## If two bulbs of power 25 W and 100 W respectively each rated at 220 V are connected in series with the supply

Q: If two bulbs of power 25 W and 100 W respectively each rated at 220 V are connected in series with the supply of 440 V. Which bulb will fuse?

(a) 25 W bulb

(b) 100 W bulb

(c) none of these

(d) both

Ans: (a)

Sol:
Resistance of one bulb, R1 = (220)2/25 Ω ;

Resistance of other bulb, R2 = (220)2/100 Ω.

Total resistance when two bulbs are in series,

R = (220)2/25 + (220)2/100 = (220)2/20

current I = V/R = 440/((220)2/20) = 2/11 amp.

Pot. diff. across 25 watt bulb
= IR1 = 2/11 × (220)2/25 = 352 V

Pot. diff. across 100 watt bulb
= IR2 = 2/11 × (220)2/100 = 88 V

Thus, the bulb 25 W will be fused, because it can tolerate only 220 V while the voltage across it is 352 V.

## The length of a Potentiometer wire is 5 metres. An electron in this wire experiences a force…

Q: The length of a Potentiometer wire is 5 metres. An electron in this wire experiences a force of 4.8 × 10-19 N , e.m.f. of the main cell used in Potentiometer is :

(a) 3 volt

(b) 15 volt

(c) 1.5 volt

(d) 5 volt

Ans: (b)

Sol: Force = electric intensity × charge

$\displaystyle Force = \frac{Potential \; difference }{distance} \times charge$

$\displaystyle 4.8 \times 10^{-19} = \frac{V}{5} \times 1.6 \times 10^{-19}$

V = 15 volt

## A and B are two points on a uniform ring of resistance R. The ∠AOB = θ…

Q: A and B are two points on a uniform ring of resistance R. The ∠AOB = θ, where O is the center of the ring. The equivalent resistance between A and B is:

(a) $\displaystyle \frac{R \theta}{2 \pi}$

(b) $\displaystyle \frac{R (2\pi - \theta)}{4 \pi}$

(c) $\displaystyle R (1- \frac{\theta}{2 \pi} )$

(d) $\displaystyle \frac{R}{4 \pi^2} (2\pi -\theta )\theta$

Ans: (d)

Sol: Let r be the radius of the ring.

Resistance per unit length of ring , ρ = R/(2 π r)

Lengths of section making angle θ = r θ

Resistance of this section $\displaystyle R_1 = \rho r \theta = \frac{R}{2 \pi r} r \theta$

$\displaystyle R_1 = \frac{R \theta}{2 \pi }$

Lengths of section making angle (2π -θ) = r (2π -θ)

Resistance of this section $\displaystyle R_2 = \rho r (2\pi -\theta) = \frac{R}{2 \pi r} r (2\pi -\theta)$

$\displaystyle R_2 = \frac{R (2\pi - \theta ) }{2 \pi }$

Here, R1 and R2 are in parallel between A and B,

$\displaystyle R_{eq} = \frac{R_1 R_2 }{R_1 + R_2}$

$\displaystyle R_{eq} = \frac{\frac{R \theta}{2 \pi }\times \frac{R (2\pi - \theta )}{2 \pi} }{\frac{R \theta}{2 \pi } + \frac{R (2\pi - \theta ) }{2 \pi }}$

$\displaystyle = \frac{R}{4 \pi^2} (2\pi -\theta )\theta$

## Resistance of a resistor at temperature t° C is Rt = R0 (1 + αt + βt^2 )…

Q: Resistance of a resistor at temperature t° C is Rt = R0 (1 + αt + βt2 ). Here R0 is the resistance at 0 °C. The temperature coefficient of resistance at temperature t °C is:

(a) $\displaystyle \frac{1+\alpha t + \beta t^2}{\alpha + 2 \beta t}$

(b) $\displaystyle \alpha + 2 \beta t$

(c) $\displaystyle \frac{\alpha + 2 \beta t}{1+\alpha t + \beta t^2}$

(d) $\displaystyle \frac{\alpha + 2 \beta t}{2(1+\alpha t + \beta t^2)}$

Ans: (c)

Sol:
Temperature coefficient of resistance $\displaystyle = \frac{1}{R_t}\frac{dR_t}{dt}$

$\displaystyle = \frac{1}{R_0(1+\alpha t + \beta t^2)}\frac{d}{dt}(R_0(1+\alpha t + \beta t^2))$

$\displaystyle = \frac{\alpha + 2 \beta t}{1+\alpha t + \beta t^2}$

## Masses of the three wires of same material are in the ratio of 1 : 2 : 3 and their lengths in the ratio…

Q: Masses of the three wires of same material are in the ratio of 1 : 2 : 3 and their lengths in the ratio of 3: 2: 1, Electrical resistance of these wires will be in the ratio of:

(a) 1 : 1: 1

(b) 1: 2: 3

(c) 9: 4: 1

(d) 27: 6: 1

Ans: (d)

Sol : Mass, M = volume × density = A l × d

A = M/ld

Resistance $\displaystyle R = \rho \frac{l}{A} = \frac{\rho l}{M/ld}$

$\displaystyle R = \rho \frac{l^2 d}{M}$

$\displaystyle R = \propto \frac{l^2 }{M}$

$\displaystyle R_1 : R_2 : R_3 = \frac{l_1^2 }{M_1} : \frac{l_2^2 }{M_2} : \frac{l_3^2 }{M_3}$

$\displaystyle = \frac{3^2 }{1} : \frac{2^2 }{2} : \frac{l^2 }{3}$

= 27 : 6 : 1