists in their investigations. 

2. Scientists make accurate measurements by using different measuring instruments like thermometer clocks, timers, rules, a spring scale, beakers, balance and other containers to measure liquids. 

3. Scientists collect, organize, display and interpret data by using tables, charts and graphs. 

4. Scientists compare and order numbers, compute with numbers shown on graphs and read the scales on thermometers, measuring cups, beakers and other tools. 

5. Good scientists apply their maths skills to help them display and interpret the data they collect.

1. Informative writing, Narrative writing, Expressive writing, Persuasive writing are used by the Scientists. 

2. In informative writing, scientists describe the observation, inferences and their conclusion. 

3. In narrative writing the scientists describe about something, give examples or tell a story. 

4. In expressive writing, they may write letters, poems, or songs.

5. In persuasive writing they write letters about important issues in science and also write about what they have learned about science which helps others to understand about their thinking.

Observe, compare, classify are the process skills.

1. Observe – Use the senses to learn about objects and events.

2. Compare – Identify characteristics of things or events to find out how they are alike and different. 

3. Classify – Group or organize objects or events in categories based on specific characteristics.

Based on the experiment and its results explaining the solution for the problem is called Generalisation.

Example:

a) The bulb did not light in the room. 

b) Identify the problem that may be defilament, fuse failure, switch problem, wire problem. 

c) Then take tester, screwdriver, wooden scale, wires, insulation tape, table and blade. 

d) Observe the filament of the bulb. 

e) If the filament of the bulb is good, then observe fuse. 

f) As the fuse is damaged, we need to replace the fuse. 

Based on this, the bulb did not lit in the room because the fuse is damaged. This is the generalisation in the above experiment.

Right

Let x be the supplement of the right angle.

Then, x + 90° = 180° 

⇒ x = 180° – 90° = 90°

True

Measure of a right angle is 90°. Then, sum of two right angles will be (90°+ 90°) = 180°. 

So, two right angles are always supplementary to each other.

Acute

The supplement of an obtuse angle is always an acute angle. As, if we subtract an obtuse angle from 180°, then result will be an acute angle, i.e. 90°.

False

If two angles are supplementary angles, then it is not necessary that they are always obtuse angles.

e.g. 60° and 120° are supplementary angles but both are not obtuse.

False

Since, sum of two complementary angles is 90°, so sum of one obtuse and one acute angles cannot make a pair of complementary angles as obtuse angle is greater than 90°.

As we know cosec(–θ) = –cosec θ

∴ cosec\((\frac{-9\pi}4)\) = -cosec\((\frac{9\pi}4)\)

-cosec\((\frac{9\pi}4)\) can be written as -cosec\((2\pi+\frac{\pi}4)\)

Also,

cosec (2π+θ) = cosec θ

∴ -cosec\((2\pi+\frac{\pi}4)\) = -cosec\((\frac{\pi}4)\)

As we know –cosec (θ) = cosec (–θ)

∴ -cosec\((\frac{\pi}4)\) = cosec\((\frac{-\pi}4)\)

Now the question becomes cosec^-1(cosec\((\frac{-\pi}4)\))

cosec^-1(cosec x) = x

Provided x ∈ \([\frac{-\pi}2,\frac{\pi}2]\) - {0}

∴ We can write cosec^-1(cosec\((\frac{-\pi}4)\)) = \(\frac{-\pi}4.\)

sec\((\frac{9\pi}5)\) can be written as sec\((2\pi-\frac{\pi}5)\)

Also, we know sec(2π - θ) = sec(θ)

∴ sec \((2\pi-\frac{\pi}5)\) = sec \((\frac \pi 5)\)

∴ Now the given equation can be written as sec^–1sec\((\frac \pi 5)\)

As sec^-1(sec x) = x

Provided x ∈ [0,π] - \(\{\frac \pi 2\}\)

∴ we can write sec^-1sec\((\frac{\pi}5)\) as \(\frac{\pi}5.\)

Correct Answer is \(\frac{\pi}{4}\)

Now, let x = \(cot^{-1}\left(cot\frac{5\pi}{4}\right)\)

⇒ cot x =cot ( \(\frac{5\pi}{4}\)) 

Here range of principle value of cot is [ \(-\frac{\pi}{2},\frac{\pi}{2}\)] 

⇒ x = \(\frac{5\pi}{4}\)∉ [ \(-\frac{\pi}{2},\frac{\pi}{2}\)] 

Hence for all values of x in range [ \(-\frac{\pi}{2},\frac{\pi}{2}\)] ,the value of 

\(cot^{-1}\left(cot\frac{5\pi}{4}\right)\) is 

⇒ cot x =cot (π+ \(\frac{\pi}{4}\)) (\(\because\) cot ( \(\frac{5\pi}{4}\))= cot (  π+ \(\frac{\pi}{4}\) ) ) 

⇒ cot x =cot ( \(\frac{\pi}{4}\)) ( \(\because\)cot (π + θ)= cot θ) 

⇒ x =\(\frac{\pi}{4}\)

Given that: cot (tan^-1  + cot^-1 )

= cot (/) (since, tan^-1 x + cot^-1 x = /2)

= cot (180°/2) ( we know that cot 90° = 0 )

= cot (90°)

= 0

Therefore, the value of cot (tan^-1 α + cot^-1 α) is 0.

(i) Given as sin^-1{(sin – 17π/8)}

As we know that – sin θ = sin (-θ)

So, (sin -17π/8) = – sin 17π/8

– sin 17π/8 = – sin (2π + π/8) [since sin (2π – θ) = sin (θ)]

It can also be written as – sin (π/8)

– sin (π/8) = sin (-π/8) [since – sin θ = sin (-θ)]

On, substituting these values in sin^-1{(sin – 17π/8)} we get,

sin^-1(sin – π/8)

As sin^-1(sin x) = x with x ∈ [-π/2, π/2]

So, sin^-1(sin -π/8) = – π/8

(ii) Given as sin^-1(sin 3)

As we know that sin^-1(sin x) = x with x ∈ [-π/2, π/2] which is approximately equal to [-1.57, 1.57]

But here x = 3, which does not lie on the above range,

So, we know that sin (π – x) = sin (x)

Thus, sin (π – 3) = sin (3) also π – 3 ∈ [-π/2, π/2]

Sin^-1(sin 3) = π – 3

(iii) Given as sin^-1(sin 4)

As we know that sin^-1(sin x) = x with x ∈ [-π/2, π/2] which is approximately equal to [-1.57, 1.57]

But here x = 4, which does not lie on the above range,

So, we know that sin (π – x) = sin (x)

Hence sin (π – 4) = sin (4) also π – 4 ∈ [-π/2, π/2]

sin^-1(sin 4) = π – 4

(iv) Given as sin^-1(sin 12)

As we know that sin^-1(sin x) = x with x ∈ [-π/2, π/2] which is approximately equal to [-1.57, 1.57]

But here x = 12, which does not lie on the above range,

So, we know that sin (2nπ – x) = sin (-x)

Hence, sin (2nπ – 12) = sin (-12)

Here n = 2 also 12 – 4π ∈ [-π/2, π/2]

sin^-1(sin 12) = 12 – 4π

(v) Given as sin^-1(sin 2)

As we know that sin^-1(sin x) = x with x ∈ [-π/2, π/2] which is approximately equal to [-1.57, 1.57]

But here x = 2, which does not lie on the above range,

So, we know that sin (π – x) = sin (x)

Thus, sin (π – 2) = sin (2) also π – 2 ∈ [-π/2, π/2]

sin^-1(sin 2) = π – 2

The given equations are: 

2x - 3y/ 4 = 3 …..(i) 

5x = 2y + 7 ……..(ii) 

On multiplying (i) by 2 and (ii) by 3/4 , we get: 

4x - 3/2 y = 6 …...(iii) 

15/4 x = 3/2 y + 21/ 4 ….(iv) 

On subtracting (iii) and (iv), we get:

 - 1/4 x = - 3/ 4 

⇒x = 3 

On substituting x = 3 in (i), we get: 

2 × 3 - 3y/4 = 3 

⇒ 3y/4 = (6 – 3) = 3 

⇒y = (3 ×4 )/3 = 4 

Hence, the solution is x = 3 and y = 4.

The value of cosec\((\frac{13\pi}6)\) is 2.

∴ The question becomes cosec^–1(2)

Let, cosec^–1(2) = y

∴ cosec y = 2

⇒ cosec\((\frac{\pi}6)\) = 2

The range of principal value of cosec^–1 is \([\frac{-\pi}2,\frac{\pi}2]\)- {0} and

cosec\((\frac{\pi}6)\) = 2

Therefore, the value of cosec^-1(cosec\((\frac{13\pi}6)\)) is \(\frac{\pi}6.\)

The given equations are: 

2x - 3y/ 4 = 3 …..(i) 

5x = 2y + 7 ……..(ii) 

On multiplying (i) by 2 and (ii) by 3/4 , we get: 

4x - 3/2 y = 6 …...(iii) 

15/4 x = 3/2 y + 21/ 4 ….(iv) 

On subtracting (iii) and (iv), we get:

 - 1/4 x = - 3/ 4 

⇒x = 3 

On substituting x = 3 in (i), we get: 

2 × 3 - 3y/4 = 3 

⇒ 3y/4 = (6 – 3) = 3 

⇒y = (3 ×4 )/3 = 4 

Hence, the solution is x = 3 and y = 4.

The given equations are: 

2x - 3/y = 9 ……..(i) 

3x + 7/y = 2 ……..(ii) 

Putting 1/y = v, we get: 

2x - 3v = 6 …….(iii) 

3x + 7v = 2 ……(iv) 

On multiplying (iii) by 7 and (iv) by 3, we get: 

14x - 21v = 63 ……..(v) 

9x + 21v = 6 ……..(vi) 

On adding (v) from (vi), we get: 

23x = 69 

⇒ x = 3 

On substituting x = 3 in (i), we get: 

2 × 3 - 3/y = 9 

⇒6 - 3/y = 9 

⇒ 3/y = -3 

⇒ y = -1 

Hence, the required solution is x = 3 and y = -1.

The given equations are: 

5/x + 6y = 13 ……..(i) 

3/x + 4y = 7 ……..(ii) 

Putting 1/x = u, we get:

5u + 6y = 13 …….(iii) 

3u + 4y = 7 ……(iv) 

On multiplying (iii) by 4 and (iv) by 6, we get: 

20u + 24y = 52 ……..(v) 

18u + 24y = 42 ……..(vi) 

On subtracting (vi) from (v), we get: 

2u = 10 ⇒ u = 5 

⇒ 1/x = 5 

⇒ x = 1/5 

On substituting x = 1/5 in (i), we get: 

5/ 1/3 ⁄ + 6y = 13 

25 + 6y = 13 

6y = (13 – 25) = -12 

y = -2 

Hence, the required solution is x = 1/5 and y = -2

The given equations are: x + 6/y = 6 ……..(i) 

3x - 8/y = 5 ……..(ii) 

Putting 1/y = v, we get: 

x + 6v = 6 …….(iii) 

3x – 8v = 5 ……(iv) 

On multiplying (iii) by 4 and (iv) by 3, we get: 

4x + 24v = 24 ……..(v) 

9x – 24v = 15 ……..(vi) 

On adding (v) from (vi), we get: 

13x = 39 

⇒ x = 3 

On substituting x = 3 in (i), we get: 

3 + 6/y = 6 

⇒ 6/y = (6 – 3) = 3 

⇒ 3y = 6 

⇒ y = 2 

Hence, the required solution is x = 3 and y = 2. 

Correct answer is (b) 47 1/4

The given equations are: 

2x – 5y = 8/3 …..(i) 

3x – 2y = 5/6 ……..(ii) 

On multiplying (i) by 2 and (ii) by 5, we get: 

4x - 10y = 16/3 …...(iii)

15x – 10y = 25/6 ……(iv) 

On adding (iii) and (iv), we get: 

19x = 57/ 6 

⇒x = 57/ (6 × 19) = 3/6 = 1/2 

On substituting x = 1/2 in (i), we get: 

2 × 1/2 + 5y = 8/3 

⇒ 5y = ( 8/3 − 1) = 5/3 

⇒y = 5/ (3 × 5) = 1/3 

Hence, the solution is x = 1/2 and y = 1/3 .

(a) [1, 2]

The domain for sin^-1 x is [0, 1]

So, √(x - 1) = 0 ⇒ x – 1 = 0 ⇒ x = 1

√(x - 1) = 1 ⇒ x – 1 = 0 ⇒ x = 2

∴ The domain is [1, 2]

The given system of equations is 

0.4x + 0.3y = 1.7 …….(i) 

0.7x – 0.2y = 0.8 …….(ii)

Multiplying (i) by 0.2 and (ii) by 0.3 and adding them, we get 

0.8x + 2.1x = 3.4 + 2.4 

⇒2.9x = 5.8 

⇒x = 5.8/2.9 = 2 

Now, substituting x = 2 in (i), we have 

0.4 × 2 + 0.3y = 1.7 

⇒0.3y = 1.7 – 0.8 

⇒y = 0.9/ 0.3 = 3 

Hence, x = 2 and y = 3. 

The given equations are: 

(7 − 4x)/ 3 = y 

⇒ 4x + 3y = 7 ……..(i) 

and 2x + 3y + 1 = 0 

⇒2x + 3y = -1 ……….(ii) 

On subtracting (ii) from (i), we get: 

2x = 8 

⇒x = 4 

On substituting x = 4 in (i), we get: 

16x + 3y = 7 

⇒3y = (7 – 16) = -9 

⇒y = -3 

Hence, the solution is x = 4 and y = -3.

Given sin as (tan^-1 24/7)

Let tan^-1 (24/7) = y

tan y = 24/7, where y ∈ [0,π/2]

Let us find

sin(tan^-1 (24/7)) = sin y

As we know that, 

1 + cot²θ = cosec²θ

1 + cot² y = cosec² y

On, substituting this trigonometry form, we get

1 + (7/24)² = cosec²y 

1 + (49/576) = 1/sin²y

By rearranging we get,

sin²y = 576/625

sin y = 24/25, where y ∈ [0,π/2]

sin(tan^-1(24/7)) = 24/25

By microscopic observation, we can find up and downs over surface of a paper. Hence paper is not a smooth fine surface. Even though sheet of white paper reflects the light rays, they do not form an image of an object.

Correct option is  : (c)

Strategy : use the mole concept where moles of HCI are equated to the volume occupied by H₂ gas at N.T.P. as given by the balanced chemical equation.

One mole of a gas at N.T.P. (0° C, 1 atm) occupies 22.4 L of volume.

Solution : 6 moles of HCI liberate 3 moles of H₂ gas. 

Convert the moles of H₂ gas to the volume occupied by H₂ gas = 3 x 22.4 L of H₂

6 moles of HCI liberate 3 x 22.4 L of H₂

1 mole of HCI will liberate \(\frac{3\times 22.4}{6}=11.2L\)

Correct option is (c) 100 mL of 0.5 M Na₂ SO₄ (formula mass 14)

C. meat as well as eggs 

Poultry gives us meat as well as eggs.

When Nitric acid is added to egg shell carbon dioxide gas is released.

I stayed there for five days.

I went there with my family.

Guru Nanak Dev Ji passed through Jalandhar and Hoshiarpur and finally reached the present Himachal Pradesh. There, Guru Sahib visited Bilaspur, Mandi, Suket, JawalaJi, Kangra, Kulu, Spiti, etc. and made many people his followers. Guru Sahib then visited Tibet, Kailash Mountain and Amarnath Cave in Kashmir. After that, Guru Sahib also visited Hassan Abdal and Sialkot. From there, Guru Sahib came back to Sultanpur Lodhi.

I visited Jaipur during my last summer vacation.

• Egg shells contain calcium carbonate. 
• When nitric acid is added to it, carbon dioxide gas is evolved. 
• The reaction can be given as:

 CaCO₃ + 2HNO₃  → Ca(NO₃)₂ + H₂O + CO