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TN 12th Computer Applications மின்னணு செலுத்தல் முறைகள் Sample Question Papers Study Material - QB365 Set A

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TN 12th Computer Applications மின் - வணிகம் Sample Question Papers Study Material - QB365 Set A

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TN 12th Computer Applications வலையமைப்பு வடமிடல் Sample Question Papers Study Material - QB365 Set A

12th Standard English Medium Physics Subject Current Electricity Book Back 1 Mark Questions with Solution Part - I Set B

Published on: 01/06/2021

Question Set B

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Std: 12th Standard - Physics

chapter: - Book back Questions

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Questions:

Multiple Choice Question

5 x 1 = 5

A piece of copper and another of germanium are cooled from room temperature to 80 K. The resistance of ______.

(a)

each of them increases

(b)

each of them decreases

(c)

copper increases and germanium decreases

(d)

copper decreases and germanium increases

The temperature coefficient of resistance of a wire is 0.00125 per °C. At 20°C, its resistance is 1 Ω. The resistance of the wire will be 2 Ω at ______.

(a)

800 °C

(b)

700 °C

(c)

850 °C

(d)

820 °C

There is a current of 1.0 A in the circuit shown below. What is the resistance of P ?

(a)

1.5 Ω

(b)

2.5 Ω

(c)

3.5 Ω

(d)

4.5 Ω

In India electricity is supplied for domestic use at 220 V. It is supplied at 110 V in USA. If the resistance of a 60 W bulb for use in India is R, the resistance of a 60 W bulb for use in USA will be ______.

(a)

(b)

(c)

\(\frac{R}{4}\)

(d)

\(\frac{R}{2}\)

Two wires of A and B with circular cross section made up of the same material with equal lengths. Suppose R_A = 3 R_B, then what is the ratio of radius of wire A to that of B?

(a)

(b)

\(\sqrt3\)

(c)

\(\frac{1}{\sqrt3}\)

(d)

\(\frac{1}{3}\)

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Answers Key:

Multiple Choice Question

5 x 1 = 5

Resistivity ∝ temperature for conductor. so, copper → decreases
Resistivity ∝\(\frac{1}{\text {temperature for semiconductor}}\)
so, germanium → increases.

R_t = R_o [1 + α(T₂ - T₁)
2 = 1[1 + 0.00125(T₂ - 293)
I = 5/4 x 10^-3(T₂ - 293)
T₂ = 1093 K
T₂ = 820 ^oC

R_s = 3 + 2.5 + P = 5.5 + P
V = 9 V, I = 1.0 A
R_s = \(\frac{V}{I}=\frac{9}{1}= 9 \Omega\)
∴ 9 = 5.5 + P
∴ P = 9 - 5.5 = 3.5 Ω

\(\mathrm{V}_1 =220 \mathrm{~V}, \quad \mathrm{P}_1=60 \mathrm{~W} \)
\(\mathrm{~V}_{\mathrm{U}} =110 \mathrm{~V}, \mathrm{P}_{\mathrm{U}}=60 \mathrm{~W} \)
\(P =\frac{V^2}{R} \Rightarrow R=\frac{V^2}{P} \)
\(\therefore R_l =\frac{V_I^2}{P_l} \text { Similarly, } \quad \mathrm{R}_U=\frac{V_U^2}{P_U} \)
\(R_l =\frac{220 \times 220}{60} \quad \mathrm{R}_{\mathrm{U}}=\frac{110 \times 110}{60} \)
\(R_I =\frac{48400}{60} \quad R_U=\frac{12100}{60} \)
\(\frac{R_U}{R_l} =\frac{12100}{60} \times \frac{60}{48400}=\frac{1}{4} \)
\(R_U =\frac{R_l}{4}=\frac{R}{4}\)

\(R \propto \frac{1}{A}, R \propto \frac{1}{r^2} \)
\(R_A \propto \frac{1}{r_A^2}, R_B \propto \frac{1}{r_B^2} \)
\(\frac{r_A}{r_B}=\left(\frac{R_B}{R_A}\right)^{1 / 2}=\left(\frac{R_B}{3 R_B}\right)^{1 / 2}=\frac{1}{3^{\frac{1}{2}}}=\frac{1}{\sqrt{3}}\)