## The current sensitivity of a moving coil galvanometer is 5 div/mA and its voltage sensitivity…

Q: The current sensitivity of a moving coil galvanometer is 5 div/mA and its voltage sensitivity (angular deflection per unit voltage applied) is 20 div/V . The resistance of galvanometer is

(a) 250 Ω

(a) 500 Ω

(a) 25 Ω

(a) 400 Ω

Ans: (a)

Sol: Current sensitivity $\displaystyle I_s = \frac{n B A}{k}$

Voltage sensitivity $\displaystyle V_s = \frac{n B A}{k R}$

$\displaystyle V_s = \frac{I_s}{R}$

$\displaystyle R = \frac{I_s}{V_s}$

$\displaystyle R = \frac{5 \times 10^3}{20}$

R = 250 Ω

## A set of n equal resistors , each of value R are connected in series to a battery of emf E and internal resistance R …

Q: A set of n equal resistors , each of value R are connected in series to a battery of emf E and internal resistance R . The current drawn is I . Now the n resistors are connected in parallel to same battery . Then the current drawn from the battery becomes 10I . The value of n is

(a) 20

(b) 9

(c) 11

(d) 10

Ans: (d)

Sol: For series combination ,

$\displaystyle I = \frac{E}{nR + R}$ …(i)

For series combination ,

$\displaystyle 10I = \frac{E}{\frac{R}{n} + R}$ …(ii)

On dividing (ii) by (i) we get

n = 10

## 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,

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

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

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