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Physics NEET Questions — Free MCQ Practice

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A fixed volume of iron is drawn into a wire of length L. The extension x produced in this wire by a constant force F is proportional to

A.

$\frac{1}{L^{2}}$

B.

$\frac{1}{L}$

C.

$L^{2}$

D.

L

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Explanation

(c) $l = \frac{F L}{A Y} = \frac{F L^{2}}{(A L) Y} = \frac{F L^{2}}{V Y}$

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On applying a stress of $20 \times 10^{8}$ N/ $m^{2}$ the length of a perfectly elastic wire is doubled. Its Young’s modulus will be

A.

$40 \times 10^{8} N / m^{2}$

B.

$20 \times 10^{8} N / m^{2}$

C.

$10 \times 10^{8} N / m^{2}$

D.

$5 \times 10^{8} N / m^{2}$

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Explanation

(b) Young's modules = $\frac{s t r e s s}{s t r a i n}$

As the length of wire get doubled therefore strain=1

$∴ Y = s t r a i n = 20 \times 10^{8} N / m^{2}$

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The length of an elastic string is a metre when the longitudinal tension is 4 N and b metre when the longitudinal tension is 5 N. The length of the string in metre when the longitudinal tension is 9 N is

A.

a - b

B.

5b - 4a

C.

2b - $\frac{1}{4} a$

D.

4a - 3b

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Explanation

(b) Let L is the original length of the wire and K is force constant of wire.

Final length = initial length + elongation

$L^{1} = L + \frac{F}{K} F o r f i r s t c o n d i t i o n α = L + \frac{4}{K} . . . . (i) F o r s e c o n d c o n d i t i o n b = L + \frac{5}{K} . . . . (i i) B y s o l v i n g (i) a n d (i i) e q u a t i o n w e g e t L = 5 a - 4 b a n d K = \frac{1}{b - a} N o w w h e n t h e l o n g i t u d i n a l t e n s i o n i s 9 N, l e n g t h o f t h e s t r i n g = L + \frac{9}{K} = 5 a - 4 b + 9 (b - a) = 5 b - 4 a .$

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How much force is required to produce an increase of 0.2% in the length of a brass wire of diameter 0.6 mm ?

(Young’s modulus for brass = $0.9 \times 10^{11} N / m^{2}$ )

A.

Nearly 17 N

B.

Nearly 34 N

C.

Nearly 51 N

D.

Nearly 68 N

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Explanation

(c) $F = \frac{Y A l}{L} = 0.9 \times 10^{11} \times π \times {(0.3 \times 10^{- 3})}^{2} \times \frac{0.2}{100} = 51 N$

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A 5 m long aluminium wire $(Y = 7 \times 10^{10} N / m^{2})$ of diameter 3 mm supports a 40 kg mass. In order to have the same elongation in a copper wire $(Y = 12 \times 10^{10} N / m^{2})$ of the same length under the same weight, the diameter should now be, in mm

A.

1.75

B.

1.5

C.

2.5

D.

5.0

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Explanation

(c)

$l = \frac{F L}{{πr}^{2} Y} \Rightarrow r^{2} \propto \frac{1}{Y} (F, L a n d l a r e c o n s \tan t) \frac{r_{2}}{r_{1}} = {(\frac{Y_{1}}{Y_{2}})}^{1 / 2} = {(\frac{7 \times 10^{10}}{12 \times 10^{10}})}^{1 / 2} \Rightarrow r_{2} = 1.5 \times {(\frac{7}{12})}^{1 / 2} = 1.145 m m ∴ d i a = 2.29 m m$

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A steel wire of 1 m long and cross section area $1 {mm}^{2}$ is hang from rigid end. When mass of 1kg is hung from it then change in length will be (given $Y = 2 \times 10^{11} N / m^{2}$ )

A.

0.5 mm

B.

0.25 mm

C.

0.05 mm

D.

5 mm

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Explanation

(c) $l = \frac{M g L}{Y A} = \frac{1 \times 10 \times 1}{2 \times 10^{11} \times 10^{- 6}} = 0.05 m m$

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A force F is applied on the wire of radius r and length L and change in the length of wire is l. If the same force F is applied on the wire of the same material and radius 2r and length 2L, Then the change in length of the other wire is

A.

l

B.

2l

C.

$\frac{l}{2}$

D.

4l

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Explanation

(c)

$l = \frac{F L}{A Y} \Rightarrow l \propto \frac{L}{r^{2}} (F a n d Y a r e c o n s \tan t) \frac{l_{2}}{l_{1}} = \frac{L_{2}}{L_{1}} \times {(\frac{r_{1}}{r_{2}})}^{2} = 2 \times {(\frac{1}{2})}^{2} = \frac{1}{2} ∴ l_{2} = \frac{l_{1}}{2}$

i.e., the change in the length of other wire is $\frac{l}{2}$

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An iron rod of length 2m and cross section area of 50 X $10^{- 6} m^{2}$ , is stretched by 0.5 mm, when a mass of 250 kg is hung from its lower end. Young's modulus of the iron rod is-

A.

$19.6 \times 10^{10} N / m^{2}$

B.

$19.6 \times 10^{15} N / m^{2}$

C.

$19.6 \times 10^{18} N / m^{2}$

D.

$19.6 \times 10^{20} N / m^{2}$

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Explanation

(a) $Y = \frac{M g L}{A l} = \frac{250 \times 9.8 \times 2}{50 \times 10^{- 6} \times 0.5 \times 10^{- 3}} = 19.6 \times 10^{10} N / m^{2}$

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In which case there is maximum extension in the wire, if same force is applied on each wire

A.

L = 500 cm, d = 0.05 mm

B.

L = 200 cm, d = 0.02 mm

C.

L = 300 cm, d = 0.03 mm

D.

L = 400 cm, d = 0.01 mm

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Explanation

(d) $l \propto \frac{L}{r^{2}} (Y a n d F a r e c o n s \tan t)$

Maximum extension takes place in that wire for which the ratio of $\frac{L}{r^{2}}$ will be maximum.

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The extension of a wire by the application of load is 3 mm. The extension in a wire of the same material and length but half the radius by the same load is -

A.

12 mm

B.

0.75 mm

C.

15 mm

D.

6 mm

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Explanation

(a) $l = \frac{F L}{A Y} \Rightarrow l \propto \frac{1}{r^{2}} (F, L a n d Y a r e c o n s \tan t) \frac{l_{2}}{l_{1}} = {(\frac{r_{1}}{r_{2}})}^{2} = {(2)}^{2} \Rightarrow l_{2} = 4 l_{1} = 4 \times 3 = 12 m m$

Physics MCQs for NEET — Practice Questions with Answers

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