A car moves from X to Y with a uniform speed $v_u$ and returns to Y with a uniform speed $v_d$. The average speed for this round trip is

 Problem 5. A car moves from X to Y with a uniform speed $v_u$ and returns to Y with a uniform speed $v_d$. The average speed for this round trip is [NEET 2007]

(a) $\sqrt{v_u v_d}$

(b) $\frac{v_d v_u}{v_d + v_u}$

(c) $\frac{v_d + v_u}{2}$

(d) $\frac{2 v_d v_u}{v_d + v_u}$



If a car at rest accelerates uniformly to a speed of 144 km/h in 20 s, it covers a distance of

 Problem 4. If a car at rest accelerates uniformly to a speed of 144 km/h in 20 s, it covers a distance of [NEET 1997]

(a) 2880 m

(b) 1440 m

(c) 400 m 

(d) 20 m












A bus travelling the first one third distance at a speed of 10 km/h, the next one third at 20 km/ h and the last one-third at 60 km/h. The average speed of the bus is

Problem 3. A bus travelling the first one third distance at a speed of 10 km/h, the next one third at 20 km/ h and the last one-third at 60 km/h. The average speed of the bus is [NEET 1991] 

(a) 9 km/h

(b) 16 km/h

(c) 18 km/h

(d) 48 km/h



A car moves a distance of 200 m. It covers the first half of the distance at speed 40 km/h and the second half of distance at speed v. The average speed is 48 km/h. Find the value of v

 Problem 2. A car moves a distance of 200 m. It covers the first half of the distance at speed 40 km/h and the second half of distance at speed v. The average speed is 48 km/h. Find the value of v [NEET 1991]

(a) 56 km/h

(b) 60 km/h

(c) 50 km/h

(d) 48 km/h









A car covers the first half of the distance between two places at 40 km/h and other half at 60 km/h. The average speed of the car is [NEET 1990]

Problem 1. A car covers the first half of the distance between two places at 40 km/h and other half at 60 km/h. The average speed of the car is [NEET 1990]

(a) 40 km/h

(b) 48 km/h

(c) 50 km/h

(d) 60 km/h





Coulomb’s law for electrostatic force between two point charges and Newton’s law for gravitational force between two stationary point masses, both have inverse-square dependence on the distance between the charges and masses respectively. (a) Compare the strength of these forces by determining the ratio of their magnitudes (i) for an electron and a proton and (ii) for two protons. (b) Estimate the accelerations of electron and proton due to the electrical force of their mutual attraction when they are 1 Ã… (= $10^{-10}$ m) apart? ($m_p$ = 1.67 × $10^{–27}$ kg, $m_e$ = 9.11 × $10^{–31}$ kg)

NCERT Example 1.3 Coulomb’s law for electrostatic force between two point charges and Newton’s law for gravitational force between two stationary point masses, both have inverse-square dependence on the distance between the charges and masses respectively. (a) Compare the strength of these forces by determining the ratio of their magnitudes (i) for an electron and a proton and (ii) for two protons. (b) Estimate the accelerations of electron and proton due to the electrical force of their mutual attraction when they are 1 Ã… (= $10^{-10}$ m) apart? ($m_p$ = 1.67 × $10^{–27}$ kg, $m_e$ = 9.11 × $10^{–31}$ kg)

If $10^9$ electrons move out of a body to another body every second, how much time is required to get a total charge of 1 C on the other body?

NCERT  Example 1.1 If $10^9$ electrons move out of a body to another body every second, how much time is required to get a total charge of 1 C on the other body? 

How much positive and negative charge is there in a cup of water?

NCERT Example 1.2 How much positive and negative charge is there in a cup of water?

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NEET Past Year Questions Practice 






To simulate car accidents, auto manufacturers study the collisions of moving cars with mounted springs of different spring constants. Consider a typical simulation with a car of mass 1000 kg moving with a speed 18.0 km/h on a smooth road and colliding with a horizontally mounted spring of spring constant 5.25 × 10^3 N m–1. What is the maximum compression of the spring ? Example 5.9 Consider Example 5.8 taking the coefficient of friction, µ, to be 0.5 and calculate the maximum compression of the spring.

Example 5.8 To simulate car accidents, auto manufacturers study the collisions of moving cars with mounted springs of different spring constants. Consider a typical simulation with a car of mass 1000 kg moving with a speed 18.0 km/h on a smooth road and colliding with a horizontally mounted spring of spring constant $5.25\times 10^3 Nm^{-1}$ . What is the maximum compression of the spring ? [NCERT Class 11 Example 5.8]


Example 5.9 Consider Example 5.8 taking the coefficient of friction, µ, to be 0.5 and calculate the maximum compression of the spring. [
NCERT Class 11 Example 5.9]

$5.25\times 10^3 Nm^{-1}$ corrected as $6.25\times 10^3 Nm^{-1}$

Three concentric spherical shells have radii a, b and c (a < b < c) and have surface charge densities σ,−σ and σ respectively. If VA, VB and VC denotes the potentials of the three shells, then for c = a + b, we have [NEET 2009]

Three concentric spherical shells have radii a, b and c (a < b < c) and have surface charge densities σ,−σ and σ respectively. If VA, VB and VC denotes the potentials of the three shells, then for c = a + b, we have [NEET 2009]

a) $V_C=V_B\neq V_A$
b) $V_C\neq V_B\neq V_A$
c) $V_C=V_B=V_A$
d) $V_C=V_A\neq V_B$

Concept: Use Van de graph principles. In concentric spheres, the voltage due to the charges on the surface and the voltage due to the inner spheres and outer spheres should be added. $V_{inside\hspace{2mm}sphere}= V_{at\hspace{2mm}surface\hspace{2mm}of\hspace{2mm}sphere}$
\[V_A=\frac{1}{4\pi\epsilon_o}\frac{q_A}{r_A}-\frac{1}{4\pi\epsilon_o}\frac{q_B}{r_B}+\frac{1}{4\pi\epsilon_o}\frac{q_C}{r_C}\]
\[=K\frac{\sigma 4\pi a^2}{a}-K\frac{\sigma 4\pi b^2}{b}+K\frac{\sigma 4\pi c^2}{c}\]
\[=K\sigma 4\pi \hspace{2mm}(a-b+c)\]
\[=K\sigma 4\pi \hspace{2mm}(a-b+a+b)\]
\[=K \sigma 4\pi \hspace{2mm} (2a) \]

A man of 50 kg mass is standing in a gravity free space at a height of 10 m above the floor. He throws a stone of 0.5 kg mass downwards with a speed of 2 m/s. When the stone reaches the floor the distance of the man above the floor will be

A man of 50 kg mass is standing in a gravity free space at a height of 10 m above the floor. He throws a stone of 0.5 kg mass downwards with a speed of 2 m/s. When the stone reaches the floor the distance of the man above the floor will be

A) 9.9 m   B) 10.1 m  C) 10 m   D) 20 m

Concepts: Conservation of Linear momentum. Similar to the recoiling of gun. When the man pushes the stone downward the man will get pushed upward. Since there is no external force acting on the system, conservation of linear momentum is valid and center of mass of the whole system remains in the same location.

12P01 Electric Charges and Fields

 Step-by-step


Step 1: Theory and Concepts

Step 2: Numerical Practice
NCERT Examples, NCERT Exercises with solutions

Step 3: Numerical Practice
Chapterwise CBSE Past papers (Any one of them- Arihant, MTG, XAM Idea etc.)

Step 4: Numerical Practice

Step 5: Numerical Practice
JEE Main past year Questions, Solutions



Two charges and Forces between them.
Plot F vs. r graph to see the shape of the graph.


How to solve Numerical problem in physics?

 

How to solve Numerical problem in physics?

“Start walking. The path will become clearer”


 Step 1: Translating from English language to Maths and Drawing.

·        Nature is written in the language of Mathematics. Drawing is the language of Engineers and Scientists.

·        Read the question line by line converting it into a simple cartoon diagram or a mathematical expression. Mark the numerical values in appropriate places in the diagram or with the mathematical notations.

 Step 2: Understand what is given and what is to be found.