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Physics NEET Questions — Free MCQ Practice

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The resistance of a wire is R ohm. If it is melted and stretched to n times its original lenght, its new resistance will be

A.

nR

B.

$\frac{R}{n}$

C.

$n^{2} R$

D.

$\frac{R}{n^{2}}$

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Explanation

(c)Thinking process volume of material remains same in streching .

As volume remains same,

$A_{1} l_{1} = A_{2} l_{2}$

Now,given $l_{2} = n l_{1}$

So, New area $A_{2} = \frac{A_{1} l_{1}}{l_{2}} = \frac{A_{1}}{n}$

Resistance of wire after stretching

$R_{2} = ρ \frac{l_{2}}{A_{2}}$

= $ρ . \frac{n l_{2}}{A_{2} \ln}$

$= (ρ \frac{l_{1}}{A_{1}}) . n^{2} = n^{2} . R$

$[S o, R = (ρ \frac{l_{1}}{A_{1}})]$

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A potentiometer is an accurate and versatile device to make electrical measurement of EMF because the method involves

A.

cells

B.

potential gradients

C.

a condition of no current flow through the galvanometer

D.

a combination of cells, galvanometer, and resistances

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Explanation

(c) When a cell is balanced against potential drop across a certain length of potentiometer wire, no current flows through the cell

$∴$ emf of cell=potential drop across balance length of potentiometer wire. So, a potentiometer is a more accurate device for measuring emf of a cell or no current flows through the cell during the measurement of emf.

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The charge following through a resistance R varies with time $t a s Q = a t - b t^{2},$ where a and b are positive constants. The total heat produced in R is

A.

$\frac{a^{3} R}{3 b}$

B.

$\frac{a^{3} R}{2 b}$

C.

$\frac{a^{3} R}{b}$

D.

$\frac{a^{3} R}{6 b}$

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Explanation

(d) Given, charge $Q = a t - b t^{2},$ ...(i)

$∵$ We know that current, $I = \frac{d Q}{d t}$

So, eq(i) can be written as

$I = \frac{d}{d t} (a t - b t^{2})$

$I = a - 2 b t$ ...(ii)

For maximum value of t, till the current exist is given by

$\Rightarrow a - 2 b t = 0$

$t = \frac{a}{2 b}$ ...(iii)

$∵$ The total heat produced (H) can be given as

$H = \int_{0}^{t} I^{2} R d t = \int_{0}^{a / 2 b} {(a - 2 b t)}^{2} R . d t \{∵ t = \frac{a}{2 b}\} = \int_{0}^{a / 2 b} (a^{2} + 4 b^{2} t^{2} - 4 a b t) R d t H = {[a^{2} t + 4 b^{2} \frac{t^{3}}{3} - \frac{4 a b t^{2}}{2}]}_{0}^{a / 2 b} R$

solving above equation we get,

$\Rightarrow H = \frac{a^{3} R}{6 b}$

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A potentiometer wire has a length 4 m and resistance 8Ω. The resistance that must be connected in series with the wire and an accumulator of emf 2V, so as to get a potential gradient 1mV per cm of the wire is

A.

32Ω

B.

40Ω

C.

44Ω

D.

48Ω

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Explanation

$P o t e n t i a l d r o p i n t h e p o t e n t i o m e t e r w i r e = 400 \times 1 \times 10^{- 3} = 0.4 V P o t e n t i a l d r o p i n t h e r e s i s \tan c e t o b e c o n n e c t e e d i n s e r i e s = 1.6 V C u r r e n t i n t h e p o t e n t i o m e t e r w i r e = \frac{0.4}{8} A R e s i s \tan c e = \frac{1.6}{0.4} \times 8 = 32 o h m$

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A potentiometer wire of length L and a resistance r are connected in series with a battery of e.m.f. $E_{o}$ and a resistance r₁. An unknown e.m.f. is balanced at a length l of the potentiometer wire. The e.m.f. E will be given by

A.

$\frac{E_{o} r L}{(r_{1}) l}$

B.

$\frac{E_{o} r l}{(r + r_{1}) L}$

C.

E_ol/L

D.

$\frac{E_{o} r L}{(r + r_{1}) l}$

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Explanation

In a potentiometer circuit, the unknown EMF E is given by the ratio of the potential drop across the length l of the potentiometer wire to the total potential drop across the entire wire. This is given by the formula: E = (E_o * r * l) / ((r + r_1) * L), where E_o is the EMF of the battery.

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Two cities are 150 km apart. Electric power is sent from one city to another city through copper wires. The fall of potential per km is 8V and the average resistance per km is 0.5 Ω. The power loss in the wire is

A.

19.2W

B.

19.2kW

C.

19.2J

D.

12.2kW

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Explanation

Potential difference (drop) between two cities=150x8=1200V

Average resistance of total wire=0.5x150=75Ω

Power loss=P=V²/R

=1200x1200/75

=19200 W

=19.2 kW

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A potentiometer circuit has been setup for finding the internal resistance of a given cell. The main battery, used across the potentiometer wire, has an emf of 2.0 V and a negligible internal resistance. The potentiometer wire itself is 4 m long. When the resistance, R, connected across the given cell, has values of

(i)infinity

(ii)9.5Ω

the 'balancing lengths', on the potentiometer wire are found to be 3m and 2.85m, respectively. The value of internal resistance of the cell is

A.

0.25Ω

B.

0.95Ω

C.

0.5Ω

D.

0.75Ω

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Explanation

Given, e=2V and l=4m

Potential drop per unit length

φ=e/l=2/4=0.5V/m

$F o r t h e f i r s t c a s e, e' = φ l_{1} . . . (i) F o r t h e s e c o n d c a s e, V = φ l_{2} . . . (i i) F r o m E q s . (i) a n d (i i), e' / V = l_{1} / l_{2} e' = l (r + R) a n d V = I R f o r t h e s e c o n d c a s e S o, r = R (l_{1} / l_{2} - 1) = 9.5 (3 / 2.85 - 1) = 9.5 (1.05 - 1) = 9.5 x 0.05 = 0.475 = 0.5 Ω$

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A wire of resistance 4Ω is stretched to twice its original length. The resistance of a stretched wire would be :

A.

2Ω

B.

4Ω

C.

8Ω

D.

16Ω

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Explanation

(d) As R'=n²R=2² x 4=4 x 4=16Ω

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The internal resistance of a 2.1V cell which gives a current of 0.2A through a resistance of 10Ω is :

A.

0.2Ω

B.

0.5Ω

C.

0.8Ω

D.

1.0Ω

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Explanation

(b) As I=E/R+r or E=I(R+r)

2.1=0.2(10+r)

10+r=2.1/2 x 10

∴ r=10.5-10=0.5Ω

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If voltage across a bulb rated 220 V-100 W drops by 2.5% of its rated value, the percentage of the rated value by which the power would decrease is

A.

20%

B.

2.5%

C.

5%

D.

10%

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Explanation

Power $P = \frac{V^{2}}{R}$

For small variation

$\frac{∆ P}{P} \times 100 % = \frac{2 \times ∆ V}{V} \times 100 %$

$= 2 \times 2.5 = 5 %$

Therefore, power would decrease by 5%.

Physics MCQs for NEET — Practice Questions with Answers

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