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Published on: 13/05/2022
QB365 provides detailed and simple solution for every Book back Questions in class 12 Physics Subject. It will helps to get more idea about question pattern in every book back questions with solution.
latest Book back QuestionsDownload Tamil Nadu 12th Standard Physics question papers, model tests, one-mark questions, important questions, and public exam papers in PDF format. Free study materials and answer keys for TN State Board students.
Questions + Answers key
Take MCQ Physics Test1.
In a large building, there are 15 bulbs of 40 W, 5 bulbs of 100 W, 5 fans of 80 W and 1 heater of 1 kW are connected. The voltage of electric mains is 220 V. The maximum capacity of the main fuse of the building will be ______.
14 A
8 A
10 A
12 A
2.
A wire connected to a power supply of 230 V has power dissipation P1. Suppose the wire is cut into two equal pieces and connected parallel to the same power supply. In this case power dissipation is P2. The ratio \(\frac{P_2}{P_1}\) is ______.
1
2
3
4
3.
What is the value of resistance of the following resistor?
100 k Ω
10 k Ω
1 k Ω
1000 k Ω
4.
A toaster operating at 240 V has a resistance of 120 Ω. The power is ______.
400 W
2 W
480 W
240 W
5.
The following graph shows current versus voltage values of some unknown conductor. What is the resistance of this conductor?
2 ohm
4 ohm
8 ohm
1 ohm
1.
Total power = 15 x 40 + 5 x 100 + 5 x 80 + 1000
= 600 + 500 + 400 + 1000
= 2500 W
P = VI, V = 220 V
\(I=\frac{P}{V}=\frac{2500}{220}=11.363 \ A\)
≃ 12 A
2.
\(\mathrm{V}=230 \mathrm{~V} \)
\(P=\frac{V^2}{R} \text { since } \mathrm{V} \text { is same } \mathrm{P} \propto \frac{1}{R} \)
\(\frac{1}{R_2}=\frac{1}{\frac{R_1}{2}}+\frac{1}{\frac{R_1}{2}}=\frac{2}{R_1}+\frac{2}{R_1}=\frac{2+2}{R_1} \)
\(\frac{1}{R_2}=\frac{4}{R_1} \)
\(\therefore R_2=\frac{R_1}{4} \)
\(R_1=4 R_2 \)
\(\therefore \frac{P_2}{P_1}=\frac{R_1}{R_2}=\frac{4 R_2}{R_2}=4\)
3.
Brown - 1
Black - 0
Yellow - 104
(∴ R = 10 x 104 Ω = 100 kΩ)
4.
\(P=\frac{V^2}{R}=\frac{240 \times 240}{120}=480 \ W\)
5.
Resistance, \(R=\frac{V}{I}=\frac{4}{2}=2 \ ohm\)
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