The curriculum for all schools in the country that follow the Central Board of Secondary Education (CBSE) is set by the National Council of Education Research and Training (NCERT). These NCERT Solutions are structured so that students learn to check the marks allotted to a question before correctly solving it.

Students would benefit from using NCERT Solutions for **CBSE Class 12** Physics to improve their problem-solving skills. These NCERT Class 12 Solutions provide detailed, step-by-step explanations of textbook problems. It would also aid students in preparing for undergraduate engineering entrance exams such as VITEEE, **JEE Main 2022, **and others. The **JEE Main** Physics syllabus is available on the Aakash website for students.

Every budding engineer aims to be one of the top **JEE Main** test scorers. Due to the vast **JEE Main **syllabus, many students find it difficult to cover concepts from all three disciplines. Physics is a tough subject for many students due to the numerical and theoretical concepts. However, it is a high-scoring subject.

Table of Contents |

Important Solved Questions for JEE Main 2022 Physics Examination |

Conclusion |

Frequently Asked Questions about NCERT Physics Revision Notes |

**Important Solved Questions for JEE Main 2022 Physics Examination**

**Q1. Two bodies, each of mass M, are kept fixed with a separation 2L. A particle of mass m is projected from the midpoint of the line joining their centres, perpendicular to the line. The gravitational constant is G. The correct statement is**

Options:

- The minimum initial velocity of the mass m to escape the gravitational field of the two bodies is 4√GML
- The minimum initial velocity of the mass m to escape the gravitational field of the two bodies is 2√GML
- The minimum initial velocity of the mass m to escape the gravitational field of the two bodies is √2GML
- The energy of the mass m is not constant.

Ans.

Explanation:

Let v is the minimum velocity.

From the energy conservation rule:

-2GMmL+12mv2= 0

v = 2√GML

Hence, option (b) is correct.

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**Q2. When a 10V potential difference is applied across a wire of length 0.2 m, the drift speed of electrons is 5 ×10****−4**** ms****−1****. If the electron density in the wire is 8 × 10****28****m****−3****, the resistivity of the material is close to :**

Options:

(a) 0.8×10−8 Ω m

(b) 0.8×10−7 Ω m

(c) 0.8×10−6 Ω m

(d) 0.8×10−5 Ω m

Ans.

Option (d) is correct

Explanation:

The current density across a wire is

J = nevd

IA= nevd

I = neAvd

where,

vd is the drift speed

and R =ρlA

By Ohm’s law

ρ = Vnelvd

108 × 10 28 × 1.6 × 10 -19 × 0.2 × 5 × 10-4

= 0.8 × 10-5 Ω m

Hence, option (d) is correct.

**Q3. Two identical capacitors each of capacitance C are charged to the same potential V and are connected in two circuits (i) and (ii) at t = 0 as shown. The charged on the capacitor at t = CR are**

- CVe, CVe
- CV , CV
- VCe, VC
- VC,VCe

In Fig. (i) the p-n junction diode is forward biassed and represents a very low resistance, the capacitor, therefore discharges itself through resistor R according to relation.

q = q0e -t/CR

and q0 = CV at t = CR

Therefore,

q = q0e-1 =CVe

In Fig. (ii), the p-n junction diode is reverse biassed, the capacitor. Therefore, hold the charge intact.

q = q0 =CV

Hence, option (c) is correct.

**Q4. A particle of mass 0.5 kg is moving in one dimension under a force that delivers a constant power 1 W to the particle. If the initial speed (in m/s) of the particle is zero, the speed (in m/s) after 9 s is: **

Options:

(a) 8

(b) 7

(c) 6

(d) 5

Answer: Power = dWdt

Total work done in 9 s is

W = 1 x 9 = 9 = KEf – KEi

Since the initial speed of the particle is zero, its initial kinetic energy will also be zero.

9 = M2 ( vf2 – vi 2 )

9 = M2 ( vf2 – 0 )

vf2 = 9 x 20.5= 36

vf = 6

Hence, option (c) is correct.

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**Q5. Bob of mass m, suspended by a string of length l****1**** is given a minimum velocity required to complete a full circle in the vertical plane. At the highest point, it collides elastically with another bob of mass m suspended by a string of length l****2****, which is initially at rest. Both the strings are massless and inextensible. If the second bob, after collision acquires the minimum speed required to complete a full circle in the vertical plane, the ratio l****1****/l****2 ****is:**

(a) 1 : 5

(b) 5 : 1

(c) 1 : 4

(d) 4 : 1

The minimum velocity required to complete a full circle in the vertical plane by a

bob of mass m, suspended by a string of length l = √5gl

Also, the speed at highest point is given as √gl

The initial speed of 1st bob (suspended by a string of length l1) is √5gl1

The speed of this bob at highest point will be √gl1

When this bob collides with the other bob there speeds will be interchanged.

√gl1 = √5gl2

l1/l2 = 5/1

Hence, option (b) is correct.

**Q6.** **A diatomic molecule is formed by two atoms which may be treated as mass points m1 and m2 joined by a massless rod of length r. Then, the moment of inertia of the molecule about an axis passing through the centre of mass and perpendicular to rod is:**

Options:

- Zero
- (m1 + m2) r2
- (m1+ m2m1m2) r2
- (m1m2m1+m2) r2

Ans.

Option (d) is correct

Let the centre of mass be situated at distance x from the atom of mass m1 and at distance (r − x) from mass m2.

By definition, we have, m1x = m2 (r−x)

x = (m2m1+m2) r2 … 1

Now, moment of inertia about the centre of mass is

l = m1x2 + m2 (r−x) 2

Putting the value of x from equation (1)

I = m1 (m2rm1m2)2 + m2 (r-m2rm1 +m2) 2

= m1 (m2rm1m2)2 + m2 (m1rm1 +m2) 2

= (1m1 +m22) (m1m22r2 + m2m12r2)

= (m1m2r2m1 +m22) (m2 + m1)

I = (m1m2r2m1 +m2)

Hence, option (d) is correct.

**Conclusion**

The Physics syllabus is vital during the preparation and final revision phases. Students should be meticulous in their revision strategy with only a few weeks until the **JEE Main Examination 2022**. To avoid wasting their time and efforts, pupils must rigorously adhere to the Physics syllabus. For every entrance test, NCERT books are essential because they cover all the fundamental concepts. JEE experts also recommend the NCERT solutions. You can study from NCERT and reference books to improve your JEE Main preparation and score a high rank!

Students can access the previous 14 years’ question papers on the Aakash website and detailed solutions.

**FAQs**

**1.The picture of an object created by a plano-convex lens at an 8-metre distance behind the lens is actually one-third the size of the item. The light inside the lens has a wavelength that is 2/3 that of light in open space. The lens’s curved surface has a radius of**

- 3m
- 10m
- 7m
- 9m

- Given that- positive of image, v = +8

magnification, m = 1/3

m = v/u = 1/3

8/u = 1/3

u = 8 * 3

u = -24

**μ = **λ air / λ medium

λ air = 1 and λ medium = ⅔

So, **μ = 3/2**

According to the lens formula, **1/****f**** = 1/****v**** + 1/****u**** **

where , **1/****f**** = (u – 1) (1/R – 1/****∞)**

Substituting all the values we get **R = 3m. **Therefore the correct answer is (a) R = 3m.

**2. The volume of a gas becomes four times if **

- Temperature becomes four times at constant pressure
- Temperature becomes one fourth at constant pressure
- Temperature becomes two times at constant pressure
- Temperature becomes half at constant pressur
**e**

Hence, the correct answer is (a) Temperature becomes four times at constant pressure.

**3. A wire’s resistance at 20°C room temperature is found to be 20W. Now, in order to increase the resistance by 20%, the wire must be heated to?**

[The temperature coefficient of resistance of the wire material is 0.002 per °C].

(a) 176°C

(b) 124°C

(c) 142°C

(d) 133°C

- R1 = R0 (1 + αt)

At the beginning R0 (1 + 20α) = 20Ω

Finally, R0 (1 + αt) = 24

At the beginning R0 (1 + 20α) = 20Ω

Finally, R0 (1 + αt) = 24

Now, 6/5 = (1 + αt) / (1 + 20α)

Substituting all the values we get:

5 + (5 x 0.002 x t) = 6 + 120 x 0.002

0.01t = 1.24

t = 1.24/0.01 = 124°C

Hence, (b) 124°C is the correct answer.

**4. Can students from other boards utilise the NCERT Class 12 Physics Solutions?**

Answer: CBSE and NCERT guidelines are followed by NCERT Solutions for **CBSE Class 12** Physics. Although NCERT books may not follow the same criteria and standards as other boards, students can still get information on the ideas and concepts included in the textbooks. The NCERT Solutions for **CBSE Class 12 **Physics might be useful to students from all boards because the majority of the syllabus in Physics is the same.

Are you looking for a coaching institute near you for revision before your **JEE Main 2022 **examination? Find an Aakash coaching institute near you today!

**5. How much time should you spend studying for JEE Physics?**

The subject of physics covers a wide range of topics. It would be beneficial if you had enough time to cover all the issues completely. If you begin your **JEE Main 2022 **preparation early, you must devote approximately 5 hours every day. However, increasing the preparation time over time would be beneficial. You should devote about 7-8 hours to studying Physics throughout the final revision period.

Is it your dream to study at an IIT? Then make sure you go through the **JEE Main** and Advanced eligibility criteria before sitting for the exam.

**6. How can I score 360 in JEE Main 2022 after studying the important Physics topics?**

Here are some tips to help you achieve a good rank in **JEE Main 2022:**

- Go through the CBSE Physics syllabus.
- The syllabus should be divided based on importance.
- Select the high weightage topics first.
- Based on the chapter’s curriculum, highlight the most important aspects.
- Students should always have a notebook with them for revision on the go.

**7. How valuable are mock exams in improving your JEE Physics score?**

- Students benefit from mock tests as they ensure that students are well-prepared for the expected paper design of the final exam. Attempting additional
**JEE Main**Physics practice examinations would aid candidates in improving their speed and time management abilities.

At Aakash, students can take a **JEE Main **mock test based on the most recent **JEE Main** pattern. Students can compete against other applicants on an all-India level to assess their performance.

**8. Is the NCERT book helpful in learning about physics concepts?**

The NCERT and CBSE curricula are heavily influenced by the JEE Physics syllabus. As a result, using NCERT books will be advantageous. It is recommended that you read through the chapters to ensure that you completely understand the subject. The book’s language is straightforward to comprehend.