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JEE Mains Physics Handwritten Notes PDF for Revision 2023

   Published Date: 20th September 2022  Leave a Comment on JEE Mains Physics Handwritten Notes PDF for Revision 2023
JEE Mains Physics Handwritten Notes PDF for Revision 2023

JEE Mains Physics Handwritten Notes PDF for Revision 2023

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JEE Mains Physics Handwritten Notes PDF for Revision 2023| Free PDFs of Topper’s Notes . If you are a student of class 11 or 12th and want to crack JEE exam, then given Physics handwritten notes are the best resources to do this. As we know that Physics is the basic subjects in JEE entrance exam which is held by NTA National test agency JEE (UG), or National Eligibility cum Entrance Test (Graduation). JEE exam is used for admission in IIT and NIT collages as well as alternative or traditional medicine and nursing in India. Download JEE Mains Physics Handwritten Notes PDF for Revision 2023 | Free PDFs of Topper’s Notes from the given links of Google drive.

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JEE Mains Physics Handwritten Notes PDF for Revision 2023 he subject of Physics will help you in preparing for your final exams. Your final exams in Physics will be held in March or April. These Physics pdf notes are short, neat and clean, colored, regularly updated, and according to the syllabus set by the CBSE board following NCERT guidelines. The notes are very useful for competitive exams like JEE, IIT, and ACT (American College Testing), etc…. Not only our Physics short notes are suitable for CBSE board students, but also for other state board students. If you want to get high marks in JEE Mains Physics Handwritten Notes then you must study from these notes. To download JEE Mains Physics Handwritten notes please read this article till the end.

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JEE Main Past Year Questions With Solutions on Kinematics

Q1: A piece of wood of mass 0.03 kg is dropped from the top of a 100 m height building. At the same time, a bullet of mass 0.02 Kg is fired vertically upward, with a velocity 100 m/s, from the ground. The bullet gets embedded in the wood. Then the maximum height to which the combined system reaches above the top of the building before falling below is (g = 10 m/s)

10 m
30 m
20 m
40 m
Solution

Suppose both collide at the point P after time t. Time taken for the particles to collide,

t = d/vrel = 100/100 = 1s

Speed of wood just before collision =gt = 10m/s

Speed of bullet just before collision

v -gt = 100 -10 = 90 m/s

Before

0.03 kg ↓ 10 m/s

0.02 kg ↑ 90 m/s

After

↑ v

0.05 kg

Now, the conservation of linear momentum just before and after the collision

-(0.03)(10) + (0.02)(90) = (0.05)v ⇒ v = 30 m/s

The maximum height reached by the body a = v2/2g

= (30)2/2(10)

= 45 m

(100 -h) = ½ gt2 = ½ x 10 x1 ⇒h = 95 m

Height above tower = 40 m

Answer: (d) 40 m

Q2: A passenger train of length 60 m travels at a speed of 80 km/hr. Another freight train of length 120 m travels at a speed of 30 km/hr. The ratio of times taken by the passenger train to completely cross the freight train when: (i) they are moving in the same direction, and (ii) in the opposite direction, is

(a) 25/11

(b) 3/2

(c) 5/2

(d) 11/5

Solution:

The total distance to be travelled by train is 60 + 120 = 180 m.

When the trains are moving in the same direction, the relative velocity is v1 – v2 = 80 – 30 = 50 km hr–1.

So time taken to cross each other, t1 = 180/(50 x 103/3600) = [(18 x 18)/25] s

When the trains are moving in opposite direction, relative velocity is |v1 – (–v2 )| = 80 + 30 = 110 km hr–1

So time taken to cross each other

t2= 180/(110 x 103/3600) = [(18 x 36)/110] s

t1/t2= [(18 x 18)/25] / [(18 x 36)/110] = 11/5

Answers: (d) 11/5

Q3: A particle has an initial velocity

and an acceleration of
. Its speed after 10s is
7 units
8.5 units
10 units
7√2 units
Solution:

v = u + at

v = 3i + 4j + (0.4 i + 0.3 j) x 10

= 3i + 4j + 4i + 3i

v = 7i + 7j

Answer: (d) 7√2 units

Q4: An automobile travelling at 40 km/h can be stopped at a distance of 40 m by applying brakes. If the same automobile is travelling at 80 km/h, the minimum stopping distance, in metres, is (assume no skidding)

(a) 100 m

(b) 75 m

(c) 160 m

(d) 150 m

Solution :

Using v2 = u2 – 2as

0 = u2 – 2as

S = u2 /2a

S1/S2 = u12/u22

S2 = (u12/u22)S1 = (2)2(40) = 160 m

Answer: (c)160 m

Q5: A body is thrown vertically upwards. Which one of the following graphs correctly represents the velocity vs time?

Velocity time graph

Solution:

Velocity at any time t is given by

v = u + at

v = v(0) + (-gt)

v = -gt

Straight line with negative slope

Answer: (1) is the correct answer.

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JEE Main Past Year Questions With Solutions on Kinematics

Q6: All the graphs below are intended to represent the same motion. One of them does it incorrectly. Pick it up.

Motion graph

Solution: In this question option (2) and (4) are the corresponding position-time graph and velocity – position graph of the option (3) and its distance-time graph is given as

Distance Time graph

Hence, the answer is given as (1)

Answer: (1)

Q7. A parachutist after bailing out falls 50 m without friction. When the parachute opens, it decelerates at 2 m/s2. He reaches the ground with a speed of 3 m/s. At what height, did he bailout?

(a) 293 m

(b) 111 m

(c) 91 m

(d) 182 m

Solution:

Initially, the parachutist falls under gravity u 2 = 2ah = 2 × 9.8 × 50 = 980 m2s –2

He reaches the ground with speed = 3 m/s,

a = –2 m s–2 ⇒ (3)2 = u 2 – 2 × 2 × h1

9 = 980 – 4 h1

h1 = 971/4

h1 = 242.75 m

Total height = 50 + 242.75 = 292.75 = 293 m.

Q8: A car, starting from rest, accelerates at the rate f through a distance s, then continues at a constant speed for time t and then decelerates at the rate f/2 to come to rest. If the total distance traversed in 15 s, then

(a) s= ½ft2

(b) s= (¼)ft2

(c) s = ft

(d) s= (1/72)ft2

Solution:

For the first part of the journey, s = s1,

s1 = ½ ft12………………………(1)

v = f t1 …………………………(2)

For second part of journey,

s2 = vt or s2 = f t1 t ……………(3)

For the third part of the journey,

s3 = ½(f/2)(2t1)2= ½ x ft12

s3 = 2s1 = 2s ………………….(4)

s1 + s2 + s3 =15s

Or s + ft1t + 2s = 15s

ft1t = 12 s

From (1) and (5) we get

(s/12 s )= ft12/(2 x ft1t)

Or t1 = t/6

Or s= ½ ft12

s= ½ f(t/6)2

s= ft2/72

Answer: (d) s= (1/72)ft2

Q9: An automobile travelling with a speed of 60 km/h, can brake to stop within a distance of 20 m. If the car is going twice as fast, i.e., 120 km/h, the stopping distance will be

(a) 20 m

(b) 40 m

(c) 60 m

(d) 80 m

Solution:

Let a be the retardation for both the vehicles.

For automobile, v 2 = u 2 – 2as

u1 2 – 2as1 = 0

u1 2 = 2as1

Similarly for car, u2 2 = 2as2

(u2/u1)2 = s2/s1 = (120/60)2 = s2/20

S2 = 80 m

Answer: (d) 80 m

Q10: A ball is released from the top of a tower of height h metre. It takes T second to reach the ground. What is the position of the ball in T/3 second?

(a) (h/9)metre from the ground

(b) (7h/9) metre from the ground

(c) (8h/9) metre from the ground

(d) (17h/18) metre from the ground

Solution:

Equation of motion

s= ut + gt2

h = 0 + ½ gT2

Or 2h = gT2………(1)

After T/3 sec, s = 0 +½ x g(T/3)2= gT2/18

18 s = gT2 …………(2)

From (1) and (2), 18 s = 2h

S = (h/9) m from top.

Height from ground = h – h/9 = (8h/9) m

Answer: (c) (8h/9) metre from the ground

Q11: Which of the following statements is false for a particle moving in a circle with a constant angular speed?

(a) The velocity vector is tangent to the circle

(b) The acceleration vector is tangent to the circle

(c) The acceleration vector points to the centre of the circle

(d) The velocity and acceleration vectors are perpendicular to each other

Answer: (b) The acceleration vector acts along the radius of the circle. The given statement is false.

Q12: From a building, two balls A and B are thrown such that A is thrown upwards and B downwards (both vertically). If vA and vB are their respective velocities on reaching the ground, then

(a) vB > vA

(b) vA = vB

(c) vA > vB

(d) their velocities depend on their masses

Solution

Ball A projected upwards with velocity u, falls back with velocity u downwards. It completes its journey to the ground under gravity.

vA 2 = u2 + 2gh …(1)

Ball B starts with downwards velocity u and reaches ground after travelling a vertical distance h vB 2 = u2 + 2gh …(2)

From (1) and (2)

vA = vB

Answer: (b) vA = vB

Q13: If a body loses half of its velocity on penetrating 3 cm in a wooden block, then how much will it penetrate more before coming to rest?

(a) 1 cm

(b) 2 cm

(c) 3 cm

(d) 4 cm

Solution

For first part of penetration, by equation of motion,

(u/2)2 = u2 -2a(3)

3u2 = 24a ⇒ u2 = 8a …(1)

For latter part of penetration,

0= (u/2)2 -2ax

or u2 = 8ax……………(2)

From (1) and (2)

8ax = 8a x = 1 cm

Answer: (a) 1 cm

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