Here are **5 essential numerical problems** from the Light and Electricity chapters of Class 10 Physics, broken down step-by-step using standard formulas.
### **Light: Reflection and Refraction**
#### **Problem 1: Concave Mirror (Mirror Formula)**
An object 2\text{ cm} high is placed at a distance of 15\text{ cm} in front of a concave mirror of focal length 10\text{ cm}. Find the position, size, and nature of the image.
**Solution:**
* **Given:**
* Height of object (h_o) = +2\text{ cm}
* Object distance (u) = -15\text{ cm} *(by sign convention)*
* Focal length (f) = -10\text{ cm} *(concave mirror)*
* **Step 1: Find Image Distance (v) using Mirror Formula**
* **Step 2: Find Image Height (h_i) using Magnification**
* **Conclusion:**
* **Position:** 30\text{ cm} in front of the mirror.
* **Nature:** Real and inverted (indicated by negative sign in h_i and v).
* **Size:** Magnified (4\text{ cm} tall).
#### **Problem 2: Convex Lens & Lens Power**
A convex lens forms a real and inverted image of a needle at a distance of 50\text{ cm} from it. Where is the needle placed in front of the convex lens if the image is equal to the size of the object? Also, find the power of the lens.
**Solution:**
* **Given:**
* Image distance (v) = +50\text{ cm} *(real image formed by convex lens)*
* Magnification (m) = -1 *(since image is real, inverted, and equal in size to object)*
* **Step 1: Find Object Distance (u)**
* **Step 2: Find Focal Length (f) using Lens Formula**
* **Step 3: Calculate Power (P)**
* **Conclusion:** The needle is placed at **50\text{ cm}** in front of the lens, and the power of the lens is **+4\text{ D}**.
### **Electricity**
#### **Problem 3: Equivalent Resistance & Circuit Current**
Three resistors of 5\ \Omega, 10\ \Omega, and 30\ \Omega are connected in parallel across a 12\text{ V} battery. Calculate:
1. Total effective resistance of the circuit.
2. Total current flowing in the circuit.
**Solution:**
* **Given:** R_1 = 5\ \Omega, R_2 = 10\ \Omega, R_3 = 30\ \Omega, V = 12\text{ V}
* **Step 1: Total Parallel Resistance (R_p)**
* **Step 2: Calculate Total Current (I) using Ohm's Law**
* **Conclusion:** Total effective resistance is **3\ \Omega** and the total current is **4\text{ A}**.
#### **Problem 4: Resistivity Formula**
A copper wire of length 2\text{ m} has a cross-sectional area of 1.7 \times 10^{-6}\text{ m}^2. If the resistivity of copper is 1.7 \times 10^{-8}\ \Omega\cdot\text{m}, calculate the resistance of the wire.
**Solution:**
* **Given:**
* Length (l) = 2\text{ m}
* Area (A) = 1.7 \times 10^{-6}\text{ m}^2
* Resistivity (\rho) = 1.7 \times 10^{-8}\ \Omega\cdot\text{m}
* **Step 1: Apply Formula R = \rho \frac{l}{A}**
* **Conclusion:** The resistance of the wire is **0.02\ \Omega** (or 2 \times 10^{-2}\ \Omega).
#### **Problem 5: Commercial Unit of Electric Energy (Joule's Heating)**
An electric refrigerator rated 400\text{ W} operates 8 hours/day. What is the cost of the energy to operate it for 30 days at ₹ 3.00 per \text{kWh}?
**Solution:**
* **Given:**
* Power (P) = 400\text{ W} = 0.4\text{ kW}
* Time per day (t_1) = 8\text{ hours}
* Total duration (D) = 30\text{ days}
* Rate = ₹ 3.00 per unit (\text{kWh})
* **Step 1: Calculate Total Hours (t)**
* **Step 2: Calculate Total Energy Consumed (E)**
* **Step 3: Calculate Total Cost**
* **Conclusion:** The cost of energy for 30 days is **₹ 288**.
