Explore topic-wise InterviewSolutions in Current Affairs.

This section includes 7 InterviewSolutions, each offering curated multiple-choice questions to sharpen your Current Affairs knowledge and support exam preparation. Choose a topic below to get started.

1.

Give two safety measures to be taken during thunderstrom​

Answer»

Answer:

❤️❤️hello DEAR here is your answer mark it BRAINLIST ok and FOLLOW me... :) ☺️☺️❤️❤️

2.

The two bodies of masses 40 kg and 80kg are at a distance of 0.15m the force of gravitation between them is 1N.calculate the constant of gravation.

Answer»

Answer:

7.03 × 10^-6 Nm²/kg

Explanation:

From GRAVITATIONAL force formula

F = GMm/d²

G = Fd² / (Mm)

= 1 N × (0.15 m)² / (40 kg × 80 kg)

= 7.03 × 10^-6 Nm²/kg²

Note: The correct Universal gravitation CONSTANT is 6.67 × 10^(-11) Nm²/kg².

3.

if the angle of projection is 60°,the height of the projectile when it was travelled a distance 3R/4 where R is the range​

Answer»

EXPLANATION:

which angle HEIGHT of projectile is maximum? ... What will the angle of projection be if the horizontal range of ... The PARTICLE travels for a TIME t = 2*V*sin A/g and the ...

4.

A model of helicopter is made in which a kind of box is introduced inside it for the sitting of passengers. The box is further connnected to a parachute which get opens during any hard situation (such as fire). Is the model accurate or it Do have any demerit ?Quick reply please!​

Answer»

HELLO MATES

YES , this can be RIGHT I.e ACCURATE...

5.

a boy press horizontally a book against the front wall such that book does not slide down . the force of friction between wall and book is

Answer»

Answer:

f=\mu mg.

Explanation:

Force of FRICTION is the force which opposes the MOTION of the object.

It always acts in the OPPOSITE direction of the motion.

When a book is press horizontally against the front wall, the force of friction is

f=\mu \times R..................(1)

Where R= Normal force of the book which is reaction of book exerted by the weight of the book.

R=mg.............(2)

where m=mass of the book, g=acceleration due to gravity

using (1) and (2) we get

So,force of friction is,

f=\mu mg.

6.

Examples for free oscillations

Answer»

Answer:

Free oscillations[edit]

When an object is in free OSCILLATION, it vibrates at its natural frequency. For example, if you strike a TUNING FORK, it will begin to vibrate for some time after you struck it, or if you HIT a PENDULUM, it will always oscillate at the same frequency no matter how hard you hit it.

7.

Give me some examples of unit vector from real life ????

Answer»

EXPLANATION:

HOPE it will HELP you .... PLEASE mark me as a BRAINEST....

8.

When we freely suspend a bar magnet,the north pole of the magnet turns to North and South pole of the magnet turns to South . But if South pole of one bar magnet is kept near other magnet 's South pole,they repel and if we keep north and south pole then they get attached to each other. WHY?

Answer»

the south gets ATTRACTED towards north . it is because like pole ATTRACT and UNLIKE pole REPEL

9.

What is the dimensonl of k in the expresion of work in Spring

Answer» K is a CONSTANT Vance it has no UNIT or DIMENSION
10.

Name the kind of frictional friction that comes into play when a book lepton a collection of cylindrical pencils, is moved by pushing​

Answer»

ANSWER:

ROLLING FRICTION MARK me as BRAINLIST

11.

.What can be the angle between P vector + Q vector and p vector - q vector

Answer»

ANSWER:

The angle between P VECTOR + Q vector is 0° while the angle between P vector - Q vector is 180°.

Since RESULTANT vector=√P^2 vector +Q^2 vector +2PQvector cos 0°. (cos 0°=1 and cos 180°=-1)

HOPE THIS HELPS..☺️

12.

What is a motion? and what is its formula for numerical​

Answer»

Answer:

As we have already discussed earlier, motion is the state of change in position of an object over time. It is described in terms of DISPLACEMENT, distance, velocity, acceleration, time and speed. Jogging, driving a car, and EVEN simply taking a walk are all everyday examples of motion. The relations between these quantities are known as the equations of motion.

In case of uniform acceleration, there are three equations of motion which are also known as the laws of constant acceleration. Hence, these equations are used to derive the COMPONENTS like displacement(s), velocity (initial and final), time(t) and acceleration(a). Therefore they can only be applied when acceleration is constant and motion is a straight line. The three equations are,

v = u + at

v² = u² + 2as

s = ut + ½at²

where, s = displacement; u = initial velocity; v = final velocity; a = acceleration; t = time of motion. These equations are REFERRED as SUVAT equations where SUVAT stands for displacement (s), initial velocity (u), final velocity (v), acceleration (a) and time (T)

DERIVATION of the Equations of Motion

v = u + at

Let us begin with the first equation, v=u+at. This equation only talks about the acceleration, time, the initial and the final velocity. Let us assume a body that has a mass “m” and initial velocity “u”. Let after time “t” its final velocity becomes “v” due to uniform acceleration “a”. Now we know that:

Acceleration = Change in velocity/Time Taken

Therefore, Acceleration = (Final Velocity-Initial Velocity) / Time Taken

Hence, a = v-u /t or at = v-u

Therefore, we have: v = u + at

v² = u² + 2as

We have, v = u + at. Hence, we can write t = (v-u)/a

Also, we know that, Distance = average velocity × Time

Therefore, for constant acceleration we can write: Average velocity = (final velocity + initial velocty)/2 = (v+u)/2

Hence, Distance (s) = [(v+u)/2] × [(v-u)/a]

or s = (v² – u²)/2a

or 2as = v² – u²

or v² = u² + 2as

s = ut + ½at²

Let the distance be “s”. We know that

Distance = Average velocity × Time. Also, Average velocity = (u+v)/2

Therefore, Distance (s) = (u+v)/2 × t

Also, from v = u + at, we have:

s = (u+u+at)/2 × t = (2u+at)/2 × t

s = (2ut+at²)/2 = 2ut/2 + at²/2

or s = ut +½ at²

13.

Please help mesub:physics Thanks in advance​

Answer»

ANSWER:

POTENTIAL= ML-2T-2

unit is JOULE

14.

suppose A=B^m C^m where A has dimension LT. B has dimensions L^2 T^-1 and C has dimensions LT^2 then for the value of m & n.​

Answer»

EXPLANATION:

SOLVE it your SELF ibuuhyvrh5giggtgtvrvrcr

15.

The value of 120 joule per second on a system thathas 10 g, 10^2 cm and 10^3 second as base unit is ____(1) 12 x 10^10 units (2) 12 x 10^12 units(3) 12 x 10^14 units (4) 12 x 10^16 units​

Answer»

here we are USING the basic formula for unit conversion through dimensional analysis

any physical QUANTITY is equal to the product of its unit with the magnitude

if you have a physical quantity X then

X = n1u1 = n2u2

where YUVAN YouTube or units of the physical quantity in two different systems and n1and n2 are magnitudes corresponding to two different systems

n2 = n1u1/u2

= 120(1kg/10g)(1m/10²cm)²(1s/10³s)-³

=120(10²)(1)²(10-³)-³

=12×10¹²

answer is (2)

Any doubts ask me through chat

16.

Please answer the question as soon as possible​

Answer»

Answer:

OPTION a.

From units and DIMENSIONS you'll STUDY that arguments of trigonometric FUNCTIONS should be dimensionless

17.

A boat which has a speed of 5 km per hour in still water crosses a river of width 1km along the shortest possible path in fifteen minutes. The velocity of the river water in km per hour is​

Answer»

Answer:

LET the velocity of the RIVER be x. the velocity of the river will be in the HORIZONTAL direction

velocity of boat in still water is 5km/hr. As the width of the river is 1 km and time taken by the boat to cross the river along the shortest path is 15 min therefore the velocity of the boat is= 1*60/15= 4km/hr

according to the figure the velocity of the river is along AB . By applying triangle law of vector addition the value of AB COMES to be 3km/hr

18.

15. A car start from rest and acquire a velocity of 54 km/h in 2 sec. Finda) the accelerationb) distance travelled by car assume motion of car is uniformc) If the mass of the car is 1000 Kg, What is the force acting on it​

Answer»

Explanation:

speed of car = 54 KMPH

= 54 X 5/18

= 15m/s

(a) V = u+ at

54 = 0+ 2a

a= 27 m/s^2

(b) s= UT + 1/2 at^2

= 27x 2

= 54M

19.

Strongest fundamental force in nature??

Answer»

ANSWER:

GRAVITY is strongest fundamental FORCES in nature

20.

Defination + formula of method of application of science​

Answer»

Explanation:

The scientific method. When conducting research, scientists use the scientific method to collect measurable, empirical evidence in an experiment related to a HYPOTHESIS (OFTEN in the form of an if/then statement), the results AIMING to support or contradict a theory.

21.

a pushing force of 50 newtons acts on a block at an angle of 30 degree with horizontal. the respicetive horizontal and vertical components of force will be ​

Answer»

HORIZONTAL COMPONENT is = 50cos30°= 50sqrt3/2N

vertical component is = 50sin30°

= 25N

22.

When a wheel of radius 50 cmmakes 2 revolutions, the lineardistance travelled by it is:A) 3.14 cmB) 6.28 cmC) 628 cmD) 12.5 cm​

Answer»

Answer:

628cm because in one revolution it will GO 2*π*r

So in TWO REV it goes 4πr

=4*(22/7)*50CM

23.

The energy associated with one gram of mass is(a) 9x10^-13 J(b) 9x10^-16 J(c) 9x10^13 J(d) 9x10^16 J​

Answer»

option C is the right answer i. e. for ONE gram of MASS it is 9*10^13 joules.

hope It help uh

24.

A stone on the edge of vertical Cliff is kicked so that its initial velocity is 9 m/s horizontally. If the Cliff is 200m high calculate time taken by stone to reach the ground and how far from the Cliff the stone will hit the ground

Answer»

ANSWER:

t=2root10

r=18root10

Explanation:

use time and range formula of OBLIQUE PROJECTILE

25.

What is the magnitude of vector A= 2i+3j-k and B=4i-j-6k

Answer»

Magnitude of VACTOR A = SQRT 2^2 + 3^2+(-1)^2 = sqrt14

magnitude of vector B = sqrt (4^2+(-1)^2+(-6)^2)= sqrt53

26.

A toy of mass 0.1kg acquires a speed of 5 m/s find the impulse​

Answer»

ANSWER:

27.

A person walks 12m towards north and then 4 m towards west and there he climbs a pole of 3 m height . The total distance travelled by the person is

Answer»

Answer:

16m in GROUND &19M from ground to above

28.

a projectile is thrown with an initial velocity u and angle 15 to the horizontal has a range R if the same projectile is thrown at an angle of 45 to the horizontal with speed 2u its range will be

Answer»

ANSWER:

R15=2u2/g(cos15×sin15)

so R 45degree=R15×2u

29.

A car is going towards north with a velocity 20 metre per second full stop after some time it starts to go towards south with velocity 20 metre per second full stop determine changes in the velocity of the car full stop

Answer»

ANSWER:

ZERO because DISPALCEMENT is zero

30.

What is Maxwell's Right hand rule?​

Answer»

Answer:

MAXWELLS RIGHT hand RULE was beam BALANCE

31.

Hiiii plzz answerfollow mee​

Answer»

ANSWER:

we think ithey are GOOD because :

1.It helps farmers to buy good SEEDS and other necessary things so they do not have to take loan on high interest

2.Farmers feel good that tere is someone who can help them

3.We can see that the suicide rate of farmers decreased after all the efforts

PLEASE MARK AS BRAINIEST

32.

The mass of the earth is 6(10)^24 kg and that of the moon is 7.4 (10)^22 kg .If the distance between the earth and the moon is 3.84 (10)5 kg .Calculate the force exerted by the earth on the moon​

Answer»

Answer:

20.2 × 10¹⁹ N

Explanation:

Given :

Mass of earth = 6 × 10²⁴ kg

Mass of moon = 7.4 × 10²² kg

Distance between them = 3.84 × 10⁵ km = 3.84  × 10⁸ m

We have value of G = 6.7 × 10⁻¹¹ N m² kg⁻²

We have to find FORCE :

We have :

F = G m₁ m₂ / r²

F = ( 6.7 × 10⁻¹¹ ) ( 6 × 10²⁴ ) ( 7.4 × 10²² ) / (  3.84 × 10⁸ )² N

F = 20.2 × 10¹⁹ N

Hence force EXERTED by the earth on the moon is 20.2 × 10¹⁹ N.

33.

For a body starting from rest the displacement in 10 sec when it acquires 4m/s in 2 sec is

Answer»

Answer:

Given, initial velocity u = 0m/s

Velocity after 2s, v = 4m/s

Acceleration a = (v - u)/t = (4-0)/2 = 2m/s^2

Under UNIFORMLY accelerated MOTION,

After t = 10s

S = UT + (1/2)at^2

S = 0 + (1/2) x 2 x (10^2) = 100m

S = 100m

34.

For 100%efficiency of a carnot engine the temperature of the source should be

Answer»

Dear Student,

◆ Answer -

For 100% EFFICIENCY of a carnot engine the temperature of the source should be infinite.

● Explaination -

Efficiency of Carnot engine is calculated by formula -

η = 1 - T2/T1

Here, for η to be 1 i.e. 100 %, EITHER T2 (temperature of the sink) should be zero or T1 (temperature of the source) should be infinite.

Therefore, for 100% efficiency of a carnot engine the temperature of the source should be infinite.

Hope this helps you...

35.

Find the recoil velocity of gun of mass 2kg from which a bullet of 20 g is fired with a velocity of 200m/s

Answer»

ANSWER:

Given m1= 2 kg, V1 =? , m2= 20 x 10∧-3 kg, v2=200m/s

        According to the LAW of linear momentum,

                 m1 x v1=m2 x v2

                  2 x v1 = 20 x 10∧-3 x 200

                   v1= 2 x 10∧-2 x 2 x 10∧2 / 2

                        v1= 2m/s

Explanation:

36.

Find the maximum horizontal distance travelle by a ball thrown with a velocity 40 m per s without hitting thr roofd

Answer»

Answer:

160M.

Explanation:

R = u²sin2∝/g

∝ = 45°

R = u²sin90°/g  = u²/g = 40²/10 = 160m.

37.

Find the magnetic field intensity at the centre o of a square of the sides equal to 5m and carrying 10a of current

Answer»

ANSWER:

I HOPE IT WILL BE HELPFUL FOR YOU.

MARK AS BRAINIEST!!!!

38.

Find the flux contained by the material when the flux density is 11.7 tesla and the area is 2 units.

Answer»

ANSWER:

It will be about 23.4 T (unit).

Explanation:

The FORMULA of magnetic flux is :

ϕ = B × A cosθ

Where B is the magnetic field and A is the area vector.

Let US assume that B is parallel to A.

That IMPLIES,

ϕ = B × A cos0°

Thus,

ϕ = B × A

ϕ = 11.7 T × 2 unit = 23.4 T (unit).

39.

Make a model of seismograph​

Answer»

ANSWER:

OK i will send it WITHIN TWO minutes

40.

A meter scale of mass 400 gm is lying horizontally onthe floor. If it is to be heldvertically with one end touching the floorwork to be done is​

Answer»

ANSWER:

here torque will PLAY a role

Explanation:

the torque required will be totally DEPENDENT on the LENGTH of the scale

(*note i didn't refer any book based on my conscience i just answered this question)

41.

Angular 5 Property 'style' does not exist on type 'Element'

Answer» EVERY ELEMENT EXIST on STYLE of the CHEMICAL
42.

A rectangular car top carrier of 2.1 ft height front to back and 4.2 ft width

Answer»

The AREA of the CAR is 8.82 FT

43.

Explain level triggering and edge trigerrinfg

Answer»

It is CONNECTED to the SEX

44.

Write a short note on bose einstein condenstate

Answer»

Hey...here your answer______________________

In 1920,indian physicist SATYENDRA Nath bose had done some calculations for a FIFTH state of MATTER .Building on his calculations,Albert EINSTEIN predicted a new state of matter --the BOSE EINSTEIN CONDENSATE【BEC】


I hope it will helpful to you

45.

Working principle of one cavity clystron

Answer» HLO FRD...I HOPE the above ans will help u
46.

Methods to analyze the energy efficiency of electrical utilities

Answer» IMPROVE POWER factor
leading pf
efficient APPLIANCES
47.

Difference between active and passive elements

Answer»

Hey ...... here your answer______________________
Active and Passive Commonest (Very Easy Explanation with Examples)
 
 
 
Active Components:
Those devices or components which required external source to their operation is called Active Components. 
For Example: Diode, Transistors, SCR etc…
Explanation and Example:As we know that Diode is an Active Components. So it is required an External Source to its operation. 
Because,   If we CONNECT a Diode in a Circuit and then connect this circuit to the  Supply voltage., then Diode will not conduct the current Until the  supply voltage REACH to 0.3(In case of Germanium) or 0.7V(In case of  Silicon). I think you got it 
 
Passive Components:
 
Those devices or components which do not required external source to their operation is called Passive Components. 
For Example: Resistor, Capacitor, Inductor etc…
 Explanation and Example:Passive Components do not require external source to their operation. 
Like  a Diode, Resistor does not require 0.3 0r 0.7 V. I.e., when we connect a  resistor to the supply voltage, it starts work automatically WITHOUT  using a specific voltage. If you UNDERSTOOD the above statement about  active Components, then you will easily get this example. 

I hope it will helpful to U

48.

Why small engines having larger RPM and larger engines having smaller RPM/quora

Answer»

Small engine have small shaft and small arm and BIG engines have big shaft and very very big arm that is why it has LESS RPM

49.

Why induction genrator in small wind energysystems

Answer»

Because there is CHANGE in MAGNETIC FLUX

50.

What is problem if fingerprint red light is not comes

Answer»

If FINGERPRINT RED LIGHT is not COMES it is MISMATCHED