Explore topic-wise InterviewSolutions in Current Affairs.

The rural seasonal migrants are mainly agricultural labourers or marginal farmers in their place of origin and mostly belong to low-income households, dalits and Adivasis.

1. This could not be considered as a seasonal migration. In accordance with our social norms wife is supposed to live permanently in the house of her husband. 

2. This could be considered as a seasonal migration. Because duration of the season is three months, which is below the 6 months norm. 

3. This could be considered as a seasonal migration as it is not more than 6 months and the migration is done every year. 

4. This could not be considered as seasonal migration as the domestic maids need to serve throughout the year.

The correct option is A) no new substances are formed..

C) Turmeric harvesters moving from one district to another district in Tamil Nadu for three months in a year.

Burning of magnesium ribbon and burning of a candle.

(c) can be reversed and cannot be reversed, respectively.

When hot glass is cooled fast, the glass cools down unevenly and therefore cause the glass to crack because inside contracts while the outside remains expanded. Glass expands when hot and contracts when cold. It’s a physical but irreversible change.

The number of seasonal migrants is underestimated in India due to limitations in the definition of the term ‘migrant’ used in national surveys.

The Emigration Act, 1983 is the Indian law governing migration and employment of Indians abroad.

The migrants from rural to urban are not able to find jobs in the organised sector and therefore there’s no job security and decent income that they were aspiring for. They continue to live as daily workers.

(a) Transpiration.
(b) On a bright sunny day.
(c) Small drops of water inside the polythene cover.
(d) The set-up must be airtight/polythene bag must be dry/ the twig must be fresh with 10-12 leaves.

D) All the above

Correct option is B) industry

(a) Parallel venation and fibrous root

Examples of Reversible Changes:
1. Melting of Ice into water. By freezing the water we can obtain ice again.
2. Folding a paper: By unfolding it, we can undo the change.
3. Hot milk to cold milk: By boiling milk, we can make it warm.
Example of Irreversible Changes:
1. Burning of candle.
2. Bursting of crackers.
3. Cutting of trees.

Activities (i), (iv), (v), (vi) and (vii) can be reversed and rest cannot be reversed.

(b) That cannot be reversed

(b) Melting of ghee

No, we cannot get fresh drawing sheet once a picture is drawn on it with paint/oil or water. However, we can reverse the change, if soft pencil is used to draw the picture.

We can help the farmers by testing the soil in their fields.

Bases like sodium hydroxide react with metals and release hydrogen gas.

D) Both A and B

If we add an acid to a base, they neutralize each other.

Cone cells are responsible for the colour determination. As the name suggests they are cone shaped photoreceptor cells on retina. They help us see and distinguish between different colours hence making colour perception possible.

Correct answer is (d) table

Mass of earth M = \(\frac{gR^2}{G}\)

g = 9.8 m/s²

R = 6.38 x 10⁶ m

G = 6.673 x 10^-11 Nm² kg^-2

∴ M = \(\frac{9.8 \times (6.38 \times 10^6)^2}{6.673 \times 10^{-11}}\)

=\(\frac {9.8 \times 6.38 \times 6.38}{6.673}\) x \(\frac{10^{12}}{10^{-11}}\) 

= 5.98 x 10²⁴ kg

(d) all the above

Acceleration due to gravity g = \(\frac{GM}{R^2}\)

Where G is gravitational constant.

M is the mass of the earth.

R is radius of the earth.

(a) Force on the floor of the helicopter by the crew and passengers 

= apparent weight of crew and passengers 

= 500(10+ 15) 

=12500 N

(b) Action of rotor of helicopter on surrounding air is Obviously vertically downwards, because helicopter rises on account of reaction of this force. Thus force of action

= (2000 + 500) (10 + 15) 

= 2500 x 25 

= 62,500 N

(c) Force on the helicopter due to surrounding air is obviously a reaction. As action and reaction are equal and opposite, therefore Force of reaction F’ = 62,500 vertically upwards.

(b) Assertion is true but Reason is false.

Correct option is (A) ₁μ₂ = \(\cfrac{sin\,i}{sin\,r}\)

(a) Inertia of passenger

Given:

Mass of bicycle m = 25 kg

Fig. I:

Net force acting in the forward direction, F = 500 – 400 = 100 N 

acceleration a = \(\frac{F}{m}\) = \(\frac{100}{25}\) = 4ms^-2

Fig. II:

Net force acting on bicycle F = 400 – 400 = 0 

∴ acceleration a = \(\frac{F}{m}\)= \(\frac{0}{25}\) = 0

Correct option is (B) λ_a > λ_w > λ_g

• Positive Acceleration: When the velocity of an object increases, the acceleration is positive. In this case, the acceleration is in the direction of velocity.
• Negative Acceleration: When the velocity of an object decreases with time, it has negative acceleration. Negative acceleration is also called deceleration. Its direction is opposite to the direction of velocity.
• Zero Acceleration: If the velocity of the object does not change with time, it has zero acceleration.

0.18 N; in the direction of motion of the body

Mass of the body, m = 3 kg

Initial speed of the body, u = 2 m/s

Final speed of the body, v = 3.5 m/s

Time, t = 25 s

Using the first equation of motion, the acceleration (a) produced in the body can be calculated as: v = u + at

a = (v - u)/t

= (3.5 - 2)/25 = 1.5/25 = 0.06 m/s² 

As per Newton’s second law of motion, force is given as:

F = ma

= 3 × 0.06 = 0.18 N

Since the application of force does not change the direction of the body, the net force acting on the body is in the direction of its motion. 

A system of co-ordinate axes with reference to which the position or motion of an object is described is called a frame of reference.

(a) zero acceleration and non zero velocity

m_gun = 20 kg, m_b = 20 g

v_gun = ? v_b = 500 ms^-1

From Law of conservation of momentum

M_gun V_gun + m_b v_b = 0

⇒ V_gun = - \(\frac {20 \times 10^{-3} \times 500}{20}\)

= – 0.5^-1

∴ Force required to hold its position

F = m \(\frac{v-u}{t}\) = 20 × \(\frac{(0.5-0)}{(\frac{1} {500})s}\)

= 5000 N.

Initially, both the gun and the bullet are at rest.

∴ The total initial momentum of the system is f zero.

If v₁ and v₂ are the final velocity of the gun and the bullet, final momentum is given by,

p_f = m₁v₁ + m₂v₂

According to the law of conservation of momentum, p_i = p_f

m₁v₁ + m₂v₂ = 0

i.e., v₁ = – m₂v₂/m₁

_{= \(\frac {-20 \times 110.2}{5 \times 1000}\)}

= – 0.44 ms^-1

∴ Recoil velocity of the gun is 0.44 ms^-1 .

Acceleration a = \(\frac{-F}{m}\) = \(\frac{50}{20}\) = -2.5 ms^-2

u = 15 ms^-1

V = 0

t = ? 

v = u + at 

0 = 15 – 2.5t

t = \(\frac{15}{2.5}\) =6s

Given: m = 20 kg, u = 15ms^-1, v = 0,

F = – 50 N (retarding force)

To find: Time (t)

Formula: v = u + at

Calculation: Since, F = ma

∴ a = \(\frac{F}{m}=\frac{50}{20}\) = -2.5 m s^-2

From formula,

0 = 15 + (-2.5) × t

∴ t = 6s

Time taken to stop the body is 6 s.

Retarding force, F = –50 N

Mass of the body, m = 20 kg

Initial velocity of the body, u = 15 m/s

Final velocity of the body, v = 0

Using Newton’s second law of motion, the acceleration (a) produced in the body can be calculated as:

F = ma 

–50 = 20 × a

a = -50/20 = -2.5 m/s² 

Using the first equation of motion, the time (t) taken by the body to come to rest can be calculated as:

v = u + at

t = -u/a = -15/-2.5 = 6s

Given; u = 15m/s; v = 0; t = ?
F = 50N, m = 20kg
∴ Retardation,
a = \(\frac{F}{m}=\frac{50}{m}=\frac{50}{20}\) = 2.5 ms^-2
Now applying the equation, v = u + at
0 = 15 – 2.5t ⇒ 2.5t = 15
∴ t = \(\frac{15}{2.5}\) = 6 sec
or t = 6 sec

Given F = – 50 N (retarding force)

m = 20 kg

u = 15 m/s. V = 0 m/s

t = ?

F = ma 

⇒ a =  \(\frac {F}{m} =\frac {-50}{20}\)

= – 2.5m/s²

but we know that

V = u + at

0 = 15 + (- 2.5) t

⇒ t = 6 s.

They do not cancel each other because they never act on the same body. Since they act on different bodies, they do not cancel each other.