Explore topic-wise InterviewSolutions in Current Affairs.

1. The Mughal Emperor Jahangir was jealous of the rising popularity of Guru Arjan Dev Ji.
2. He was perturbed (troubled) by the fact that like the Hindus, many Muslims were coming under the influence of Guru Arjan Dev Ji. Thus, the Mughal Emperor Jahangir wanted to put to death Guru Arjan Dev Ji.

1. (T) 

2. (F) 

3. (F) 

4. (T)

The residue of fractional distillation of petroleum is Coal tar.

Guru Hargobind Ji put on two swords and called them ‘Miri’ and ‘Piri’. The ‘Miri’ sword represented the leadership of Guru Hargobind Ji over the Sikhs in their worldly affairs. The sword ‘Piri’ indicated the leadership of Guru Sahib in spiritual matters of the Sikhs.

It is mainly useful as fuel in kitchen as cooking gas and automobiles.

1. A long process and time is needed to form oxide oil or natural fuels.

2. Their deposits are limited only for next 100 years. ‘ 

3. They cause air pollution.

4. They help in global warming.

Suggestions to control pollution due to use of petroleum:

1. Petroleum should be used economically and judiciously. 

2. Drive vehicles at a constant and moderate speed as far as possible. 

3. Switch off the engine at traffic lights or at a place where you have to wait. 

4. Ensure correct pressure in tyres. 

5. Ensure regular maintenance of the vehicle. 

6. Sustainable use of petroleum products is a must so that future generations may also use them. 

7. Use unleaded petrol, CNG, LPG to minimise pollution.

When RBCs are piled on top of each other, this formation is known as Rouleaux. It occurs when blood flows slowly.

Vitamin K is essential for blood clotting because it is necessary for the synthesis of prothrombin in the liver.

Role of haemoglobin in the transport of of respiratory gases are : 

(i) Oxygen binds with haemoglobin to form oxyhaemoglobin. 1 molecule of haemoglobin carries 4 molecules of oxygen. 

(ii) CO₂ binds with haemoglobin to form carbamino- haemoglobin. CO₂ is carried as carbamino-haemoglobin by blood.

Formed elements or blood corpuscles constitute nearly 45% of the blood and are of the following three types : Erythrocytes, Leucocytes and thrombocytes.

The spindle cells help in clotting of blood like the blood platelets of mammals. These are found in pisces, amphibians, reptiles and birds.

There is great difference in respects of efficiency of carrying the oxygen from respiratory organs to the body tissue as follows :

Monocytes are the largest of all types of leucocytes they are phagocytic in nature and engulf bacteria and cellular debris.

Lymph nodes are small oval or bean shaped structures located along the length of lymphatic vessels.

(i) Lymph acts as a "middle man" which transports oxygen, food materials, hormones, etc., to the body cells and brings carbon dioxide and other metabolic wastes, from the body cells to blood and then finally pours the same into the venous system. Body cells are kept moist by the lymph. 

(ii) Lymph capillaries present in the intestinal villi are called lacteals which are associated with absorption and transportation of fat and fat soluble vitamins.

1. (iv) double of the volume of gas collected at the anode

2. (iii) Positive, negative

3. (i) Cathode

4. (iv) All of the above

5. (iii) 3

(i) At Cathode: hydrogen gas (H₂)

At Anode: oxygen gas (O₂)

(ii) Since 2H₂O + Electric current → 2H₂+O₂

2 molecules of H₂ combine with 1 mol of O₂ to form H₂O, so the volume of H₂, liberated is double that of O₂.

(iii) When a burning splinter is brought near the mouth of the liberated gases, the burning splinter extinguishes near the H₂ gas while the burning splinter keeps burning more near the O₂  gas.

(i) The flow of deoxygenated blood from the right ventricle to the lungs and the return of oxygenated blood from the lungs to the left atrium is called the pulmonary circulation. 

(ii) The pulmonary trunk arises from the right ventricle and then divides into the right pulmonary artery and left pulmonary artery which supply deoxygenated blood to the right and left lung respectively. 

(iii) Exchange of gases takes place in the lungs. Two pulmonary veins from each lung transport the oxygenated blood to the left atrium.

Valves in heart: (i) In the heart there are a number of openings through which blood passes from one chamber to another. 

(ii) These openings are guarded by certain valves which regulate the flow of blood in desired direction. The bicuspid valves and tricuspid valves are present guarding the auriculoventricular apertures in between the auricles and ventricles on the left and right sides respectively. These valves open into the ventricles and prevent the back flow of the blood. 

(iii) A set of semilunar valves are also present at the base of aorta in the ventricle. These allows the blood to flow only in direction of an aorta, when the left ventricle contracts.

Veins are blood vessels that carry blood towards the heart. For example, arteries are more muscular than veins, veins are often closer to the skin, and veins contain valves to help keep blood flowing toward the heart, while arteries do not have valves and carry blood away from the heart.

Differences between arteries and veins:

The correct answer is 27 1/4 kg

  correct answer is  (A) 12√6.

The correct answer is 14 5/8 kg

Answer is (A)

Circumference of a circle = Area of the circle. 

2πr = πr² 

2 × π × r = π × r × r 

∴ 2 = r 

∴ r = 2 units.

The correct answer is (A) 12m²

Perimeter of a rectangle room = 250 m 

= 2(l + b) 

Cost of Painting = Rs. 15000 

For Rs. 10, 1 sq.m. 

For Rs.15000 … ? = \(\frac{15000}{10}\)

= 1500m2 

Area of four walls = Lateral surface area 

Lateral surface area of rectangle = 1500 m² 

2h(l + b) = 1500

h × 2(l + b) = 1500 

h × 250 = 1500

h = \(\frac{1500}{250}\)

∴ h = 6 m

(a) In the first 20 s, the ice block have no change in temperature.

(b) There was a change in temperature from 20 s to 50 s, i.e. 50-20 = 30s.

(c) Observing the graph, we see that after 50 s of heating the temperature became constant.

(d) 20°C was the temperature after 25 s.

(e) Since, the temperature became constant at 100°C after 50 s heating, so the temperature will be 100°C even after 1.5 min.

(a) (i) After 2 weeks, the height of plant A was 7 cm.

(ii) After 3 weeks, the height of plant A was 9 cm.

(b) (i) After 2 weeks, the height of plant B was 7 cm.

(ii) After 3 weeks, the height of plant B was 10 cm.

(c) Growth of plant A during 3rd week = 9 cm − 7 cm = 2 cm

(d) Growth of plant B from the end of the 2nd week to the end of the 3rd week = 10 cm − 7 cm = 3 cm

(e) Growth of plant A during 1st week = 2 cm − 0 cm = 2 cm

(f) Growth of plant B during 1st week = 1 cm − 0 cm = 1 cm Growth of plant B during 2nd week = 7 cm − 1 cm = 6 cm Growth of plant B during 3rd week = 10 cm − 7 cm = 3 cm Therefore, plant B grew the least, i.e. 1 cm, during the 1st week.

(g) At the end of the 2nd week, the heights of both plants were same.

Observing the graph carefully, we conclude that

(a) The X-axis represents subjects and the Y-axis represents the marks obtained by Sonia.

(b) In Maths, she scored the highest in Test I.

(c) In English and Hindi, she scored the least in Test II.

(d) In Hindi and Maths, she scored the same marks in both tests.

(e) She scored 6 marks in English in Test II.

(f) Same performance in both tests.

(g) Test I in Maths, she scored full marks i.e. 10 marks.

Observing the graph carefully, we conclude that

(a) At 10 AM, the person start from his home.

(b) In first 4 h (i.e. till 2PM), he travelled 16 km.

(c) He was taking rest from 3 PM to 5 PM.

(d) The total distance covered by the person throughout the day was 40 km, i.e. 20 km from A to Sand then 20 km from C to D.

(e) The distance covered by him in the first 8 h i.e. from 10 AM to 6 PM was 24 km,

(f) He covered 16 km of his journey at 2 PM,

(g) The total distance covered from A to S=20 km
and the time taken to travel from A to B = 5 h

.^.. Average speed of the man from A to B =20/5 = 4 km/h and average speed from S to C = 0/2 = 0 km/h

(h) He returned home at 10 PM.

A) From the graph we can say that, Sneha made the least progress from the interval 25 minutes to 40 minutes.

We see that initially , she is travelling approximately 0.5 km in 10 min , 0.25 km in 5 min , till the first 25 mins. But after 25 mins, her speed decreased as she travelled 0.25 km in 15 min.

We can say that because of her speed in that interval i.e.0.25 km/15 mins = 0.25 km/0.25 hr = 1 km/hr which the lowest among all.

B) As we know that ,

Average speed = total distance travelled/ total time 

= 2 km/ 55 min 

= 2 km/ 55/60 hr 

= 2/55 x 60 

= 2.18 km/hr.

∴ Average speed = 2.18 km/hr.

Analysing the graph carefully, we observe that

(a) He started his journey at 5:30 AM and end at 6 PM.

(b) Total duration of journey was 12:30 h.

(c) His forward journey is of duration 8:30 h and return journey is of duration 4 h. Forward journey was of longer duration.

(d) He did not move from 6:30 AM to 9:30 AM and 10 AM to 1 PM.

So, he did not move for 6 h.

(e) He have the fastest speed at 1 PM.

(a) False, the older car is 8 i.e. 8 valuable more than car A.

(b) True, in the second graph 8 is larger car having greater speed.

(c) True, larger car is 8 which is older than A

(d) True, as 8 is faster as well as older than A.

(e) False, as 8 is more valuable but not slower.

(a) On the basis of given graph, the cyclist II rest at 8 : 45 AM for 15 min.

(b) Cyclist II is cycling faster after rest as he has covered a distance of 20 km in 1 h.

(c) Both cyclists meet at 9:00 AM.

(d) The cyclist II had travelled 20 km, when he met cyclist I.

(e) When cyclist II reached town A, the cyclist I was 10 km for from town B.

(a) the X-axis represents days of a particular week and the X-axis represents the maximum temperature (in °C) recorded.

(b) Observing the graph, we see that in the first week temperature was higher on most of
the days.

(c) The temperature was same on Wednesday in both the weeks.

(d) The difference in temperatures was the maximum on Friday for both the weeks.

(e) The temperature for the first week on Thursday was 37°C and the temperature for the second week on the same day was 34°C.

(f) On Sunday, the temperature was 35° for the first week.

(g) On Wednesday, the temperature was highest for the second week.

It is given that, the cost of one note book is Rs.10

(a) Hence, the cost of 7 note books is 7×10 = 70 rs

(b) To find the number of note books that can be purchased for Rs. 50 is

= 50/10

= 5

Hence, the number of note books for Rs. 50 is 5.

We know that, the ordinate represents the y coordinate in the ordered pair.

For the given points, the ordinate are as follows:

(a) (3, 5) – 5 is the ordinate

(b) (4, 0) – 0 is the ordinate

(c) (2, 7) – 7 is the ordinate

We know that, abscissa represents the x coordinate in the ordered pair.

For the given points, the abscissa are as follows:

(a) (7, 3) – 7 is the abscissa

(b) (5, 7) – 5 is the abscissa

(c) (0, 5) – 0 is the abscissa

True

Explanation:

For any y coordinate point, when it lies on the x axis, then the y coordinate should be zero. For example, (5, 0) and (7, 0), when these points are plotted in the graph, both the points will lie on the x axis.

The coordinates of the origin are (0, 0).

Origin is intersection of x-axis and y-axis. This means that coordinates of the origin will be intersection of lines y = 0 and x = 0.

Hence, coordinates of origin are (0, 0).

∴ Given statement is true.

True

Explanation:

Both the axis such as x axis and y axis intersects at each other when the coordinates of x and y are 0 and 0 respectively.

origin

The X-axis and y-axis intersect each other at the point representing the origin. Coordinates of origin are (0,0).

True

In the ordered pair (2,3), the second number is called the y-coordinate or ordinate of the number.

Hence, 3 denotes the y-coordinate of the point (2,3).

y-coordinate

As we have discussed in the above question, that the second number denotes the y-coordinate, also known as the ordinate.

(5,4)

To denote a point in 2-D, we use two numbers viz. the x-coordinate, the y-coordinate.
In the ordered pair, the x-coordinate is written in the first slot and the y-coordinate in the second slot separated by comma. Hence, the required point is (5,4).

Einstein’s photoelectric equation is given by 

K_max = hν - ϕ₀ 

where Kmax = Maximum kinetic energy 

ϕ₀ = Work function 

h = plank's constant 

ν = Frequency of incident radiation 

λ = 800 pm = 800 x 10^-12 = 8 x 10^-10 m

Mass of neutron, m = 1.67 x 10^-27 kg

We have, λ = h/mv

or, 8 x 10^-10 = {6.626 x 10^-34}/{1.67 x 10^-27 x v}

or, 8 x 10^-10 x 1.67 x 10^-27 x v = 6.626 x 10^-34

or, v = {6.626 x 10^-34}/{8 x 10^-10 x 1.67 x 10^-27}

= 4.95 x 10² m/s.