﻿ Medical Entrance Exam Question and Answers | Current Electricity - Zigya

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# Current Electricity

#### Multiple Choice Questions

1.

An electric bulb marked 40 W and 200 V, is used in a circuit of supply voltage 100 V. Now its power is

• 100 W

• 20 W

• 40 W

• 10 W

D.

10 W

Actual power of the bulb (P1) = 40 W

Actual voltage of bulb ( V1 ) = 200 V

Supply voltage ( V2 ) = 100 V

Power (P) =$\frac{{\mathrm{V}}^{2}}{\mathrm{R}}$∝ V2

Therefore

$\frac{{\mathrm{P}}_{1}}{{\mathrm{P}}_{2}}=\frac{{\mathrm{V}}_{1}^{2}}{{\mathrm{V}}_{2}^{2}}$

$\frac{40}{{\mathrm{P}}_{2}}=\frac{{\left(200\right)}^{2}}{{\left(100\right)}^{2}}$= 4

⇒ P2 =$\frac{40}{4}$

⇒ P2 = 10W

( wheere P2 = power when voltage is 100 V )

2.

Dimension of resistivity is

• [ M L2 T-2 A-1 ]

• [ M L3 T-3 A-2 ]

• [ M L3 T-2 A-1 ]

• [ M L2 T-2 A-2 ]

C.

[ M L3 T-2 A-1 ]

By definition

R =ρ$\frac{\mathrm{l}}{\mathrm{A}}$

⇒ρ =$\frac{\mathrm{R}\mathrm{A}}{\mathrm{l}}$

=$\frac{\frac{\mathrm{V}}{\mathrm{I}}\mathrm{A}}{\mathrm{l}}$

∴ [ρ ] =$\frac{\left[\mathrm{V}\right]\left[\mathrm{A}\right]}{\left[\mathrm{l}\right]\left[\mathrm{l}\right]}$

=$\frac{\left[\mathrm{M}{\mathrm{L}}^{2}{\mathrm{T}}^{-2}{\mathrm{A}}^{2}\right]}{\left[\mathrm{A}\mathrm{L}\right]}$

[ρ ] = [ M L3 T-2 A-1 ]

3.

Assertion:In a simple battery circuit, the pointof the lowest potential is positive terminal of thebattery.

Reason: The current flows towards the point ofthe higher potential, as it does in such a circuitfrom the negative to the positive terminal.

• If both the assertion and reason are true and reason is a correct explanation of the assertion.

• If both the assertion and reason are true and reason is not a correct explanation of the assertion.

• If the assertion is true but the reason is false.

• If both assertion and reason are false.

D.

If both assertion and reason are false.

In a battery circuit, the point of lowestpotential is the negative terminal of the battery. Andthe current flows from higher potential to lowerpotential.

4.

Two bulbs A and B are connected in parallel. Bulb A will glow more than bulb B. If their resistances are RA and RB respectively. Then :

• R$<$ RB

• R$=$ RB

• R$>$ RB

• none of these

A.

R$<$ RB

Since,bulbs A and B are connected in parallel,therefore the potential across both the bulbs is same i.e,

V = iARA= iBRB

But bulb A glow more than bulb B

$\therefore$ iA$>$iB

Hence RA$<$RB

5.

In a neon discharge tube 2.9 x 10 Ne18ions move to be the right per second while 1.2 x 108 electron move to the left persecond electric charge is 1.6 x 10-19C. The current in discharge tube is

• 0.66A towards left

• 0.66 A towards left

• 1A towards right

• Zero

B.

0.66 A towards left

Currentdue to both types of ions are in the same direction towards the right so

i=i1+i2

$\mathrm{i}=2.9×{10}^{18}×1.6×{10}^{-19}+1.2×{10}^{18}×1.6×{10}^{-19}\phantom{\rule{0ex}{0ex}}\mathrm{i}=0.66\mathrm{A}$

6.

When a wire is stretched and its radius becomesr/2, then its resistance will be

• 16 R

• 2R

• 4R

• 0

A.

16 R

Initial radius if wire r1 = r

Since volume of the wire after stretch remains constant,

therefore

l1A1 = l2A2

$\frac{{\mathrm{l}}_{1}}{{\mathrm{l}}_{2}}=\frac{{\mathrm{A}}_{2}}{{\mathrm{A}}_{1}}$

=$\frac{{\mathrm{r}}_{2}^{2}}{{\mathrm{r}}_{1}^{2}}$

=${\left(\frac{0.5\mathrm{r}}{\mathrm{r}}\right)}^{2}$

$\frac{{\mathrm{l}}_{1}}{{\mathrm{l}}_{2}}=\frac{1}{4}$

The resistance

(R) =ρ$\frac{\mathrm{l}}{\mathrm{A}}\propto \frac{\mathrm{l}}{\mathrm{A}}$

Therefore

$\frac{{\mathrm{R}}_{1}}{{\mathrm{R}}_{2}}=\frac{{\mathrm{I}}_{1}}{{\mathrm{I}}_{2}}×\frac{{\mathrm{A}}_{2}}{{\mathrm{A}}_{1}}$

=$\frac{1}{4}×\frac{1}{4}$

$\frac{{\mathrm{R}}_{1}}{{\mathrm{R}}_{2}}=\frac{1}{16}$

⇒ R2 = 16 R1

R2 = 16 R

7.

If a current is flowing in a spring, then it:

• compress

• swing

• expand

• remain unaffected

A.

compress

When current is made to flow in the spring it flows in each turn in parallel order and an attractive force is produced between consecutive turns and therefore, it is compressed.

8.

When a voltmeter connected across theterminals of cell, measures 5 V and anammeter connected measures 10 A. Aresistance of 2 Ω is connected across the terminal of the cell. The current flowingthrough this resistance is

• 7.5amp

• 5.0amp

• 2.5amp

• 2.0amp

D.

2.0amp

Internal resistanceof the cell is

$\mathrm{r}=\frac{\mathrm{E}}{\mathrm{i}}=\frac{\mathrm{E}}{10}=0.5\mathrm{\Omega }\phantom{\rule{0ex}{0ex}}\mathrm{Current}\mathrm{i}=\frac{\mathrm{E}}{\mathrm{R}+\mathrm{r}}=\frac{5}{2+0.5}=2\mathrm{A}$

9.

A rectangular loop carrying a current i1, is situated near a long straight carrying a straight current i2. The wire is parallel to one of the sides of the loop as shown in the figure. Then the current loop will • move away from the wire

• move towards the wire

• remain stationary

• rotate about axis parallel to wire

B.

move towards the wire

As two like currents attract and unlike currentsrepel each other, the wire will attract the side A of theloop near the wire and parallel to it and the far sideof the loop again parallel to the wire will be repelled.But the force of attraction will be more as that sideof the loop is closer to the wire and so effectively thecurrent loop will more towards the wire.

10.

A wire of length Lis drawn such that its diameteris reduced to half of its original diameter. If theinitial resistance of the wire were 10 Ω, its newresistance would be

• 40 Ω

• 80 Ω

• 120 Ω

• 160 Ω

D.

160 Ω

Let the original diameter of the wire be D.

Therefore the new diameter is D/2.

Original area of cross-section is

$\frac{\mathrm{\pi }{\mathrm{D}}^{2}}{4}$

where D is diameter

and the final area of cross-section is

$\frac{\mathrm{\pi }{\left(\mathrm{D}}{2}\right)}^{2}}{4}$

$\frac{\mathrm{\pi }{\mathrm{D}}^{2}}{16}$

The new length of the wire is given by

$\frac{\mathrm{\pi }{\mathrm{D}}^{2}}{4}$= L'×$\frac{\mathrm{\pi }{\mathrm{D}}^{2}}{16}$

⇒ L' =$\frac{16}{4}$L

L' = 4L

Now, we know that the resistance is given by

R =ρ$\frac{\mathrm{L}}{\mathrm{A}}$

∴ R' =ρ$\frac{\mathrm{L}\text{'}}{\mathrm{A}\text{'}}$

$\frac{4\mathrm{L}}{\mathrm{A}}{4}}$

R' = 16R

$\left[\because \mathrm{A}\text{'}=\frac{\mathrm{\pi }{\mathrm{D}}^{2}}{16}=\frac{\mathrm{A}}{4}\right]$

∴ R' = 16× 10

R' = 160Ω