PHYSICSNEET and JEE Main 2020 : 45 Days Crash Course

Problem Solving Class(Electrostatics, Gravitation and Radiation)

By,Ritesh Agarwal, B. Tech. IIT Bombay

P-Q1416

P-Q1416-Solution

Ans [D]

Fe = ������ ������

When dielectric insert it reduce the force by dielectric constant time �� = �4

P-Q1418

Two similar spheres having +� and −� charge are kept at a certain distance. � force

acts between the two. If in the middle of two spheres, another similar sphere having +� charge is kept, then it experience a force in magnitude and direction as

(a) Zero having no direction (b) 8F towards + � charge

(c) 8F towards − � charge (d) 4F towards + � charge

P-Q1418-Solution

Force formula btw two charge

Direction of force from + to –ve

Ans [C]

P-Q1419

P-Q1419-Solution

For maximum force

�� − 2���� = 0Solving the equation

Force btw 2 chargeAns [A]

P-Q1420

P-Q1420-Solution

2

1

2

2

1

2

1

÷÷ø

öççè

æ=Þ

µ

rr

FF

rF

NF

F

25.11

06.004.05

2

2

2

==

÷øö

çèæ=

Force is inversely proportional to ��

Both moving 0.01m so new distance btw then is 0.04

Ans [B]

P-Q1421

P-Q1421-Solution

Ans [B]

Force formula btw 2 charge

Electrostatic force is vector Quantity so apply vector sum

P-Q1422

P-Q1422-Solution

Charge distribution on surfaces

Net charge on each surface

Ans [A]

P-Q1426

A) 2.36 × 10�� B) 4.72 × 10�� C) 2.16 × 10�� D) 1.23 × 10��

P-Q1426-Solution

Gravitational force

Electrostatic force btw charges

Ans [A]

P-Q1427

A charge q is placed at the Centre of the line joining two equal charges Q. The system

of the three charges will be in equilibrium, if q is equal to

P-Q1427-Solution

Ans [B]Suppose in the following figure, equilibrium of charge B is considered

At equilibrium net force is zero

P-Q1428

P-Q1428-Solution

Relation btw Electric potential and electric field

Ans [A]

P-Q1429

P-Q1429-Solution

Since potential inside the hollow sphere is same as that on the surface.

Ans [B]

Potential inside is 10V

P-Q1430

P-Q1430-Solution

Ans [D]On the equipotential surface, electric field is normal to the charged

surface (where potential exists) so that no work will be done.

Equipotential Surface à same potential across surface

P-Q1431

P-Q1431-Solution

Electric lines force due to negative charge are radially inward.

Field lines come toward negative charge

Ans [C]

P-Q1432

P-Q1432-Solution

Potential due to all four charges Ans [B]

4

P-Q1433

P-Q1433-Solution

In non-uniform electric field. Intensity is more, where the

lines are more denser

More the lines à charge moreMore denser à electric field

Ans [A]

P-Q1434

P-Q1434-Solution

29

20

1094

1rq

rq

E ´´=´=pe

39

26

9

2

100833.2109

)5.2(103109

-´=´

´´=

´´

=rE

q

q should be less than 2.0833 ´ 10–3. In the given set of options 2 ´ 10–3

is the maximum charge which is smaller than 2.0833 ´ 10–3.

Electric field due to charge at distance r

Ans [B]

P-Q1436

P-Q1436-Solution

Ans [D]

Remember this formula

Charge distribute depends on radius

P-Q940

If a body is released from a great distance from the centre of the earth find the velocity

when it strikes the surface of the earth . Take R=6400km

(A)7.2km/s (B)11.8km/s(C) 7.8km/� (D) 11.2km/s

P-Q940-Solution

Ans [D]

It will collide the earth with escape velocity or 11.2km/s

Above result can also be found by taking initial Total energy zero.

Energy conservation 0 = 12��� − ���� Energy on the surface of earth, when body hits

�� = 2���� = 2��� Putting the value of all constants

= 11.2��/�

P-Q943

(A)

(C)

(B)

(D)

P-Q943-Solution

Ans [C]

Centripetal force is provided by the gravitational force of attraction between two particles

P-Q944

(A)

(C)

(B)

(D)

P-Q944-Solution

Ans [D]

Earth revolving around sun so there is centripetal force

P-Q947

(A)

(C)

(B)

(D)

P-Q947-Solution

Ans [B]

Variation of g with height and

depth respectively

P-Q949

(A)

(C)

(B)

(D)

P-Q949-Solution

Ans [B]

Gravitation on planet

Gravitation on Earth

P-Q952

(A)

(C)

(B)

(D)

P-Q952-Solution

Ans [B]

Gravitation varies with depth like this

Gravitation decreases as we go depth

P-Q954

(A)

(C)

(B)

(D)

P-Q954-Solution

Ans [B]

Given that gravity is only 25% of that on earth

P-Q958

(A)

(C)

(B)

(D)

P-Q958-Solution

Ans [B]

So the weight of the body at this height will become one fourth.

P-Q2635

We consider the radiation emitted by the human body. Which of the following statements

is true

(A) The radiation is emitted only during the day

(B) The radiation is emitted during the summers and absorbed during the winters

(C) The radiation emitted lies in the ultraviolet region and hence is not visible

(D) The radiation emitted is in the infra-red region

P-Q2635-Solution

Ans [D]The radiation emitted by the human body is in the infra-red region.

Every body at all time, at all temperatures emits radiation (except at T = 0 K)

T = 0 K is known as absolute zero temperature

P-Q2636

Good absorbers of heat are :

(A) Poor emitters (B) Non-emitters

(C) Good emitters (D) Highly polished

P-Q2636-Solution

Ans [C]Good absorbers are always good emitters of heat

Sand is a good absorber of heat but also good emitter of heat that is why Days are hotter

and nights are colder in deserts as compare to the other places

P-Q2656

Liquid is filled in a vessel which is kept in a room with temperature 20°C . When the

temperature of the liquid is 80°C, then it loses heat at the rate of 60 cal/sec. . What will

be the rate of loss of heat when the temperature of the liquid is 40°C.

(A) 180 cal/sec (B) 40 cal/sec

(C) 30 cal/sec (D) 20 cal/sec

P-Q2656-Solution

Ans [D]

Rate of loss of heat ∆�� temperature difference Dq

P-Q2658

A body takes 5 minute to cool from 80°C to 50°C. How much time it will take to cool from 60°C

to 30°C, if room temperature is 20°C.

(A) 40 minute (B) 9 minute

(C) 30 minute (D) 20 minute

P-Q2658-Solution

Ans [B]

Dividing equation (i) by (ii) we

get

According to Newton’s law of cooling.