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.
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