IIT JEE, NEET Physics Problems in Gravitation and Ray Optics concepts And Their Solutions with Detailed Step by Step Explanation.
This section includes many important IIT JEE and NEET Physics problems in in Gravitation and Ray Optics topics that are considered to be very important from IITJEE and NEET exams point of view. Here the solutions for these problems are very detailed with explanation for each step. This section is very helpful for students to get very high scores in IIT JEE and NEET exams.
Gravitation
🟩 Problem 1:
Earth has mass 8 times and radius 2 times that of a planet. If the escape velocity from the Earth is 11.2 km/s, the escape velocity (in km/s) from the planet will be:
(1) 11.2  (2) 5.6  (3) 2.8  (4) 8.4
✅ Solution:
Step 1: Escape Velocity Formula
The escape velocity from a planet is given by:
$$ v_e = \sqrt{\frac{2GM}{R}} $$
- ve = escape velocity
- G = universal gravitational constant
- M = mass of the planet
- R = radius of the planet
Step 2: What’s Given?
- Earth’s escape velocity: \( v_E = 11.2 \text{ km/s} \)
- Earth’s mass: \( M_E = 8M_P \Rightarrow M_P = \frac{M_E}{8} \)
- Earth’s radius: \( R_E = 2R_P \Rightarrow R_P = \frac{R_E}{2} \)
- Mp = Mass of Planet
- Rp = Radius of Planet
Step 3: Escape Velocity of Earth and Planet
Escape velocity of Earth:
$$ v_E = \sqrt{\frac{2GM_E}{R_E}} $$
Escape velocity of Planet:
$$ v_P = \sqrt{\frac{2GM_P}{R_P}} = \sqrt{\frac{2G \cdot \frac{M_E}{8}}{\frac{R_E}{2}}} $$
Step 4: Simplify
$$ v_P = \sqrt{\frac{2G M_E}{8 \cdot \frac{R_E}{2}}} = \sqrt{\frac{2G M_E}{4 R_E}} = \sqrt{\frac{1}{4} \cdot \frac{2G M_E}{R_E}} = \frac{1}{2} \cdot v_E $$
Step 5: Final Calculation
$$ v_P = \frac{1}{2} \cdot 11.2 = \boxed{5.6 \text{ km/s}} $$
✅ Final Answer: (2) 5.6 km/s
Ray Optics
🟩 Problem 1:
A concave mirror produces an image of an object such that the distance between the object and image is 20 cm. If the magnification of the image is -3, then the magnitude of the radius of curvature of the mirror is:
Options:
(1) 3.75 cm (2) 30 cm (3) 7.5 cm (4) 15 cm
✅ Solution:
Step 1: Use magnification formula
Magnification for mirrors is given by:
$$ m = \frac{-v}{u} $$
- \( m \) = magnification
- \( v \) = image distance (from mirror pole)
- \( u \) = object distance (negative for real object)
Given: \( m = -3 \Rightarrow -3 = \frac{-v}{u} \Rightarrow v = 3u \tag{1} \)
Step 2: Use object-image distance
Given: \( |v – u| = 20 \text{ cm} \tag{2} \)
From (1): \( v = 3u \), substitute into (2):
$$ |3u – u| = 20 \Rightarrow 2u = 20 \Rightarrow u = \pm10 $$
Since object is real: \( u = -10 \text{ cm} \), so \( v = 3u = -30 \text{ cm} \)
Step 3: Use mirror formula
$$ \frac{1}{f} = \frac{1}{v} + \frac{1}{u} $$
Substitute: $$ \frac{1}{f} = \frac{1}{-30} + \frac{1}{-10} = -\left( \frac{1}{30} + \frac{1}{10} \right) = -\frac{4}{30} \Rightarrow f = -7.5 \text{ cm} $$
Step 4: Find Radius of Curvature
$$ R = 2f = 2 \times (-7.5) = -15 \text{ cm} $$
Magnitude: $$ |R| = \boxed{15 \text{ cm}} $$